
1.
Explain briefly how you would decide which of the following two events is the more unusual:
a. A 90 degree day in Vermont. b. A 100 degree day in Florida.
Answer: One would examine previous weather records and note the relative fre quency of 90+ degree days in Vermont, and 100+ degree days in Florida. Thus, relative frequencies are one method of estimating the probability of each event. (Note: The smallest frequency is the more unusual event.)
2.
The Getrich Tire Company is having a tire sale on tires salvaged from a train wreck. Of the 15 tires offered in the sale five tires have suffered internal damage and the remaining ten are damage free. You are planning on purchasing two of these tires. In finding the proba bility that the two tires selected at random from the 15 will be damage free, the probability distribution to use is:
(a) Normal (c) Hypergeometric (b) Poisson (d) Binomial
Answer: (c) Hypergeometric
Since two tires are selected without replacement, it is Hypergeometric.
3.
As a seamstress you have observed that flaws in a certain type of material occur on the average of 0.2 per yard. The distribution to find the probability of no more than one flaw occurring in a dress requiring four yards of this material would be:
(a) Normal (c) Hypergeometric (b) Poisson (d) Binomial
Answer: (b) Poisson; with LAMBDA = .2
Any random phenomenon for which a count of some sort is of interest is a candidate for modeling by assuming a Poisson distribution.
4.
Suppose that a neutron passing through plutonium is equally likely to release 1, 2, or 3 other neutrons, and suppose that these second generation neutrons are in turn each likely to release 1, 2, or 3 third generation neutrons. What is the probability distribution of the number of third generation neutrons? What is the mean of this distribution?
Answer: To find the probability distribution use a tree diagram and count.
N=n ^ 1 2 3 4 5 6 7 8 9  P(N=n) ^ 1/9 4/27 16/81 12/81 12/81 10/81 2/27 1/27 1/81
Mean = E(N) = (1)(1/9) + (2)(4/27) + ... + (9)(1/81) = 4
5.
The lengths of Frank's 24 inch franks are normally distributed with mean of 2 feet and variance of 0.03 feet. If you purchase 3 of Frank's franks for your family, what is the probability that you will have a total length of hot dogs in excess of 6.60 feet?
a. .008 b. .014 c. .019 d. .023 e. .029
Answer: d. .023
Variance = 3*(.03) = .09 Z = (6.60  6)/SQRT(.09) = .60/.3 = 2
Area beyond Z = .0228 == .023
6.
If the life of wild pheasants follows a normal distribution with a mean of 9 months and a variance of 9, what percent of the population will be less than 11 months of age? (Note that MU = 9 and SIGMA(X)**2 = 9.)
(a) 34.13 (c) 74.86 (b) 84.13 (d) 62.93
Answer: (c) 74.86
Z = (11  9)/3 = .67 P(Z < .67) = (.2486) + (.5000) = .7486 = 74.86%
7.
The distribution of lifetimes for a certain type of light bulb is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the 33rd percentile of the distribution of lifetimes.
a. 560 b. 330 c. 1044 d. 1440 e. none of these
Answer: e. none of these P(z=?) = .33 z = .44 .44 = (x1000)/(100) x = 44 + 1000 = 956
8.
In testing a new rifle, the new rifle and a standard rifle are fired a large and equal number of times under similar conditions. The new rifle scored 53 hits while the old rifle scored 47 hits. For a comparison of the two rifles, the total number of hits by the two rifles may be re garded as equivalent to 100 flips of a coin and a hit by the new rifle as a head. Consider the null hypothesis that the two rifles are equally good (prob. of head = 1/2) against the alternative that the new rifle is better. Answer the following three questions without using the continuity correction.
A. The Zvalue is
1) .5 2) .6 3) .6 4) 4.3 5) none of the above
B. The significance level is
1) 0.011 2) 0.274 3) 0.726 4) less than 0.001 5) none of the above
C. The null hypothesis should
1) be rejected at 5% but not at 1% level 2) be rejected at 1% but not at 5% level 3) be rejected at either 5% or 1% level 4) not be rejected at either 1% or 5% level 5) not be continued or rejected without further information
Answer: A. (2) .6 XBAR = np = 100*.5 = 50 SIGMA = SQRT(npq) = SQRT(100 * .5 * .5) = 5
Z = (53  50)/5 = .6
B. (2) 0.274
C. (4) not to be rejected at either 1% or 5% level.
9.
Molybdenum rods produced on a production line are supposed to average 2.2 inches in length. It is desired to check whether the process is in control. Let X = length of such a rod. Assume X is approximately nor mally distributed with mean = MU and variance = SIGMA**2, where the mean and the variance are unknown.
Suppose a sample of n = 400 rods is taken and yields a sample average length of XBAR = 2 inches, and SUM((X  XBAR)**2) = 399.
To test H(0): MU = 2.2 vs. H(1): MU =/= 2.2 at level ALPHA = 8%, one would use a _____ confidence interval for MU and hence a table value of _____.
a) 92%, 1.67 b) 92%, 1.41 c) 92%, 1.75 d) 96%, 2.06 e) 96%, 1.75
10.
A rod from a production line has length X where X is normally distributed with mean = 2 and variance = 1/2.
Draw two rods X(1) and X(2) and place them end to end. The sum of their lengths is X(1) + X(2).
P(X(1) + X(2) < 3.6) = P(XBAR < 1.8) since (X(1) + X(2))/2 = XBAR, for sample size n = 2. Hence P(X(1) + X(2) < 3.6) is expressible in Z terms as
a) P(Z < SQRT(2)/5) b) P(Z < (2/5)) c) P(Z < (4/5)) d) P(Z < (1/5)) e) P(Z < (2*SQRT(2)/5))
Answer: b) P(Z < (2/5))
SIGMA(XBAR) = SQRT((SIGMA**2)/n) = SQRT((1/2)/2) = 1/2
Z = (XBAR  MU)/(SIGMA(XBAR)) = (1.8  2)/(1/2) = (1/5)/(1/2) = (2/5)
11.
Rods produced by G&R Company are normally distributed with a mean of 66 cm. and a standard deviation of 2 cm. Rods are too long to be useable if they are longer than 68.5 cm. What percentage of these rods are too long?
a) 0.1056 b) 0.1151 c) 0.3849 d) 0.3944 e) None of the above are correct.
Answer: a) 0.1056
Z = [X  MU!/[SIGMA! = [68.5  66!/[2! = 1.25
Prob.(Z>1.25) = .1056
12.
A particular type of bolt is produced having diameters with mean 0.500 inches and standard deviation 0.005 inches. Nuts are also produced having inside diameters with mean 0.505 inches and standard deviation 0.005 inches. If a nut and a bolt are chosen at random, what is the probability that the bolt will fit inside the nut?
Answer: Mean for the distribution of differences = .005 Standard deviation = SQRT((.005)**2/1 + (.005)**2/1) = .007071
Z = value of interest  mean of distribution (of differences) / standard error of the distribution of differences
Z = 0  .005/.007071 = .71
We want all the area to the right of .71
= .7611 or 76%.
13.
It is known that the lengths of a particular manufactured item are normally distributed with a mean of 6 and a standard deviation of 3. If one item is selected at random, what is the probability that it wil fall between 5.7 and 7.5?
Answer: P(5.7 < Y < 7.5) = P((5.76)/3 < Z < (7.56)/3) = P(.1 < Z < .5) = .0398 + .1915 = .2313
14.
A company manufactures cylinders that have a mean 2 inches in diameter. The standard deviation of the diameters of the cylinders is .10 inches. The diameters of a sample of 4 cylinders are measured every hour. The sample mean is used to decide whether or not the manufacturing process is operating satisfactorily. The following decision rule is applied: If the mean diameter for the sample of 4 cylinders is equal to 2.15 inches or more, or equal to 1.85 inches or less, stop the process. If the mean diameter is more than 1.85 inches and less then 2.15 inches, leave the process alone. a. What is the probability of stopping the process if the process average MU, remains at 2.00 inches? b. What is the probability of stopping the process if the process mean were to shift to MU = 2.10 inches? c. What is the probability of leaving the process alone if the process mean were to shift to MU = 2.15 inches? To MU = 2.30 inches?
Answer: a. Z = (1.85  2.00)/(.10/SQRT(4)) or (2.15  2.00)/(.10/SQRT(4)) = .15/.05 or = +.15/.05 = 3 = +3 P(Z<3 or Z>+3) = .0013 + .0013 = .0026 b. Z = (1.85  2.10)/.05 or (2.15  2.10)/.05 = 2.5/.05 .05/.05 = 5 1 P(Z<5 or Z>1) = .00000 + .1587 = .1587 c. Using MU = 2.15: Z = (1.85  2.15)/.05 or (2.15  2.15)/.05 = .30/.05 0/.05 = 6 0 P(6Using MU = 2.30: Z = (1.85  2.30)/.05 or (2.15  2.30)/.05 = .45/.05 .15/.05 = 9 3 P(9
15.
A company manufactures rope. From a large number of tests over a long period of time, they have found a mean breaking strength of 300 lbs. and a standard deviation of 24 lbs. Assume that these values are MU and SIGMA.
It is believed that by a newly developed process, the mean breaking strength 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 = (YBARMU)/(SIGMA/SQRT(n)) 2.33 = (YBAR300)/(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 = (307310)/(24/SQRT(64)) = 1.00 Area = 0.1587 or 15.87%
P(type II error) = 0.1587
16.
A certain kind of automobile battery is known to have a length of life which is normally distributed with a mean of 1200 days and standard deviation 100 days. How long should the guarantee be if the manufacturer wants to replace only 10% of the batteries which are sold?
Answer: Z = 1.28 for 10 percent failure
1.28 = (X  1200)/100
X = 1072 days for guarantee
17.
It is known from past experience that when a certain type of farm machine is used, the length of time it will run before needing an overhaul is approximately normally distributed with MU=455 hours and SIGMA=50 hours. When running the output is 100 bushels per hour.
a. What is the probability that such a machine will process at least 40,000 bushels before needing an overhaul?
b. If a large number of such machines are put into service about 25% will be running after X hours. Calculate X.
c. If 25 machines are put into service, what is the probability that their AVERAGE life will be at least 445 hours?
Answer: a. Z = (40,000455*100)/(50*100) = 1.1 prob. = .5 + .3643 = .8643
b. prob. = .25 Z = .68 X = 455*100 + 50*100*.68 = 48,900
c. S = 50/SQRT(25) = 10 Z = (445455)/10 = 1
prob. = .5 + .3413 = .84
18.
