1.

`If 2% of the fuses produced are defective, the probability that in arandomly selected sample of six there are two defectives is:a.  (6C2)((.02)**2)((.98)**4)b.  ((.02)**4)((.98)**2)c.  (6C4)((.02)**4)((.98)**2)d.  ((.98)**4)((.02)**2)e.  none of the above `
`Answer: a.  (6C2)((.02)**2)((.98)**4)Use the binomial probability formula. `

2.

`A company engaged in recruiting wishes to develop a questionnaire thatspreads out applicants (shows greater variation in scores than doesits standard form).  A new, longer form is developed and tested byhaving 16 randomly selected applicants from the current applicantpool fill it out.  (Another group of 16 randomly and independentlyselected applicants fills out the standard form.)Results obtained were:Variance - new form        S**2(1) = 40.2Variance - standard form   S**2(2) = 12.3Regarding scores as normally distributed -a.  Test at the 5% level the claim that the variance of the new form    is greater than that of the old form.b.  Sketch the relevant F distribution and indicate the rejection    region. `
`Answer: a.  F(calculated) = 40.2/12.3 = 3.268    F(critical)   = 2.4 at the 95% confidence level with df=15,15.    Since F(calculated) is greater than F(critical), we will reject    the null hypothesis that the new variance is less than or equal    to the old variance.b.  If available, consult file of graphs and diagrams that could not    be computerized for accompanying diagram. `

3.

`It is desired to see if there is a relationship in tastes for anexpensive car and owning a tri-maran.  A survey of 200 upper-upper class potential purchasers of cars and tri-marans gavethese responses:                  want expensive car    do not want expensive    totals                                                carwant tri-maran          100                      40                140don't want               20                      40                 60tri-marantotals                  120                      80                200Specifically, it is desired to test H(O):  the desire for a tri-maranis independent of a desire for an expensive car vs. H(1):  thereis a relationship, at level ALPHA = 10%.The correct table value (cutoff point) to use for this problem isclosest to:a)  2.71      d)  9.21b)  4.61      e)  6.25c)  6.63 `
`Answer: a)  2.71    CHISQUARE(critical, df = 1, ALPHA = .10) = 2.70554 `

4.

`The thickness of the individual cards produced by a certainplaying card manufacturer is normally distributed with mean =0.01 inches and variance = 0.000052.  What is the probabilitythat a deck of 52 cards is more than 0.65 inches in thickness?A.  .001      B.  .006      C.  .023      D.  .036E.  .067      F.  .087      G.  .159      H.  .184 `
`Answer: B.  .006MU (deck) = 52 * .01 = .52Var(deck) = 52 * .000052 = .002704Z = (.65 - .52)/SQRT(.002704) = .13/.052  = 2.5P(Z > 2.5) = .0062 `

5.

`A particular type of bolt is produced having diameters with mean 0.500inches and standard deviation 0.005 inches.  Nuts are also producedhaving inside diameters with mean 0.505 inches and standard deviation0.005 inches.  If a nut and a bolt are chosen at random, what is theprobability that the bolt will fit inside the nut? `
`Answer: Mean for the distribution of differences = .005Standard deviation = SQRT((.005)**2/1 + (.005)**2/1) = .007071Z = value of interest - mean of distribution (of differences) /    standard error of the distribution of differencesZ = 0 - .005/.007071 = -.71We want all the area to the right of -.71= .7611 or 76%. `

6.

`It is known that the lengths of a particular manufactured item arenormally distributed with a mean of 6 and a standard deviation  of3.  If one item is selected at random, what is the probability thatit wil fall between 5.7 and 7.5? `
`Answer: P(5.7 < Y < 7.5) = P((5.7-6)/3 < Z < (7.5-6)/3)                 = P(-.1 < Z < .5)                 = .0398 + .1915                 = .2313 `

7.

`Suppose the length of life of certain kinds of batteries is normallydistributed with MU = 36 months, SIGMA = 4 months.  The company guar-antees the battery to last 30 months.  What proportion of the batter-ies will they have to make an adjustment on? `
`Answer: P(X < 30) = P(Z < (30 - 36)/4)          = P(Z < -1.5)          = .0668 or 6.68% `

8.

`It is found that the gyre swivit, manufactured on any given dayat Gozornenplatz, Inc.'s Swat City plant, has the following cha-racteristics with respect to length:      Normally distributed with MU = 3.5"      and SIGMA = .2".Draw a picture for each of the following, and show your work.a.  What percent of a day's output lies within two standard    deviations of the mean?b.  Of a day's output of 2500 gyre swivits, how many will    measure less than 3.966"?c.  95% of a day's output, centered around the average, will    measure between _______ and _______.d.  What percent of a day's output lies between 1 and 2.58    standard deviations above the mean? `
`Answer: If available, consult file of graphs and diagrams that could not becomputerized.a.  95.4%    P(3.3 <= X <= 3.7) = P(-2 <= Z <= 2)                       = 2(P(0 <= Z <= 2))                       = 2 * (.4772)                       = .9544b.  2475    Z = (X - MU)/SIGMA      = (3.966 - 3.5)/.2      = (.466/.2)      = 2.33    P(Z < 2.33) = .99    99% of 2500 = (.99)(2500) = 2475c.  3.108, 3.892    P(0 <= Z <= ?)  = .475       Therefore ?  = 1.96    Lower Boundary  = MU - (Z)(SIGMA)                    = 3.5 - (1.96)(.2)                    = 3.108    Higher Boundary = MU + (Z)(SIGMA)                    = 3.5 + (1.96)(.2)                    = 3.892d.  15.4%    P(1 <= Z <= 2.58)        = P(0 <= Z <= 2.58) - P(0 <= Z <= 1)        = .4951 - .3413        = .1538        = 15.4% `

9.

`The U.S. Department of Commerce has just completed a sample survey ofweekly food expenditures.  A simple random sample of 100 families wastaken.  The average weekly food expenditure was \$70.00 per week, witha standard deviation of \$8.00.  You may assume expenditures in thepopulation to be normally distributed.a.  What proportion of the families spent \$85.00 or more per week    on food?  Be sure to diagram your problem solution]b.  Using the information above, find the expenditure value above    which 80% of the families lie. `
`Answer: If available, consult file of graphs and diagrams that could not becomputerized.a)  Z = (X - MU)/SIGMA      = (85 - 70)/8      = 1.875    Area beyond this Z value is .0301, so 3.01% of the families spent    85 dollars or more per week.b)  A cumulative Z value such that 80% lies above it or 20% lies below    it is -.84.       Z = (X - MU)/SIGMA    -.84 = (X - 70)/8       X = 63.28    Therefore, 80% of the families lie above the expenditure value of    \$63.28/week. `

10.

`Suppose a floor manager of a large department store isstudying buying habits of their customers.a)  If he is willing to assume that monthly income of these customers    is distributed normally, what proportion of the income should he    expect to fall in the interval determined by MU +/- 1.2(SIGMA)?b)  What proportion of the income should he expect to be greater    than MU + SIGMA?c)  Still assuming normality, what is the probability that a    customer selected at random will have an income exceeding the    population mean by 3*SIGMA? `
`Answer: a)  P(MU <= X <= 1.2SIGMA) = .3849    P(X is in interval MU +/- 1.2SIGMA) = 2(.3849) = .7698b)  P(X > (MU + SIGMA)) = (.5 - .3413) = .1587c)  P(X > MU + (3*SIGMA)) = (.5 - .4987) = .0013 `

11.

`Suppose a floor manager of a large department store is studyingthe buying habits of the store's customers.a)  If he is willing to assume that monthly income of these    customers is distributed normally and SIGMA = \$500, find    the proportion of customers exceeding the population mean    by \$375.b)  Find the proportion of customers within \$125 of the    population mean. `
`Answer: a)  Z = 375/500 = .75    P(Z > .75) = (0.5 - .2734) = .2266b)  Z = 125/500 = .25    P(-.25 <= Z <= .25) = 2(.0987) = .1974 `

12.

`A floor manager of a large department store is studying the buyinghabits of the store's customers.  Suppose the manager has  someonetell him that monthly income of these customers is distributed nor-mally with a population mean of \$600 and standard deviation of \$500.a)  What proportion of the customers should he expect to have    incomes less than \$600?b)  What proportion should he expect to have incomes less    than \$725? `
`Answer: a)  .5b)  Z = (725 - 600)/500 = .25    P(Z < .25) = .5 + .0987 = .5987 `

13.

`A company manufactures rope.  From a large number of tests over a longperiod of time, they have found a mean breaking strength of 300 lbs.and a standard deviation of 24 lbs.  Assume that these values areMU and SIGMA.It is believed that by a newly developed process, the mean breakingstrength can be increased.(a)  Design a decision rule for rejecting the old process with an     ALPHA error of 0.01 if it is agreed to test 64 ropes.(b)  Under the decision rule adopted in (a), what is the probability     of accepting the old process when in fact the new process has     increased the mean breaking strength to 310 lbs.?  Assume SIGMA     is still 24 lbs.  Use a diagram to illustrate what you have done,     i.e., draw the reference distributions. `
`Answer: a.  One tail test at ALPHA = .01, therefore Z = 2.33.    Z = (YBAR-MU)/(SIGMA/SQRT(n))    2.33 = (YBAR-300)/(24/SQRT(64))    YBAR = 307    Decision Rule:  If the mean strength of 64 ropes tested is 307                    lbs. or more, we reject the hypothesis of no im-                    provement, i.e., we continue that the new process                    is better.b.  If available, consult file of graphs and diagrams that could not    be computerized for reference distributions.    Z = (307-310)/(24/SQRT(64)) = 1.00    Area = 0.1587 or 15.87%    P(type II error) = 0.1587 `

14.

`Suppose X is the price  that a certain stock will be exactly 6 monthsfrom today.  Assume that X is normally distributed with a mean of \$30and a standard deviation of \$5.a.  Find the probability that X will be at least \$30.b.  Find the probability that X will be greater than \$40.c.  Find the probability that X will be between \$20 and \$35.d.  How many standard deviations is \$38 from the mean?e.  If you paid \$29 for the stock today, what is the probability that    you will make a profit if you sell the stock exactly 6 months from    today? `
`Answer: a.  P(X >= 30) = P(Z >= 0) = 1/2, where Z = (30 - 30)/5b.  Z = (40 - 30)/5 = 2;   P(X > 40) = .5 - .4772 = .0228c.  Z(1) = (20 - 30)/5 = -2    Z(2) = (35 - 30)/5 = 1    Prob(20 < X < 35) = .4772 + .3413 = .8185d.  8/5 = 1.6 SD'se.  Prob(X > 29) = .5 + .0793 = .5793, where Z = (29 - 30)/5 = -.2 `

15.

`Distribution of the I.Q.'s of 4,500 employees of a company isroughly normal with mean 104 and standard deviation 15. Findthe number of employees whose I.Q. is:     a.  greater than or equal to 110     b.  between 95 and 110. `
`Answer: a.   Z = (110 - 104)/(15) = .4    NO. = (4500)(.5 - .1554) = 1550b.   Z = (95 - 104)/(15) = -.6    NO. = (4500)(.1554) + (4500)(.2258) = 1715 `

16.

`A certain kind of automobile battery is known to  have  a  length  oflife  which  is  normally  distributed  with  a mean of 1200 days andstandard deviation 100 days.  How long should the guarantee be if themanufacturer wants to replace only 10%  of  the  batteries which  aresold? `
`Answer: Z = -1.28 for 10 percent failure-1.28 = (X - 1200)/100X = 1072 days for guarantee `

17.

`A floor manager of a large department store is studying the buyinghabits of the store's customers.  Suppose he assumes that the monthlyincome of these customers is normally distributed with a standard de-viation of 500.  If he were to draw a random sample of N = 100 custo-mers and determine their income:a)  What is the probability that the sample mean of incomes will    differ from the population mean by more than \$25?b)  What is the probability that the sample mean is larger than    the population mean?c)  Could you provide a reasonable answer to (a) and (b) if    the population of incomes were not normal?  Explain. `
`Answer: a)  SIGMA(XBAR) = SIGMA/SQRT(n)                = 500/SQRT(100)                = 50    Z = (XBAR - MU)/SIGMA(XBAR)    Z = 25/50 = .5    P(Z < -.5 or Z > +.5) = 2(.5 - .1915) = .6170b)  .5c)  Yes, the central limit theorem assures us that the    distribution of means for n = 100 is symmetrical andapproximately normal. `

18.

`Suppose that you work for a brewery as a clerk to receive barleyshipments.  As part of your job you are to decide whether to keepor return new shipments of barley.  The criteria used for making yourdecision is an estimation of the moisture content of the shipment.If the moisture level is too high (above 17.5%) the shipment has agood possibility of rotting before use and, therefore, a loss ofmoney to the company.  You know from past experience that the variancefor all barley shipments is 36 and that your staff can process at themost one sample of 9 moisture readings per shipment.a.  Propose a rule for accepting and rejecting grain shipments on the    basis of sample means where the null claim is a shipment has a    mean moisture content of 17.5% or less (H(0):  MU <= 17.5%).    Let the probability of Type I error be .10.b.  When will you make incorrect decisions about a grain shipment    having MU = 17.4?  What will be the probability of such an    error?c.  When will you make incorrect decisions about a grain shipment    having MU = 19?  What will be the probability of such errors?d.  When will you make incorrect decisions about a grain shipment    having MU = 21?  What will be the probability of such errors? `
`Answer: SIGMA**2 = 36Take a sample, n = 9SIGMA(XBAR) = SIGMA/SQRT(n) = 6/3 = 2a.  H(0):  MU <= 17.5    H(1):  MU >  17.5    ALPHA = .10 implies Z = 1.28    Z = XBAR - MU/SIGMA(XBAR)    1.28 = XBAR - 17.5/2    2.56 = XBAR - 17.5    XBAR = 20.06    Reject H(0) when XBAR > 20.06.b.  I am rejecting H(0) when XBAR > 20.06, so when MU is REALLY 17.4,    I make incorrect decisions whenever XBAR > 20.06.    Z = 20.06 - 17.4/2    Z = 1.33    Area beyond Z = 1.33 is .0918.    The probability of an incorrect decision is .0918.c.  I am rejecting H(0) when XBAR > 20.06, so when MU is REALLY 19,    I make incorrect decisions whenever XBAR <= 20.06.    Z = 20.06 - 19/2    Z = .53    Area between mean and Z = .2019.    The probability of an incorrect decision is .5 + .2019 = .7019.d.  I am rejecting H(0) when XBAR > 20.06, so when MU is REALLY 21,    I make incorrect decisions whenever XBAR < 20.06.    Z = 20.06 - 21/2    Z = -.47    Area beyond Z = -.47 is .3912.    The probability of an incorrect decision is .3912. `

19.

`If the number of complaints which a laundry receive per day is arandom variable having Poisson distribution with LAMBDA = 4, findthe probabilities that on a given day the laundry will receive:     a.  no complaints,     b.  exactly 2 complaints. `
`Answer: a.  P(x=0) = (4**0)(e** -4)/0] = .018b.  P(x=2) = (4**2)(e** -4)/2] = .147 `

20.

`As a quality control inspector you have observed that  wooden  wheelswhich are bored off-center occur about three percent of the time.  Ifsix of these wheels are to be used on each toy truck produced by AcmeToy  Company,  the  probability  that a given truck has no wheels offcenter would be obtained by using which distribution?(a)  Normal                  (c)  Hypergeometric(b)  Poisson                 (d)  Binomial `
`Answer: (d)  Binomial `

21.

`Suppose that 2% of the fuses produced by a machine are defective.If we take a sample of 6 from the machine's output, the probabilitythat the first four fuses are good and the last two defective is:a.  (6C4)((.98)**4)((.02)**2)b.  ((.02)**4)((.98)**2)c.  (6C4)((.02)**4)((.98)**2)d.  ((.98)**4)((.02)**2)e.  none of these `
`Answer: d.  ((.98)**4)((.02)**2)The combination (6C4) is not necessary because the order GGGGDDis distinct. `

22.

`An accounting firm processed 1000 balance sheets for its clientlast year.  If 20% of these are known to contain errors, whatis the probability of finding at least one error in asample of 4 balance sheets chosen at random with replacement?a.  .4906b.  .0016c.  .5904d.  .9984e.  none of these `
`Answer: c.  Let x = # balance sheets containing errors        P(x>=1) = 1-P(x=0);        P(x=0) = b(0;4,.20) = (4C0)(.20**0)(.80**4)               = .4096        P(x>=1) = 1 - .4096 = .5904 `

23.

`We have a manufacturing process which produces good itemswith probability .9.  We select a sample of 15 items.  Assume abinomial experiment.What is the probability that there is at least one good item inthe sample?a)  15C1(.9)**1b)  15C0(.9)**0(.1)**15c)  1 - (15C0(.9)**0(.1)**15)d)  1 - (15C1(.9)**1(.1)**14)e)  none of the above `
`Answer: c)  1 - (15C0(.9)**0(.1)**15) `

24.

`The following triangle test is sometimes used to identify tasteexperts.  In the case of wine tasting, a test subject is presentedwith three glasses of wine, two of one kind and a third glass ofanother wine.  The test subject is asked to identify the singleglass of wine.  A test subject who merely guesses has a 1 chancein 3 of identifying the single glass correctly.  An expert winetaster should be able to do much better.  Let K stand for the num-ber of correct identifications made by a test subject in 10 inde-pendent triangle tests.A test subject makes at least 5 correct identifications (k >= 5).  Thedescriptive level associated with this result is:    a.  .076            c.  .213    b.  .137            d.  .057 `
`Answer: c.  .213    Descriptive level = P(5 or more correct identifications)                      = P(5) + P(6) + P(7) + P(8) + P(9) + P(10)                      = .1366+.0569+.0162+.0030+.0003+.0000                      = .213 `

25.

`In a dispute over the proportion of defects in  a  large  shipment  abuyer  claims  there  are  20% defective while the seller claims only10%. To settle the dispute it is decided to take a  sample  of  size100  from  the shipment and if there are less than 15 defectives foundto rule in favor of the seller. (Note: the shipment is so large  thatsampling can be considered to be with replacement.)      a.  What is the probability of ruling in favor of the seller if          he is correct?      b.  What is the probability of ruling in favor of the seller if          the buyer is correct? `
`Answer: a.  If the seller is correct -    Using normal approximation:  p = .1, np = 10, SIGMA = SQRT(npq) = 3    Prob(Z < (14.5 - 10)/3) = .9322    Using binomial with n = 100 and p = .1:    P(X < 15) = .9274b.  If the buyer is correct -    Using normal approximation:  p = .2, np = 20, SIGMA = 4    Prob(Z < (14.5 - 20)/4) = .0845    Using binomial with n = 100 and p = .2:    P(X < 15) = .0804 `

26.

`Suppose that it is known that one out of  ten  undergraduate  collegetextbooks  is  an  outstanding  financial  success.   A publisher hasselected  four  new  text  books  for  publication.   What   is   theprobability that:a.  Exactly one will be an outstanding financial success?b.  at least one?c.  at least two? `
`Answer: a.  P = (4C1)(.1**1)(.9**3) = .2916b.  P = 1 - (4C0)(.9**4) = .3439c.  P = 1 - .6561 - .2916 = .0523 `

27.

`The Liddalol Airline Company runs an airline from New York toBoston.  Its planes carry a maximum of 90 passengers.  Knowingthat not all persons who reserve seats will actually use them,they accept 100 reservations for each flight.  The company hasdetermined that 80% of the persons who make reservations actu-ally use them.  Assuming that 100 reservations are made for aparticular flight, find the probability that some passengerswill not get seats.  What two assumptions do you have to make? `
`Answer: p = .81 - p = .2    n = 100Using normal approximation to binomial:XBAR =  np = 100*.8 = 80S**2 = npq = 100*.8*.2 = 16   S = 4P(X > 90) = P(Z > (90 - 80)/4)          = P(Z > 2.5)          = .0062The two assumptions are:1.  That the decisions for individuals to use their reservations    are independent2.  That the probability that a person who makes a reservation    actually uses it remains constant from person to person. `

28.

`The Noglow automatic cigarette lighter is claimed to light 80% ofthe time when the button is pushed.  If this is true, and if thelighter is tried 25 times:a.  What is the probability of getting exactly 20 lights?b.  What is the probability of getting fewer than 17 lights?c.  What is the probability of getting no lights on the first    4 trials? `
`Answer: a.  P(X = 20) = (25C20)(.8**20)(.2**5)              = .196b.  P(X < 17) = .046 (use table)c.  (.2**4) = .0016 `

29.

`A floor manager of a large department store is studying habits of theircustomers.  One aspect of this research pertains to residence locationof customers.a)  If 1/2 of the customers live outside the city, what is the proba-    bility that 4 customers selected at random will all live inside    the city?b)  Continuing to suppose that 1/2 live outside, what is the probability    that 3 or fewer in a random sample of size n = 10 will live outside?c)  If 1/2 live outside, the probability is 0.10 that the random sample    of size n = 100 will contain ____ or fewer persons living outside. `
`Answer: a)  (1/2)**4 = 1/16b)  Let X   = number chosen who live outside    P(X<=3) = b(0; 10, .5) + b(1; 10, .5) + b(2; 10, .5) + b(3; 10, .5)            = .001 + .0098 + .0439 + .1173            = 0.172c)  P(X<=?) = .10 = b(?; 100, 1/2)    Using the fact that for large n and p and q not too close to zero,    the binomial distribution can be closely approximated by a normal    distribution where                Z = (X - np)/SQRT(npq)    Therefore      X = (Z*SQRT(npq)) + np        = (-1.28*SQRT(100*.5*.5) + (100*.5)        = (-1.28 * 5) + 50        = -6.4 + 50        = 43.6      X == 44 `

30.

`A salesman has found that, on the average, the probability of a  saleon  a  single  contact is .3.  If the salesman contacts 50 customers,what is the probability that at least 10 will buy?   Write  an  exactexpression  for  the  probability  and  then  obtain  an  approximatenumerical value using the normal approximation. `
`Answer: Exact Expression:  p = .3,  q = 1 - p = .7,  n = 50p(X >= 10) = SUM(X = 10, 50)((50CX)(.3**X)(.7**[n - X!)) = .9598Using normal approximation: XBAR = n*p = 50*.3 = 15SIGMA = SQRT(npq) = SQRT(50*.3*.7) = 3.24            Z = (10 - 15)/3.24 = -1.54P(Z >= -1.54) = .4382 + .5000              = .9382If you use the correction factor, Z = -1.70 and P(Z >= -1.70) = .9554. `

31.