Suppose the hour life lengths, X(1) and X(2), of two brands of electronic tubes, say T(1) and T(2), are:
MU(1) = 100 MU(2) = 102 SIGMA(1)**2 = 36 SIGMA(2)**2 = 9
a. Find the value of K such that P(X(1) > K) = .93319.
b. If a tube is needed for a 106 hour time period, which brand should be selected? Why?
c. If one tube is selected at random from brand T(1), find the probability that its life will exceed 100 hours.
d. Find P(X(2)X(1) > 0).
Answer: a. P(Z > Y) = .9334 or P(Z < Y) = .0666 Therefore, Y = 1.5
Using the formula Z = (X  MU)/SIGMA: X = (Z*SIGMA) + MU K = (1.5)(6) + 100 = 9 + 100 = 91
b. Z = (106  100)/6 Z = (106  102)/3 = 1 = 1.3 P(Z > 1) = .1587 P(Z > 1.3) = .0918
T(1) should be selected because 15.87% of T(1) tubes last for 106 hours or more, but only 9.18% of T(2) tubes last that long.
c. Z = (100  100)/6 = 0/6 = 0 P(Z > 0) = .5000
d. Since the variances are known, the standard error of differences between elements equals:
SIGMA(X(2)  X(1)) = SQRT[(SIGMA(2)**2) + (SIGMA(1)**2)! = SQRT(36 + 9) = 6.71
MU(X(2)  X(1)) = 2
Therefore, Z = (0  2)/6.71 = .3 P(Z > .3) = .1179 + .5000 = .6179 or 61.79%.
19.
Suppose that you work for a brewery as a clerk to receive barley shipments. As part of your job you are to decide whether to keep or return new shipments of barley. The criteria used for making your decision is an estimation of the moisture content of the shipment. If the moisture level is too high (above 17.5%) the shipment has a good possibility of rotting before use and, therefore, a loss of money to the company. You know from past experience that the variance for all barley shipments is 36 and that your staff can process at the most 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 = 36 Take a sample, n = 9 SIGMA(XBAR) = SIGMA/SQRT(n) = 6/3 = 2
a. 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.
20.
A man purchases 100 boxes of nails, each box containing 1000 nails. If, on the average, one out of every 500 nails is rusty, how many of the 100 boxes would you expect to contain less than 2 rusty nails?
a. 18 b. 27 c. 41 d. 49 e. 68
Answer: c. 41
f(X) = (LAMBDA**X)*(exp(LAMBDA))/X], where LAMBDA = np = 2 100*(f(0) + f(1)) = 100 * (exp(2) + 2exp(2)) = 13.5 + 27.0 == 41
21.
Failures of electron tubes in airborne applications have been found to follow closely the Poisson Distribution. A receiver with sixteen tubes suffers a tubefailure on the average of once every 50 hours of operat ing time. Find the probability of more than one failure on an eight hour mission.
Answer: Using the Poisson distribution where:
P(Y) = ((e**  LAMBDA)(LAMBDA**Y))/(Y])
LAMBDA = 8/50 = .16
P(Y > 1) = 1  P(0)  P(1) = 1  ((e**(.16))*(.16**0))/0]  ((e**(.16))*(.16**1))/1] = .1152
22.
Suppose the weather forecaster is either right or wrong with his daily forecast and that the probability he is wrong on any day is .4. Assume his performance is to be evaluated on 18 randomly selected days such that his performance is independent from day to day. Let A be the event that he is wrong on less than 5 of the days.
a. Find the exact value of P(A). b. Find the approximate value of P(A), based on the Poisson approximation. c. Is the approximation in (b.) valid? Why or why not? d. Find the approximate value of P(A), based on the Central Limit Theorem. (Hint: SIGMA**2 = np(1  p)) e. Is the approximation in (d.) valid? Why or why not?
Answer: a. P = P(4 wrong) + P(3 wrong) + P(2 wrong) + P(1 wrong) + P(0 wrong) = (18C4)(.4**4)(.6**14) + ... + (18C0)(.4**0)(.6**18) = .061 + .025 + .007 + .001 + .000 = .094
b. LAMBDA = np = (18)(.4) = 7.2 P = (7.2**4)(e**7.2)/4] + (7.2**3)(e**7.2)/3] + (7.2**2)(e**7.2)/2] + (7.2)(e**7.2) + (e**7.2) = .084 + .046 + .019 + .005 + .001 = .115
c. No, because n is too small and P is too large.
d. p = .4 SIGMA = SQRT(npq) = 2.08 MU = np = 7.2 4.5 in standard units is 1.30 = (4.5  7.2)/2.08. P(Z < 1.30) = .5  .4032 = .0968
e. Yes, because np and nq are greater than 5, (or p is not close to 0 or 1 and n is at least moderate size).
23.
The probability of a snow storm on any given day during January is equal to P.
a) What is the probability of at least one snow storm during January (the month has 31 days)? Set this up in general since you have no values for P.
b) If p = 1/10, what is the probability of exactly three storms during the period beginning with January 10 and ending with January 21? Set this up but do not evaluate.
c) Use the normal approximation to evaluate the above probability in part b.
Answer: a) P(at least 1 storm) = 1P(no storm) = 1Q**31 where P = 1Q
b) P(3) = (12C3)*(.1**3)*(.9**9) = .085
c) p=.1 mean=np=1.2 SIGMA=SQRT(npq)=1.04 P(3) == P(2.5
standard score = (2.51.2)/1.04 = 1.25 standard score = (3.51.2)/1.04 = 2.21
prob. = .4864  .3944 = .092
24.
Seventy five percent of the Ford autos made in 1976 are falling apart. Determine the probability distribution of the number of Fords in a sample of 4 that are falling apart. Draw a histogram of the distribution. 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 X
mean = np = 4*.75 = 3 variance = npq = 4*.75*.25 = .75
25.
The following results were obtained from life tests on miniature bearings. Each datum represents the hours to failure in a particular turbine.
Time to Failure  Turbine No. 1 2 3 4 5 6 7 8 9 10           
Bearing 110 116 670 530 260 190 116 254 150 99 Runs 600 1130 525 242 336 414 300 213 769 140 350 90 194 112 78 558 308 280 123 108 330 930 320 41 96 690 925 92 122 260
a) Plot a frequency histogram with cell interval of 50 hours. Does the distribution look normal?
b) Plot a conventional % relative cumulative frequency curve on normal probability paper. What do you conclude?
c) Plot a similar curve on lognormal probability paper. What do you conclude?
d) Calculate the proper estimate of central tendency and dispersion.
Answer: LIFE TEST DATA
Time to Failure Frequency % Relative Cumulative Frequency    41 1 2.5 78 1 5.0 90 1 7.5 92 1 10.0 96 1 12.5 99 1 15.0 108 1 17.5 110 1 20.0 112 1 22.5 116 2 27.5 122 1 30.0 123 1 32.5 140 1 35.0 150 1 37.5 190 1 40.0 194 1 42.5 213 1 45.0 242 1 47.5 254 1 50.0 260 2 55.0 280 1 57.5 300 1 60.0 308 1 62.5 320 1 65.0 330 1 67.5 336 1 70.0 350 1 72.5 414 1 75.0 525 1 77.5 530 1 80.0 558 1 82.5 600 1 85.0 670 1 87.5 690 1 90.0 769 1 92.0 925 1 95.0 930 1 97.5 1130 1 100.0
a) No, the distribution does not look normal. (If available, consult file of graphs and diagrams that could not be computerized.)
b) This plot does not form a straight line, and thus appears to be non normal. (If available, consult file of graphs and diagrams that could not be computerized.)
c) This plot does form a straight line, and thus the log of the data appears to form a normal distribution. (If available, consult file of graphs and diagrams that could not be computerized.)
d) Arithmetic mean is 329.275 hours. Standard deviation is 269.96.
Arithmetic mean of logs is 2.382. Antilog is 241 hours. As expected, the geometric mean is less than the arithmetic mean. The geomtric standard deviation is 0.352 as is the log, and the anti log of this number is 2.25 hours.
26.
The strengths of elevator cables are to be measured. Let X = strength of a cable, and assume X is normal with mean MU and variance SIGMA**2, both unknown. A sample of 89 cables is taken, with results XBAR = 31 and S**2 = 89.
A 93% confidence interval for MU uses a table value closest to:
(a) 1.60 (b) 2.11 (c) 1.32 (d) 1.12 (e) 1.81
Answer: (e) 1.81
Use Z value because sample size is large, although t distribution would ordinarily be used when SIGMA**2 is unknown.
27.
Rods from a production line have a length X which is distributed normally with a mean of 2 and a variance of 1/2. Draw two rods X(1), X(2) and place them end to end. The sum of their lengths is X(1) + X(2).
P[(X(1) + X(2)) < 3.6! = P(XBAR < 1.8) has a value closest to:
(a) .1554 (d) .3446 (b) .2157 (e) .7843 (c) .2843
Answer: (d) .3446
Z = (X  MU)/SQRT(Variance/n)
Z = (1.8  2)/SQRT(.5/2) = 4
Area beyond Z of .4 = .3446
Therefore, the probability that the sum of the two rods will be < 3.6 is .3446.
28.
You, as a manufacturer, can use a particular part only if its diameter is between .14 and .20 inches. Two companies, A and B, can supply you with these parts at comparable costs. Supplier A produces parts whose mean is .17 and whose standard deviataion is .015 inches. However, supplier B produces parts whose mean is .16 inches and whose standard deviation is .012. The diameters of the parts from each company are normally distributed. Which company should you buy from and why?
Answer: For Supplier A:
Z = (X  MU)/SIGMA = (.14  .17)/.015 = 2
and Z = (.20  .17)/.015 = 2
Area 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 = 3.33
Area 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.
29.
A lightbulb is selected randomly from a factory's monthly production. The bulb's lifetime (total hours of illumination) is a random variable with 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
30.
Suppose that the duration of a storm on a tropical island is expo nentially distributed with mean value of THETA = 5 minutes. What is the probability that a storm on the island will last at least two minutes more, given that it has already lasted for 5 minutes?
Answer: The distribution for the duration of a storm f(X) is:
f(X) = (1/5) * (e**(X/5)) X > 0 = 0 elsewhere
P(rain will last at least 2 minutes morelasted for 5 min.) = P(X >= 7X >= 5) = (INT(INFNTY/7)((1/5)(e**(X/5))))/ (INT(INFNTY/5)((1/5)(e**(X/5)))) = (e**(7/5))/(e**1) = (e**(2/5)) = .67032
31.