`A machine produces bolts in a length (in inches) found toobey a normal probability law with mean MU = 5 and standarddeviation SIGMA = 0.1.  The specifications for a bolt call foritems with a length (in inches) equal to 5 +/- 0.15.A bolt not meeting these specifications is called defective.a.  What is the probability that a bolt produced by this    machine will be defective?b.  If a sample of 10 bolts is chosen at random what is    the probability that there will be at least two    defective bolts? `
`Answer: X = length of bolta.  P(defective) = 1 - P(4.85 < X < 5.15)                 = 1-P((4.85 - 5)/.1 < (X - 5)/.1 < (5.15 - 5)/.1)                 = 1 - P(-1.5 < Z < 1.5)                 = 1 - .8664                 = .1336b.  Y = number of defectives in a sample of size 10    Then Y has a binomial distribution with parameters n = 10,    P = .1336.    Hence, P(at least 2 are defective) = P(Y >= 2)           = 1 - P(Y = 0) - P(Y = 1)           = 1 - (10C0)*(P**0)*((l - P)**10) - (10C1)*P*((1-P)**9)           = 1 - (.8664**10) - 10*(.1336)*(.8664**9)           = .394 `

32.

`A factory finds that, on the average, 20% of the bolts produced by agiven machine are defective.  If 10 bolts are selected at random fromthe day's production, find the probability that:a)  exactly 2 will be defective.b)  2 or more will be defective. `
`Answer: a)  P(X = 2) = (nCX)(p**X)(q**(n - X))             = (10C2)(.2**2)(.8**(10 - 2))             = .3020b)  P(X >= 2) = 1 - P(X <= 1)              = 1 - (P(1) + P(0))              = 1 - (((10C1)(.2**1)(.8**(10 - 1)))                + ((10C0)(.2**0)(.8**(10 - 0))))              = 1 - (.2684 + .1074)              = .6242 `

33.

`Seventy  five  percent  of  the  Ford  autos made in 1976 are fallingapart. Determine the probability distribution of the number of  Fordsin  a  sample  of  4 that are falling apart.  Draw a histogram of thedistribution. What is the mean and variance of the distribution? `
`Answer: Let X = the number of Fords falling apart in a sample of four.probability distribution: (binomial distribution with n=4 and p=.75)        X   ^   p(X)     -------^----------        0   ^  0.0039     = (4C0)(.75**0)(.25**4)        1   ^  0.0469     = (4C1)(.75**1)(.25**3)        2   ^  0.2109     = (4C2)(.75**2)(.25**2)        3   ^  0.4219     = (4C3)(.75**3)(.25**1)        4   ^  0.3164     = (4C4)(.75**4)(.25**0)               ^          P(X) ^          ^          ^          ^          ^               ^          ^          ^          ^          ^               ^          ^          ^          ^          ^               ^          ^          ^          ^          ^               ^          ^          ^          ^          ^           0.6 ^----------^----------^----------^----------^               ^          ^          ^          ^          ^               ^          ^          ^          ^          ^           0.5 ^----------^----------^----------^----------^               ^          ^          ^          ^          ^               ^          ^          ^     ----------      ^           0.4 ^----------^----------^----^          ^-----^               ^          ^          ^    ^          ^     ^               ^          ^          ^    ^          ^----------           0.3 ^----------^----------^----^          ^          ^               ^          ^          ^    ^          ^          ^               ^          ^     ----------^          ^          ^           0.2 ^----------^----^          ^          ^          ^               ^          ^    ^          ^          ^          ^               ^          ^    ^          ^          ^          ^           0.1 ^----------^----^          ^          ^          ^               ^          ^    ^          ^          ^          ^               ^     ----------^          ^          ^          ^               ^----^----------^----------^----------^----------^----->               0          1          2          3          4      Xmean = np = 4*.75 = 3variance = npq = 4*.75*.25 = .75 `

34.

`The probability that a particular kind of machine, used in production,breaks down during a one week period is 0.2.  If a company has 10 ofthese machines, what is the probability of having:        a.  at least two breakdowns during a given week?        b.  3,4, or 5 breakdowns during a given week?        c.  How many breakdowns should the company expect to have over            a one month (4 weeks) time period?        d.  Now suppose that during a particular week six machines            break down.  Do you have reason to believe that the break-            down rate may have increased above the 0.20 rate?  State            reasoning.        e.  What assumptions are necessary in making the above calcula-            tions of probability? `
`Answer: x = # of breakdowns during a given week        x has a binomial distribution with n=10, p=.2a.  P(x>=2) = 1 - [b(0;10,.2) + b(1;10,.2)!            = 1 - [.376!            = .624b.  P(3 <= x <= 5) = b(3;10,.2) + b(4;10,.2) + b(5;10,.2)                   = .2013 + .0881 + .0264                   = .3158c.  E[x! = np = 10(.2) = 2 for 1 week    for 4 weeks = 4(2) = 8d.  The probability of 6 or more breakdowns given p = .2, n = 10    is .0064.  Since the probability that this event occurred by    chance is so small, there appears to be some indication that    the breakdown rate has increased above .2.e.  1)  The breakdown of one machine is not related to the condi-        tion of any of the other machines;  in other words, machine        breakdowns are independent.    2)  The probability of a breakdown is the same for each of the        machines. `

35.

`A large TV retailer in San Francisco claims that 80 percent of allservice calls on color television sets are concerned with the smallreceiving tube.  Test this claim against the alternative PI =/= 0.80at ALPHA = 0.05 if a random sample of 222 calls on color televisionsets included 167 which were concerned with the small receiving tube. `
`Answer: Continue the null hypothesis that the proportion is .80.Two methods of solution:  1. Using the binomial distribution:         H(0): PI = .80         H(A): PI =/= .80                Z = ((167/222) - .80)/SQRT(PI*Q/n)                  = (.752 - .80)/SQRT(.00072)                  = -1.779  2. Using the normal approximation to the binomial:              Mean = n*PI = 222 * .8 = 177.6          Variance = n*PI*Q = 222*.8*.2 = 35.52         H(0): MU = 177.6         H(A): MU =/= 177.6                Z = (167 - 177.6)/SQRT(35.52)                  = -10.6/5.9599                  = -1.779The critical Z value for both of these cases is:                Z(ALPHA/2 = .025) = 1.96Since the two Z calculated values are less than the critical Z value,we continue the null hypothesis that the proportion is equal to .80. `

36.

`You, as a manufacturer,  can  use  a  particular  part  only  if  itsdiameter  is between .14 and .20 inches.  Two companies, A and B, cansupply you with these parts at comparable costs.  Supplier A producesparts whose mean is .17 and whose standard deviataion is .015 inches.However, supplier B produces parts whose mean is .16 inches and whosestandard  deviation  is  .012.   The diameters of the parts from eachcompany are normally distributed.  Which company should you buy fromand why? `
`Answer: For Supplier A:     Z = (X - MU)/SIGMA       = (.14 - .17)/.015       = -2and Z = (.20 - .17)/.015       = 2Area between Z = 2 and Z = -2 under the normal curve is .9544.  There-fore, 95.44% of the parts would be within .14 in. and .20 in.For Supplier B:     Z = (.14 - .16)/.012       = -1.67 and Z = (.20 - .16)/.012       = 3.33Area between Z = 3.33 and Z = -1.67 under the normal curve is .9520.Therefore, 95.20% of the parts would be within .14 in. and .20 in.Conclusion:  I would choose Supplier A by a hair. `

37.

`A lightbulb is selected randomly from a factory's monthly production.The bulb's lifetime (total hours of illumination) is a random variablewith exponential density function       f(x) = (1/MU)*(e**[-x/MU!)   if x >= 0            = 0                     if x < 0,where the fixed parameter MU is the mean of this distribution (MU > 0).a)  Derive the cumulative distribution function F(x).    Show that a random lifetime X exceeds x hours (x > 0) with    probability                  P(X > x) = e**(-1/MU)b)  Let M denote the smallest value in a random sample of n bulb    lifetimes  X(1), X(2), ..., X(n).    Show that P(M > x) = P(X(1) > nx).    HINT:  M > x if and only if X(1) > x and X(2) > x and ...           and X(n) > x.c)  Assume the mean lifetime MU = 700 hours.    Use a) and a table of the exponential function to evaluate    numerically    i)  the median lifetime x(.50),    ii)  P(X <= 70),    iii)  P(70 < X <= 700). `
`Answer: a)  F(X) = INT(X/0)((1/MU)*(e**[-t/MU!)dt)                       X         = -e**(-t/MU)!                       0         = 1/0 - [e**-X/MU)!    F(X) = [ 0;  x < 0           [ 1.0 - [e**(-x/MU)!; x >= 0    Prob (X>x) = 1.0 - F(X)               = 1.0 - [1.0 - [e**(-x/MU)!!               = e**(-x/MU)b)  Prob(M > x) = [Prob(X(1)>x)!*[Prob(X(2)>x)!*...*[Prob(X(n)>x)!                = [e**(-x/MU)!**n                = e**(-xn/MU)                = [Prob(X(1)>xn)!c)  i)  0.50 = Prob(X <= Median)             = F(x)             = 1.0 - [e**(-x/700)!        0.50 = e**(-x/700)        using a table of the exponential function        x/700 == .693        x == 485.1 hours    ii)  Prob(X<=70) = F(X=70)                     = 1.0 - [e**-70/700)!                     = 1.0 - 0.90484                     = 0.09516    iii)  Prob(70 < x <= 700) = F(x=700) - F(x=70)                              = [1.0-[e**(-700/700)!!-[1.0-[e**(-70/700)                              = [1.0 - .36788! - [0.09516!                              = 0.53696 `

38.

`A lightbulb is selected randomly from a factory's monthly production.The bulb's lifetime (total hours of illumination) is a random variablewith exponential density function      f(x) = [(1/MU)*(e**[-x/MU!)  if x >= 0             [ 0                   if x < 0,where the fixed parameter MU is the mean of this distribution (MU>0).a)  For an exponential distribution the standard deviation SIGMA = MU.    Let XBAR = (1/n)(X(1)+X(2)+...+X(n)) denote the average value in    a random sample of n bulb lifetimes.  Express E[XBAR! and VAR[XBAR!    in terms of MU.  If the mean MU = 700 hours and sample size n = 100,    then the statistic Z=(XBAR-700)/70 has approximately a normal    distribution with what mean and variance?b)  Describe a test of the null hypothesis H(0):  MU <= 700 against the    alternative hypothesis H(1):  MU > 700, using only the sample mean    XBAR.  If the desired significance level is ALPHA = .05 and sample    size n = 100, then indicate which numerical values of XBAR corre-    spond to this test rejecting H(0).    (Use the table of the standard normal distribution.)c)  If mean MU = 700 hours, then P(X > 2100) = .04979.  If instead    MU > 700, is P(X > 2100) larger or smaller than .04979? `
`Answer: a)  E[XBAR! = E[(1/n)*(X(1)+X(2)+...+X(n))!            = (1/n)*[E[X(1)+E[X(2)!+...+E[X(n)!!            = (1/n)*[n*E[X!!            = E[X!            = INT(INFNTY/0)(X*(1/MU)*e**[-x/MU!)dx)              (Integrating by parts, with                u = x       dv = (1/MU)(e**[-x/MU!)dx               du = dx       v = -e**[-x/MU!                             INFNTY            = -x*(e**[-x/MU!)!  - INT(INFNTY/0)(-e**[-x/MU!dx)                             0                   INFNTY            = -MU * e**[-x/MU!!                              0            = MU    E[x**2! = INT(INFNTY/0)((x**2)*(1/MU)*(e**[-x/MU!)dx)              by parts with,                u = (x**2)   dv = (1/MU)(e**[-x/MU!)dx               du = 2x dx     v = -e**[-x/MU!                                   INFNTY            = (x**2)*(-e**[-x/MU!)! -INT(INFNTY/0)((2x)*(-e**[-x/MU!)dx)                                   0            = -2*INT(INFNTY/0)((x*(-e**[-x/MU)dx)              by parts with                u = x      dv = -e**[-x/MU!dx               du = dx      v = mu*(e**[-x/MU!)                                   INFNTY            = -2*[x*MU*(e**[-x/MU!)!  - INT(MU*(e**[-x/MU!)dx)!                                   0                                    INFNTY            = -2(MU**2)*(e**[-x/MU!)!                                    0            = 2(MU**2)    VAR[XBAR! = VAR[(1/n)*(X(1)+X(2)+...+X(n))!              = [(1/n)**2!*[VAR[X(1)!+VAR[X(2)+...+VAR[X(n)!!              = [(1/N)**2!*[n*VAR[X!!              = (1/n)*(VAR[X!)              = (1/n)*[E[X**2!-(E[X!**2)!              = (1/n)*[2(MU**2)-(MU**2)!              = (MU**2)/n       Z = (XBAR-700)/70    E[Z! = (E[XBAR!-700)/70         = (MU-700)/70         = (700-700)/70         = 0/70         = 0    VAR[Z! = VAR[(XBAR-700)/70!           = [(1/70)**2! * VAR(XBAR)           = [(1/70)**2! * [(MU**2)/n!           = [1/4900! * [(700**2)/100!           = 1b)  test statistic:  Z = [XBAR-700!/[700/SQRT(n)!    critical region:  Any value of Z(calc) that lies beyond the Z(crit)           which is found in the standard normal table with ALPHA per           cent of the distribution beyond it.    with n = 100 and ALPHA = .05, Z(crit) = 1.645    Thus in order to reject H(0),    [XBAR-700!/[700/SQRT(100)! >= 1.645    XBAR >= (1.645*70) + 700    XBAR >= 815.15c)  It can be shown that a random lifetime X exceeds x hours (X>0)    with probability    P(X > x) = e**(-x/MU)    Therefore,    P(X > 2100) = e**(-2100/700)                = e**(-3)    Now if MU > 700, the exponent of e becomes less and looking at a    table of the exponential function it is evident that the probability    becomes smaller. `

39.

`In a given business venture a man can make a profit of \$1000 orsuffer a loss of \$500.  The probability of a profit is 0.6.  Whatis the expected profit (or loss) in that venture? `
`Answer: p = .6Expected profit = (.6*1000) - (.4*500)                = 600 - 200 = 400 `

40.

`The following probability distribution applies to the value of a stockduring the coming year:        VALUE        P(VALUE)         100           .46         150           .04         200           .20         250           .20         300           .10Compute the expected value of the stock.  What interpretation wouldyou give to this value? `
`Answer: E(V) = (100 * .46) + (150 * .04) + (200 * .20) + (250 * .20)        + (300 * .10)     = 46 + 6 + 40 + 50 + 30     = 172The \$172 is the average, or mean, of the distribution of stock values. `

41.

`Suppose that the probability that a salesman makes a sale to anycustomer is .4.  If each sale is worth \$100 in commissions and theevents of making a sale to two different customers are independent,what is his expected commission if he sees two customers on a parti-cular day? `
`Answer: If we let X = commissions that day, the probability distributionfor X is:        X  ^  p(X)    _________________        0  ^  .6 * .6 = .36     \$100  ^  (.6 * .4) + (.4 * .6) = .48     \$200  ^  (.4 * .4) = .16E(X) = (0 * .36) + (100 * .48) + (200 * .16)     = 48 + 32     = \$80 `

42.

`An investor wishes to buy a stock and sell it three months later.After much investigation he has narrowed his possible choices tofive different stocks and decides to pick one of these at random.The first stock has one chance in four of losing value.  Thesecond stock has one chance in three of losing value.  The thirdstock has two chances out of nine of losing value.  The fourthand the fifth stocks both have three chances out of ten of losingvalue.  What is the probability that the investor loses money onhis investment? `
`Answer: P(picking a particular stock) = .2Given:  First Stock:  P(losing) = .25       Second Stock:  P(losing) = .33        Third Stock:  P(losing) = .22       Fourth Stock:  P(losing) = .3        Fifth Stock:  P(losing) = .3Probability that the investor loses money on hisinvestment = E[P(losing)!E[P(losing)! = (.25*.2) + (.33*.2) + (.22*.2) + (.3*.2) + (.3*.2)             = .28 `

43.

`Suppose an investor buys a stock for \$100 per share with intentionsof selling it three months later.  At the end of three months he hasone chance in four of selling for \$80 per share, one chance in fourof selling for \$100 per share and one chance in two of selling for\$140 per share.  How much per share can the investor expect to makeon this stock when he sells it? `
`Answer: Let X = amount made by investor when he sells stock.  x ^ p(x)_____________-20 ^ .25  0 ^ .25 40 ^ .5E(X) = (-20*.25) + (0*.25) + (40*.5)     = -5 + 0 + 10     = 5The investor can expect to make \$5 on this stock when he sells it. `

44.

`Joe Pennyworth has a very rare 1919S-VDB penny.   He  is  consideringaccepting  a  firm  offer  of  \$8  for the penny or putting it up forauction at the local numismatic club.  His possible actions are:       a(1):  put the penny up for auction       a(2):  accept \$8 for the pennyAn estimate of the probability distribution for the sales price at theauction is given to Joe as:       Sales Price                Probability       -----------                -----------          \$ 6                         .10            7                         .20            8                         .30            9                         .30           10                         .10Joe has determined his utility function as:       Dollar Value                  Utility       ------------                  -------          \$ 6                          1.0            7                          2.5            8                          4.0            9                          4.5           10                          5.0a.  If Joe evaluates the problem by considering expected monetary values    what is his decision?b.  If Joe evaluates the problem by considering expected utilities, what    is his decision?c.  Plot the utility function.d.  Is Joe a risk preferrer? `
`Answer: a.  Expected profit if the penny is put up for auction:       E[a(1)! = [6*.10!+[7*.20!+[8*.30!+[9*.30!+[10*.10!               = .60 + 1.40 + 2.40 + 2.70 + 1.00               = \$8.10    Expected profit if he accepts the offer:       E[a(2)! = [8.00*1.00!               = \$8.00    Since E[a(1)! > E[a(2)!, he should put the penny up for auction.b.  Expected utility of putting the penny up for auction:       Exp. Ut. [a(1)! = [1*.10!+[2.5*.20!+[4*.30!+[4.5*.30!+[5*.10!                       = 3.65    Expected utility of accepting offer:       Exp. Ut. [a(2)! = [4*1.00!                       = 4    Since Exp. Ut.[a(2)! > Exp. Ut.[a(1)!, he should accept the offer of    \$8 for the penny.c.           ^     Utility ^             ^           5 +                                       *             ^                                   *           4 +                                *             ^           3 +             ^                           *           2 +                                   (NOTE:  Connect *'s             ^                                    with a smooth curve.)           1 +                       *             ^             ----+---+---+---+---+---+---+---+---+---+------>                 1   2   3   4   5   6   7   8   9  10                             Dollar Valued.  No, a risk avoider. `

45.

`A lot containing 12 parts among which 3 are defective is put on  sale"as  is"  at  \$10.00  per  part  with  no  inspection possible.  If adefective part represents a complete loss of the \$10.00 to the  buyerand  the good parts can be resold at \$14.50 each, is it worthwhile tobuy one of these parts and select it at random? `
`Answer: Expected return value of part = .75*(14.50) + .25(0) = 10.875Therefore, you expect to gain approximately \$.87 on each part you buy,and it is worthwhile to buy one selected at random. `

46.

`The Connecticut Daily Numbers game uses a selection procedurevery similar to the one described in the following paragraphfor the selection of random numbers.    There are 10 identical ping pong balls on which the    digits 0, 1, ..., 9 have been written.  After mixing    the balls thoroughly in a box, one is selected with-    out looking.  The digit written on the ball is record-    ed, and then the ball is put back in the box.  This    whole process of mixing, selecting, writing down a    digit, and returning the ball to the box is repeated    again and again.The first 100 digits selected by the Connecticut Lottery showedthe following distribution:digit:         0   1   2   3   4   5   6   7   8   9number ofoccurrences:  23   7   5   8  12  11   8   9   7  10Test the hypothesis that all 10 digits are equally likely.The descriptive level, DELTA, for the test satisfies:a.  DELTA < .01                c.  .025 < DELTA < .10b.  .01 < DELTA < .025         d.  .10 < DELTA `
`Answer: a.  DELTA < .01    H(O):  All digits are equally likely to occur.  (This defines    a Goodness of Fit Test for a uniform distribution.)    Expected frequency for each cell is 10.    CHISQ(calc) = [[(23-10)**2!+[(7-10)**2!+[(5-10)**2!+[(8-10)**2!+                  [(12-10)**2!+[(11-10)**2!+[(8-10)**2!+[(9-10)**2!+                  [(7-10)**2!+[(10-10)**2!!/10                = 22.6    The probability of obtaining CHISQ(calculated) = 22.6 is less than    .01.  (This DELTA value was obtained from Table of CHISQUARE dis-    tribution with 9 df.) `

47.

`Among twenty-five articles, nine are defective, six having only minordefects and three having major defects.   Determine  the  probabilitythat  an  article  selected at random has major defects given that ithas defects.a.  1/3b.  .25c.  .24d.  .08 `
`Answer: a.  1/3    P[MD/D! = (3/25)/(9/25)            = 3/9            = 1/3 `

48.

`Among twenty-five articles eight are defective, six having onlyminor defects and two having major defects.  Determine the pro-bability that an article selected at random has major defectsgiven that it has defects.(a)  .08              (c)  1/3(b)  .25              (d)  .24 `
`Answer: (b)  .25     P(MDD) = P(MD and D)/P(D)             = (2/25)/(8/25)             = .25 `

49.

`The following table shows the composition of employees at Dwinal'sInn.                    (FT)    ^     (PT)    ^                  Full-time ^  Part-time  ^  TOTAL              ------------------------------------   Waiters (W)        20    ^             ^    30              -------------------------------------Bartenders (B)              ^             ^              -------------------------------------     Cooks (C)        10    ^             ^    15              -------------------------------------     TOTAL                  ^      15     ^    50a)  Complete the above table.b)  From this table structure (form) another table showing all    the marginal and joint probabilities.c)  Find the following conditional probabilities:      i)  P(PTW) = ?     ii)  P(BFT) = ?    iii)  P(BC)  = ?     iv)  P(CPT) = ? `
`Answer: a)                    (FT)    ^     (PT)    ^                  Full-time ^  Part-time  ^  TOTAL              ------------------------------------   Waiters (W)      20      ^      10     ^   30              ------------------------------------Bartenders (B)       5      ^       0     ^    5              ------------------------------------     Cooks (C)      10      ^       5     ^   15              ------------------------------------     TOTAL          35      ^      15     ^   50b)                Joint      Marginal              ----------------------              ^  .4  ^  .2  ^  .6  ^              ----------------------       Joint  ^  .1  ^ 0.00 ^  .1  ^              ----------------------              ^  .2  ^  .1  ^  .3  ^              ----------------------    Marginal  ^  .7  ^  .3  ^      ^              ----------------------c)    i)  P(PTW) = P(PT INTRSCT W)/P(W)                  = .2/.6                  = .33     ii)  P(BFT) = P(B INTRSCT FT)/P(FT)                  = .1/.7                  = .143    iii)  P(BC)  = P(B INTRSCT C)/P(C)                  = 0.0/.3                  = 0     iv)  P(CPT) = P(C INTRSCT PT)/P(PT)                  = .1/.3                  = .33 `

50.

`A certain kind of job opening can be filled by hiring only  either  ahigh  school graduate or a college graduate.  In the past all hiringsfor the job have resulted in 80%  of  the  hirings  being  successfulaccording  to  the company's evaluation.  It is also known that amongall failures at the job, 40% have been high school  graduates,  whileamong  all successes only 30% have been high school graduates.  Usingthe percentages as probabilities, find the probability  that  if  thejob is given to a high school graduate it will be successful. `
`Answer: Using Bayes' Law:P(success) = .8      P(HSF) = .4      P(HSS) = .3   P(SHS) = [P(S)*P(HSS)!/[P(S)*P(HSS) + P(F)*P(HSF)!           = ((.8)(.3))/((.8)(.3) + (.2)(.4)) = .75or the population breaks down as follows:                           Successful     Failure                          ------------------------    High school graduate  ^    24%    ^      8%  ^    32%                          ------------------------    College graduate      ^    56%    ^     12%  ^    68%                          ------------------------                               80%          20%Therefore, Prob(successfulhigh school graduate) = 24/32                                                 = .75 `

51.

`Suppose in a 500 mile race there are 12 entries. 4 cars are to beplaced in each of three rows to start the race.  How many ways canthe first row of cars in the race be formed? `
`Answer: 4]*12C4 = 11880or12P4 = 11880 `

52.

`An ice cream store has three sizes of ice cream cones, small, medium,and large.  If four cones are randomly selected, one at a time,  whatis the probability that a small cone will be selected before threelarge cones are selected?a.  1/3      b.  2/3      c.  64/81      d.  9/16      e.  61/64 `
`Answer: c.  64/81    1 - P(no S selected) - P(LLLS) = ((2/3)**4) - ((1/3)**4)                                   = 1 - (16/81) - (1/81)                                   = 64/81    OR:    Sxxx    1/3                       27/81  !    -Sxx    2/3*1/3                   18/81  !    --Sx    2/3*2/3*1/3               12/81  !  64/81    MMMS    ((1/3)**4)                 1/81  !    MMLS    3*((1/3)**2)((1/3)**2)     3/81  !    MLLS    3*((1/3)**2)((1/3)**2)     3/81  ! `

53.

`A certain assembly consists of two sections, A and B, which are boltedtogether.  In a bin of 100 assemblies, 12 have only section A defective,10 have only section B defective, and 2 have both section A and sectionB defective.  What is the probability of choosing, without replacement,2 assemblies from the bin which have neither section A nor section Bdefective?a.  (76)**2/(100)**2b.  (98)**2/(100)**2c.  98(97)/[100(99)!d.  76(75)/[100(99)!e.  none of these `
`Answer: d.  76(75)/[100(99)!    # of sections without defectives = 100 - (12 + 10 + 2)                                    = 100 - 24 = 76    P(of no defectives) = (76/100)*(75/99) `

54.

`Given that each of three identical devices operating independentlyhas probability 3/4 of operating successfully, determine the pro-bability that exactly two of the three fail.(a)  3/64                    (c)  4/64(b)  9/64                    (d)  27/64 `
`Answer: (b)  9/64     P(2 failures and 1 success) = 3*(1/4)*(1/4)*(3/4)                                 = 3*(3/64)                                 = 9/64 `

55.

`The  life in months of service before failure of the color televisionpicture tube in 8 television sets manufactured by Firm A and  8  setsmanufactured by Firm B are as follows (arranged according to size):     Firm A:  25,  29,  31,  32,  35,  37,  39,  40     Firm B:  34,  36,  41,  43,  44,  45,  47,  48Let ETA(A) and ETA(B) denote the median service life of picture tubesproduced by the 2 firms. A confidence interval for ETA(B) - ETA(A) isbounded by the dth smallest and the dth largest of all differences ofB-  and  A-observations.  For confidence coefficient .99, we take dequal to:     (a)  9     (b)  14     (c)  15      (d)  17 `
`Answer: (a)  9 `

56.

`Prices of shares on the stock market are recorded to 1/8th of a dollar.We might then expect to find stocks selling at prices ending in:    0    1/8    1/4    3/8    1/2    5/8    3/4    7/8with about equal frequency.  On a certain day, 120 stocks showed thefollowing frequencies:    26    8    15    9    22    12    19    9.H(O):  P(0) = P(1/8) = P(1/4) = P(3/8) = P(1/2) = P(5/8) = P(3/4) =       P(7/8), or the distribution of final eighths is uniform.H(A):  the distribution of final eighths is not uniform.At significance level ALPHA = .10, the hypothesis being tested isrejected provided the test statistic is:a)  greater than  1.28.           d)  greater than 12.0.b)  greater than 14.7.            e)  smaller than  2.83.c)  smaller than  4.17. `
`Answer: d)  greater than 12.0.    CHISQUARE(critical, df=7, ALPHA=.10) = 12.0 `

57.

`Prices of shares on the stock market are recorded to 1/8th of a dollar.We might then expect to find stocks selling at prices ending in:    0    1/8    1/4    3/8    1/2    5/8    3/4    7/8with about equal frequency.  On a certain day, 120 stocks showed thefollowing frequencies:    26    8    15    9    22    12    19    9.H(O):  P(0) = P(1/8) = P(1/4) = P(3/8) = P(1/2) = P(5/8) = P(3/4) =       P(7/8), or the distribution of final eighths is uniform.H(A):  the distribution of final eighths is not uniform.In testing the above hypothesis, all expected frequencies equal:a)  12      b)  60      c)  125      d)  500      e)  none of these `
`Answer: e)  none of these    e = 120 * (1/8)      = 15 `

58.

`A manufacturer of floor polish conducted a consumer-preference experi-ment to determine which of the five different floor polishes wassuperior.  A sample of 100 housewives viewed five patches of flooringwhich received the five polishes.  Each housewife indicated the patchthat she considered superior in appearance.  The lighting, background,etc., were approximately the same for all five patches.  The resultof the survey was as follows:        Polish    A   B   C   D   E   TOTAL        Frequency 27  17  15  22  19   100a.  State the hypothesis of "no preference" in statistical    terminology.b.  State the test statistic used.c.  Test the hypothesis at ALPHA = .10 and draw the conclusion. `
`Answer: a.)  H(O):  P(A) = P(B) = P(C) = P(D) = P(E) = 1/5     H(A):  P(A) =/= P(B) =/= P(C) =/= P(D) =/= P(E)b.)  Use CHISQUARE with 4 dfc.)  CHISQUARE (calculated) = Sum [((O-E)**2)/E!                  A      B      C       D      E    TOTAL      O          27     17     15      22     19    100  E = n*p(i)     20     20     20      20     20    100    O-E           7     -3     -5       2     -1      0   (O-E)**2      49      9     25       4      1((O-E)**2)/E    2.45   .45    1.25    .20     .05   4.40CHI SQUARE (calculated) = 4.40CHI SQUARE (critical, df = 4, ALPHA = .10) = 7.78Therefore the data supports the null hypothesis at the .10level and a conclusion that no significant consumer-preference for floor polish has been found. `

59.

`A  market  research  firm  was  hired to test consumer preference fordifferent packages for some  soap.   Two  hundred  randomly  selectedhousewives  were  given  a  package  of  soap  wrapped in each of thefollowing colors:  red, white, blue, green.  After a month  in  whichthey  could  use  the  soap, they were given a free case of the colorpackage of their choice.  There were no markings to differentiate thepackages - just color - and the soap itself was the same.  Is there asignificant difference in the colors they selected?        Color Package        No. Housewives Choosing        -------------        -----------------------             red                         50             white                       75             blue                        30             green                       45 `
`Answer: Contingency table:    0     50     75     30     45    E     50     50     50     50CHISQ = ((50 - 50)**2)/50 + ((75 - 50)**2)/50 + ((30 - 50)**2)/50 +        ((45 - 50)**2)/50      = 0 + 12.5 + 8 + .5      = 21   df = (K - 1) = 3P(CHISQ(3) >= 21) < .001Reject H(O) at ALPHA = .10, .05, or .01.Conclude that there is a significant difference in the colors chosen. `

60.

`Horse-racing fans often insist that in a race around a circular trackthe  horses  in  certain  post positions have significant advantages.Post position 1 is nearest to the inside rail and post position 8  isfarthest to the outside.  Suppose we observed the results of races forone month of racing  at  a  track.   (Horses were randomly assigned topost positions.)  Results were as follows:     post position    1    2    3    4    5    6    7    8     no. of wins     29   15   18   25   17   10   15   11Use CHISQUARE to test whether there is any difference in number of wins.State the null hypothesis you are testing. `
`Answer: H(O):  No difference in number of wins per post position (uniform       distribution).Total number of wins = 140.Expected number of wins/position = 140/8 = 17.5CHISQUARE(calculated) = ((29 - 17.5)**2)/17.5 + ((15 - 17.5)**2)/17.5 +                        ((18 - 17.5)**2)/17.5 + ((25 - 17.5)**2)/17.5 +                        ((17 - 17.5)**2)/17.5 + ((10 - 17.5)**2)/17.5 +                        ((15 - 17.5)**2)/17.5 + ((11 - 17.5)**2)/17.5                      = 17.143ALPHA = .05 (or .10 or .01) usually;  df = 8 - 1 = 7.01 <= P(CHISQ(7)=17.143) <= .05Therefore, the conclusion is to reject H(O) at ALPHA = .05 or .10,and continue H(O) at ALPHA = .01. `

61.

`New York State Thruway Commission is examining lane usage on the bridgeleading to the Big Apple (Tappan Zee Bridge).  It is hypothesized that,during rush hours, traffic in vehicles/hour in the rightmost four lanesis in the ratio:          Lane      1      2      3     4                ^--------------------------^          LEFT  ^  11  ^  12  ^  10  ^  7  ^  RIGHT                ^__________________________^In essence this means that of the total traffic for the four lanes:     11/40 took lane 1     12/40 took lane 2     10/40 took lane 3      7/40 took lane 4A sampling of lane traffic for one day is as follows:          ^-----------------------------------^          ^  8200  ^  9000  ^  7350  ^  5200  ^          ^___________________________________^Can we conclude at ALPHA = .05 that traffic lane usage occurred inthe hypothesized ratios?(a)  Pick the most appropriate hypothesis test.  What is it?(b)  State the null and alternative hypotheses.(c)  Compute a test statistic.(d)  Indicate the critical value or values.(e)  Do you continue or reject H(O)?  What is your conclusion     relative to the question posed above? `
`Answer: (a)  CHISQUARE goodness of fit(b)  H(O): The lane usage occurs in the ratio of 11:12:10:7.     H(A): The lane usage is other than 11:12:10:7.(c)  observed  ^  8200  ^  9000  ^  7350  ^  5200  ^  29750     __________^________^________^________^________^_______     expected  ^8181.25 ^  8925  ^ 7437.5 ^5206.25 ^     CHISQUARE(calc) = .042 + .630 + 1.029 + .008 = 1.709(d)  CHISQUARE(.05,3) = 7.815(e)  Do not reject H(O) since CHISQUARE(calc) < CHISQUARE(crit).     Conclude that lane usage occurs in hypothesized ratios. `

62.

`The following is the number of cars produced in an auto plant.    MON    TUE    WED    THU    FRI    -------------------------------     20     25     25     20     10Test the null hypothesis at ALPHA = .01 that production does not dependon the day of the week. `
`Answer: n(i)    20    25    25    20    10-------------------------------------E[n(i)!    20    20    20    20    20CHISQUARE(calculated) = 7.5CHISQUARE(critical, df=4, ALPHA=.01) = 13.3Since CHISQUARE(calculated) < CHISQUARE(critical), we cannot rejectthe null hypothesis at ALPHA = .01. `

63.

`It is desired to see whether there is a relationship in tastes for anexpensive car and owning a tri-maran.   A  survey of  200 upper-classpotential purchasers of cars and tri-marans gave these responses:                       Want Expensive    Do Not Want Expensive    Totals                            Car                  CarWant Tri-maran              100                   40                140Don't Want Tri-maran         20                   40                 60Totals                      120                   80                200Specifically, it is desired to test H(O):  the desire for a tri-maranis independent of a desire for an expensive car.If H(O) were true, the estimate of the expected number of those who donot want either a tri-maran or an expensive car would be:a)  (140*120)/200        d)  (140*80)/200b)  (140*60)/200         e)  (120*60)/200c)  (60*80)/200 `
`Answer: c)  (60*80)/200    Expected Value = (80/200)(60/200)(200)                   = (80*60)/200 `

64.

`It is desired to see if there is a relationship in tastes for anexpensive car and owning a tri-maran.  A survey of 200 upper-upperclass potential purchasers of cars and tri-marans gave theseresponses:                  want expensive    don't want expensive    totals                       car                 carwant tri-maran         100                  40                140don't want              20                  40                 60tri-marantotals                 120                  80                200Specifically, it is desired to test H(O):  the desire for a tri-maran is independent of the desire for an expensive car.The contribution to the Chi-square statistic of the term:  desire tri-maran - and desire expensive car is:a)  ((100-120)**2)/120      d)  ((100-78)**2)/78b)  ((100-140)**2)/140      e)  ((100-80)**2)/80c)  ((100-84)**2)/84 `
`Answer: c)  ((100-84)**2)/84    Expected Value = (120/200)(140/200)(200)                   = 84    Contribution   = ((100-84)**2)/84 `

65.

`It is desired to see if there is a relationship in tastes for anexpensive car and owning a tri-maran.  A survey of 200 upper-upperclass potential buyers of cars and tri-marans gave these results:                  want expensive    don't want expensive    total                       car                 carwant tri-maran         100                  40                140don't want              20                  40                 60tri-marantotals                 120                  80                200Specifically, it is desired to test H(O):  the desire for a tri-maranis independent of the desire for an expensive car vs. H(1):  there isa relationship at level ALPHA = .20.Given that the appropriate normalized statistic is greater than 23and less than 30, one should ______ H(O):  independence since thevalue ______ is ______ than the correct cutoff point.a)  reject, 30, bigger      d)  continue, 23, biggerb)  reject, 23, bigger      e)  continue, 23, smallerc)  continue, 30, bigger `
`Answer: b)  reject, 23, bigger    CHISQUARE(critical, df = 1, ALPHA = .20) = 1.64    and if CHISQUARE(calculated) is in the interval (23,30)    CHISQUARE(calculated) > CHISQuARE(critical), which implies    that H(O) should be rejected. `

66.

`The life in months of service before failure of the color televisionpicture tube in 8 television sets manufactured by Firm B are as follows(arranged according to size):     Firm B:  34,  36,  41,  43,  44,  45,  47,  48Let ETA(B) denote the median service life of picture tubes produced bythe firm.  To test the hypothesis ETA(B) = 38.5 against the alternativeETA(B) =/= 38.5, the value of CHISQ(calculated) for the median testequals:     (a)  8     (b)  6     (c)  4      (d)  2 `
`Answer: (d)  2              ^ Above 38.5 ^ Below 38.5     -----------------------------------     observed ^      2     ^     6     ^     -----------------------------------     expected ^      4     ^     4     ^     -----------------------------------CHISQ = [[(2 - 4)**2! + [(6 - 4)**2!!/4 = 2 `

67.

`A car rental agency is in the process of deciding the brand  of  tireto  purchase  as  standard  equipment for their fleet. As part of thedecision process, they are interested in studying  the  treadlife  offive  competing  brands.  Based  on  testing, the research departmentdetermined that each  of  10  tires  of  each  brand  will  last  thefollowing  number of miles (in 1000's to the nearest 1000). Compute aCHISQ median test. Test the null hypothesis H(O): no difference amongtires, with ALPHA = .05.                Tire Brands               -------------      A      B      C      D      E     40     45     30     35     28     42     40     32     40     32     45     40     31     42     34     38     44     35     36     28     40     42     28     38     32     41     44     29     34     26     43     41     31     41     29     43     41     30     41     31     37     43     34     35     25     40     41     27     37     31 `
`Answer: MD(overall) = 37Observed:                  A     B     C     D     E                  -     -     -     -     -     above MD     9    10     0     5     0     below MD     0     0    10     4    10Expected:                 4.5    5     5    4.5    5                 4.5    5     5    4.5    5CHISQ(calculated) = 41.34CHISQ(ALPHA=.10, df=4) = 7.779CHISQ(calculated) > CHISQ(critical), therefore reject H(O) andconclude the samples are from populations with different medians. `

68.

`Test that there is no relationship between  performance  in  a company'straining program and ultimate success in the job. Use ALPHA = 0.01.  Thefollowing data is obtained from 400 samples of a company.                    PERFORMANCE IN TRAINING PROGRAM                         A          B          C                    ----------------------------------     SUCCESS    A   ^    63    ^    49    ^     9    ^     IN JOB         ----------------------------------                B   ^    60    ^    79    ^    28    ^                    ----------------------------------                C   ^    29    ^    60    ^    23    ^                    ---------------------------------- `
`Answer: H(O):  There is no relationship between performance in the training       program and success in the job.H(A):  There is a relationship between performance in the training       program and success in the job.                            A             B             C                    -------------------------------------------                A   ^  63(45.98)  ^  49(56.87)  ^   9(18.15)  ^  121                    -------------------------------------------                B   ^  60(63.46)  ^  79(78.49)  ^  28(25.05)  ^  167                    -------------------------------------------                C   ^  29(42.56)  ^  60(52.64)  ^  23(16.8)   ^  112                    -------------------------------------------                         152            188            60        400CHISQUARE = ((63 - 45.98)**2)/45.98 + ... + ((23 - 16.8)**2)/16.8          = 20.18Critical value = 13.3df = (3 - 1)(3 - 1) = 4Since CHISQUARE(critical) < CHISQUARE(calculated), reject the nullhypothesis and conclude that there is a relationship. `

69.

`In order to find out how viewing preferences of TV viewers changeover the years, networks conduct viewer surveys.  In such a survey,viewers of sports events were asked to name their favorite sport.The following table gives responses for the years 1960 and 1970.                  1960      1970                  ----      ----      football     150       250      baseball     250       150    basketball     100       100The null hypothesis tested by an appropriate CHISQUARE test is:a)  1970 viewers prefer football to baseball.b)  1960 viewers prefer baseball to football.c)  There have been no changes in viewing preferences between 1960    and 1970.d)  Viewing habits of TV watchers have changed between 1960 and 1970.e)  The number of sports viewers has remained the same over the years. `
`Answer: c)  There have been no changes in viewing preferences between 1960    and 1970. `

70.

`In order to find out how viewing preferences of TV viewers changeover the years, networks conduct viewer surveys.  In such a survey,viewers of sports events were asked their favorite sport.  The fol-lowing table gives responses for the years 1960 and 1970.                  1960      1970                  ----      ----      football     150       250      baseball     250       150    basketball     100       100Using the null hypothesis that there have been no changes in viewingpreferences between 1960 and 1970, the value of CHISQUARE for the giventable is:a)  less than 2.            d)  between 25 and 45.b)  between 2 and 10.       e)  greater than 45.c)  between 10 and 25. `
`Answer: e)  greater than 45.    Expected values:                  1960      1970                  ----      ----      football     200       200      baseball     200       200    basketball     100       100    CHISQUARE(calc) = SUM([(O-E)**2!/E)                    = ([(150-200)**2!/200) + ([(250-200)**2!/200) +                      ([(100-100)**2!/100) + ([(250-200)**2!/200) +                      ([(150-200)**2!/200) + ([(100-100)**2!/100)                    = (2500/200) + (2500/200) + 0 + (2500/200) +                      (2500/200) + 0                    = 10000/200                    = 50 `

71.

`In order to find out how viewing preferences of TV viewers changeover the years, networks conduct viewer surveys.  In such a survey,viewers of sports events were asked their favorite sport.  The fol-lowing table gives responses for the years 1960 and 1970.                  1960      1970                  ----      ----      football     150       250      baseball     250       150    basketball     100       100The network was interested in testing the null hypothesis that therehave been no changes in viewing preferences between 1960 and 1970.If the correct value of CHISQUARE is sufficiently high to rejectthe hypothesis being tested, then we can conclude that:a)  viewing habits have not changed over the 10-year span.b)  basketball is more popular in 1970 than in 1960.c)  both football and baseball have become more popular in 1970.d)  the appeal of football has increased and that of baseball has    decreased between 1960 and 1970.e)  the appeal of baseball has increased and that of football has    decreased between 1960 and 1970. `
`Answer: d)  the appeal of football has increased and that of baseball has    decreased between 1960 and 1970.    Since we reject H(O):  that there has been no change in viewing    habits, and from the table, we can see that more people preferred    football in 1970 than in 1960, that fewer people preferred base-    ball in 1970 than in 1960, and that there was no change in pre-    ference with regards to basketball, the above conclusion is    appropriate. `

72.

`Given the following data matrix:                        AUTOMOBILES                                                         CHEV.             CORVETTE     MUSTANG II     VW RABBIT     MONTE CARLO             --------    -----------     ---------     -----------OWNER'S AGEless than 40    21            143            36           28       ^ 228greater thanor equal to40              26             61            35           41       ^ 163               ----           ----          ----         ----        ---                47            204            71           69       ^ 391Test at ALPHA = .05  if  the  populations  of  cars  have  differentdistributions  of ages of persons owning them. If the null hypothesisis  rejected,  construct  confidence  intervals  about  the  proportiondifferences as a post-hoc test procedure. `
`Answer: a.  Expected Frequencies:              27.41     118.96     41.40     40.24              19.59      85.04     29.60     28.76    CHISQ(calculated) = 21.88    CHISQ(ALPHA=.05, df=3) = 7.815    CHISQ(calculated) > CHISQ(critical), therefore reject H(O) and con-    clude that the distributions are different.b.  P(1)     P(2)     P(3)     P(4)    ----     ----     ----     ----    .4468    .7010    .5070    .4058    P(i) - P(j) +/- SQRT(CHISQ(critical) * (p(i)q(i)/n +  p(j)q(j)/n))     Pairs     C.I.     -----     ----     1 - 2     -.2542 +/- .2216 *     1 - 3     -.0602 +/- .2619     1 - 4     -.0041 +/- .2615     2 - 3      .1940 +/- .1885 *     2 - 4      .2952 +/- .1879 *     3 - 4      .1012 +/- .1751    Conclude:  P(1) - P(2)               P(2) - P(3)  caused rejection of H(O)               P(2) - P(4) `

73.