A lightbulb is selected randomly from a factory's monthly production. The bulb's lifetime (total hours of illumination) is a random variable with 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=(XBAR700)/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 = (XBAR700)/70 E[Z! = (E[XBAR!700)/70 = (MU700)/70 = (700700)/70 = 0/70 = 0
VAR[Z! = VAR[(XBAR700)/70! = [(1/70)**2! * VAR(XBAR) = [(1/70)**2! * [(MU**2)/n! = [1/4900! * [(700**2)/100! = 1
b) test statistic: Z = [XBAR700!/[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), [XBAR700!/[700/SQRT(100)! >= 1.645 XBAR >= (1.645*70) + 700 XBAR >= 815.15
c) 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.
32.
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 a defective part represents a complete loss of the $10.00 to the buyer and the good parts can be resold at $14.50 each, is it worthwhile to buy one of these parts and select it at random?
Answer: Expected return value of part = .75*(14.50) + .25(0) = 10.875
Therefore, you expect to gain approximately $.87 on each part you buy, and it is worthwhile to buy one selected at random.
33.
Usually when we make use of a random numbers table we wish to arrange things so that each each event has an equal probability of occurring. If we were interested in locating 5 corn trials in a region having 48 corn farms and we wanted each farm to have an equal likelihood of being selected (in contrast to the common practice of locating trials on the farms of the growers most friendly to the local extension agent), describe a method using the random numbers table that could be used to make the selection. Indicate the five farms selected using your method.
Answer: To use a random numbers table one must do the following: 1. Make up a rule for converting digits from the table into sample identification numbers. The rule used ordinarily should make selections of each population item equally likely. It should also indicate if the same element can be counted more than once. 2. Find a starting point in the table in a manner that will not always lead to the same starting point or a small set of starting points. 3. Translate the digits that follow the starting point into sample identification numbers. In this case we will use sampling without replacement meaning that a population element can only appear once in a sample. It is also assumed that the I.D. numbers 1 to 48, have been assigned to the farms. a. The rule for converting digits is: beginning at the starting point and going left to right take a pair of digits and use those if they are in the range 1 to 48 otherwise discard. Continue this process until you get five. b. To arrive at the starting point, haphazardly put your finger on a group of digits, use the first two digits (that fit the table) to get a row number and the next two to get a column. Using this process I get row 44 and column 04 as my starting point. Starting from there I get the following pairs: 76, 54, 91, 40, 69, 90, 67, 24, 56, 83, 50, 82, 94, 81, 13, 98, 42, 87, 88, 02 Therefore, the sample would contain the following farms: 40, 24, 13, 42, 2
34.
Electron tubes made by two factories, A and B, are installed at random in single tube units. Thirty percent of the tubes are from factory B. The probability that a factory B tube will fail in the first week of operation is .1, and the probability that a factory A tube will fail is .3. If a particular unit fails in the first 100 hours of continuous operation, what is the probability that it had a tube from factory A installed? From factory B?
Answer: a. P(Afailure) = [P(A)*P(failureA)!/ [P(A)*P(failureA) + P(B)*P(failureB)! = (.7*.3)/[(.7*.3) + (.3*.1)! = .875 = 7/8
b. P(Bfailure) = 1  P(Afailure) = 1  .875 = .125 = 1/8
35.
Suppose that two of the six spark plugs on a sixcylinder automobile engine require replacement. If the mechanic removes two plugs at random, what is the probability that he will select the two defective plugs? At least one of the two defective plugs?
Answer: a. prob = 1/(6C2) = 1/15 b. prob = 1(  (4C2))/(6C2) = 9/15 = 3/5
36.
A certain assembly consists of two sections, A and B, which are bolted together. 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 section B defective. What is the probability of choosing, without replacement, 2 assemblies from the bin which have neither section A nor section B defective?
a. (76)**2/(100)**2 b. (98)**2/(100)**2 c. 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)
37.
Suppose that the probability is 0.1 that the weather (being either sun shine or rain) does not change from one day to the next. The sun is shining today. What is the probability that it will rain the day after tomorrow?
Answer: S(today)  R(tomorrow)  R(day after)
P(SRR) = (.9)(.1) = .09
S(today)  S(tomorrow)  R(day after)
P(SSR) = (.1)(.9) = .09
P(rain day after tomorrow) = .09 + .09 = .18
38.
   ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^    ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^    ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^   
I II III
Three fields have been divided into plots as in the above figure. Define an edge plot as one that is on the outside of the field (this includes corner plots).
A farmer selects at random one plot from each field, with no relation between choices from different fields. What is the probability he ends up with an edge plot from I and II but not from III? Give answer as simplified fraction.
Answer: P(edge 1) = 8/9 P(edge 2) = 10/12 = 5/6 P(not edge 3) = 3/15 = 1/5 P(edge 1, edge 2, not edge 3) = (8/9)(5/6)(1/5) = 4/27
39.
Suppose that the probability is 0.7 that the weather (sunshine or rain) is different for any given day than it was on the preceding day. If it is a sunshine day today, what is the probability that it will be rain ing the day after tomorrow?
Answer: PROB = P(sun, sun, rain) + P(sun, rain, rain) = (.3)(.7) + (.7)(.3) = .42
40.
The following table gives data which have been rounded from an actual federal report on the subject.
DISTRIBUTION OF ADMINISTRATORS FOR NURSING AND PERSONAL CARE HOMES, BY LENGTH OF TOTAL WORK EXPERIENCE AND SIZE OF THE HOME, UNITED STATES, EXCLUDING ALASKA AND HAWAII, JUNEAUGUST 1969.
 Length of ^ Under ^ ^ ^ ^ ^ ^ Size total work ^ 1 ^ 14 ^ 59 ^ 1019 ^ 20+ ^ Total ^ of home experience ^ year ^ years ^ years ^ years ^ years ^ ^ _____________________^_______^_______^_______^_______^_______^_______^ ^ ^ ^ ^ ^ ^ ^ Under 25 beds ^ 200 ^ 600 ^ 850 ^ 1100 ^ 550 ^ 3300 ^ ^ ^ ^ ^ ^ ^ ^ 2549 beds ^ 200 ^ 750 ^ 500 ^ 600 ^ 350 ^ 2400 ^ ^ ^ ^ ^ ^ ^ ^ 5099 beds ^ 250 ^ 700 ^ 550 ^ 450 ^ 250 ^ 2200 ^ ^ ^ ^ ^ ^ ^ ^ 100299 beds ^ 100 ^ 300 ^ 250 ^ 200 ^ 150 ^ 1000 ^ ^ ^ ^ ^ ^ ^ ^ 300 beds and over ^ 0 ^ 20 ^ 30 ^ 30 ^ 20 ^ 100 ^ ______________________^_______^_______^_______^_______^_______^_______^ ^ ^ ^ ^ ^ ^ ^ Total ^ 750 ^ 2370 ^ 2180 ^ 2380 ^ 1320 ^ 9000 ^ ______________________^_______^_______^_______^_______^_______^_______^
If administrator's experience were independent of size of home, find:
A. the probability that an administrator chosen at random is adminis tering a home with 25 beds or more, given that he/she has at least 10 years experience.
B. the probability that (for an administratorhome pair selected at random) the home will have 99 or fewer beds and the administrator will have a work experience of 1 to 9 years inclusive.
C. the probability of an administrator with experience of 1 to 9 years with a nursing home of 300 beds or over.
D. the number of administrators you would expect to have from 1 to 9 years experience and work in a home with 99 or fewer beds.
E. the number of administrators with 1 to 9 years experience in a home of 300 or over beds.
Answer: A. (note: If events A and B are independent,then P(AB) = P(A).)
P(25 beds+ 10 yrs.+) = P(25 beds+) = (2400+2200+1000+100)/9000 = 57/90 = .633
B. P(99 or fewer beds INTERSECTION 1  9 yrs) = [(2370 + 2180)/9000! * [(3300 + 2400 + 2200)/9000! = [4550/9000! * [7900/9000! = 0.4438
C. P(300+ beds INTERSECTION 19 yrs) = [(2370 + 2180)/9000! * [100/9000! = .0056
D. Using the probability found in part B. Expected number = 9000*.4438 = 3994.2
E. Using the probability found in part C. Expected number = 9000*.0056 = 50.4
41.
A population of 160 communities is arranged according to death rate and air pollution level as follows:
AIR POLLUTION LEVEL
Low Medium High  Low 2 6 8 ^ 16 DEATH  Medium 14 42 56 ^ 112 RATE  High 4 12 16 ^ 32  20 60 80 ^ 160
How many communities would you expect to have a low death rate and a high air pollution level if death rate and air pollution level are independent (i.e. are not associated)?
a. (16*80)/(160**2) b. (20*32)/(160**2) c. 8/160 d. 4 e. none of these
Answer: e. none of these
P(high pollution) = 80/160 = 1/2 = .5 P(low death rate) = 16/160 = 1/10 = .1 P(both) = .05
Number of communities with low death rate and high pollution = (.05)*(160) = 8
42.
A population of 160 communities is arranged according to death rate and air pollution level as follows:
AIR POLLUTION LEVEL
Low Medium High  Low 2 6 8 ^ 16 DEATH  Medium 14 42 56 ^ 112 RATE  High 4 12 16 ^ 32  20 60 80 ^ 160
This is the entire population not a sample. In view of this 
Which of the following statements is correct about the two events "low death rate" and "high air pollution level":
a. they are independent b. they are mutually exclusive c. they are exhaustive d. they are opposite e. none of these
Answer: a. they are independent
P(high pollution)*P(low death) = .5 * .1 = .05 8/160 = .05
43.
A random sample of 160 communities is distributed according to death rate and air pollution level as follows:
AIR POLLUTION LEVEL
Low Medium High  Low 2 6 8 ^ 16 DEATH  Medium 14 42 56 ^ 112 RATE  High 4 12 16 ^ 32  20 60 80 ^ 160
Which of the following statements is correct?
a. There is no evidence that air pollution level and death rate are related.
b. Death rate and air pollution level are dependent variables.
c. The CHISQUARE index for the table is very large.
d. The probability that a randomly selected community will have a high death rate will vary as the air pollution level of the community varies.
e. None of these.
Answer: a. There is no evidence that air pollution level and death rate are related.
CHISQUARE(calculated) = 0, so continue the notion in H(O) of independence.
44.
A special steel alloy has an average tensile strength of 25,800 psi. The numerical value of the variance is 1,500,000. The units assoc iated with this variance would be:
(a) (psi)**2 (c) SQRT(psi) (b) psi (d) unknown
45.
The life in months of service before failure of the color television picture tube in 8 television sets manufactured by Firm A and 8 sets manufactured 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, 48
Let ETA(A) and ETA(B) denote the median service life of picture tubes produced by the 2 firms. A confidence interval for ETA(B)  ETA(A) is bounded by the dth smallest and the dth largest of all differences of B and Aobservations. For confidence coefficient .99, we take d equal to:
(a) 9 (b) 14 (c) 15 (d) 17
46.
A Physicist comes to you (as Associate Professor of Statistics) for your help. He has a special electronic circuit consisting of Anodes and Diodes linked together in a special way. This is an experimental piece of equipment and for the 4 components (i.e. Anodes and Diodes) he knows the following probability distribution from previous experiments.
____________________________________________________________ ^ Anodes or Diodes failing per circuit ^ 0 ^ 1 ^ 2 ^ 3 ^ 4 ^ ^^^^^^^ ^ Probability that many fail ^0.1^0.2^0.3^0.3^ 0.1^ 
He creates a special circuit consisting of the same Anodes and Diodes and wonders whether the distribution has remained the same. He uses 500 of the circuits and counts the number of Diodes and Anodes that fail. He finds the following:
___________________________________________________ ^ Anodes or Diodes failing ^ 0 ^ 1 ^ 2 ^ 3 ^ 4 ^ ^^^^^^^ ^ Frequency ^ 50^105^145^155^ 45 ^ 
Could the Physicist conclude the new circuit had the same characteristics as the previous circuits? Be careful to state the Level of Significance used.
Answer: Here we need to see how well the data we have fits the theoretical (past) distribution. This is a Chi square goodness of fit problem.
H(O): The data fits the past distribution. H(A): The data does not fit the past distribution.
CHISQ = SUM([(OE)**2!/[E!)
where O = observed value E = expected value = probability*total
df= km1, k = no. of categories, m = no. of estimated parameters
Table:
No. failing ^ O ^ E ^ (OE) ^ [(OE)**2! ^ [(OE)**2!/[E! ^^^^^ 0 ^ 50^ 50^ 0 ^ 0 ^ 0.000 1 ^105^100^ 5 ^ 25 ^ 0.250 2 ^145^150^ 5 ^ 25 ^ 0.167 3 ^155^150^ 5 ^ 25 ^ 0.167 4 ^ 45^ 50^ 5 ^ 25 ^ 0.500 ^^^^^ TOTAL ^500^500^ 0 ^ 100 ^ 1.084
CHISQ (calc.) = 1.084
From tables:
CHISQ(crit., df=4, ALPHA=.05) = 9.49 CHISQ(crit., df=4, ALPHA=.10) = 7.78
Since CHISQ(calc.) < CHISQ(crit.), we shall continue (not reject) H(O) with ALPHA = 10%. It seems most likely that the characteris tics of the new circuit are the same as the previous circuits.
47.
The works known to be written by a famous author have been thoroughly analyzed as to sentence length. A newly found manuscript is claimed to have been written by the same author. The data below are taken from a sample of 2000 sentences in this new manuscript. Use CHISQUARE to decide whether the new manuscript is by the same author.
proportion of sentences _______________________ no. words in sentence known author new manuscript ____________ _________________________________ 3 or less .010 .007 45 .030 .024 68 .041 .031 912 .102 .034 1316 .263 .250 1720 .279 .203 2124 .118 .198 2527 .105 .156 2829 .042 .081 30 or more .010 .016
Answer: Number of sentences  0 14 48 62 68 500 406 396 312 162 32 E 20 60 82 204 526 558 236 210 84 20
CHISQ = (1420)**2/20 + (4860)**2/60 + (6282)**2/82 + (68204)**2/204 + (500526)**2/526 + (406558)**2/558 + (396236)**2/236 + (16284)**2/84 + (312210)**2/210 + (3220)**2/20 = 380.081
d.f. = (k1) = 9
P(CHISQ(9) >= 380.081) < .001
Reject H(0) that new manuscript by same author at ALPHA = .10, .05 or .01.
48.
On the basis of the data presented below, do we have reason to believe the geneticist who says offspring with characteristics A, B, C, and D should occur with relative frequency 1:2:4:8 in the experiment? Use ALPHA = .005.
Characteristic A B C D Number 28 60 208 304
Answer: There are a total of 600 offspring. If the geneticist is correct, then (1/(1+2+4+8))(600)=(1/15)(600) = 40 offspring are expected to have Cha racteristic A. We can compute other expected values as follows.
Characteristic A B C D
Expected Number 40 80 160 320 Observed Number 28 60 208 304 Difference (O(i)E(i)) 12 20 48 16
Now these differences look very large, which implies that the geneticist is probably wrong. A statistical test can be made by computing:
W = SUM(i=1,4)(([O(i)E(i)!**2)/E(i)),
which is the CHISQUARE test with 3 degrees of freedom.
W = (12**2)/40 + (20**2)/80 + (48**2)/160 + (16**2)/320 = 23.8
CHISQUARE(critical, ALPHA=.005, df=3) = 12.8381
Since 23.8 > 12.8381, reject the geneticist's claim.
49.
A random sample of 160 communities is distributed according to death rate and air pollution level as follows:
AIR POLLUTION LEVEL
Low Medium High  Low 2 6 8 ^ 16 DEATH  Medium 14 42 56 ^ 112 RATE  High 4 12 16 ^ 32  20 60 80 ^ 160
What would you estimate from the above table to be the probability that a randomly sampled community will have a low death rate and a high air pollution level?
a. (16*80)/160 b. (20*32)/(160**2) c. 8/160 d. 4/160 e. none of these
50.
The following table gives data which have been rounded from an actual federal report on the subject.
DISTRIBUTION OF ADMINISTRATORS FOR NURSING AND PERSONAL CARE HOMES, BY LENGTH OF TOTAL WORK EXPERIENCE AND SIZE OF THE HOME, UNITED STATES, EXCLUDING ALASKA AND HAWAII, JUNEAUGUST 1969.
 Length of ^ Under ^ ^ ^ ^ ^ ^ Size total work ^ 1 ^ 14 ^ 59 ^ 1019 ^ 20+ ^ Total ^ of home experience ^ year ^ years ^ years ^ years ^ years ^ ^ __________ ___________^_______^_______^_______^_______^_______^_______^ ^ ^ ^ ^ ^ ^ ^ Under 25 beds ^ 200 ^ 600 ^ 850 ^ 1100 ^ 550 ^ 3300 ^ ^ ^ ^ ^ ^ ^ ^ 2549 beds ^ 200 ^ 750 ^ 500 ^ 600 ^ 350 ^ 2400 ^ ^ ^ ^ ^ ^ ^ ^ 5099 beds ^ 250 ^ 700 ^ 550 ^ 450 ^ 250 ^ 2200 ^ ^ ^ ^ ^ ^ ^ ^ 100299 beds ^ 100 ^ 300 ^ 250 ^ 200 ^ 150 ^ 1000 ^ ^ ^ ^ ^ ^ ^ ^ 300 beds and over ^ 0 ^ 20 ^ 30 ^ 30 ^ 20 ^ 100 ^ ______________________^_______^_______^_______^_______^_______^_______^ ^ ^ ^ ^ ^ ^ ^ Total ^ 750 ^ 2370 ^ 2180 ^ 2380 ^ 1320 ^ 9000 ^ ______________________^_______^_______^_______^_______^_______^_______^
Decide whether work experience is associated with size of home in which the administrator works. Explain your decision.
Answer: Using the CHI SQUARE statistic to test the hypotheses:
H(0): Work experience and size of home are independent. H(A): Work experience and size of home are dependent.
CHI SQUARE = SUM[((0E)**2)/E! = 344.73
CHI SQUARE (d.f. = 16, ALPHA = .05) = 26.29
Therefore we have very strong evidence to reject the null hypothesis at the .05 ALPHA level.
51.
Brand of Tire A B C D E      151 157 135 147 146 143 158 146 174 171 159 150 142 179 167 152 142 129 163 145 156 140 139 148 147 165 166
The data in the above table give stopping distance for five brands of tires. You want to test the hypothesis that brands D and E do not differ with respect to stopping ability. This hypothesis can be tested using
a. either the sign test or the Wilcoxon signed rank test. b. the sign test, but not the Wilcoxon signed rank test. c. either the median test or the Wilcoxon twosample test. d. the median test, but not the Wilcoxon twosample test. e. the test of homogeneity.
Answer: c. either the median test or the Wilcoxon twosample test.
52.
The life in months of service before failure of the color television picture 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, 48
Let ETA(B) denote the median service life of picture tubes produced by the firm. To test the hypothesis ETA(B) = 38.5 against the alternative ETA(B) =/= 38.5, the value of CHISQ(calculated) for the median test equals:
(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
53.
Five cars are entered in a race:
starting order: 1 2 3 4 5 finishing order: 2 1 4 3 5
The Kendall rank correlation coefficient between starting order and finishing order equals
a. .4 b. .2 c. .6 d. .2 e. .4
Answer: c. .6
N(C) N(D) X Y (# concordant pairs) (# discordant pairs) 1 2 3 1 2 1 3 0 3 4 1 1 4 3 1 0 5 5 0 0   8 2
# of pairs in data = n = 5
T = [N(C)  N(D)!/[n(n1)/2! = [8  2! / [5(4)/2! = .6
54.
The observed life, in months of service, before failure for the color television picture 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 48
Let ETA(B) denote the median service life of picture tubes produced by the 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.
55.
The life in months of service before failure of the color television picture tubes in 8 television sets manufactured by Firm A and 8 sets manufactured 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 48
Let ETA(A) and ETA(B) denote the median service life of picture tubes produced by the two firms.
The Sinterval with confidence coefficient .71 for ETA(A) is bounded by:
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 dfactors for Sign Test and Confidence Intervals for the median, d = 3. The confidence interval is bounded by the dsmallest and dlargest sample observations. Thus, the Sinter val about the median is bounded by the third smallest and third largest sample observations, or 31 and 37.
56.
The life in months of service before failure of the color television picture tube in 8 television sets manufactured by Firm A and 8 sets manufactured 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 48
Let ETA(A) and ETA(B) denote the median service life of picture tubes produced by the two firms.
The Winterval with confidence coefficient .98 for ETA(A) is bounded by:
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 Winterval 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
Winterval is 27 and 39.5.
57.