`Frequency of repairs are being examined for two populations of cars,foreign and domestic.  Given the sample data below, can we concludethat the population distributions are the same at ALPHA = .10?         Frequency of Repairs/Year                  0     1 - 2     3 - 5     More than 5                  -     -----     -----     -----------Foreign Autos     6  ^    11   ^    11   ^       7Domestic Autos  100  ^    50   ^    22   ^      17 `
`Answer: Expected frequencies:                  0     1 - 2     3 - 5     More than 5                  -     -----     -----     -----------Foreign Autos   16.56 ^   9.53  ^  5.16  ^    3.75Domestic Autos  89.44 ^  51.47  ^ 27.84  ^   20.25CHISQUARE(calculated) = 19.439CHISQUARE(critical, ALPHA=.10, df=3) = 6.251Since CHISQUARE(calculated) > CHISQUARE(critical), reject H(0)and conclude that the distributions are not the same. `

74.

`Below are the results of an insurance survey to relate amount ofinsurance to income.          Amount of Insurance         IncomeFamily      (in Thousand \$)      (in Thousand \$)------    -------------------    ---------------   A              9                     10   B             20                     14   C             22                     15   D             15                     14   E             17                     14   F             30                     25   G             18                     12   H             25                     16   I             10                     12   J             20                     15Find RHO and TAU and test each for significance.(Note:  data has ties.) `
`Answer: a.  R(X)    R(Y)    (R(X)-R(Y)**2    ----    ----    -------------     1       1          0     2       2.5         .25     3       5          4     4       5          1     5       2.5        6.25     6.5     5          2.25     6.5     7.5        1     8       7.5         .25     9       9          0    10      10          0                       -----                        15    H(O):  X and Y are independent    H(A):  X and Y are correlated    RHO(crit) = .6364    RHO(calc) = 1 - ((6*15)/(10*99))              = .9091    Since RHO(calc) > RHO(crit), reject H(O) and conclude that    X and Y are correlated.b.  N(C)    N(D)    #Neither    ----    ----    --------      9       0        0      7       0        1      4       1        2      4       1        1      5       0        0      4       0        0      2       0        1      2       0        0      1       0        0      0       0        0     --       -        -     38       2        5    U(X) = (1/2) * (2*1) = 1    U(Y) = (1/2) * (2*1 + 3*2 + 2*1) = 5    TAU = (38 - 2)/SQRT(44)*SQRT(40)        = 36/41.952        = .8581    H(O):  X and Y are independent    H(A):  X and Y are correlated    T(calc) = 38 - 2 = 36    T(crit) = 21    Since T(calc) > T(crit), reject H(O) and conclude that    X and Y are correlated. `

75.

`The observed life, in months of service, before failure for the colortelevision picture tube in 8 television sets manufactured by Firm B areas follows (arranged according to size):Firm B:    34   36   41   43   44   45   47   48Let ETA(B) denote the median service life of picture tubes produced bythe firm.The point estimate of ETA(B) equals:a.  35        b.  43.5        c.  44        d.  33.5 `
`Answer: b.  43.5    n = 8    Therefore, the median equals the average of the two middle values.    Median = (43 + 44)/2 = 43.5    or any number between 43 and 44. `

76.

`The life in months of service before failure of the color televisionpicture tubes in 8 television sets manufactured by Firm A and 8 setsmanufactured by Firm B are as follows (arranged according to size):Firm A:    25   29   31   32   35   37   39   40Firm B:    34   36   41   43   44   45   47   48Let ETA(A) and ETA(B) denote the median service life of picture tubesproduced by the two firms.The S-interval with confidence coefficient .71 for ETA(A) is boundedby:a.  29 and 39      b.  36 and 47      c.  31 and 37      d.  41 and 45 `
`Answer: c.  31 and 37    GAMMA = .71        n = 8    From the Table of d-factors for Sign Test and Confidence Intervals    for the median, d = 3.  The confidence interval is bounded by the    d-smallest and d-largest sample observations.  Thus, the S-inter-    val about the median is bounded by the third smallest and third    largest sample observations, or 31 and 37. `

77.

`The life in months of service before failure of the color televisionpicture tube in 8 television sets manufactured by Firm A and 8 setsmanufactured by Firm B are as follows (arranged according to size):Firm A:   25   29   31   32   35   37   39   40Firm B:   34   36   41   43   44   45   47   48Let ETA(A) and ETA(B) denote the median service life of picture tubesproduced by the two firms.The W-interval with confidence coefficient .98 for ETA(A) is boundedby:a.  29 and 39    b.  36 and 47    c.  35 and 47.5    d.  27 and 39.5 `
`Answer: d.  27 and 39.5    n = 8    Using a table of critical values for the W-interval with ALPHA=.02,    d=2, the table of averages:       ^  25   29   31   32   35   37   39   40    --------------------------------------------    25 ^  25  [27!  28    29 ^       29   30    31 ^            31    32 ^    35 ^    37 ^                           37   38   38.5    39 ^                                39  [39.5!    40 ^                                     40    W-interval is 27 and 39.5. `

78.

`Explain which measure of central tendency is most useful when reportingan average income for persons employed by Beech Aircraft. `
`Answer: The median would be the most useful measure of central tendencywhen reporting an average income.  The distributionof income is positively skewed since there are relativelyfew people who earn a substantially high income.  Theseextreme values would affect the mean by inflating it.  Themedian, which simply indicates the point where half theobservations are above and half are below, wouldnot be affected by such extreme values and in this sensewould more accurately convey the "average" income. `

79.

`A college athlete, equally talented in baseball and football, comparesthe income potential in the two sports before choosing to specialize inone of them.  The data available for annual income from all sources isbelow:                     Mean          Median          90th percentileFootball players:    \$25,000       \$20,000         \$70,000Baseball players:    \$23,000       \$28,000         \$50,000a.  Give a one-sentence interpretation of the mean which indicates how    it can be used to help him to decide between the two sports.b.  Do the same for the median and the same for the 90th percentile.c.  Based on the data above, which sport would you suggest he choose?    Indicate why. `
`Answer: a.  The mean is the average dollar income (or expected income) and in-    dicates football is the  better  choice,  though  the  difference    between the two means is not great.b.  The median is the midpoint in terms of rankings and is substantially    higher for baseball.  The 90th percentile is the dollar value that    exceeds 90 percent of the salaries and indicates  football  is  the    best choice.c.  The distribution of salaries in football is skewed right, indicating    higher potential salary extremes -- a  riskier  but  higher - payoff    choice.  A risk-averse athlete might choose baseball, whose  left-    skewed salaries suggest higher "typical" salaries. `

80.

`The life in months of service before failure of the color televisionpicture tube in 8 television sets manufactured by Firm A and 8 setsmanufactured by Firm B are as follows (arranged according to size):Firm A:    25   29   31   32   35   37   39   40Firm B:    34   36   41   43   44   45   47   48Let ETA(A) and ETA(B) denote the median service life of picture tubesproduced by the two firms.You want to test the hypothesis ETA(A) = 38 against the alternativeETA(A) < 38.  The correct sign test statistic and its value is:a.  S(+) = 2    b.  S(-) = 2    c.  S(+) = 3    d.  S(-) = 3 `
`Answer: a.  S(+) = 2    Since we have H(A):  ETA(A) < 38, we expect fewer observa-    tions to be larger than the median, and the correct test    statistic is S(+).  Its value is:        S(+) = # observations > 38 = 2. `

81.

`A student organization surveyed food prices at 4 local food stores:                          Stores  Item                      Weight/volume       A      B      C      D  ----------------------------------------------------------------------  Apples                    per lb              .30    .30    .33    .45  Lettuce                   one head            .39    .25    .25    .39  Milk, homogenized         1/2 gal container   .84    .76    .81    .76  Eggs: fresh, grade A,     1 doz               .89    .83    .69    .93    large  Hamburger                 per lb             1.29    .99    .99   1.09  Frying chicken            cut up, per lb      .65    .46    .59    .69  Chicken noodle soup       10 3/4 oz can       .22    .19    .22    .19  White bread               1 lb loaf           .48    .59    .48    .33  Raviolios with meat       15 oz               .45    .41    .43    .35    sauce  Soda                      qt bottle           .38    .40    .37    .39  Coffee                    4 oz               1.39   1.31   1.29   1.23  Peanut butter             28 oz jar          1.19   1.16   1.17   1.09  Laundry soap              3 lb 1 oz           .89    .85    .81    .80You  may  want  to  compare  prices at stores C and D. An appropriatetwo-sample test can be based on either:     a.  the sign test or the median test.     b.  the Wilcoxon one- or two- sample test.     c.  the sign test or Wilcoxon one-sample test.     d.  the median test or Wilcoxon two-sample test. `
`Answer: c.  the sign test or Wilcoxon one-sample test. `

82.

`The observed life, in months of service, before failure for the colortelevision picture tube in 8 television sets manufactured by Firm B areas follows (arranged according to size):     Firm B:  34,  36,  41,  43,  44,  45,  47,  48Let ETA(B) denote the median service life of picture tubes produced bythe firm and assume the lifetimes have symmetric distributions.  Youwant to test the hypothesis ETA(B) = 38.5 against the alternativeETA(B) =/= 38.5  using the Wilcoxon signed rank test.  From thefollowing list, select the most reasonable test statistic:     (a)  W(+) = 2     (b)  W(+) = 5     (c)  W(-) = 5     (d)  W(-) = 2 `
`Answer: (c)  W(-) = 5      X(i)     D(i)     ]D(i)]     Rank     -----    -----     ------     ----      34      -4.5       4.5       3.5      36      -2.5       2.5       1.5      41       2.5       2.5       1.5      43       4.5       4.5       3.5      44       5.5       5.5       5      45       6.5       6.5       6      47       8.5       8.5       7      48       9.5       9.5       8W(-) = SUM(R(-)) = 3.5 + 1.5 = 5 `

83.

`Ten randomly selected cars of a specific year, make, and model andwith similar equipment, are subjected to an EPA gasoline mileagetest.  The resulting miles/gallon are:    24.6, 30.0, 28.2, 27.4, 26.8,    23.9, 22.2, 26.4, 32.6, 28.8Using the Wilcoxon Median Test, test the hypothesis that the populationmedian is 30 miles/gallon at the ALPHA = .10 level.  Construct a 90%confidence interval for the median. `
`Answer: Measurement     D(i)     ]D(i)]     Rank-----------     ----     ------     -----   24.6         -5.4       5.4        7   30.0          0         0          -   28.2         -1.8       1.8        2   27.4         -2.6       2.6        3.5   26.8         -3.2       3.2        5   23.9         -6.1       6.1        8   22.2         -7.8       7.8        9   26.4         -3.6       3.6        6   32.6          2.6       2.6        3.5   28.8         -1.2       1.2        1R+ =  3.5          ---> T = 3.5R- = 41.5Lower w = 9Upper w = (9*10)/2 - 9 = 36Since (T=3.5) < 9, we reject H(0):  median = 30.For the confidence interval, we need the 11th largest and smallestvalues, to be obtained from the following table:      ^ 32.6  30.0  28.8  28.2  27.4  26.8  26.4  24.6  23.9   22.2-------------------------------------------------------------------- 32.6 ^ 32.6  31.3  30.7  30.4  30.3  29.7  29.5  28.6  28.25  27.4 30.0 ^       30.0  29.4  29.1  28.7  28.4   --    --    --     -- 28.8 ^            [28.8! 28.5  28.1  27.8   --    --    --     -- 28.2 ^                    --    --    --    --    --   26.05 [25.2! 27.4 ^                          --    --    --   26.0  25.65  24.8 26.8 ^                                --    --   25.7  25.35  24.5 26.4 ^                                     26.4  25.5  25.15  24.3 24.6 ^                                           24.6  24.25  23.4 23.9 ^                                                 23.9   23.05 22.2 ^                                                        22.2Therefore, 90% C.I.:  from 25.2 to 28.8. `

84.

`The  life in months of service before failure of the color televisionpicture tube in 8 television sets manufactured by Firm A and  8  setsmanufactured by Firm B are as follows (arranged according to size):     Firm A:  25,  29,  31,  32,  35,  37,  39,  40     Firm B:  34,  36,  41,  43,  44,  45,  47,  48Against the two-sided alternative, the Wilcoxon (Mann Whitney) two-sample test has descriptive level:     (a)  .050     (b)  .010     (c)  .007     (d)  .004 `
`Answer: (c)  .007U(A) = 0 + 0 + 0 + 0 + 1 + 2 + 2 + 2     = 7U(B) = 64 - 7     = 57P(U(A) <= 7) = .007 `

85.

`The  life in months of service before failure of the color televisionpicture tube in 8 television sets manufactured by Firm A and  8  setsmanufactured by Firm B are as follows (arranged according to size):     Firm A:  25,  29,  31,  32,  35,  37,  39,  40     Firm B:  34,  36,  41,  43,  44,  45,  47,  48Suppose the data is ranked as one combined set.  The sum of the ranksR(B) for the B-observations equals:     (a) 36     (b)  43     (c)  57     (d)  93 `
`Answer: (d)  93          Table of Ranks:Firm A:   1     2      3     4     6     8     9     10Firm B:   5     7     11    12    13    14    15     16SUM(R(B)) = 5 + 7 + 11 + 12 + 13 + 14 + 15 + 16          = 93 `

86.

`An expert gave the following subjective ratings of the driving abilitiesof a group of two subjects.  Test the hypothesis  that according  to theexpert's ratings, women are better  drivers  than men.  (Use a non-para-metric test with ALPHA = .05.)   (NOTE:  higher  scores indicate  betterdrivers.)                Expert's Ratings     Male       7,  4,  2,  3,  12,  1,  14,  10,  10     Female     6,  13,  12,  10,  14,  7,  3,  11 `
`Answer: H(O):  Female drivers are worse or as good as male driversH(A):  Female drivers are better than male driversUsing the Mann Whitney-Wilcoxon test:U(M) = 2.5 + 1 + 0 +.5 + 5.5 + 0 + 7.5 + 3.5 + 3.5     = 24U(critical, onetail, ALPHA=.05,9,8) = 19Since U(M) > U(critical) continue H(O). Therefore sample evidence wasnot strong enough to indicate that females are  better  drivers  thanmales.Using Wilcoxon version of test:Ranks:Ratings  ^    1    2    3    4    6    7    10    11    12    13    14-----------------------------------------------------------------------M or F   ^    M    M   M,F   M    F   M,F  M,M,F   F    M,F    F    M,FRank     ^    1    2   3.5   5    6   7.5   10    12   13.5   15   16.5Sum of ranks for females = 3.5 + 6 + 7.5 + 10 + 12 + 13.5 + 15 + 16.5                         = 84 = T(F)  Sum of ranks for males = [(17)(18)/2! - 84                         = 69 = T(M)Converting to U statistic:                    U(M) = T(M) - [.5 * n(M) * (n(M) + 1)!                         = 69 - [.5 * 9 * 10!                         = 24Conclusion reached is the same as above. `

87.

`A large consulting firm hires a west coast university to provide anMBA program for its employees.  The basic statistics course is taughtat two locations of the firm.  After completion of the course, stan-dardized tests are given to the participating employees at each loca-tion.  Assume the distributions of test scores are symmetric for bothgroups.  The results are shown below.    Observation      Location A      Location B    -----------      ----------      ----------         1               65              60         2               74              72         3               77              66         4               82              75         5               70              78         6               78              65         7               84              --Test the hypothesis at ALPHA = .10 that the two samples came from thesame population, or equivalently that the populations have the samemedian scores. `
`Answer: H(O):  The two samples came from the same population.       (median(1) = median(2))H(A):  The two samples came from different populations.       (median(1) =/= median(2))Using Wilcoxon-Mann Whitney test statistic we find:          A      Rank       B      Rank         --      ----      --      ----         84      13        78      10.5         82      12        75       8         78      10.5      72       6         77       9        66       4         74       7        65       2.5         70       5        60       1         65       2.5                 ----              ----Ranked Sums:     59.0              32.0Smaller Rank Sum = 32;  (Note that some books give critical values                         for this sum.)T = 32 - (6 * 7)/2   ;  (Transforming to the Mann-Whitney U Statis-  = 11                   tic.)Critical Values:  lower U = 9                  upper U = (6 * 7) - 9                          = 33Since 9 <= (T=11) < 33, do not reject H(O).  We do not have sufficientevidence to claim a difference between the two populations. `

88.

`New employees of the ABC corporation are given a training program  toacquaint them with business procedures and principles.  Two groups often each are selected randomly from a large set of new employees. Thefirst  group  is  trained  using  Method  A,  and the second group istrained using Method B.  At the end  of  the  training  period,  eachgroup  is  given  the same test to determine how much information hasbeen assimilated.  The data are:     Method A        Method B     --------        --------        55              50        70              91        70              90        65              62        62              75        81              88        72              84        58              78        67              82        50              80Use ALPHA = .05 to test that the two training methods result in thesame amount of assimilated information. `
`Answer: X(i)         Rank    ----         ----     50          1.5     50*         1.5*     55*         3  *     58*         4  *     62*         5.5*     62          5.5     65*         7  *     67*         8  *     70*         9.5*     70*         9.5*     72*        11  *     75         12     78         13     80         14     81*        15  *     82         16     84         17     88         18     90         19     91         20 (* indicates Method A) S = SUM(R(X(i))) = 1.5 + 3 + 4 + 5.5 + 7 + 8 + 9.5 + 9.5 + 11 + 15                  = 74                T = S - n*(n - 1)/2 = 74 - 10*11/2                  = 19          lower w = 24          upper w = 10*10 - 24 = 76T < 24, reject H(0); conclude methods result in different scores. `

89.

`Eight names are selected at random from the subscriber list ofMagazine A, and eight additional names from the list of MagazineB.  The ages of the subscribers are determined and listed below(fictitious data):A:    18,  24,  35,  19,  20,  20,  40,  17B:    20,  30,  45,  38,  42,  34,  50,  22a.  Using appropriate 5-year age groups, prepare a stem & leaf    plot OR a histogram for each group.b.  Is there evidence (at the 5% level of significance) that the    two magazines appeal to different age groups?c.  Give two possible reasons for choosing the test you chose for    part b instead of some other test. `
`Answer: a.  STEM & LEAF PLOT:                    Magazines                     A     B                1 ^ 897 ^                2 ^ 400 ^ 02                2 ^     ^    Age         3 ^     ^ 04    Groups      3 ^ 5   ^ 8                4 ^ 0   ^ 2                4 ^     ^ 5                5 ^     ^ 0    HISTOGRAMS:                ^            Magazine A                ^              5 +                ^              4 +    Frequency   ^              3 +           _________                ^           ^   ^   ^              2 +           ^   ^   ^                ^           ^   ^   ^              1 +           ^   ^   ^       ---------                ^           ^   ^   ^       ^   ^   ^                ----+---+---+---+---+---+---+---+---+---+---+--->                    5  10  15  20  25  30  35  40  45  50  55                                      Age                ^            Magazine B                ^      5 +                ^              4 +    Frequency   ^              3 +                ^              2 +               -----   -----                ^               ^   ^   ^   ^              1 +               ^   ^   ^   ^----------------                ^               ^   ^   ^   ^   ^   ^   ^   ^                ----+---+---+---+---+---+---+---+---+---+---+--->                    5  10  15  20  25  30  35  40  45  50  55                                      Ageb.  Using the Mann-Whitney U Test:    U(A) = 0 + 2 + 4 + 0 + .5 + .5 + 5 + 0         = 12    Using a table of critical values for this test at 5% level    of significance, U(critical) = 14    Since U(observed) < U(critical), there is evidence that the    two magazines appeal to different age groups.c.  1)  Normality cannot be assumed.    2)  The sample sizes are not large enough to avoid necessity        for normality. `

90.

`Suppose  you  run  a  warehouse  that  stocks  replacement  parts forappliances.  You randomly sample orders  for  parts  for  replacementburner  units  for  two  brands (A and B) of electric stoves.  Over aperiod of 50 weeks you observe the following:        weekly demand            number of weeks         for burners           Brand A     Brand B        -------------          -------------------              0                    28         22              1                    15         21              2                     6          7              3                     1          0           over 3                   0          0Use the Kolmogorov-Smirnov test to determine if these two distributionsare the same. `
`Answer: F(A)      F(B)        D    ----      ----        -    .56       .44        .12    .86       .86       0    .98      1.00        .02   1.00      1.00       0   1.00      1.00       0P(D(50) >= .12) > .2Do not reject H(O) at ALPHA = .10, .05 or .01. `

91.

`Suppose we examine a random sample of subjects to see whether there is apreference for certain colors.  The colors  selected  are  green,  blue,brown, yellow, and black.  Theory suggests that people may prefer  thosecolors that are most commonly found in nature, so the colors  have  beenranked from least commonly  found  in  nature (black)  to most  commonlyfound (green).  We have recorded the number of people (n = 50) who ratedeach color as their favorite.  Our null hypothesis is one of no  specialpreference.  Perform a  Kolmogorov-Smirnov  test  to see whether the ob-served preferences fit our expected uniform distribution.     color          expected          observed     -----          --------          --------     black             10                 0     yellow            10                 5     brown             10                 0     blue              10                25     green             10                20 `
`Answer: S(n)(X)     F(0)(X)     D     -------     -------     -        0          .2       .2       .1          .4       .3       .1          .6       .5       .6          .8       .2      1.0         1.0       0Prob.(D(n=50) >= .5) < .01Reject H(O), that all colors are uniformly preferred, at ALPHA = .10,.05 and .01. `

92.