The coded values for a measure of brightness in paper (light reflectivity), prepared by two different processes, are as follows for samples of size 9 drawn randomly from each of the two processes:
A B ___ ___
6.1 9.1 9.2 8.2 8.7 8.6 8.9 6.9 7.6 7.5 7.1 7.9 9.5 8.3 8.3 7.8 9.0 8.9
Do the data present sufficient evidence (ALPHA = .10) to indicate a difference in the populations of brightness measurements for the two processes?
a. Use the sign test. b. Use the MannWhitney rank test.
Answer: H(O): Brightness has the same distribution under both processes
H(A): Brightness has different distributions under the two processes
a. Sign test:
The signs associated with the differences are: ,+,+,+,+,,+,+,+.
The smaller number of like signs is 2. With 9 pairs, 1 or fewer signs are required for significance at the .10 level, therefore we continue the null hypothesis of no difference.
b. MannWhitney rank test:
A B
6.1 1 9.1 16 9.2 17 8.2 9 8.7 13 8.6 12 8.9 14 6.9 2 7.6 5 7.5 4 7.1 3 7.9 7 9.5 18 8.3 10.5 8.3 10.5 7.8 6 9.0 15 8.1 8   96.5 74.5
With n(1) = 9 and n(2) = 9, a value of 63 or less for the smaller sum of ranks should lead to rejection at the .05 level. Therefore we will continue H(O) at both the .05 and .10 levels.
58.
Model ^ G F C  Compacts ^ 20.3 25.6 24.0 Intermediate 6s ^ 21.2 24.7 23.1 Intermediate 8s ^ 18.2 19.3 20.6 Fullsize 8s ^ 18.6 19.3 19.8 Sport Cars ^ 18.5 20.7 21.4
The data in the above table give gasoline mileage for various types of cars produced by three different manufacturers. You want to compare cars produced by manufacturers G and F. The hypothesis that gasoline mileage does not differ for the two manufacturers can be tested using
a. either the sign test or the Wilcoxon signed rank test. b. the sign test, but not the Wilcoxon signed rank test. c. either the median test or the Wilcoxon twosample test. d. the median test, but not the Wilcoxon twosample test. e. the test of homogeneity.
Answer: a. either the sign test or the Wilcoxon signed rank test.
59.
The life in months of service before failure of the color television picture tube in 8 television sets manufactured by Firm A and 8 sets manufactured 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 48
Let ETA(A) and ETA(B) denote the median service life of picture tubes produced by the two firms.
You want to test the hypothesis ETA(A) = 38 against the alternative ETA(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.
60.
An experiment was designed to compare the durability of two highway paints named type A and type B. An "A" strip and a "B" strip were painted across a highway at each of 30 locations. At the end of the test period the following results were observed: at 6 locations type A showed the least wear, at 15 locations type B showed the least wear, and at 9 locations both had the same amount of wear. Use the sign test at the 5% level to test that both paints have equal durability.
Answer: H(O): P(A < B) = P(B < A) = .5
(NOTE: All tied cases are dropped from the analysis for the sign test.)
n = 21 X = the number of fewer signs = 6
Using appropriate table: P(X <= 3) = .001 Therefore, reject the null hypothesis.
61.
The observed life, in months of service, before failure for the color television picture 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, 48
Let ETA(B) denote the median service life of picture tubes produced by the firm and assume the lifetimes have symmetric distributions. You want to test the hypothesis ETA(B) = 38.5 against the alternative ETA(B) =/= 38.5 using the Wilcoxon signed rank test. From the following 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 8
W() = SUM(R()) = 3.5 + 1.5 = 5
62.
The state highway department is collecting data to determine whether a highway's repair priorities should be raised, lowered, or should remain the same. The decision will be made in the following manner. If the population median of traffic flow = 100 cars per day, the priority will remain the same. If the population median of traffic flow > 100 cars per day, raise the priority. If the population me dian of traffic flow < 100 cars per day, decrease the priority. Data for nine randomly selected days is as follows:
Traffic Flow: 88, 91, 89, 101, 93, 86, 95, 98, 92
Can we conclude at ALPHA = .05 that the median number of cars per day is 100?
(a) Pick the most appropriate nonparametric procedure. (b) State null and alternative hypotheses. (c) Compute a test statistic. (d) Indicate your critical values. (e) Do you or do you not reject H(0)? What is your conclusion? What happens to the road in question?
Answer: (a) Use Wilcoxon test.
(b) H(0): Md = 100 H(1): Md =/= 100
(c) X D(i) = X  100 ABS(D(i)) Rank    
86 14 14 9 88 12 12 8 89 11 11 7 91 9 9 6 92 8 8 5 93 7 7 4 95 5 5 3 98 2 2 2 101 +1 1 1
T = 1 R+ = 1 (Sum of positive ranks) R = 44 (Sum of negative ranks)
(d) lower W = 6 upper W = 9(10)/2  6 = 39
(e) (T=1) < 6, therefore reject H(O). Conclude that the median is less than 100 cars per day, and decrease the priority.
63.
Ten randomly selected cars of a specific year, make, and model and with similar equipment, are subjected to an EPA gasoline mileage test. The resulting miles/gallon are:
24.6, 30.0, 28.2, 27.4, 26.8, 23.9, 22.2, 26.4, 32.6, 28.8
Using the Wilcoxon Median Test, test the hypothesis that the population median 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 1
R+ = 3.5 > T = 3.5 R = 41.5
Lower w = 9 Upper w = (9*10)/2  9 = 36
Since (T=3.5) < 9, we reject H(0): median = 30.
For the confidence interval, we need the 11th largest and smallest values, 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 ^ 22.2
Therefore, 90% C.I.: from 25.2 to 28.8.
64.
On eight occasions of cloud seeding, the following amounts of rainfall were observed: .74, .54, 1.25, .27, .76, 1.01, .49, .70. On six control occasions (when no cloud seeding took place), the following amounts of rainfall were measured: .25, .36, .42, .16, .59, .66.
We test (using the Wilcoxon  Mann  Whitney test) the hypothesis that cloud seeding does not increase amount of rainfall against the alter native that it does. The descriptive level for the given data equals
a. .00 b. .01 c. .02 d. .04 e. .05
Answer: c. .02
U(S) = number of times seeded observations are larger than control observations U(S) = 6 + 4 + 6 + 2 + 6 + 6 + 4 + 6 = 40 U(NS) = (8*6)  40 = 48  40 = 8
from table: P(U(NS) < 9) = .021
65.
The life in months of service before failure of the color television picture tube in 8 television sets manufactured by Firm A and 8 sets manufactured 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, 48
Against the twosided alternative, the Wilcoxon (Mann Whitney) two sample test has descriptive level:
(a) .050 (b) .010 (c) .007 (d) .004
Answer: (c) .007
U(A) = 0 + 0 + 0 + 0 + 1 + 2 + 2 + 2 = 7 U(B) = 64  7 = 57 P(U(A) <= 7) = .007
66.
The life in months of service before failure of the color television picture tube in 8 television sets manufactured by Firm A and 8 sets manufactured 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, 48
Suppose the data is ranked as one combined set. The sum of the ranks R(B) for the Bobservations equals:
(a) 36 (b) 43 (c) 57 (d) 93
Answer: (d) 93
Table of Ranks:
Firm A: 1 2 3 4 6 8 9 10 Firm B: 5 7 11 12 13 14 15 16
SUM(R(B)) = 5 + 7 + 11 + 12 + 13 + 14 + 15 + 16 = 93
67.
Question type 1 2 3 4 An investigator is interested in teachers' use of various Teacher types of questions in teaching A 9 1 9 2 mathematics. He identifies 4 B 4 6 7 0 types of questions which de C 8 2 5 1 mand responses of different D 6 9 2 3 levels of complexity. He re E 7 5 6 2 cords the number of questions F 7 3 4 1 of each type asked by each G 8 5 2 5 teacher in a random sample of H 8 9 7 1 10 teachers. The frequencies I 6 5 8 4 are reported for teacher and J 7 2 5 1 question type.
The most appropriate nonparametric test for these data would be:
A. MannWhitney Test B. Friedman Test C. Wilcoxon Test D. CHISQUARE Test of Homogeneity
Answer: B. Friedman Test
We are interested in comparing the average ranks for the four question types. Friedman will do this directly.
68.
Using the list of designs below, indicate which type of design is most descriptive of the following study:
a. one shot case study b. factorial design c. time series design d. nonequivalent control group e. corelational study f. one group pretest posttest g. equivalent time series h. patched up design i. posttest only control group design j. criterion group k. pretestposttest control l. separate sample pretestposttest
A company making electric drills has kept accurate monthly records of the number of faulty drills sold as indicated by the number of them that have been returned to the factory. Because this number has been increasing, the company has instituted a training program to im prove the skills of its inspectors in the hopes that fewer faulty drills will be distributed. The company plans to assess the effects of this training by continuing to examine monthly records of how many drills are returned to the factory for repairs after the training pro gram has been completed.
Answer: c. time series
This study consists first of repeated observation, then conduct of training program, and then continued observation.
69.
Regarding the testing of ammunition using a 16inch gun subject to a linear trend for wear, W.J. Youden makes these comments on a testing sequence
AAAA/BBBB/..../EEEE where A, ..., E are brands of shells
"Nothing good comes from this work. The averages are worthless. Each average depends on its position in the firing order. The estimate of the experimental error based on repeat rounds fired in succession ob viously has no applicability for judging differences between ammunitions not fired in immediate succession."
a. Why are the averages worthless? b. What is wrong with the estimate of experimental error?
Answer: a. The averages are worthless because differences between ammunition means are mixed up with differences in firing order where firing or der is an important source of variation. If we let, say, RHO(1) re present the effect of testing during the first four firings, RHO(2) represent the effect of testing during the second four firings, etc. and let TAU(A) represent the effect of Brand A ammunition, TAU(B) represent the effect of Brand B ammunition, etc., then the differ ence YBAR(A)  YBAR(B) = (TAU(A) + RHO(1))  (TAU(B) + RHO(2)). The proper difference is distorted by an amount (RHO(1)  RHO(2)).
b. Repeat rounds fired in succession will tend to agree with each other much more than rounds scattered randomly through the firing se quence. They will underestimate the variance that should be used in comparing brands of ammunition.
70.
An Experiment is to be conducted in a greenhouse to compare three treatments. The 9 experimental units to be used will be arranged as below:
Units
Greenhouse X X X
Heat X X X
Source(Radiator) X X X
Since the plants used in the experiment are sensitive to heat, it is expected that the closer an experimental unit is to the heat source, the poorer the response that will be obtained. Otherwise the experimental units are considered to be about the same.
a. How would you assign treatments to experimental units? Explain. b. What, if any, experimental design is defined by this method of assignment?
Answer: a. I would assign treatments such that each treatment would occur in each of the distances from the radiator because we suspect that the distrance from the heat will affect the response. b. This is the randomized complete block design, with distance from the heat source used as the blocking factor.
71.
Suppose that you have been appointed energy czar of New Hampshire and have been instructed to provide guidance to consumers on the relation of speed to miles per gallon for various makes of cars. Disregarding cost considerations, which of the following static test schemes would you prefer? Why?
a.) 5 tests all at 25 mph b.) 2 tests at 25 mph, 3 tests at 55 mph c.) 1 test at 25, 1 at 35, 2 at 45, and 1 at 55 mph d.) 1 test each at 25, 35, 45, 55, and 65 mph
Answer: d.) Because I would be able to say something about speeds of 25 and 65 without extrapolating, and I would be able to say more about the whole range because it is well covered using this scheme.
72.
Suppose that you are a member of a garden club that has 30 members. The club has fallen into controversy as to whether or not planting garlic next to Baby's Breath is an effective way to reduce insect attacks on Baby's Breath.
One club member proposes that the best gardener in the club plant various parts of his garden either with Baby's Breath alone or with Baby's Breath next to garlic. The proposal is that random selection be used so that half of 20 areas are planted one way and half are planted the other way.
Another club member disagrees. He suggests that each club member plant 2 areas  one with Baby's Breath alone and one with Baby's Breath plus garlic (again randomly assigning treatment to area). Which scheme do you favor? Why?
Answer: I favor the second scheme because the scope of inference will be much broader. After the study is finished, the results would apply to many types of soil conditions, environmental conditions and gardener's abilities.
73.
Suppose that you are an entomologist and wish to test 4 compounds that are said to attract a certain insect. You have 16 insect traps in a square (4X4) arrangement in a field that contains a growing crop. You think, but are not sure, that either wind direction or distance from insect source may affect number trapped. Your traps are like this:
Wind >
X X X X
X X X X
X X X X
X X X X
Insect Source /] ]
a. How many df will be associated with experimental error if the design used for this situation is i) completely random? ii) randomized blocks (with one block consisting of the traps clos est to the insect source, ..., another block consisting of the traps furthest from the source)? iii) latin square?
b. What model terms have to be important if it's to be worthwhile to have used the latin square?
c. Which model terms (or influences) have to be unimportant if a CR design is to be a good choice?
Answer: a. i) t(r1) = 4(41) = 4(3) = 12 ii) (r1)(t1) = (41)(41) = (3)(3) = 9 iii) (t1)(t2) = (41)(42) = (3)(2) = 6 where t = number of treatments r = number of blocks or repetitions
b. Both wind and distance from insect source have to be important in order for the latin square design to be worthwhile, therefore, both RHO(j) and KAPPA(k) for all i and k must be important.
c. Both wind and distance from insect source, or RHO(j) and KAPPA(k) for all j and k, must be unimportant for the completely random design to be appropriate.
74.
A computer user wishes to compare two programs in terms of amount of computer time used. Both programs perform the same analysis and use the same data set. The computer user randomly selects 10 time periods during the time when the computer ordinarily is used. For each of these periods, both programs are run. Each time a random method is used to decide which program should be run first. What experimental design has been used?
Answer: A randomized complete block (RCB) design has been used to compare the two programs. The implied blocking factor is time period, since each program must run within each time period. Each treatment (program) occurs randomly within each time period.
75.
A computer user is charged for the amount of computer time that he uses. He has at his disposal two programs that perform the same analysis. He sets up a standard data set and wishes to compare programs in terms of computer time used. He randomly chooses 20 time periods from those times when he ordinarily would use the computer. Programs are randomly assigned to these 20 periods subject only to the requirement that each program be run 10 times. What experimental design has been used?
Answer: The completely random design has been used since the treatments (programs) have been randomly assigned to the experimental units without any restriction on randomization except that each treatment be assigned 10 times.
76.
An experiment has been conducted in which two computer programs were compared in terms of computer time needed to perform the same analysis of the same data.
Results obtained included:
ANOVA Source of Variation df SS M.S. Total 20   Mean 1   Corrected Total 19   Treatments 1   Error 18  
Means Program 1 20 Program 2 15
LSD(at ALPHA .05) = t * S(dBAR) = 2
a. Which computer program would you use in the future? Why? b. What design is suggested by this report?
Answer: a. I would use program 2 because the difference in mean times between the two programs is 5 which is larger that the LSD of 2, which indicates that the observed difference was unlikely (at the .05 level) to have occurred by chance alone. Thus, the sample data indicates a significant difference between the two programs. b. A completely random design is indicated by the ANOVA table because no degrees of freedom have been subtracted from the total for any blocking factors. The total loss in degrees of freedom is attributable to the mean and one treatment factor.
77.
An imaginary experiment was conducted to compare length of life of batteries sold be 4 manufacturers.
Results obtained included:
ANOVA Source of Variation df SS M.S.
Total 16   Mean 1   Corrected Total 15   Type of flashlight 3   Month of testing 3   Brands 3   Error 6  
Means Brand c 4.0 Brand b 3.9 Brand a 3.5 Brand d 3.3
LSD(.05) = .2
a. What brand(s) would you suggest buying if all prices were the same? Why? b. What design was used? c. What does the ANOVA table tell you about how the data was obtained?
Answer: a. I would suggest buying brands c or b because although they are not significantly different from each other, both of them are significantly different from the other two brands, based on the LSD at a .05 significance level. b. A latin square design is implied by the ANOVA table. Two blocking factors are indicated; type of flashlight and month of testing, each having four levels. c. The ANOVA table indicated an LSQ design with the two restrictions on randomization of treatments to experimental units. Each brand of battery was required to occur once within each month and once with each type of flashlight.
78.
Suppose that you are in charge of product testing for a chemical company. You are persuaded that your company had a new product that is a promising way of relieving insomnia.
a. Suppose that the resources available only permit testing one other treatment in addition to the new compound. What will that treatment be? Why?
b. Suppose that available resources permit testing two treatments besides the new product. What will be your choice of treatments? Why?
c. What will be your choice if five additional treatments can be tested (in addition to the new product)? Explain you choice.
Answer: a. The other treatment would be a control so that there will be some real basis for comparison to see if the product does any good at all b. Then I would choose a control and a product which is believed to be effective at relieving insomnia. c. If I could test 5 additional treatments, they would be a control and four treatments for insomnia. If four were much more widely used or more interesting that others, they would be the four tested. If all insomnia preparations were about equally interesting, then I would choose four randomly.
79.
Sominex commercials repeatedly present endorsements of the form: I take Sominex and sleep something fierce. Some skeptics would suggest that many, if not all, of the Sominex endorsers could take an inert pill and sleep something fierce. Why would an inert pill be a better "control" for testing the effectiveness of a sleeping potion that no pill at all? (e.g. test subjects might report not being sleepy, then they might randomly receive either Sominex or no pill at all.)
Answer: The problem associated with not taking any pill at all to compare with taking a Sominex pill is that such a design has not controlled for the possibility of an effect on response of taking "any" pill. That is, some people may take a pill to sleep, and just the act of taking a pill, which they think works will in fact have an effect on their ability to sleep. By comparing Sominex to an inert pill, or control, one can control for this possible effect and look at just the effect of Sominex. Since both groups are going through the process of "taking pills", this effect when comparing the two has been controlled and the differences between the two groups will be due to the ingredients of the pill alone, rather than including effects of "taking a pill".
80.
Suppose that you have identified three pocket calculator models that you regard as suitable for your work and comparable in price. Suppose that you will make your decision on which one to buy on the basis of the time that it takes for you to perform a particular set of calculations. You feel that you are equally familiar with all 3 models, but suspect that if you repeat the same set of calculations over and over again, you will become slower and slower. Suppose that it is reasonable to repeat the calculations six times on each machine. Which design will you use? Why?
i) Completely Random ii) Two 3X3 Latin Squares iii) Randomized Block
Answer: Choose two 3X3 Latin Squares because the position in the testing process appears to have a significant effect on the response, and, therefore, this effect should be balanced out.
81.
The time required for a computer user to finish running a program to perform a particular analysis depends on many things such as Language used by the program, Number of other jobs being processed on the system, Amount of data being summarized, etc.
Suppose that 3 programs were available to perform the same analysis where the chief difference among programs was the language used for writing them. Suppose that these languages were:
1. FORTRAN 2. BASIC 3. Assembler
Suppose further that 3 terminals were available and all 3 versions could be run at the same time. (A different person at each terminal). Which of the following designs would you use? Why?
i) Randomized Block ii) Latin Square iii) Completely Random
Answer: I would use a Randomized Block design so that each program would run at the same time. This assumes that there will be no important effect of person or terminal on the amount of time needed to run a program.
82.
An investigator wished to study the effect of an operator on the performance of a machine. He could arrange to have each of four operators run the machine five times. A response measurement could be recorded each time the machine was used. How many experimental units will 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.
83.
An investigator plans to conduct an experiment to evaluate three different methods for measuring a chemical. An experimental unit will involve the activities of a technician during 12 time periods over six days. When asked what pattern of variation he would expect if the technician used the same method in each of those time periods, the investigator responded that all measurements would be about the same, that he expected no re gular pattern of variation. Which of the following experimental designs do you recommend? Why? i) Latin Square ii) Randomized Block iii) Completely Random Design
Answer: I recommend iii, Completely Random Design because there are no expected influencing factors for which we should balance the effects.
84.
Suppose that we wished to compare two drivers in terms of miles travelled per gallon of gas used. Suppose that an experimental unit consists of independently driving over a single prescribed course involving about the same traffic and both city and open road driving within a prescribed time range. In order to broaden the scope of this comparison, it is desired to use: a Cadillac Ford Volkswagen Datsun Z Jeep Mazda a. How would you set up this trial to balance the effects of kind of car on driver performance? What design would you use? b. How would you balance both the effects of kind of car and day of travel?
Answer: a. I would use a randomized block design and use kind of car as a blocking factor, having each driver drive in each of the cars. b. If length of time required to complete a run were short enough so that two comparable runs could be finished on the same day, I would define a block as 2 runs on the same day involving the same kind of car. Drivers would be randomly assigned to runs each day. If it was not reasonable to regard 2 runs on the same day as comparable, then a scheme involving a 2 X 2 latin square for each brand of car might be useable.