`You wish to compare four methods of displaying apples sold insupermarkets.  The question to be answered is:Does one of these methods (A,B,C, or D) provide greater daily applesales than another?In order to evaluate methods, a single display is used in a store for afull day.  Displays cannot be changed during the working day but can bechanged between days.  A store owner agrees to let you set up and meas-ure sales on Monday, Tuesday, Wednesday, and Thursday for each of 4consecutive weeks during October and November (i.e. a total of 16selling days).  Prior to this test period you obtain the followinginformation on apple sales where the same display has been usedeach day:Units of Apples Sold    Monday     Tuesday     Wednesday     ThursdayFirst Week                100        105           70           100Second Week               125        120           85           120Design an experiment to compare A,B,C, and D.  Explain your choice ofdesign. `
`Answer: The information available on apple sales using the same display (uniformtreatment) indicates one should not assume uniform results, since both1.  Day of week (Wednesday is noticiably worse) and2.  Week (second week had more sales)pear to affect the response.  Therefore, a Latin Square Design isappropriate to restrict randomization so that each display is testedan equal number of times on each day of the week and an equal numberof times during each week.  In this way, each display is equallyexposed to effects of week and day.The following assignment of treatments (from the program LSQPLN***) isone possible assignment since each display occurs on each day of theweek as well as once during each week.                Day of WeekWeek            1       2       3       4 1              C       D       B       A 2              D       C       A       B 3              A       B       D       C 4              B       A       C       D `

93.

`Suppose that you have been instructed to test 2 chemicals that aresaid to be mosquito repellants.  You are to compare these 4 treat-ment combinations:               Amount of Chemical A        Amount of Chemical B               ____________________        ____________________     T(11)              0                           0     T(12)              0                           +     T(21)              +                           0     T(22)              +                           ++ indicates the recommended rate of application for each chemical.An experimental unit consists of an arm of a subject.  Each subjectprovides 2 arms that are considered comparable, but subjects maydiffer in mosquito appeal (i.e., experimental units are homogeneouswithin incomplete blocks of size 2).          ^----^----^          ^----^----^          ^    ^    ^          ^    ^    ^          ^    ^    ^          ^    ^    ^          ^    ^    ^          ^    ^    ^          ^----^----^          ^----^----^     Arm    1    2               1    2            Person 1             Person 2A)  How would you  group treatments so  that main effect differences    between rates for chemical  A  would be measured precisely, while    main effect differences between rates for B are measured less pre-    cisely?  (Indicate clearly which treatments must be applied to the    same person).B)  How would you group treatments so that  A  and B main effects are    measured most precisely while AB interaction is estimated less pre-    cisely? `
`Answer: A)  To obtain high precision  for  A  at the expense of low precision    for B, let B be the basis for grouping (B is main plot factor) ==>    Group 1:  T(11), T(21); Group 2:  T(12), T(22).  Both group members    are assigned to the same person.B)  Let AB be the basis for grouping, i.e., group members will consist    of treatments receiving (+) signs for AB interaction, or those re-    ceiving (-) signs.          T(11)    T(12)    T(21)    T(22)   A        +        +        -        -   B        +        -        +        -  AB        +        -        -        + ==>  Group 1:  T(11) and T(22)                                              Group 2:  T(12) and T(21) `

94.

`An investigator wished to study the effect of an operator on theperformance of a machine.  He could arrange to have each of fouroperators run the machine five times.  A response measurement could berecorded each time the machine was used.  How many experimental unitswill he have if-a.  He randomly selects an operator, has him run the machine five times,    then selects another operator, etc.?b.  He identifies 20 turns for running the machine and randomly assigns    operators to turns subject to the requirement that each operator    perform five times?c.  He forms five groups of four turns and randomly and independently    assigns operators within each group of four? `
`Answer: a.  4 : an experimental unit is a set of 5 turns or time of running the    machine.b.  20: an experimental unit is a turn.c.  20: an experimental unit is a turn even though turns have been    arranged in groups. `

95.

`An investigator suspected that the time required to pour a mold in afoundry was longer after lunch than before lunch.  He proposed comparingthese two conditions or treatments by measuring times needed to pour amold.Which of the following schemes meets the requirement of independence ofresponse among experimental units?  Why or why not?a.  The foundry was visited one day and four times were recorded before    lunch and four times after lunch.b.  The foundry was visited on four days.  On each day one time was    recorded before lunch and one after lunch. `
`Answer: Scheme b comes close to meeting the requirement of independence ofresponse among experimental units.  With this scheme other factors thatmay affect time to pour a mold, either before or after lunch such as:the weather conditions or the handing out of paychecks during lunchetc. would be balanced, and, therefore, the responses would not all betied together with that common influencing factor. `

96.

`A company is interested in adopting a new type of machine.  Sinceit is an expensive model they are not willing to adopt it unlessthey are fairly positive it will decrease the production time perunit.  If MU(S) is the mean production time per unit under the stan-dard machine and MU(n) is the mean production time per unit underthe new machine, the appropriate pair of hypotheses to test is:(a)  H(O): MU(S) = MU(n) vs. H(A): MU(S) < MU(n)(b)  H(O): MU(S) >= MU(n) vs. H(A): MU(S) < MU(n)(c)  H(O): MU(S) = MU(n) vs. H(A): MU(S) =/= MU(n)(d)  H(O): MU(S) = MU(n) vs. H(A): MU(S) > MU(n) `
`Answer: (d)  H(O): MU(S) = MU(n) vs. H(A): MU(S) > MU(n)     The hypothesis we do not wish to reject unduly is MU(S) = MU(n).     This we call H(O).  The alternative we wish to investigate and     not accept unduly is MU(S) > MU(n). `

97.

`A home owner claims that the current market value of his house is atleast \$40,000.  Sixty real estate agents were asked independently toestimate the house's value.  The hypothesis test that followed endedwith a decision of "reject H(O)".  Which of the following statementsaccurately states the conclusion?a)  The home owner is right, the house is worth \$40,000.b)  The home owner is right, the house is worth less than \$40,000.c)  The home owner is wrong, the house is worth less than \$40,000.d)  The home owner is wrong, the house is worth more than \$40,000.e)  The home owner is wrong, he should not sell his home. `
`Answer: c)  The home owner is wrong, the house is worth less than \$40,000. `

98.

`In an experiment to determine the effect of a utilization review (UR)procedure on the length of hospitalization, patients were paired by ageand sex and one member of each pair was randomly assigned to a ward thathad UR, the other to a regular ward.  The results of the 30 pairs aresummarized below.  The hospital wants to know at the 5% significancelevel if the length of hospitalization is different for those experien-cing a utilization review.                       Regular Ward    UR Ward    Paired DifferenceMean Length of Stay        5.64          4.29          1.35S.D.                       2.75          2.41          3.41a.  Set up the appropriate confidence interval to evaluate this    experiment.b.  How can the hospital use the confidence interval to make a    decision about the effectiveness of the procedure? `
`Answer: a.  C.I. = DBAR +/- [t*S(D)/SQRT(n)!         = 1.35 +/- [2.045*3.41/SQRT(30)!         = 1.35 +/- 1.27    .08 <= MU(D) <= 2.62b.  Because MU = 0 is not included in this interval, we know the    null hypothesis would be rejected in a significance test at    the 5% level.  Therefore, we can conclude that the procedure    is effective. `

99.

`A shirt manufacturer  is  considering  the  purchase  of  new  sewingmachines.  If  MU(1) is the average number of shirts made per hour byhis old machines and MU(2) is the  corresponding  average  number  ofshirts  per  hour  for  the  new  machine,  he wants to test the nullhypothesis MU(1) = MU(2) against a suitable alternative.a.  What alternative hypothesis should he use if he does not want to    buy the new machine unless it is proven superior?b.  What alternative hypothesis should the manufacturer use if he    wants to buy the new machine (which has nice features) unless    the old machines are actually superior? `
`Answer: a.  MU(2) > MU(1)b.  MU(1) > MU(2) `

100.

`A manufacturer who produces auto tires wished to compare the wearingqualities of two types of tires, A and B.  To make the comparison, atire of Type A and one of Type B were randomly assigned and mountedon the rear wheels of each of 5 automobiles.  The automobiles werethen operated for a specific number of miles and the amount of wearwas then recorded for each tire.     Auto        A          B     ___________________________      1         10.6       10.2      2          9.8        9.4      3         12.3       11.8      4          9.7        9.1      5          8.8        8.3Test H(0):  MU(A) = MU(B) against H(A):  MU(A) =/= MU(B) withALPHA = .05:     a)  using 2-sample test.     b)  using paired-sample test.What are your conclusions?  Explain. `
`Answer: a)  XBARA = 51.2/5 = 10.24    XBARB = 48.8/5 = 9.76    S(A)**2 = (SUMX(A)**2 - ((SUMX(A))**2)/n)/(n - 1)            = (531.22 - (51.2**2)/5)/4            = 1.73    Similarly:  S(B)**2 = 2.44    (NOTE:  The following is the same as pooling since the sample sizes            are equal.  However, the proper df = 5 + 5 - 2 = 8.)    S(XBARA - XBARB) = SQRT((S(A)**2)/n(A) + (S(B)**2)/n(B))                     = SQRT((1.73/5) + (2.44/5))                     = .913    Critical values of:     (XBARA - XBARB)   = (MU(A) - MU(B)) +/- t(crit)*S(XBARA - XBARB)                       = 0 +/- (2.31*.913)                       = +/- 2.109    DM(CALC) = XBARA - XBARB             = 10.24 - 9.76             = .48    Since .48 is neither less than -2.109 nor more than 2.109, we    cannot reject H(O).b)    A        B        D      D - DBAR = d    d**2    10.6      10.2      .4         -.08       .0064     9.8       9.4      .4         -.08       .0064    12.3      11.8      .5          .02       .0004     9.7       9.1      .6          .12       .0144     8.8       8.3      .5          .02       .0004                       ----                   -------                       2.4                    .0152    DBAR = 2.4/5 = .48    S(D) = SQRT((SUM(d)**2)/n - 1)         = SQRT(.0152/4)         = .0616    S(DBAR) = S(D)/SQRT(n)            = .0616/SQRT(5)            = .0276    t(calc) = DBAR/S(DBAR)            = .48/.0276            = 17.41    t(crit) = 2.776 for n - 1 = 4 df    Since 17.41 > 2.776, we reject H(O) and conclude that the means    are different.    H(O) was rejected in (b) but not in (a) since the test for related    samples is stronger than that for independent samples. `

101.

`We  are  interested in the wearing capabilitites of tires.  We obtainGood-day and Good-poor Tires and 9 racing cars (and  also  the  trackused  for  the  Indianapolis  500  Race).   We  put  Good-day  on theleft-hand side of the car (front  and  rear)  and  Good-poor  on  theright-hand  side  of the car (front and rear). We then allow the carsto complete the 500 miles at  a  (relatively)  safe  speed  and  thenmeasure the wear (in millimeters) per tire.      Car No.        Good-day        Good-poor      -------        --------        ---------        77              17              16        82              18              19        92              17              12        41              16              13        17              15              14        22              14              12        18              10              10        23              18              15        43              17              13a.  All the advertising literature claims equality between Good-day and    Good-poor.  Can you present evidence to disprove this claim?  Use a    significance level of 5%.b.  Comment on the validity of this experimental set-up. `
`Answer: a.  Let d be the difference in wear between tires on the left-hand side    compared to tires on the right-hand side.  We are interested in    testing the hypothesis that the mean (dBAR) of such different scores    is zero.         H(0):  MU(dBAR)  =  0         H(1):  MU(dBAR) =/= 0    The problem is obviously a paired experiment set-up and therefore we    perform a t-test on the difference.        Car No.         GD         GP        d(i)        d(i)**2        -------         --         --        ----        -------          77            17         16          1            1          82            18         19         -1            1          92            17         12          5           25          41            16         13          3            9          17            15         14          1            1          22            14         12          2            4          18            10         10          0            0          23            18         15          3            9          43            17         13          4           16                                             ----        -------  SUM                                  18           66           dBAR = [SUM(d(i))!/[9!                = 18/9                = 2        S(d)**2 = [SUM([d(i)-dBAR!**2)!/[n-1!                = [[SUM(d(i)**2)!-[n*(dBAR**2)!!/[n-1!                = [[66!-[9*4!!/[8!                = 3.75       t(calc.) = [dBAR-0!/[SQRT([S(d)**2!/n)!                = [2-0!/[SQRT([3.75!/9)!                = 3.098       t(critical, .05, two-tailed, 8 df) = 2.306    Since t(calculated) > t(critical), reject H(0).   Therefore we can    claim on the basis of this test that the tires are not equal.b.  The Indianapolis race track has an oval shape with highly-banked    curves.  Since the cars travel in only one direction, only the    inner tires would wear appreciably.  There are many other drawbacks    to the design, but this one is catastrophic. `

102.

`A random sample of 625 boxes taken from the output of a box makingmachine was inspected for flaws.  It was found that 500 of the boxeswere free from flaws.  To three decimals, what is the upper limit ofthe 0.99 confidence interval estimate of the proportion ofacceptable boxes being produced?a.  .8 + 1.96*SQRT(.16/625)b.  .8 + 2.576*(.16/625)c.  .8 + 1.96*(.16/625)d.  .8 + 2.576*SQRT(.16/625) `
`Answer: d.  .8 + 2.576*SQRT(.16/625)    C.I. = p +/- Z(ALPHA/2)*SQRT(pq/n)            p = 500/625 = .8,  q = 125/625 = .2    C.I. = .8 +/- Z(.005)*SQRT(.8*.2/625)    Upper limit = .8 + 2.576*SQRT(.16/625) `

103.

`Suppose that I'm interested in a herd of Unicorns where  mean  weightis  unknown,  but  the  population  variance  is known to be 100. TheDirector of Unicorns has given me \$1200 for a year and directed me tosubmit monthly reports on the average weight of this herd. It costs \$5per unicorn to weigh a beast  and  the  Director  will  let  me  havewhatever money is left over after I pay expenses. He won't tolerate aconfidence interval greater than +/- 5 and becomes very upset with aninterval that doesn't contain the true mean. (He has a special infor-mant who reports intervals  that  don't  contain  the  true populationmean.)  What shall I do?  Why?  Will I make any money?  Will I survivethe year? `
`Answer: The steps taken would probably depend upon several factors, including:(1) What action the Director would take if you failed to    produce a confidence interval that contains the mean,(2) What chance of failure you are willing to risk, and(3) How you expect the mean weight of the herd to change through-    out the year.If you feel it is necessary to estimate the mean every month with  asmuch  confidence  as  possible and forego making any money, you woulduse a sample size = 20 and find a confidence interval  with  a  97.5%confidence level (see calculations below).     n = (1200/12)/5       = 20     Z = 5/(10/SQRT(20))       = 2.24     P(-2.24 < Z < 2.24) = .975In order to make money, you could  either  decrease  your  confidencelevel, not make a new estimate every month, or ask the director for araise. `

104.

`A random sample of 500 accounts receivable is selected from the 4,032accounts that a firm has, and the sample mean is found to be\$242.30.  The sample standard deviation is computed to be \$3.20.Set up a .99 confidence interval estimate of the population mean.How do you interpret the meaning of this interval? `
`Answer: Using t with ALPHA = .01 and df = 499,     C.I. = XBAR +/- (t) (S/SQRT(n))          = 242.30 +/- (2.576) (3.20/SQRT(500))          = 241.93 to 242.67.99% of the time that this procedure is used to calculate an interval,the resulting interval will contain MU.  This interval may or may notinclude MU. `

105.

`A survey on consumer finances reports that 33 per cent of a sampleof 2,600 spending units expected good times during the next 12 months.Assume that a simple random sample was used in the study.  Set up a.95 confidence interval estimate of the population proportion ofspending units expecting good times. `
`Answer: p = .33n = 2600Z(ALPHA=.025) = 1.96Stand. error of proportion = SQRT(pq/n)                           = SQRT((.33*.67)/2600)                           = .009C.I. = .33 +/- (1.96 * .009)     = from .312 to .348 `

106.

`In a random sample of 200 television viewers in a certain area, 95had seen a certain controversial program.  Construct a 0.99 confi-dence interval for the actual percentage of television viewers inthat area who saw the program. `
`Answer: .475 +/- 2.58 SQRT((.475*.525)/200) = .475 +/- .091 `

107.

`Taking a random sample from its very extensive files, a water companyfinds that the amount owed in 16 delinquent accounts have  a mean  of\$16.35 and a standard deviation of \$4.56.a.  Use these values to construct a .98 confidence interval for the    average amount owed on all delinquent accounts.b.  If Mr. Blackwater, the company president, claims the delinquent    accounts have a population mean of \$19.01, how could you quickly    respond to him based on part a above (also after explaining that    you were using a 2% ALPHA level)? `
`Answer: a.  C.I. = XBAR +/- [t*S(XBAR)!; t(ALPHA=.01, one-tail, df=15) = 2.602         = 16.35 +/- [2.602*(4.56/SQRT(16))!         = from 13.384 to 19.316b.  According  to  the confidence  interval  found  in part a, Mr.    Blackwater's  estimate  of  \$19.01  is a possible estimate for    the population mean.  It should be pointed out that we are 98%    confident  that  such  a confidence interval would contain the    population mean. `

108.

`A floor manager of a large department store is studying the buyinghabits of the store's customers.  Suppose he assumes that monthlyincome of these customers is normally distributed with a standarddeviation of 500.  If he draws a random sample of size N = 100 andobtains a sample mean of YBAR = 800,a)  Find a .95 confidence interval for the true population mean.b)  Do you think that it would be quite unreasonable for    the true population mean to be \$600?  Explain. `
`Answer: a)  C.I. = YBAR +/- Z*SIGMA(YBAR)         = 800 +/- 1.96*(500/SQRT(100))         = 800 +/- (1.96*50)         = 800 +/- 98    Therefore, 702 < MU < 898b)  Yes, based on the above confidence interval, we would reject    the hypothesis that MU = 700 (at ALPHA = .05). `

109.

`A cigarette manufacturer tests tobaccos of two different brands of cig-arettes for nicotine content and obtains the following results:Brand A:  4  6  5  2  3Brand B:  7  8  5  9  6a.  Using ALPHA = .01, would you say that there is a difference in    the averages?b.  Set up 99% confidence limits on the difference.  Does this    answer agree with your answer in part a?  Why or why not? `
`Answer: a.  For supplier A:        XBARA = 20/5 = 4        S(A)**2 = (SUM(X**2))-((SUM(X))**2)(n)/(n-1)                = (90-(400/5))/4 = 2.5    For supplier B:        XBARB = 35/5 = 7        S(B)**2 = (225-(1225/5))/4 = 2.5    S(XBARA-XBARB) = SQRT((S(A)**2/n(A)) + (S(B)**2/n(B)))                   = SQRT((6.25/5) + (6.25/5))                   = 1.58    (XBARA-XBARB)(crit) = MU +/- t(crit)*S(XBARA-XBARB)                        = 0 +/- (3.36)*1.58                        = +/- 5.31    (XBARA-XBARB)(calc) = 4 - 7 = -3    Since -3 is neither less then -5.31 nor more than 5.31, we cannot    reject H(0).b.  (XBARA - XBARB) - t(crit)*S(XBARA - XBARB) < MU(A) - MU(B)        < (XBARA-XBARB) + t(crit)*S(XBARA-XBARB)    -3 - 5.31 < MU(A) - MU(B) < -3 + 5.31        -8.31 < MU(A) - MU(B) < 2.31    Note that H(O):  MU(A) - MU(B) is included in this interval, and    thus we are led to the  same  conclusion,  (i.e. continuation of    H(O)), as in part (a). `

110.

`A brewery producing beer has a number of specifications for quality.Among these standards is the requirement that the degree of hop likeflavor should be a value of 8.0.The production of the brewery consists of a large number of batches.It's possible for differences to arise  between batches,  so we willregard each batch as a different population.   We will  consider thehoppiness of each batch as a normally distributed variable with meanand variance unknown.From each batch you  can remove 6 samples for hoppiness.   For  eachbatch you are to:    a.  set confidence limits for the batch (population) mean, MU;    b.  determine if these limits are consistent with the require-        ment that hoppiness is a value of 8.0.1.  Outline the procedure to be followed in setting confidence limits    where the probability of the interval calculated including MU is:        a.  90%        b.  99%2.  Apply the procedure outlined to this set of sample values:  13,    11, 9, 14, 8, 11.  Is this sample data consistent with the speci-    fication of hoppiness = 8.0 when the probability level used is:        a.  90%        b.  99%3.  Do these results suggest any weakness in the procedure used?  If    so, what? `
`Answer: 1.  a.  To set 90% confidence limits including MU, we need XBAR, the        sample standard deviation, s, the sample size, n, and a t        value for ALPHA = .1.        Use the formula:            XBAR +/- (t, ALPHA/2)(s/SQRT(n)) with t(ALPHA/2) = 2.015.    b.  Use the same procedure as in 1a,  but use t(ALPHA/2) = 4.032.2.  XBAR = 11    Standard deviation = 2.28    n = 6    a.  11 +/- (2.015)(2.28/SQRT(6))            = 11 +/- (2.015)(.93)            = 11 +/- 1.876            = 9.124 to 12.876        This is inconsistent with the specification of hoppiness = 8.0.    b.  11 +/- (4.032)(2.28/SQRT(6))            = 11 +/- (4.032)(.93)            = 11 +/- 3.753            = 7.247 to 14.753    This is consistent with the specification of hoppiness = 8.0.3.  Ths basic weakness seems to be in using a procedure that produces a    confidence interval consistent with a wide range  of values,  which    makes it difficult to detect departures from MU = 8.  This situation    is exaggerated when the 99% level is used.  In addition, the sample    size used seems small relative to the variability measured. `

111.

`Willy the Waiter claims that the amount in tips that he receives percustomer on any given day is normally distributed, but that theaverage and variability change from day to day (in response to changesin Willy, the weather, the menu, etc.).  So far today, Willy hasreceived the following amounts in tips (in dollars):1, 2, .5, 1, 1.5, 0.a.  Write a model for Willy's tips for today.  Define all terms.b.  Set 90% confidence limits for the next tip that he will receive. `
`Answer: a.  Y(J) = MU + EPSILON(J)    where:    Y(J)      :  tips received from customer J    MU        :  mean value for tips for the day    EPSILON(J):  deviation of customer J's tips from the mean value,                 assumed to be a value of a normally distributed ran-                dom variable with a mean of zero and a variance of                 SIGMA**2.b.  Since the mean and variance of the population have been estimated    the variance of the predicted, or future, value involves both the    variance of the individuals (S**2) and the variance involved in    estimating the mean ((S**2)/n).    S(Predicted value)**2 = S**2 + (S**2)/n                          = (S**2) + (1 + 1/n)    Finding XBAR = 1 and S = .707,    we get:    S(Pred. Value) = SQRT(.5(1 + 1/6))                   = SQRT (.583)                   = .764    Using ALPHA = .10 and df = 5, t = 2.132    C.I. = 1 +/- 2.015(.764)         = -.54 to 2.54 `

112.

`The following triangle test is sometimes used to identify tasteexperts.  In the case of wine tasting, a test subject is presentedwith three glasses of wine, two of one kind and a third glass ofanother wine.  The test subject is asked to identify the singleglass of wine.  A test subject who merely guesses has a 1 chancein 3 of identifying the single glass correctly.  An expert winetaster should be able to do much better.  Let K stand for the num-ber of correct identifications made by a test subject in 10 inde-pendent triangle tests.Assume that a test subject is accorded the title "expert winetaster" if the number K of correct identifications is suffi-ciently high to reject the hypothesis P = 1/3 at significancelevel ALPHA = .02 i)  A Type II error has the consequence that:    a.  an experienced wine taster is accorded the title.    b.  a person who is guessing is not accorded the title.    c.  an experienced wine taster is not accorded the title.    d.  a person who is guessing is accorded the title.ii)  The power of the test when P = .8 equals:    a.  .38      b.  .61      c.  .88      d.  .97 `
`Answer: i)  c.  an experienced wine taster is not accorded the title.ii)  c.  .88    Under H(O):  P = 1/3    P(X >= 7) = P(7) + P(8) + P(9) + P(10)              = .016 + .003 + 0 + 0             == .02    Therefore, X = 7 is the critical value.    Under H(A):  P = .8    Power = 1 - BETA;  (BETA = P(X < 7))          = P(X >= 7)          = .201 + .302 + .268 + .107          = .878 `

113.

`A toaster manufacturer produces two models, A and B.  Experienceindicates that 3% of the customers buying model A will make a claimon their warranty.  In a sample of 400 owners of model B (whosewarranties have expired), 16 made a claim on their warranty.  Themanufacturer wishes to determine if the models differ in the numberof claims.(a)  Determine the value of the test statistic.  What are your     conclusions?(b)  Let PI(B) be the probability that a buyer of model B will make a     claim on the warranty.  For what values of the test statistic     would you reject H(0) when testing H(0): PI(B) = .03 against     H(A): PI(B) =/= .03?  (Let ALPHA = .10). `
`Answer: (a)  For A:  PI(A) = .03      For B:  n = 400, p(B) = 16/400 = .04     H(0):  PI(A) = PI(B) = .03, or PI(B) - PI(A) = 0     H(A):  PI(B) =/= .03     Z(calculated) = (.04 - .03)/SQRT(.03*.97/400)                   = .01/.00853                   = 1.1724     Z(critical, twotail, ALPHA = .05) = +/- 1.96     Therefore continue H(0), and assume at the 95% confidence level     that the models are the same in the number of claims.(b)  Using ALPHA = .10:     Z(critical, twotail) = +/- 1.645 `

114.

`Past experience shows that, if a certain machine is adjusted properly, 5percent of the items turned out by the machine are defective.  Each daythe first 25 items produced by the machine are inspected for defects.If three or fewer defects are found, production is continued withoutinterruption.  If four or more items are found to be defective, produc-tion is interrupted and an engineer is asked to adjust the machine.After adjustments have been made, production is resumed.  This proce-dure can be viewed as a test of the hypothesis p = .05 against thealternative p > .05, p being the probability that the machine turnsout a defective item.  In test terminology, the engineer is asked tomake adjustments only when the hypothesis is rejected.Interpret the quality control procedure described above as a test ofthe indicated hypothesis.  A Type I error results in:a.  a justified production stoppage to carry out machine adjustments.b.  an unnecessary interruption of production.c.  the continued production of an excess of defective items.d.  the continued production, without interruption, of items that    satisfy the accepted standard. `
`Answer: b.  an unnecessary interruption of production. `

115.

`The workers in a large plant  have  complained  through  their  unionnegotiators    that    they   are   being  underpaid.    Both   sides(labor-management) agree that the mean wage for plant workers in thisindustry is about \$3.75 per hour with a standard deviation of \$.84 perhour. (i)  Does the fact that a random sample of 49 workers from this plant      gave mean wage \$3.54 provide sufficient evidence to indicate the      plant is paying an inferior wage?  Use ALPHA = .05.(ii)  State what a Type I and a Type II error would be for this problem. `
`Answer: (i)  H(O):  MU >= 3.75      H(A):  MU <  3.75      n = 49      XBAR = 3.54      MU = 3.75      SIGMA = .84      SIGMA(XBAR) = [.84!/[SQRT(49)!                  = .12      Z(calc) = [3.54 - 3.75!/[.12!              = -1.75      Z(crit, ALPHA=.05, one-tailed) = -1.645      Since Z(calc) < Z(crit), reject H(O).  Therefore, sample evidence      is strong enough to suggest that workers are being underpaid.(ii)  Type I Error:  A type I error will occur when the null hypothesis                     is rejected on the basis of the sample information                     and in reality the null hypothesis  is  true.   In                     this case, the conclusion  based  on   the  random                     sample would be that the workers are being  under-                     paid when actually they are not.  So, the workers'                     complaint would be erroneously supported.     Type II Error:  A type II error will occur when the null hypothesis                     is not rejected on the basis of the sample informa-                     tion and in reality the null hypothesis is  false.                     In this case, the conclusion  based  on the random                     sample would be that the workers are not being un-                     derpaid when actually they are being underpaid.  So                     the employers' position of just wages would be er-                     roneously supported. `

116.

`The daily yield of a chemical manufactured in a chemical plant,recorded for n = 49 days, produced a mean and standard deviationequal to XBAR = 870 tons and s = 21 tons, respectively.Test H(0):  MU = 880 against H(A):  MU < 880, using ALPHA = .05.Calculate BETA for H(A):  MU = 870. `
`Answer: S(M) = S/SQRT(n) = 21/7 = 3XBAR(crit) = MU(M) + Z(crit)S(M)           = 880 + ((-1.65)*3)           = 875.05Since 870 < 875.05, we reject H(0) and conclude that MU < 880.BETA is the probability of committing a type II  error.  Using theabove decision rule and given H(A), it is the probability that XBARis greater than XBAR(crit) = 875.05 when MU = 870.BETA(H(A): MU = 870) = P(XBAR > 875.05);    Z = (875.05 - 870)/3                     = P(Z > 1.683)    ;      = 1.683                     = .046 `

117.

`We are interested in finding the linear relation between the numberof widgets purchased at one time and the cost per widget.  Thefollowing data has been obtained:X:  Number of widgets purchased-- 1   3   6  10   15Y:  Cost per widget(in dollars)--55  52  46  32   25Suppose the regression line is YHAT = -2.5X + 60.  We compute theaverage price per widget if 30 are purchased and observe:a.  YHAT = -15 dollars; obviously, we are mistaken; the prediction    YHAT is actually +15 dollars.b.  YHAT = 15 dollars, which seems reasonable judging by the data.c.  YHAT = -15 dollars, which is obvious nonsense.  The regression    line must be incorrect.d.  YHAT = -15 dollars, which is obvious nonsense.  This reminds us    that predicting Y outside the range of X values in our data is a    very poor practice. `
`Answer: d.  YHAT = -15 dollars, which is obvious nonsense.  This reminds us    that predicting Y outside the range of X values in our data is a    very poor practice. `

118.

`A  management  analyst  is  studying  production  in  an   electroniccomponent assembly factory.  Workers individually assemble componentsinto final products.  Each worker is given 100 sets of components  toassemble  each  day.    Employees  clock  out at the time they finishassembling the 100 sets into final products.  The analyst has averagehourly  production  rates  for  each  individual worker.  Which meanshould be used to calculate the overall average production per laborhour?a.  arithmetic meanb.  geometric meanc.  harmonic mean `
`Answer: c.  harmonic mean    The harmonic mean is properly used since the numerator in each    worker's average production is 100 units and the denominator,    hours worked, varies. `

119.

`A   management  analyst  is  studying  production  in  an  electroniccomponent assembly factory.  Workers individually assemble componentsinto  final  products.  Workers assemble as many units as they can inan eight hour day.  The analyst has average hourly  production  ratesfor  each  individual  worker.   To  calculate  the factory's overallaverage hourly production per worker, which mean should be used?a.  arithmetic meanb.  geometric meanc.  harmonic mean `
`Answer: a.  arithmetic mean    The arithmetic mean of individual average hourly production rates    is the same  as  total production divided by total hours worked,    since individual rates are daily production divided by eight for    every employee. `

120.

`Nelly finds that 30 out of 100 randomly selected persons  walking  indowntown  Cincinnati  believe  that  the government should spend moremoney on health care; a similar survey in the suburbs  shows  that  20out  of 100 persons believe in more government spending. At the ALPHA= 0.05 level, are these data evidence that  the  people  in  downtownCincinnati  believe  to a different extent than people in the suburbsthat the government should spend more money on health care? `
`Answer: Z = (P(1) - P(2) -0)/(S)   where S is the standard error for the difference of proportions   S = SQRT((.3(.7)/100) + (.2(.8)/100)) = .06   Z(calc.) = ((.3 - .2) - 0)/.06 = 1.67   Z(crit.) = +/- 1.96 for ALPHA = .05Since 1.67 < 1.96; Continue the H(O). `

121.

`You are to conduct an opinion poll to determine  the  opinions  of resi-dents of a given community about a projected industrial development pro-gram.  How large a sample should you select to estimate  the  proportionof adult residents favoring the projected development?  Make all assump-tions necessary to determine the sample size, and justify these  assump-tions. `
`Answer: Necessary assumptions are:    Level of significance = .05, then Z = 1.96    Tolerable error, e = .05    Assume X has a binomial distribution with a population of size N       and PI = .50, where X is the number of adult residents favor-       ing the projected development.The first two assumptions are arbitrary values that will depend uponthe preference of the researcher.  The choice of the proportion  hasbeen set at maximum variability  since  no other  information on theproportion was available.    n = (Z**2)(PI)(1 - PI)/(e**2)      = (1.96**2)(.5)(.5)/(.05**2)      = 384.16     == 385 `

122.

`a.  For each of the samples listed below obtain:    1.  a mean    2.  a variance, and    3.  a standard deviation    Each sample was randomly obtained from the production of the hot    dog manufacturer listed.           Company               Dog Length(inches)             A                    5,5,5,5,5             B                    6,5,5,5,4             C                    9,9,5,1,1             D                    9,5,5,5,1             E                    9,5,5,5,5,5,5,5,5,1             F                    9,9,9,4,4,3,3,3,3,3b.  Given that the price per hot dog is the same for all manufacturers,    whose hot dogs would you buy?  Why? `
`Answer: a.  Company          Mean          Variance          St. Dev.         (SUM X(i))/n=XBAR   S**2=(SUM(X-XBAR)**2)/n-1     S=SQRT(VAR.)       A           25/5=5              0                 0       B           25/5=5         2/4=1/2=.5         SQRT( 1/2 )=.707       C           25/5=5         64/4=16            SQRT( 16 )=4       D           25/5=5         32/4=8             SQRT(8)=2.83       E           50/10=5        32/9=3.55          SQRT( 3.55)=1.89       F           50/10=5        70/9=7.78           SQRT(7.78)=2.79b.  This question may have a variety of answers.  The decision woulddepend on the purpose.  If it was important to have as littlevariability as possible when selling 5 inch hot dogs, company Awould be best since it has the least variability.  However, if youcould profit from selling hot dogs in a variety of lengths, companyF might prove best since it shows a lot of variability and produces hotdogs ranging from 9 to 3 inches in length. `

123.

`Two workers on the same job show the following results over a longperiod of time.                                           Worker       Worker                                              A            B-------------------------------------------------------------------Mean time of completing the job (minutes)    30           25Standard deviation (minutes)                  6            4a.  Which worker appears to be more consistent in the time he    requires to complete the job?  Explain.b.  Which worker appears to be faster in completing the job?    Explain. `
`Answer: a.  Worker B appears to be more consistent in the time he requires    to complete the job, since he has a smaller variance.b.  Worker B appears to be faster in completing the job, since he    has a smaller mean.  (You could actually test this.) `

124.

`Suppose the manager of a plant is concerned with the totalnumber of man-hours lost due to accidents for the past 12months.  The company statistician has reported the mean numberof man-hours lost per month but did not keep a record of thetotal sum.  Should the manager order the study repeated toobtain the desired information?  Explain your answer clearly. `
`Answer: No--the estimate that he would get using the mean number permonth would most likely be accurate enough, without having togo to the extra expense of another study.  Presumably the meannumber of hours lost per month is equal to the total numberof hours lost divided by 12, so it's not difficult tocalculate the total. `

125.

`A large health screening program that will have 36 clinics needs topurchase scales for the clinics.  A manufacturing firm has available36 scales on which the same 180 pound man was weighed.  The variancein his weight on the 36 scales was .07 (lb**2).  The screening programwill buy the scales if the variance is not significantly greater than.05 at the 1% significance level.a.  What test statistic would you use to test the null hypothesis that    the true variance in weights on the new scale is .05?  Set up the    computations.b.  What are the null hypothesis, alternative hypothesis, and critical    region of such a test? `
`Answer: a.  CHISQUARE = (n-1)*(S**2)/(SIGMA**2)b.  H(O):  SIGMA**2 = .05    H(A):  SIGMA**2 > .05    Critical Region:  CHISQUARE(df=35, ALPHA=.01, one-tail) = 57.34 `

126.

`Suppose that the variable measured using a random sample is annualincome.  (Suppose that it and all other items were measured accurately.)Explain what it is that these two models have to say about income.1.  Y(I) = MU + EPSILON(I)2.  Y(I,J) = MU(J) + EPSILON(I,J)    Where J = 1 indicates a Democrat                                           J = 2 indicates a Republican `
`Answer: Model 1 states that annual income can be described by a single popula-tion having a single mean and standard deviation.  Individual incomesconsist of a common mean plus random variation.Model 2 states that description of annual income may require 2 popula-tions, one for Democrats and one for Republicans.  It provides for thepossibility of different population means for income and may also pro-vide for different standard deviations.NOTE:  This is a case where it would probably not be advisable to       assume that EPSILONS are normally distributed. `

127.

`A manufacturing company operates 12 plants that are regarded asabout the same in all important respects.  This company decidesto try a new safety program.  The new program is randomlyassigned to 6 plants while the old program is continued at theother 6.  Number of man-hours lost per plant per month, weremeasured in each plant following completion of the safetyprograms.  Results were:New Program:  46, 41, 16, 11, 58, 61Old Program:  92, 65, 10, 24, 46, 51a.  Write a model for number of man hours lost.b.  Fill out an ANOVA table corresponding to this model.c.  Was there a change in accident rate that was detectable atthe 10% level? `
`Answer: a.  Y(I,J) = MU(I) + EPSILON(I,J) or MU + TAU(I) + EPSILON(I,J)Where:Y(I,J) is manhours lost in plant J under program IMU(I) is population mean for time lost under program I.orMU is population mean andTAU(I) is used to indicate effect of program I defined as adeviation from MU.and EPSILON(I,J) indicates a random element associated with the Jthplant using program I.  These random elements are normallydistributed with mean = 0 and variance = SIGMA**2b.  Source        df          Source       df    Total          6          Total         6    Mean                      General     Program 1     1    or     mean         1    Mean                      Corrected     Program 2     1           total        5    Error          4          Programs      1                              Error         4c.  Old Program:          New Program:    XBAR(1) = 48              XBAR(2) = 38.83    S(1) = 29.18              S(2) = 21.03    n = 6                     n = 6Testing:H(0):  XBAR(1) = XBAR(2)H(A):  XBAR(1) =/= XBAR(2)We first must test for homogeneity of variance orH(0):  SIGMA(1) = SIGMA(2)H(A):  SIGMA(1) =/= SIGMA(2)F(calculated) = [S(1)**2!/[S(2)**2! = 851.6/442.17                 = 1.926 with 5 EPSILON 5 dfF(critical) 5.05 with ALPHA = .05, df = 5 EPSILON 5Since F(calculated) is less than F(tabled), there is evidence at the 5%level to continue the null hypothesis of homogeneity of variance.The variance should now be pooled:S(p)**2 = ((5) (851.6) + (5) (442.17))/10        = 646.89and finally find the standard error of the difference between meansS(XBAR(1) - XBAR(2)) = SQRT ((S(p)**2)(1/n(1) + 1/n(2)))                     = 14.68Now using the two-tailed t test with ALPHA = .10, df = 10 we testthe null hypothesis about the meanst(calculated) = ((48-38.83)-0)/14.68 = .6245t(criticals)  = -1.812 EPSILON 1.812Continue the null hypothesis that there was no change in the accidentrate.  Since t(calculated) is greater than the smaller t(critical) butless than the larger t(critical) at the 90% confidence level. `

128.

`In the attached Table 1, results for the routine measurement ofnickel in a steel standard are reported.  This determination was madedaily over a long period of time to establish a quality controlprogram.In Table 2, the data have been plotted as a tally sheet ofindividual values.  Clearly, a grouped tally sheet would be moreeffective in revealing the pattern of variation in these data.Perform the following --(a)  Set up a grouped tally sheet and histogram.  A cell interval of     0.05% is recommended.  List the frequency, cumulative frequency     and relative cumulative frequency for each cell.(b)  Calculate the mean and standard deviation (use coding) by both     the ungrouped and the grouped procedures.  Compare results.(c)  What is the mode -- comment -- is it meaningful?(d)  What is the median?(e)  Calculate the standard deviation of the mean.(f)  Plot an ogive.  Plot the data on normal probability paper.  Is it     reasonable to assume a normal distribution?  If so, estimate the     standard deviation and mean and compare wih the calculated values.     Estimate the percentage of values outside of the limits 4.88 to     5.21 and compare with the actual percentage.Table 1.  Results of Daily Determination of Nickel in a Nickel                           Steel StandardDate     % Ni        Date     % Ni        Date     % NiMar. 6   4.95        Apr. 17  4.96        May  29  5.03     7   5.02             18  4.79             30  5.08     8   5.17             19  5.06             31  5.20     9   5.08             20  5.03        June 1   5.11     10  4.92             21  4.95             2   4.95     11  4.94             22  5.10             3   4.95     13  5.22             24  5.05             5   5.00     14  4.96             25  5.30             6   4.92     15  5.05             26  5.24             7   5.16     16  5.02             27  5.00             8   5.14     17  5.14             28  5.08             9   5.02     18  5.00             29  5.04             10  5.14     20  5.07        May  1   4.97             12  5.02     21  4.83             2   4.86             13  4.97     22  5.11             3   5.07             14  4.96     23  4.99             4   4.90             15  5.26     24  4.98             5   5.22             16  5.11     25  5.26             6   5.07             17  5.15     27  4.88             8   5.31             19  4.98     28  5.01             9   5.05             20  5.15     29  4.98             10  5.16             21  5.00     30  5.21             11  5.02             22  5.14     31  5.15             12  5.18             23  4.98Apr. 1   5.00             13  4.90             24  5.03     3   5.00             15  5.20             26  5.01     4   5.10             16  5.08             27  4.97     5   5.03             17  5.19             28  5.12     6   4.97             18  5.16             29  4.98     7   4.89             19  4.88     8   5.12             20  4.99     10  5.27             22  4.92     11  5.09             23  5.17     12  5.13             24  5.01     13  4.93             25  5.02     14  4.93             26  5.06     15  5.04             27  5.03Table 2.  Frequency Table and Tally Sheet for the Data                       in Table 1Ni Conc.,    Tally    Frequency      Ni Conc.,    Tally    Frequency  % (y)      Marks       (f)           % (y)      Marks       (f)  4.79       X            1            5.05       XXX          3  4.80                                 5.06       XX           2  4.81                                 5.07       XXX          3  4.82                                 5.08       XXXX         4  4.83       X            1            5.09       X            1  4.84                                 5.10       XX           2  4.85                                 5.11       XXX          3  4.86       X            1            5.12       XX           2  4.87                                 5.13       X            1  4.88       XX           2            5.14       XXXX         4  4.89       X            1            5.15       XXX          3  4.90       XX           2            5.16       XXX          3  4.91                                 5.17       XX           2  4.92       XXX          3            5.18       X            1  4.93       XX           2            5.19       X            1  4.94       X            1            5.20       XX           2  4.95       XXXX         4            5.21       X            1  4.96       XXX          3            5.22       XX           2  4.97       XXXX         4            5.23  4.98       XXXXX        5            5.24       X            1  4.99       XX           2            5.25  5.00       XXXXXX       6            5.26       XX           2  5.01       XXX          3            5.27       X            1  5.02       XXXXXX       6            5.28  5.03       XXXXX        5            5.29  5.04       XX           2            5.30       X            1                                       5.31       X            1 `
`Answer: a)  (If available, consult file of graphs and charts that could not be     be computerized.)  Cell          Cell            Cum    Rel CumMidpoints    Boundaries    f     f        f                4.775  4.80                     1     1       0.01                4.825  4.85                     2     3       0.03                4.875  4.90                     8    11       0.11                4.925  4.95                    14    25       0.25                4.975  5.00                    22    47       0.47                5.025  5.05                    15    62       0.62                5.075  5.10                    12    74       0.74                5.125  5.15                    13    87       0.87                5.175  5.20                     7    94       0.94                5.225  5.25                     4    98       0.98                5.275  5.30                     2   100       1.00                5.325                         ___                         100b)  ungrouped YBAR = 504.99/100 = 5.0499 == 5.05    ungrouped S(Y) = SQRT[(2551.3039 - 2550.1490)/99!                   = SQRT(0.01166)                   = 0.108 == 0.11    Grouped and coded by:  Y = 0.05d + 5.05      Cell    Midpoint      d      f      f*d      f(d**2)      4.80       -5      1      -5         25      4.85       -4      2      -8         32      4.90       -3      8      -24        72      4.95       -2     14      -28        56      5.00       -1     22      -22        22      5.05        0     15       0          0      5.10       +1     12      +12        12      5.15       +2     13      +26        52      5.20       +3      7      +21        63      5.25       +4      4      +16        64      5.30       +5      2      +10        50                                ___       ___                       sum(fd) = -2    sum(f*d**2) =                                          448    dBAR = (sum(fd))/n = -2/100 = -.02    YBAR = (0.05)(-.02) + 5.05 = 5.049 == 5.05    S(d) = SQRT[((448 - 2**2)/100) / 99! = SQRT(4.525) = 2.127    S(Y) = (2.127)(0.05) = 0.106 == 0.11c)  5.00 or 5.02 - not meaningful because no single value occurs    with sufficient frequency.d)  Median is average of 50th and 51st observations -          (5.03 + 5.03)/2 = 5.03e)  S(YBAR) = S(Y)/SQRT(n) = 0.108/SQRT(100) = 0.0108 == 0.011f)  Estimates graphically should compare closely.    (If available, consult file of graphs and charts that could not be     computerized.)    Actual percentage outside = 11%.    Graphical estimate should be within about 2% of this. `

129.