85.
You now have at your disposal one 16inch gun. You are to develop a testing scheme to test shell velocity for five brands of ammunition. Because of the amount of explosive required for each round, each supplier will provide just four rounds.
While the gun to be used for testing has a new barrel, you know from previous experience that gun barrel wear is pronounced. In fact, you are willing to assume that the only consequential uncontrolled environmental factor is gun wear and that order of firing has a linear effect on shell velocity. (Velocity of shell declines from firing 1 to firing 20.)
a. Define experimental unit. b. Propose a testing scheme and assign brands to experimental units. c. Will your testing scheme avoid distortions due to firing order if the influence of firing order is I = 50  2T (I: influence, T: time of firing, T = 1, 2, ..., 20)?
Answer: a. An experimental unit is one firing of the gun, and all the preparation that goes with it. b. I propose that the randomized complete block design be used. I would assign the treatments to the experimental units within four blocks of five sequential units. The following is an example of an assignment of treatments to experimental units.
Block 1 2 3 4
Treatment 1 2 6 15 18 2 5 9 11 19 3 1 10 13 20 4 3 7 14 17 5 4 8 12 16
c. This scheme will not entirely avoid distortions due to firing order. The randomized complete block goes a long way in balancing these distortions.
If you had five rounds of each brand of ammunition you could use a Latin Square design and eliminate the distortions altogether. (It's also possible to use 4 columns of a 5 X 5 Latin Square, but it's not expected that this option would be proposed in an introductory class.)
Influence values for the above randomization (e.g. for firing #1. I = 50  2 = 48 firing #2. I = 50  4 = 46)
Treatment Total Influence 1 46 38 20 14 118 2 40 32 28 12 112 3 48 30 24 10 112 4 44 36 22 16 118 5 42 34 26 18 120
Treatment 5 had the largest advantage due to firing order, but RCB randomization largely balanced firing order effects.
86.
An investigator wishes to explore the effectivess of four coagulants (named A, B, C, and D) in removing suspended material from water. He proposes to try these coagulants on six different water samples where each sample is large enough to provide an aliquot for testing each coagulant. The proposed procedure consists of
* Preparing enough of each coagulant for Sample 1 where the order of preparing coagulants is random * Treating aliquots of Sample 1 in the same order and putting samples on the stirrer in that order * Repeating the same process for each of the other five water samples (using a different randomization for each sample).
a. How would you conduct a uniformity trial for the above situation? b. What would it tell you?
Answer: a. You would conduct a uniformity trial by running the whole experi ment but using only one of the coagulants throughout.
b. It would show you any patterns of variation in response among the experimental units independent of the treatments. It would also give you an estimate of the background variation.
87.
The average number of hours an electric circuit lasts before failing is 100 hours. An engineer claims that he can develop a circuit that increases the average life of the circuit. It is desired to test H(0): MU = 100 against the appropriate al ternative hypothesis. The alternative hypothesis is best rep resented as:
(a) H(A): MU =/= 100 (c) H(A): MU < 100 (b) H(A): MU <= 100 (d) H(A): MU > 100
Answer: (d) H(A): MU > 100
88.
We want to compare two machines for production line speed of beer bottle manufacturing. At the end of n(1) = 9 days, the number of bottles pro duced by machine 1 yield XBAR(1) = 19, S(1)**2 = 4. For machine 2, we have n(2) = 6 days, XBAR(2) = 17, and S(2)**2 = 9. Assume independence of samples, normality, and equality of unknown variances.
In testing H(0): MU(1) = MU(2) vs. H(1): MU(1) =/= MU(2), the observed value of our statistic is:
(A) (2)/(SQRT(77/13))(SQRT((1/9) + (1/6))) (B) (2)/SQRT((4/9) + (9/6)) (C) (2)/(SQRT(90/13))(SQRT((1/9) + (1/6))) (D) (2)/SQRT((16/9) + (81/6)) (E) (2)/(SQRT(36/13))(SQRT((1/9) + (1/6)))
Answer: (A) (2)/(SQRT(77/13))(SQRT((1/9) + (1/6)))
First we must find the pooled variance: S(P)**2 = ((8)(4) + (5)(9))/(9 + 6  2) = 77/13
Now standard error of the difference between means: S(XBAR(1)  XBAR(2)) = (SQRT(77/13))(SQRT((1/9) + (1/6)))
The observed tvalue is: t(calculated) = ((19  17)  0)/S(XBAR(1)  XBAR(2)) = 2/S(XBAR(1)  XBAR(2))
89.
We want to compare two machines for production line speed of beer bottle manufacturing. At the end of each of n(1) = 9 days, the number of bottles produced by machine 1 yield XBAR(1) = 19, S(1)**2 = 4. For ma chine 2, we have n(2) = 6 days, XBAR(2) = 17, and S(2)**2 = 9. Assume independence of samples, normality, and equality of unknown variances.
Suppose we test H(0): MU(1) = MU(2) vs. H(1): MU(1) =/= MU(2) at ALPHA=.05, and the value of the test statistic is 2.20. Then we should:
(A) Do not reject (continue) H(0) (B) Reject H(0) (C) Do not reject (continue) H(1) (D) Both (B) and (C) (E) Both (A) and (C)
Answer: (D) Both (B) and (C)
Given that: t(calculated) = 2.20 and find : t(critical, ALPHA=.05, twotailed, df=13) = +/ 2.160.
Since t(calculated) falls in the area of rejection, we would reject H(0) and would not reject (continue) H(1).
90.
Molybdenum rods are produced by a production line setup. It is desir able to check whether the process is in control. Let X = length of such a rod. Assume X is approximately normally distributed with mean = MU and variance = SIGMA**2, where the mean and variance are unknown.
Take n = 400 sample rods, with sample average length XBAR = 2 inches, and SUM((X  XBAR)**2) = 399.
In testing H(0): MU = 2.2 vs. H(1): MU =/= 2.2 at level ALPHA = 8%, one should _____ the H(0) since the value _____ lies _____ the confidence interval.
a) continue, 2.2, within b) reject, 2, outside of c) reject, 2.2, outside of d) continue, 2, within e) either b or c
Answer: c) reject, 2.2, outside of
S**2 = 399/399 = 1 S(XBAR) = SQRT(S**2/n) = .05
If you center the confidence interval on the sample mean the confidence interval = 2 +/ (1.75)(.05) = from 1.9125 to 2.0875
which does not contain the hypothesized value, 2.2.
91.
Molybdenum rods are produced by a production line setup. It is desired to check whether the process is in control. Let X = length of such a rod. Assume X is approximately normally distributed with mean = MU and variance = SIGMA**2, where the mean and variance are unknown.
Take n = 400 sample rods, with sample average length XBAR = 2 inches and SUM((X  XBAR)**2) = 399.
If one were testing H(0): MU = 1 vs. H(1): MU =/= 1 at level ALPHA = _____, one should _____ the H(0) since the value 1 lies _____ the confidence interval.
a) 16%, not reject (continue), within b) 8%, not reject (continue), within c) 4%, not reject (continue), within d) 4%, not reject (continue), to the left of e) 4%, reject, to the left of
Answer: e) 4%, reject, to the left of
S**2 = 399/399 = 1; S(XBAR) = SQRT(S**2/n) = .05; C.I. = 2 +/ Z(ALPHA/2) * .05;
Z(16%/2) = 1.41, Z(8%/2) = 1.75, Z(4%/2) = 2.05
C.I.(ALPHA=16%) = 2 +/ (1.41)(.05) = from 1.93 to 2.07
C.I.(ALPHA= 8%) = 2 +/ (1.75)(.05) = from 1.91 to 2.09
C.I.(ALPHA= 4%) = 2 +/ (2.05)(.05) = from 1.90 to 2.10
1 is not included in any of the confidence intervals, so H(0) should be rejected in all cases.
92.
It is known that long, thin titanium rods lengthen with increasing temperature. A sample of n=20 identical titanium rods is selected. Each is subjected to a particular uniform temperature for a specified time. Let Y denote the change in length. The readings are (X(1),Y(1)),...,(X(20),Y(20)), with data XBAR=2 (in hundreds of degrees F), YBAR=3 (in milliinches), SUM(XXBAR)**2 = 10, SUM(YYBAR)**2 = 40, and SUM(XXBAR)(Y YBAR) = 16.
In testing H(0): RHO = 0 vs H(1): RHO =/= 0 (RHO = population value for Pearson correlation coefficient) at level ALPHA = 5%, one should ___ H(0) since the statistic r/SQRT((1r**2)/(n2)) = ____ is ____ than the correct table value of ____.
(a) reject, 5.7, greater, 2.086 (b) reject, 5.7, greater, 1.734 (c) reject, 5.7, greater, 2.093 (d) reject, 5.7, greater, 2.101 (e) continue, 1.6, less, 2.086
Answer: (d) reject, 5.7, greater, 2.101
93.
In order to compare two brands of tires, Nader's Raiders selected five tires of each brand, measuring the mileage for which each tire gave adequate service. The result of the test (expressed in thousands of miles) were:
BRAND A BRAND B   28.2 25.5 24.9 25.4 23.0 25.3 21.8 25.0 28.1 24.8   126.0 126.0
Assuming both brands sell for the same price, which brand of tire would you say is the better buy? (Do not compute standard deviations.)
Answer: Under the no computation requirement, it appears Brand B is the better buy, since the mileage figures are more consistent and the rank ordering (highest to lowest) AABBBBABAA seems to favor B.
94.
Two types of paint are to be tested. Paint I is somewhat cheaper than paint II. The test consists of giving scores to the paints, after they have been exposed to certain weather conditions for a period of 6 months. Five samples of each type of paint are scored as follows:
Paint I ^ 26 16 20 25 23 ________________________________ Paint II ^ 20 28 32 25 25
We should like to adopt paint I, the cheaper one, unless we have definite reason to believe that paint II is better. Test the hypothesis that MU(2) <= MU(1) at level of signifi cance ALPHA = 10.
A. State your test statistic and critical region. B. Perform your calculations. C. State your conclusions.
Answer: Assumptions: (a) Both populations are normal and independent. (b) Populations have the common variance SIGMA**2.