`A coffee dispensing machine provides servings that have a populationmean of 6 ounces and a population standard deviation of .3 ounces.If the difference is measured between randomly chosen cups (e.g.the 7th minus the 15th, the 22nd minus the 29th, etc.), thedistribution of differences will have a mean of ______ and astandard deviation of ______. `
`Answer: a.  MU = 0b.  SIGMA = SQRT(.09/1 + .09/1) = .424 `

130.

`The closing prices of two common stocks were recorded for a periodof 15 days.  The means and variances were:       Y(1)BAR = 40.33,          Y(2)BAR = 42.54,       S(1)**2 = 1.54,           S(2)**2 = 2.96Do these data present sufficient evidence to indicate a difference invariability of the two stocks for the populations associated with thetwo samples?  [Assume stock 1 is normally distributed with mean = MU(1)and variance = SIGMA(1)**2 and stock 2 is normally distributed with mean= MU(2) and variance = SIGMA(2)**2; ALPHA = 5% and S(i)**2 =SUM(j=1,n(i))([(Y(ij)-Y(i)BAR)**2!/[n(i)-1!), i=1,2.! `
`Answer: H(0): SIGMA(1)**2  =  SIGMA(2)**2H(A): SIGMA(1)**2 =/= SIGMA(2)**2F(calc) = [larger variance! / [smaller variance!        = [2.96! / [1.54! = 1.922 F(crit., df=14,14, ALPHA= .05, one-tail) = 2.48Since our calculated F value is less than our tabled F value, we do notreject (continue) the null hypothesis that  the  variances for  the twopopulations are the same. `

131.

`Once upon a time there was a king who proclaimed that a properkingdom should not have great differences in wealth.  One day heinstructed his wizard to randomly sample his kingdom so that hecould assess the distribution of wealth.So the wizard did this -1.  He randomly selected 100 people, found their income, and wrote    down the mean for the group of 100;2.  He randomly and independently repeated this process over and    over again;3.  He truthfully reported to the king, "I have repeatedly taken    average wealth in the kingdom of 100 subjects and find that the    average wealth is 10 units and variance of those averages is 1    unit.  Further, those means are normally distributed."a.  What is mean wealth of individuals in the kingdom?b.  What is the variance for individual wealth in the kingdom?c.  Why did the wizard report on means based on samples of 100? `
`Answer: a.  10 unitsb.  SIGMA**2 = (n) (SIGMA(XBAR)**2)             = 100 * 1             = 100c.  To conceal the variability that would be obvious if he reported    on individuals. `

132.

`HEADLINE:   MPG for Gas Guzzler skyrockets over                MPG for Econ Scooter]Data: in miles per gallon      Gas      Econ      Guzzler  Scooter      -------  -------1964     4       251968     5       301972     8       351976    16       40        100]                       GPercent    ]Increase   ]Over     75]Prior      ]Time       ]             GPeriod   50]           ]           ]         25]    G           ]    E           ]             E        E           -------------------------------------------              1968     1972     1976        (To complete the graph connect the three G points with         straight lines to relate the performance of Gas Guzzler.         Similarly, connect the three E points to show the trend         for Econ Scooter.)Even though the above graph is correct, explain how it has led tothe misleading headline. `
`Answer: The headline is misleading in the sense that it implies that mpg isbeing compared for the two vehicles.  Only upon inspection of the datacan one see that Econ Scooters have a substantially higher mpg, whiletheir rate of increasing mpg has not been as great.  The graph accurate-ly indicates the rate of increase in mpg, but the headline is comparingactual mpg, which is quite different. `

133.

`Suppose that a report contains this graph:                  ^                  ^    Annual Income ^    (thousands of ^     \$ per year)  ^                  ^               50 +                             *                  ^                   *                  ^                  ^                  ^                  ^                  ^                  ^                  ^         *                  ^               25 +                  ^                  ^                  ^                  ^                  *                  ^                  ^                  ^                  ^                  ----------+---------+---------+---------->                           10        20        30                        Years of Experience in Trade                   (Note:  to complete graph, connect the *'s                           with a smooth curve.)a.  What does the graph indicate as annual income for someone with no    experience in the trade?b.  Describe the relation between income and experience over the inter-    val from 0 to 20.c.  Describe the relation between income and experience over the inter-    val 20 to 30.d.  Describe the overall graph. `
`Answer: a.  Around 12,500 dollars per year.b.  There appears to be approximately a straight line relation in which    income increases with experience over the interval  from  0  to 20.    (There seems to be some curvature or flattening for experience near    20.)  The change in income in this range  is  from  around 12.5  to    around 48, so the rate of increased income is roughly \$35,500/20   =    \$1775 per year.c.  The relation between experience and income  for experience  between    20 and 30 years also appears to be roughly a straight line,  but  a    flat straight line, indicating that income stays  roughly  constant    at a little less than \$50,000 per year.d.  The overall graph indicates income initially around \$12,500 (no ex-    perience), increasing income in the range from 0 to 20 years exper-    ience, approaching a limit that seems to be a little below \$50,000.    That limit seems to be reached sometime between 10 and 25 years.    (Income seems to remain about constant afterward.) `

134.

`If a random sample of 18 homes south of Center Street in Provo showedthe average selling price to be \$15,000 with s**2 = \$2400 and a randomsample of 18 homes north of Center Street revealed an average sellingprice of \$16000 with s**2 = \$4800, can you conclude that there is astatistically significant difference (ALPHA = .05) between the sellingprice of homes in these areas of Provo? `
`Answer: H(0):  MU(north) - MU(south)  =  0H(A):  MU(north) - MU(south) =/= 0s**2 = ((18-1)(2400) + (18-1)(4800)) / (18+18-2) = 3600   s = 60Before pooling the sample variances, we will test to see if thepopulation variances are equal:H(0):  SIGMA(north)**2  =  SIGMA(south)**2H(A):  SIGMA(north)**2 =/= SIGMA(south)**2F(calc.) = 4800/2400         = 2F(crit, df=17,17, ALPHA=.05) = 2.29So do not reject (continue) H(0), and pool s**2's, the following isequivalent to pooling when the sample sizes are equal:t(calculated) = (15000-16000)/((60)*SQRT(1/18+1/18)) = -50t(crit., ALPHA=.05, df=35, two-tailed) = +/- 2.03conclusion:  there is a significant difference `

135.

`A machine is supposed to produce Zorkel fingers having a thickness of.050 inches.  To test if the machine is working properly, a randomsample of 16 Zorkel fingers is selected randomly from the day's out-put.  The mean thickness of the sample is .053 inches with S = .003.We wish to determine if the machine is in proper working order withALPHA = .01.  Use a two-tailed test. `
`Answer: Hypothetical Population:  Set of all Zorkel fingers.Sample                 :  The 16 randomly selected fingers.H(O):  MU  =  .05.  The mean Zorkel finger thickness is .05.H(A):  MU =/= .05.  The mean Zorkel finger thickness is other than .05.MU(M) = .05 by H(O)S(M)  = S/SQRT(n)      = (.003)/4      = .0007497M(crit) = MU(M) +/- t(crit)*S(M)        = .05 +/- (2.95)*(.0007497)        = .052 to .048Since M = .053 is greater than .052, we reject H(O) and conclude thatZorkel finger thickness is other than .05.OR, using a t-test:t(calc) = [XBAR - MU! / [S(M)!        = [.053 - .050! / [.0007497!        = 4.0016t(crit, df=15, ALPHA=.01, two-tailed) = 2.947Since t(calc) > t(crit), we reach the same conclusion as above. `

136.

`An investigator was interested in studying relations between a numberof  factors  and  salary  in  a  university.   One  of the factors ofinterest was tenure status.  After much agonizing,  the  investigatordecided  to  use  the  following  variable for persons having facultyappointments.Variable called T coded as:    1 for non-tenure track people such as administrators      with faculty appointments    2 for faculty with less than one year in service in      tenure track positions    3 for faculty with one to three years on tenure track    .    .    7 for faculty having tenure and more than 20 years      serviceThe  investigator  then carried out a multiple regression analysis inwhich one of the variables fitted was T.If  you  accept  his  seven  tenure  classes as a reasonable groupingscheme, would you use this approach? Why or why not? `
`Answer: If  I  regarded  the  seven  classes  specified as a good way to formgroups based on tenure, I would want to see what would  happen  if  Iused  six  independent  variables  instead  of  just one to representtenure effects.  Using T alone might work  if  the  relation  betweensalary  and  the code values for T were linear.  But, the code valuesfor T don't appear  to  be  either  well  ordered  (the  first  classincluded  administrators who are apt to have higher salaries than thefollowing classes which do seem to  provide  sort  of  an  increasingorder)  or  equally  spaced.  (Is  the  amount of "tenure" the samebetween classes 1 and 2, and 2 and 3?).I would not use this approach. `

137.

`A report on the effect of  sex  on  faculty  salaries  at  a  WesternUniversity  states that all ranks and departments have been surveyed.It states that:   A simple regression with salary as dependent variable and sex   as independent variable had a regression coefficient for sex   equal \$5000.   A multiple regression  with  the same dependent variable but   additional variables for  rank  (full  professor, associate,   assistant, instructor), department,  tenure  status,  length   of employment, etc.  had  a  regression coefficient  for  sex   equal \$100.(In  both  cases,  the independent variable used for sex was coded sothat the  regression  coefficient  for  sex  represented  the  salaryadvantage of males over females.)Which of these values would you use to represent the effect of sex onsalary?  Explain your answer. `
`Answer: I  would  expect  that  value  \$100  would  be  more  informative  ortrustworthy.  When all important factors affecting response have beenheld constant except for  a  single  independent  variable  and  thatvariable is related to response by a straight line relation, a simpleregression coefficient  can  provide  a  good  measure  of  how  thatvariable  affects  response.    But,  if response is affected by manyvariables and they are not constant in the data set being examined, asimple  regression coefficient can be very misleading.  In this case,the fact that the regression coefficient for sex changed from 5000 to100  when  other  variables  were  included  in the fitted regressionindicates that much of the apparent influence of sex on salary reallywas  the  result of treating variables like rank, department, etc. asconstant or unimportant  when in fact some of them were important andnot constant. `

138.

`Attached is a table relating current food prices and prices from3 months ago for a certain supermarket in the area.  Perform thefollowing:a.  Plot current price vs. price 3 months ago.b.  Propose a model relating current price and price 3 months ago.    Define all terms and estimate all parameters.      Produce                             Price                               3 months ago   Current-----------------------------------------------------Milk (1 gallon)                 1.39            1.39Cheese (sliced 12 oz.)           .89             .93Eggs (1 doz. large)              .89             .81Bologna (12 oz.)                 .85             .89White Tuna Fish (7 oz.)          .75             .79Soup (chicken noodle)            .22             .24Green Beans (1 lb. can)          .33             .39Ground Beef (1 lb.)              .95             .98Corn Flakes (12 oz.)             .49             .49Spaghetti (2 lbs.)               .89             .95Sauce (w/o meat, 16 oz.)         .59             .59Coffee (6 oz.)                  1.57            1.57Bread (1 oz.)                    .45             .52Lettuce (1 head)                 .49             .33Potatoes (10 lbs.)               .49             .69Fruit Cocktail (18 oz.)          .49             .49Peanut Butter (18 oz.)           .87             .99Yogurt (8 oz.)                   .37             .37Rice (2 lbs.)                    .71            1.09Cottage Cheese (1 lb.)           .65             .67Total                          14.33           15.17 `
`Answer: a.  If available, consult file of graphs and diagrams that could not be    computerized for graph.b.  The model I propose is:  Y = B(1)*X + EPSILON    where:        Y is the response, current price;               B(1) is the estimated effect of X on Y;                  X is the independent variable, price 3 months age;            EPSILON is a random error term.    The fitted equation is:  YHAT = 1.046 * X    I forced this regression through the origin because, with a regres-    sion not through the origin, (the  intercept  equalled .052),  the    intercept was not significant at  the  5% level.   Also, the usual    method of describing inflation would not include adding a constant    to some computed number.    The t test for the regression coefficient and F test for the regres-    sion mean square are significant.    ANOVA    Source               df        SS           M.Sq.    Uncorrected total    20      13.8443    Regression            1      13.6213      13.6213    Pooled Error         19       0.2230        .01739    R**2, adjusted for matched X error = .9894 `

139.

`A small mail-order house uses the weight of incoming mail to determinehow many of their employees are to be assigned to filling orders  on agiven day.  Assume a linear regression model,  given  X = weight (lbs)of mail on hand at 7:00 a.m., and Y = no. of 8-hour shifts required tofill the orders of that day.  The calculated  results  from some  dataare given below:        n = 8        SUM(X**2) = 524   SUM(X) = 56         SUM(XY) = 364   SUM(Y) = 40       SUM(Y**2) = 256        YHAT = .52 + .84X(a)  Test the hypothesis that the slope of the regression line is zero     at ALPHA = .05.(b)  Find a 90% confidence interval for the number of eight hour shifts     required if there are 10 lbs. of mail on hand at 7 a.m. on a par-     ticular day.(c)  In analyzing the fitted regression model, explain what the values     for b(0) and b(1) mean.  Is there anything inconsistent about your     fitted values from a practical standpoint? `
`Answer: (a)  H(O):  BETA  =  0     H(A):  BETA =/= 0     SSE = [256-[[40**2!/8!! - [([364-(56*40)/8!**2)/(524-[[56**2!/8!)!         = [56! - [(84**2)/(132)!         = 2.5454     MSE = [2.5454!/[6!         = 0.4242     S(b)**2 = 0.4242/132             = 0.0032     S(b) = 0.0567     t(calc) = [0.84 - 0!/[.0567!             = 14.817     t(crit, ALPHA=.05, two-tailed, df=6) = +/- 2.447     Since t(calc) < +t(crit), reject H(O).  Therefore, based on this     sample evidence, conclude that the regression coefficient is dif-     ferent from zero.(b)  S(YAT) = SQRT([4242! * [(1/8) + ([10**2!/[132!)!)            = 0.61     C.I. = YHAT +/- [t * S(YHAT)!          = [.52 + (.84*10)! +/- [1.943 * 0.61!          = 8.92 +/- 1.189          = from 7.73 to 10.11(c)  b(0) is the estimated value for BETA(0) which is the intercept     value on the Y-axis for the regression line.     b(1) is the estimated value for BETA(1) which is the slope of the     regression line.  This indicates the ratio of the change in the Y-     variable with respect to the change in the X-variable for the par-     ticular line.     The fitted value that might cause some concern from a practical     standpoint is b(0), which implies that approximately a four hour     shift is needed even when no mail is on hand. `

140.

`A report for an organization states that a simple regression was usedto relate salary to sex.  The independent variable for sex was codedso that the regression coefficient for sex represented the salary ad-vantage of males over females.  The result of fitting over 700 pairsof values was a regression coefficient of 5000 (advantage for malesof \$5000).  "A test of the regression coefficient at the 1% level wassignificant.  The correlation coefficient r was .24."What action would you take on the basis of this report?  Explain. `
`Answer: Send the report writers back to re-examine their data.1)  The only time that a simple regression would be a good way to    estimate the effect of sex on salary would be when all other    important factors affecting response have been held constant    or nearly constant.  That seems unlikely in most organizations.    (There should also be a straight line relation between salary    and sex.  That should be a good bet unless there are more than    two sexes in the organization.)2)  For this data set, sex has only accounted for around 6% (.24**2)    of the variation in salary.  It would take someone bolder than I    am to take action on a description that leaves 94% of the varia-    tion in salary unexplained.  (The test of significance says that    the evidence at hand is consistent with the claim that the regres-    sion coefficient is not zero.  It offers guarantees neither that    the model fitted is reasonable nor that a worthwhile amount of var-    iation in response has been accounted for.) `

141.

`The following data illustrates  the  relationship  between  income  andeducation for a sample of nine U.S. workers.  Education (X(j) in years)           Income (Y(j) in thousands of \$)             0                                      5             6                                      6             8                                      7            10                                      9            12                                      8            12                                     10            12                                     12            14                                     11            16                                     12a.  Obtain a scattergram for the data.b.  Perform a regression analysis using the model:              Y(j) = a + b*X(j) + e(j)c.  Draw the regression line on the scattergram.d.  What income, in thousands of dollars, would you predict for a    single U.S. worker with 10 years of education?e.  Find the correlation coefficient for the data.f.  What proportion of the variance in income is "explained" by the    regression equation?g.  Based on this sample, how much extra income would an additional    year of education be worth to a person with less than 16 years    of education? `
`Answer: a.   Y ^       ^                      Connect points A & B to form the       ^                      graph of the regression line.    15 +       ^       ^                                   B       ^                       *       *       ^                           *    10 +                       *       ^                   *       ^                       *       ^               *       ^           *     5 *       A       ^       ^       ^       --+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+------> X                 5        10        15b.  YHAT = 4.1063 + .47826(X)c.  Refer to scatter diagram above.d.  Prediction for a worker with 10 years of education:               YHAT = 4.1063 + .47826(10)               YHAT = 8.889    Therefore, I predict an income of \$8,889 when education = 10 years.e.  r = .892f.  r**2 = .796, so 79.6% of the variation in income has been explained    by the model.g.  An additional year of education  is  worth  \$478,  since b(1),  the    regression coefficient, indicates the change in Y for a unit  change    in X. `

142.

`An  experiment was conducted in a supermarket to observe the relationbetween the amount of display  space  allocated  to  Petrushka  brandcoffee and its weekly sales.  The data for the five time periods arebelow.    Space Allocated (sq. yds.) - X:  1   2   3   4   5    Weekly Sales (cases)       - Y:  2   4   5   6   8After gathering  the  data,  Clark  Kent,  the  SUPERmarket  MANager,discovered  that  the slope of the least squares line is 1.40 and theintercept is +0.80.a)  Plot the least squares line and the data points.  Comment on the    fit.b)  What would you predict the weekly sales would be if the manager    allocated 4.5 sq. yd.?c)  How much of an increase in sales can he expect for every extra    sq. yd. of display space?d)  What does LEAST SQUARES refer to?e)  Why are there two equations for the confidence intervals for a    future Y value, and when would you use each one? `
`Answer: a)            Y              ^              ^            9 +              ^            8 +                        *              ^            7 +              ^            6 +                   *    Weekly    ^     Sales  5 +              B   (cases)    ^            4 +         *                   (NOTE:  To plot an appro-              ^                              ximation to the regression            3 +                              line, connect points A & B.              ^                              Also note that point B is a            2 +    *                         data point.)              ^            1 A              ^              -----+----+----+----+----+----+----+------>  X                   1    2    3    4    5                   Space Allocated (sq. yds.)    The least squares line appears to fit the data very well.b)  Sales = (1.4*4.5) + .8          = 7.1c)  1.4 sq. yd.d)  Least squares refers to minimizing the sum of squares of the dis-    tance between the regression line and the data points.e)  One equation is used to arrive at confidence intervals for predict-    ing the mean response,  while  the other is used  for  predicting a    particular response. `

143.

`A study was conducted in which typing speed (number of words per minute)was measured each day after the beginning of a period of practicetyping.  Part of the results of fitting a series of polynomial modelsappear below.                df for Error    S**2    R**2Linear               9          28.40   .85Quadratic            8           4.51   .98Cubic                7            .91   .99On the basis of this information, which model would you choose?  Why? `
`Answer: SSE(Quad.) = 4.51*8 = 36.08SSE(Cubic) = 0.91*7 =  6.37Diff. SS. = 29.71 with 1 dfF(CALC) = Mean Sq. Diff./Error Mean Sq. (Containing Model)        = 29.71/.91 = 32.65 with 1 and 7 df.F(CRITICAL = ALPHA = .05, df = 1,7) = 5.59Since F(CALC) is greater than F(CRITICAL), we reject the nullhypothesis that the coefficient of the cubic term equals zero.I would, therefore, choose the cubic model. `

144.

`What would you guess the value of the correlation coefficient to be forthe pair  of variables:   "number of man-hours worked" and  "number  ofunits of work completed"?a)  Approximately 0.9b)  Approximately 0.4c)  Approximately 0.0d)  Approximately -0.4e)  Approximately -0.9 `
`Answer: a)  Approximtely 0.9 `

145.

`The results of an imaginary investigation of the effect on sales ofdifferent methods of displaying peaches included:ANOVASource of Variation     df      SS      M.S.Total                   25Mean                     1Corrected Total         24Day of the week          4      400     100Fruit Market             4     1000     250Display                  4      200      50Error                    9      225      25Using the information contained in this table perform appropriatetests to decide if background variation accounts for the effect onsales of:a.  Displayb.  Day of Weekc.  Fruit Market `
`Answer: a.  F(calculated) = 50/25 = 2    F(critical, df = 4, 9, ALPHA = .05) = 3.63Therefore, retain the null hypothesis that the effect of display equalszero.b.  F(calculated) = 100/25 = 4    F(critical) = 3.63Therefore, reject the null hypothesis that the effect of the day of theweek equals zero.c.  F(calculated) = 250/25 = 4    F(critical) = 3.63Therefore, reject the null hypothesis that the effect of fruit marketequals zero.From these F tests we can see that background variation accounts for theeffect on sales of display only. `

146.

`To test the hypothesis that shelf placement influences sales, amarketing researcher has collected data on sales in a random sampleof 15 comparable supermarkets with 3 different shelving policiesfor an identical brand of soup.  The data is weekly sales figures(in tens of cans).  Perform the appropriate test at the 5% level.If you reject, which shelving policies are different?  (Note:1/SQRT(.4) = 1.6.)                  bottom shelf    middle shelf    top shelf                      sales           sales         sales                  ------------    ------------    ---------                       10              25            10                        5              20            10                       10              25            20                       10              30            20                       15              50            40Sums                   50             150           100Sums of  squared scores      550            5050          2600    8200 `
`Answer: OVERALL MEAN(SALES)         20STANDARD DEVIATION          10COEFFICIENT OF VARIATION    50ANALYSIS OF VARIANCE:---------------------SOURCE OF VARIATION         DF          SS      MEAN SQUARE    F(CALC.)UNCORRECTED TOTAL           15      8200.0000CORRECT'N FOR MEAN           1      6000.0000CORRECTED TOTAL             14      2200.0000SHELF                        2      1000.0000   500.00000        5.00EXPERIMENTAL ERROR          12      1200.0000   100.00000MEANS FOR SHELFTREATMENT     MEAN(SALES)MIDDLE         30TOP            20BOTTOM         10PROBABILITY LEVEL FOR COMPARING MEANS = .05VALUE FOR STUDENT'S t (DF=12,ALPHA=.05,TWO-TAILED) = 2.179LSD FOR ABOVE MEANS IS 13.7812 at PROB.LEVEL .05(Note:  LSD means Least Significant Difference.)------------------------------------------------------------------------F(critical, df=2,12, ALPHA=.05) = 3.88Therefore, reject the null hypothesis that shelving policy does notinfluence sales.Based on the LSD given above, it appears that there is a significantdifference between the middle shelf and the bottom shelf. `

147.

`Suppose that you wish to test 4 brands of tires for length ofusefulness and that you have available 4 car-driver combinations.Thus you have available 16 experimental units if you considereach tire position on a car as an experimental unit: i.e.         Front-right   Front-left   Rear-right   Rear-left         ___________   __________   __________   _________Car 1       Unit 1       Unit 2       Unit 3      Unit 4Car 2       Unit 5       Unit 6       Unit 7      Unit 8Car 3       Unit 9       Unit 10      Unit 11     Unit 12Car 4       Unit 13      Unit 14      Unit 15     Unit 16A.  Assign Brands (A,B,C,D) randomly to experimental units    (i.e., Use a procedure appropriate to a completely random,    CR, design).  Show how you used the random numbers table.    Do you see any dangers in using a CR design for this kind    of experiment?B.  Assign brands to experimental units subject to the restric-    tion that each brand must be tested once in each tire posi-    tion (i.e., Use a procedure appropriate to a randomized com-    plete block, RCB, design where tire position is used in form-    ing blocks).  Show how you used the random numbers table.  Do    you see any dangers in using an RCB design for this kind of    experiment?C.  Suggest a way of testing tires in this situation that might    overcome the dangers of using either a CR or RCB design. `
`Answer: A.  Generating randomly 16 numbers using a computer or using    random numbers table, will assign brands randomly to ex-    perimental units such that each brand appears 4 times in    the experiment, i.e.      2, 4, 8, 14, 13, 12, 7, 1, 16, 6, 15, 3, 9, 5, 10, 11                 e.g. Brand A to unit 2                        "   B "    "  4                        "   C "    "  8                        "   D "    " 14                        "   A "    " 13                        "   B "    " 12                        "   C "    "  7                        "   D "    "  1                       etc.     (1) There is a possibility that one brand might         appear on only one tire position (e.g. A to 1,5,         9, and 13)     (2) Or one brand(s) might appear on one car (e.g.         B to 1,2,3, and 4).B.  Problem (1) will be solved since in RCB designs each brand    will be applied once to each tire position. e.g. In Front-    right (say as block I) randomly assign brands to units    1,5,9 and 13 (e.g A to 5, B to 13, C to 1 and D to 9).    This scheme will eliminate the position to position vari-    ation.  But one would expect with this scheme a car to    car variation (row-wise).C.  A Latin square design will resolve (2) as opposed to RCB    and resolve (1) and (2) as opposed to CR.  In this scheme    each brand will appear once and only once in each position    and each car. One way is:                     _________________________                     ^  A  ^  B  ^  C  ^  D  ^                     ^_______________________^                     ^  B  ^  C  ^  D  ^  A  ^                     ^_______________________^                     ^  C  ^  D  ^  A  ^  B  ^                     ^_______________________^                     ^  D  ^  A  ^  B  ^  C  ^                     ^_______________________^ `

148.

`The manager of a department store wished to compare the influence ofbackground music on the volume of sales in the shoe department.He wished to test:T1.  Waltzes,T2.  Marches,T3.  Acid Rock,T4.  PolkasHe decided to use the same treatment for a sales period where eachweek provided four periods.P1.  Friday   - 10 a.m. to 3 p.m.P2.  Friday   -  3 p.m. to 8 p.m.P3.  Saturday - 10 a.m. to 3 p.m.P4.  Saturday -  3 p.m. to 8 p.m.He also decided to use one month (4 weeks) for testing.  When askedwhat, if any, differences would he find if the same background musicwere used during all test periods he answered:a.  Sales would be greatest during P2 and P4.  P3 would be better    than P1.b.  Sales would be best during the first week of the month, next    best during the 3rd week, and about equally poor during the 2nd    and 4th week.A.  Define and illustrate experimental unit in terms of this problem.B.  Would you elect to conduct this inquiry as indicated?C.  Suppose that you have no alternative but to conduct an experiment    under the conditions decreed by the manager.  Which of the common    designs discussed in the course would you use?D.  Write the model for the design you have chosen.  Define all terms    carefully.  (Be sure that your definitions of terms is relevant    to this particular problem.) `
`Answer: A.  An experimental unit is one of the four time periods during a    certain week, such as:  Saturday from 10 a.m. to 3 p.m. during    the first week.B.  Problems:        1.  The treatment set doesn't include a treatment with no            music.  Yet that would seem to be a reasonable stan-            dard for comparison.  This treatment set only allows            comparisons among conditions involving background music.        2.  If a randomized block or latin square design is to            be used it requires the assumption that there is no            interaction between treatments and the blocking factor.            It seems questionable that the difference in sales be-            tween, say, acid rock and waltzes would be the same for            10 a.m. to 3 p.m. on Friday and 3 p.m. to 8 p.m. on            Saturday.C.  I would use the Latin Square design with time of day and week    of month as my blocking factors.D.  Y(I,J,K) = MU + TAU(I) + RHO(J) + KAPPA(K) + EPSILON(I,J,K)    with I = 1, 2, 3, 4         J = 1, 2, 3, 4         K = 1, 2, 3, 4    where Y(I,J,K) is the response        MU is the overall mean        TAU(I) are the treatment (type of music) effects        RHO(J) are the effects of the time period        KAPPA(K) are the effects of the weeks of the month        EPSILON is the random error `

149.

`A test was conducted to compare the relative effectiveness of threewaterproofing compounds, (A,B,C).  A strip of cloth was subdividedinto nine pieces - - -        Left                  Center                 Right_____  _____  _____    _____  _____  _____    _____  _____  __________  _____  _____    _____  _____  _____    _____  _____  _____Each piece was considered to be an experimental unit, but it wassuspected that the pieces differed systematically from left toright in capacity to become waterproofed.  Accordingly, therandom assignments of compounds to experimental units was res-tricted so that: I.  Each compound was tested once in each set of three pieces (sets     are left, center, and right);  andII.  Each compound was tested once in each of the positions within a     set of three (once furthest left in a section, once in the cen-     ter of a section, and once on the right of a section).a.  Write a model appropriate to such a trial.b.  Analyze and interpret the following results for such a randomization     scheme:       Left                   Center                 Right_____  _____  _____    _____  _____  _____    _____  _____  _____B, 12  A, 15  C, 16    A, 11  C, 17  B, 10    C, 10  B, 12  A, 14_____  _____  _____    _____  _____  _____    _____  _____  _____(consider higher numbers as better) `
`Answer: a.  This is an LSQ design where the model is:    Y(I,J,K) = MU + TAU(I) + RHO(J) + KAPPA(K) + EPSILON(I,J,K)    Y is response, degree of waterproofing    MU is an overall mean for waterproofing    TAU(I) are the treatment effects    RHO(J) are the column effects, or piece position on cloth    KAPPA(K) are the row effects, or the position within the piece    EPSILON is the random error, assumed to be normally distributed             with mean = 0 and variance = SIGMA**2    Estimates of parameters        SIGMA**2 = 5.333        MU      13              RHO(1)  -2              KAPPA(1)  1.333        TAU(1)    .333          RHO(2)   1.667          KAPPA(2) - .333        TAU(2) - 1.667          RHO(3)    .333          KAPPA(3) -1        TAU(3)   1.333    Treatment means were:        C = 14.333        A = 13.333        B = 11.333b.  None of the differences among treatment means appear to be signi-    ficant;  they are all less than the LSD of 18.7148 (ALPHA = .01).    The F test for treatments (alternative test with higher Type II    error rate):        H(0):  TAU(1) = TAU(2) = TAU(3) = 0        F(calculated) = 1.3125        F(table, ALPHA = .01, df = 2,2) = 99,    also does not allow one to reject H(0).  In conclusion, it appears    that none of the compounds are significantly different from any    other at ALPHA = .01. `

150.

`A test has been conducted in which four tire brands have been testedusing 12 experimental units where an experimental unit consisted of onetire position on one car.  The random assignment of brands to experi-mental units was restricted so that each brand was tested once on eachcar.  Results (in amount of wear) were:         Front Right      Front Left       Rear Right       Rear LeftCar 1    D, 7.17          A, 7.62          B, 8.14          C, 7.76Car 2    B, 8.15          A, 8.00          D, 7.57          C, 7.73Car 3    C, 7.74          B, 7.87          A, 7.93          D, 7.80a.  Write a model appropriate to this trial and estimate all parameters.b.  Do any of the assumptions for this design make you uneasy?  Explain.c.  Analyze and interpret these results. `
`Answer: a.  The model is Y(I,J) = MU + TAU(I) + RHO(J) + EPSILON(I,J)    where Y is the response, tread wear          TAU(I) are the treatment effects, effects of tire brand          RHO(J) are the block effects, effects of car          EPSILON is the random error term with mean = 0 and                  variance = SIGMA**2          MU is the overall mean    Estimates of parameters:    MU(HAT)    = 7.79    TAU(A,HAT) = .0599 = .06    TAU(B,HAT) = .2633    TAU(C,HAT) = -.04667 = -.047    TAU(D,HAT) = -.27667 = -.277    RHO(1,HAT) = -.1175    RHO(2,HAT) = .0725    RHO(3,HAT) = .045    SIGMA**2 = .0419 with 6 df.b.  Using a randomized block (RCB) design makes me uneasy since I would    expect wheel position on car to also affect tread wear.  Therefore,    I would also block on wheel position as well as car and use a Latin    Square design.c.  Treatments means are:  B = 8.053,  A = 7.85,  C = 7.743,  D = 7.513    Only one difference is significant at the .05 level.  Tires B and    D are different since their difference is greater than the LSD.    (B - D) +/- LSD    .54 +/- .409    Interval is from .131 to .949    Since the interval does not include zero, we reject the null hypo-    thesis that the true difference is zero.    The F test for treatments fails.  This is the case where the LSD    indicates a significant difference while the F test of treatments    doesn't.  These procedures usually are different and usually have    different properties regarding Type I and Type II error rates.    Here, the LSD is more exposed to Type I errors and the F test is    more exposed to Type II errors. `

151.

`Write out the sources of variation and the degrees of freedom for thefollowing industrial experiment.  Mention also the name of the design.Three  machines  were  used  to produce parts made from four kinds ofmetal. Each machine made one part from each type of metal.  The orderwith  which  the metals were assigned to the machines was establishedthrough a randomization procedure. `
`Answer: Source of Variation         df-------------------         --Total                       12Mean                         1Metals                       3Machines                     2Residual                     6 (Metal x Machine)This is a randomized block experiment with metals playing the role ofblocks. `

152.

`The Crapi Cable Company #35 cable has a mean breaking strength of 1800pounds with a standard deviation of 100 pounds.  A new material is usedwhich, it is claimed, increases the breaking strength.  To test thisclaim a random sample of 50 cables, manufactured with the new material,is tested.  It is found that the sample has a mean breaking strengthof 1850 pounds.  Test this claim using ALPHA = .01. `
`Answer: Hypothetical population:  All Crapi #35 cables made with the new                          material.                 Sample:  The 50 cables randomly selected.H(O):  MU = 1800.  The mean breaking strength of the new cable is                   1800 lb.H(A):  MU > 1800.  The mean breaking strength of the new cable is                   more than 1800 lb.MU(XBAR) = 1800 by H(O)SIGMA(XBAR) = SIGMA/SQRT(n)            = 100/SQRT(50)            = 14.142XBAR(crit)  = MU(XBAR) + Z(crit)*SIGMA(XBAR)            = 1800 + (2.33)*(14.142)            = 1832.951Since the sample mean breaking strength is 1850, which is greater than1832.51, we must reject H(O)  and  conclude  that  the  mean  breakingstrength of the new cable is significantly more than 1800 lb. `

153.

`In the past a chemical fertilizer plant has produced  an  average  of1100 pounds of fertilizer per day. The record for the past year basedon 256 operating days shows the following:      XBAR = 1060 lbs/day         S =  320 lbs/daywhere  XBAR  and  S  have  the  usual  meaning. It is desired to testwhether or not the average daily production has dropped significantlyover  the  past  year.  Suppose  that  in this kind of operation, thetraditionally acceptable level of significance has been .05. But  theplant manager, in his report to his bosses, uses level of significance.01. Analyze the data at both levels after setting up appropriatehypotheses, and comment. `
`Answer: H(O):  MU = 1100H(A):  MU < 1100Since n   = 256, use Z to approximate t.S(XBAR) = 320/SQRT(256)        = 320/16        = 20Z(calculated) = (1060 - 1100)/20              = -40/20              = -2Z(critical, ALPHA=.05, one-tailed) = 1.645Z(critical, ALPHA=.01, one-tailed) = 2.33Therefore,  H(0)  is rejected at ALPHA=.05 but continued at ALPHA=.01.It appears that the manager is trying to pull  a  fast  one  on  hisbosses  by  using  ALPHA=.01  and saying production has not dropped.However, if the traditional level of significance is used,  ALPHA=.05,there is evidence that indicates a drop in production. `

154.

`The Pfft Light Bulb Company claims that the mean life of its 2 wattbulbs is 1300 hours.  Suspecting that the claim is too high, NalphRader gathered a random sample of 64 bulbs and tested each.  He foundthe average life to be 1295 hours with s = 20 hours.  Test the com-pany's claim using ALPHA = .01. `
`Answer: Hypothetical population:  All Pfft 2 watt bulbs.                 Sample:  The 64 randomly selected bulbs.H(O):  MU = 1300.  The mean life of 2 watt bulbs is 1300 hours.H(A):  MU < 1300.  The mean life of 2 watt bulbs is less than 1300                   hours.MU(XBAR)   = 1300 by H(O) S(XBAR)   = S/SQRT(n)           = 20/8           = 2.5XBAR(crit) = MU(XBAR) + Z(crit)*S(XBAR)           = 1300 - 2.33*2.5           = 1294.18Since 1295 is not less than 1294.18, we cannot reject H(O).  There isnot enough evidence to conclude that the mean life of the 2 watt bulbsis significantly less than 1300 hours. `

155.

`The Rickety Railroad Company claims that .5 of the trains on its FoggyBottom branch run on time.  An Interstate Commerce Commission investi-gator doubts the claim but is uncertain about whether the true frac-tion is less than or greater than the claim.  A random sample of 64trains was checked and he found that 21 were on time.  Test the com-pany's claim using ALPHA = .05. `
`Answer: Hypothetical population:  All trains on the Foggy Bottom branch.                 Sample:  The 64 randomly selected trains.H(O):  PI  =  .5  50% of the Foggy Bottom branch trains run on time.H(A):  PI =/= .5  Other than 50% of the Foggy Bottom branch trains                  run on time.We can use the normal approximation since:      n*PI = 64 * .5 = 32 > 5;  andn*(1 - PI) = 64 * (1 - .5) = 32 > 5.     MU(p) = .5 by H(O)  SIGMA(p) = SQRT(PI*(1 - PI)/n)           = SQRT(.5*(1 - .5)/64)           = .0625   p(crit) = MU(p) +/- Z(crit)*SIGMA(p)           = .5 +/- 1.96*.0625           = .6225, .3775In this case, p = 21/64 = .328, which is less than .3775, so wereject H(O).  Other than 50% of the Foggy Bottom branch trainsrun on time.OR:Z(calc) = (p - PI)/SIGMA(p)        = ((21/64) - .5)/(.0625)        = -2.75Since -2.75 < -1.96 (Z-crit), we reject H(O) and reach the sameconclusion as above. `

156.

`Define the term "stratified sample" and explainwhy it would be useful in the following situation.  Acompany is composed of many small plants located through-out the United States.  A Vice President of the companywants to determine the opinions of the employees onthe vacation policy. `
`Answer: Answer - A stratified sample is one which has beenobtained by a procedure in which the frame is divided intonon overlapping categories (strata).  Sampling units arethen selected at random from each stratum thus assuringthat all strata are represented in the sample.     For the given problem, I would suggest a type ofstratified sampling procedure.  Specifically, I wouldrecommend that each plant be considered a stratum and arandom sample obtained within each stratum to insure thatall plants are represented in the sample.  I wouldfurther suggest that the sample from each stratum representthe proportional size of that stratum.  For example,if plant A employs 25% of the total company's employees,then the sample from plant A should represent 25% of thetotal sample obtained. `

157.

`A company wants to estimate with a degree of confidence of  0.95  andwith an absolute error not greather than \$4.00  the  true mean dollarsize of orders for a particular item.   How large a sample should thecompany take from its very extensive records to meet this requirement,if SIGMA is assumed to equal \$20.00? `
`Answer: (2 * SIGMA)/SQRT(n) = d   (2 * 20)/SQRT(n) = 4n = 100Note: 1.96 is  a more accurate estimate of the critical value, but      complicates the computation. `

158.

`A manufacturer wishes to determine the average weight of a certain typeof product in order to design the proper package.  What size sample isrequired so that the risk of exceeding an error of .20 pounds is .010?(Note:  Errors can be positive or negative.)  Assume SIGMA is1.10 pounds. `
`Answer: Using Z = (XBAR - MU)/(SIGMA/SQRT(n))we get n = ((Z) (SIGMA)/(XBAR - MU))**2       n = ((2.576)(1.1)/.2)**2       n = 200.73Therefore, a sample size of 201 is required. `

159.

`An aircraft parts manufacturer wishes to determine the average shearingstrength of a certain type of weld in order to submit a bid for a con-tract to produce these parts.  What size sample is required so that therisk of exceeding an error of 20 pounds or more is .005?  Assume thatSIGMA is 100 pounds. `
`Answer: Using a significance level = .005, Z = 2.576n = (Z**2)(SIGMA**2)/(e**2)  = (2.576**2)(100**2)/(20**2)  = 165.9 == 166 `

160.

`In a random sample of flashlight batteries, the average useful life was22 hours and the sample standard deviation was 5 hours.  How largeshould the sample size be if you want the mean of your sample to bewithin 1 hour of MU 99 times out of 100 in repeated sampling? `
`Answer: If significance level = .01, then Z = 2.576SIGMA(HAT) = 5 hoursTolerable error = 1 hourn = (Z**2)(SIGMA**2)/(error**2)  = (2.576**2)(5**2)/(1**2)  = 165.89  = 166 `

161.

`A floor manager of a large department store is studying the buyinghabits of his customers.  Suppose he has good reason to believe that anestimate of \$600 for the population mean for the amount spent in hisstore each year is wrong.  He makes preparation to draw a sample butlacks the funds to draw N=100 as he had planned.  How large a sampleneed he draw in order to estimate the population mean within \$100 ofthe true value with probability of 0.95?  (Assume SIGMA = \$500.) `
`Answer: n = [Z(ALPHA/2) * SIGMA/e!**2;  e = tolerable error  = [(1.96) * 500/100!**2  = 96 `

162.

`The variance of average family income in New York State is known to beabout the same as it is in Delaware.  The mean family income is to beestimated by a sample survey in each state.  It is desired to have thesampling error equal in both states.  If the recommended sample size forDelaware is 2500 families:a)  What size sample would you take in New York State?b)  What statistical formula supports your answer to part (a)? `
`Answer: a)  2500b)  Variance of mean = SIGMA**2/n    Quantity estimated in each state is a mean.  SIGMA is the same for    each state, so equal sampling error can be achieved by using equal    sample size. `

163.

`Should a sample survey be considered before a complete count(census)?  If so, why?  Be brief. `
`Answer: Yes, most of the time a sample survey should be considered beforea census because it has the following advantages:    1)  Greater accuracy advantage - it is a curious fact that the        results from a carefully planned and well executed sample        survey are expected to be more accurate than those from a        complete census.    2)  Cost advantage - if data are secured from a small fraction        of the population, expenditures are smaller than if a com-        plete count is attempted.  Low cost permits  expansion  of        the statistical program and expanded usefulness.    3)  Time advantage - in many research problems, time is a criti-        cal factor.  A sample of the data can be collected, coded,        tabulated and analyzed more quickly than a complete count.    4)  Destructive nature of the test - in order to make observations        in some problems, particularly those dealing with manufactured        products, the elementary units being observed must be destroyed        or weakened.  To test all units in the population would result        in damage to all units. `

164.

`What are the major sources of uncertainty (error) in sample survey data?Describe them and give an example of each. `
`Answer: All data, whether obtained by a census or sample, are subject to varioustypes of uncertainties.  There are three types of uncertainties:1.  Structural Limitations - are defects that are built into the survey    procedures.  The following are some examples:    a.  Failure to obtain observations which would be useful;    b.  Unclear or biased wording of the questionnaire;    c.  Poor selection of the measuring or testing instrument;    d.  Too large a gap between the frame and population;    e.  Poor choice of survey data;    f.  Incorrect usage of statistical formulas for calculating        estimates.The best control  and  method for avoiding structural limitation isachieved by detailed  planning  and design, through testing, reviewof the literature, and prior studies.2.  Operational Blemishes and Blunders - originate in the execution of    the work.  The following are some examples:    a.  Failure to ask some of the questions;    b.  Asking questions not on the questionnaire;    c.  Mistakes in reading the measuring instrument;    d.  Nonresponse or refusal;    e.  Keypunch errors.The detection, control and measurement  of  operational blemishes andblunders can be achieved through a repetition or audit of the sample,thereby enabling evaluation of their impact on the estimates.3.  Random Variation - is measured by the standard error of estimate.    The first source  of  random variation is simply the variability    (spread, dispersion) among the sampling units in the frame.  The    second source of random variation is from inherent, uncorrelated    or nonpersistent, accidental variations of the cancelling nature    that arise from inherent variability, perhaps on an hourly basis,    of the investigators, supervisors, editors, coders, keypunchers,    and other workers. `