A. Test statistic: t = [XBAR(1)  XBAR(2)!/[S(XBAR(1)  XBAR(2))! t(critical, ALPHA=.1, df=8) = 1.397
B. t = [XBAR(1)  XBAR(2)!/[S(XBAR(1)  XBAR(2))!
where: n(1) = n(2) = 5 S(XBAR(1)  XBAR(2)) = SQRT((S(1)**2/n(1)) + (S(2)**2/n(2)))
XBAR(2) = 26 XBAR(1) = 22 S(2)**2 = 19.5 S(1)**2 = 16.5
If t(calc) <= t(crit), we reject H(0).
t(crit) = 1.4 t(calc) = (2226)/2.68 = 1.491
C. Since t(calculated) < t(critical), we reject H(0): MU(2) <= MU(1) and adopt paint II.
95.
Five parallel determinations of zinc in an organic substance have been obtained. The results arranged in order are: 16.84%, 16.86%, 16.91%, 16.93%, and 17.08%. The initial reaction is a desire to discard the highest value, which seems to be an outlier.
a) Briefly describe the considerations which should arise in the mind of the experimenter in deciding how to treat the data.
b) Perform a statistical test to determine if the datum should be rejected. What is the inherent weakness of such tests?
Answer: a) 1. Are there any physical reasons on which to base a rejection? (i.e., dirty glassware, spill, etc.) 2. Are data normally distributed? 3. What are requirements of the analyses?
b) Using the first 4 measurements to calculate XBAR and S(X):
XBAR = 16.885 S(X) = 0.0420
t(calc) = (17.08  16.885)/(.042*SQRT(1+(1/4))) = 4.153
t(crit, ALPHA=.05, df=3, twotailed) = 3.182
Since t(calc) > t(crit), the datum does appear suspect and should be considered for deletion.
NOTE: In this case,
VAR(X(5)XBAR) = VAR(X(5)  [[X(1)+X(2)+X(3)+X(4)!/4!) = SIGMA**2 + [1/16![4*(SIGMA**2)!  2COV(X(5),XBAR) = (SIGMA**2)(1 + (1/4))
(COV(X(5),XBAR) = 0, since X(5) and XBAR are independent.)
96.
A report on the effect of a nuclear power plant on number of fish per unit area in nearby waters states that the hypothesis under test was: H(0): MU(1)  MU(2) >= 20 where
MU(1) is mean number of fish during the year before the plant was constructed; and
MU(2) is mean number of fish during the year after the plant began operation.
a. Is this a one tailed or a two tailed test?
b. Will a sample difference of 19 ever result in rejection of this claim? (i.e. will a reduction of 19 ever lead to rejection of this claim.)
Answer: a. one tailed test
b. yes
97.
a. The information sheet of an insecticide company carried this statement:
"Differences in control between our material and the current standard material were not statistically significant at the 99% confidence level."
How do you interpret this statement? What, if any, additional information would you like to have in order to make a choice of which material to use?
b. Suppose that the statement read that differences were significant at the 99% confidence level. How would you interpret this state ment? What, if any, additional information would you like to have in order to make a choice?
(In answering this question, it may be helpful to picture an experiment as a way of sampling a population of differences in response to these 2 chemicals. Commonly, it's hypothesized that such a population of differences has a mean of zero, i.e., H(0): MU = 0. Such a hypothesis is readily tested by using sample dif ferences to set confidence limits.)
Answer: a. As the statement stands, it simply means that the new material is neither better, nor worse, at controlling insects than the current standard material. However, no information is available on the res pective means and the corresponding measures of variation; what size sample was used to make the comparison; was a one or two tailed test made on the difference or a confidence interval created around the observed difference; what was the standard error of the difference?
b. My first reaction to the statement of significant differences at the .99 level is what was the observed difference, i.e., which product controls insects better] However, it still lacks the necessary in formation to support such a statement. Once again, knowledge of the size of the observed difference, the corresponding tvalue or confi dence interval, and measures of variation are needed to judge the merit of the concluding statement. Statistically significant dif ferences can be observed simply due to large sample sizes or dif ferences that are just barely significant may indicate the need for further experimentation or replication.
98.
In a study of learning ability, six boys and six girls were chosen at random from a kindergarten class. Scores were obtained as measure ments of their ability to learn nonsense syllables. Students were paired on the basis of IQ (that is, the boy with the lowest IQ was paired with the girl with the lowest IQ, etc.). The data are presented below.
SCORES Pair Number Boys Girls    1 10 13 2 14 8 3 16 10 4 13 13 5 13 12 6 15 13
a. Estimate the difference between the population mean scores for boys and that for girls.
b. Find a 99% confidence interval for the difference in population means, doing your calculations very roughly, so as to find which of the following is closest to the answer. Circle the corres ponding number.
1. 12 to 16 2. 4 to 8 3. 1 to 5 4. 0 to 4
(If you don't like any of these, show your work for partial credit.)
c. The primary purpose of the pairing was in hopes of reducing (circle one):
1. skewness in the data 2. the standard error of the difference in sample means 3. the degrees of freedom in the appropriate ttest 4. heterogeneity of the variances 5. correlation between boys and girls
Answer: a. Estimated difference between the population mean scores for boys and girls = XBAR(boys)  XBAR(girls) = 13.5  11.5 = 2
b. 2. 4 to 8
Since it is a paired test, we will compute S(DBAR) in order to compute the 99% confidence interval for the difference in popu lation means.
DBAR = XBAR(boys)  XBAR(girls) = 2 d(i) = (D(i)  DBAR), where D(i) is the difference between the ith pair.
S(D) = SQRT[SUM(d(i)**2)/(n1)! = SQRT(62/5) = 3.52
S(DBAR) = S(D)/SQRT(n) = 3.52/SQRT(6) = 1.44
99% confidence interval: = DBAR +/ [t(ALPHA=.01, twotailed, df=5)*S(DBAR)! = 2 +/ [(4.032)*(1.44)! = 2 +/ 5.8 = from 3.8 to 7.8
c. 2. the standard error of the difference in sample means
99.
A chemist studies two treatments applied to a chemical which he must prepare in small quantity because of cost and variability of its pro perties. The first time he runs the experiment he applies treatment A to half of the first batch and treatment B to the other half of the first batch. He experiments with six batches and on the basis of six pairs of observations he declares the means of the two treatment pop ulations to be just barely significant. The next time he runs a sim ilar experiment involving the same chemical and two treatments. He intends to run a two group (unpaired) experiment.
(1) State some advantages (things in favor of) a two group experiment for the chemist.
(2) State some reasons why the chemist might again prefer a paired de sign.
Answer: (1) A two group experiment allows for more degrees of freedom for mak ing a test. The group design has twice as many degrees of freedom and in addition it allows for the possibility of getting three or more experimental units out of one test chemical preparation batch. With the two group experiment there is no restriction with regard to equal sample sizes.
(2) With a paired design, unwanted variability can be handled. That is, the variance of difference SIGMA(D)**2 will in general be less than the variance of observations. There is no need to assume nor mality of each population and only differences need to be assumed independent with homogeneous variance when a ttest is employed.
100.
We are interested in the wearing capabilitites of tires. We obtain Goodday and Goodpoor Tires and 9 racing cars (and also the track used for the Indianapolis 500 Race). We put Goodday on the lefthand side of the car (front and rear) and Goodpoor on the righthand side of the car (front and rear). We then allow the cars to complete the 500 miles at a (relatively) safe speed and then measure the wear (in millimeters) per tire.
Car No. Goodday Goodpoor    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 13
a. All the advertising literature claims equality between Goodday and Goodpoor. Can you present evidence to disprove this claim? Use a significance level of 5%.
b. Comment on the validity of this experimental setup.
Answer: a. Let d be the difference in wear between tires on the lefthand side compared to tires on the righthand 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 setup and therefore we perform a ttest 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)!/[n1! = [[SUM(d(i)**2)![n*(dBAR**2)!!/[n1! = [[66![9*4!!/[8! = 3.75
t(calc.) = [dBAR0!/[SQRT([S(d)**2!/n)! = [20!/[SQRT([3.75!/9)! = 3.098
t(critical, .05, twotailed, 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 highlybanked 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.
101.
A computer programmer was concerned about the length of time that would be required to print exam questions and answers at a terminal (a printing device like a typewriter). The programmer knew: that the terminal was capable of printing 30 characters per second;
the number of lines (maximum length: 80 characters) of text that made up each question and answer;
the identity of the questions to be printed at any particular time; and
that question selection for a single exam was a random process.
Accordingly, the programmer estimated the time required to print a set of questions and answers by multiplying the total number of lines to be printed by 80 and then dividing the result by 30 to arrive at an estimated print time in seconds. Later the programmer was informed that his estimates for print time were always too high. What do you recommend that the programmer do to improve his estimation scheme?
Answer: A variety of suggestions should be offered in response to this question. Some should focus on the distribution of characters per line and the use of something other than the maximum length as a basis for estimation. (This assumes that the printing device recognizes short lines and does not always attempt to print 80 characters.) My preference would be to form a series of exams covering the anticipated range of useage and using random selection as much as possible. I would print these exams on the terminal and hope that a regression involving:
Y: Print time; and X: Total number of lines
would provide an adequate approximation of estimating print time. If not, I would then be inclined to examine use of such variables as question length, answer length, computer load, etc.
102.
A test was performed to determine intensity settings for a certain type of filter. In 20 separate runs of the test, the results for filter 1 are as follows:
96, 83, 97, 93, 99, 95, 97, 91, 100, 92, 88, 89, 85, 94, 90, 92, 91, 78, 77, 93.
Use this data to answer the following. (You may assume a normal distri bution.)
a. Write a model to describe an individual measurement from the popula tion in terms of an overall mean and a random element. Define all terms completely. (You may assume the intensity settings to be determined by a large number of factors that operate independently. Assume that each factor makes a small contribution to intensity set ting and that contributions are additive.)
b. Obtain sample estimates for the parameters and set 90% confidence limits for MU.
c. How would you modify your answer to part b if you knew that SIGMA**2 was 36?
d. Verify that the sample variance, S**2, is equal to:
S**2 = SUM(i=1,n)(e(i)**2) where e(i) is a deviation from the  mean and P is the number of para (n  P) meters estimated other than the population variance.
Answer: a. Population form of model: Y(i) = MU + EPSILON(i), i = 1, 2, ...
where Y(i): intensity setting at which oprator i can first detect an image using filter 1.
MU: the population average intensity setting for an indefinitely large population of operators.
EPSILON(i): a random element representing the difference in intensity setting between that required by a par ticular operator (Operator i) and the population mean. The usual assumption is that the EPSILON (i)'s are normally and independently distributed with a mean of 0 and a variance of SIGMA**2.
Sample form of model: Y(i) = MU(HAT) + e(i), i = 1, 2, ...
where Y(i): same as above.
MUHAT or YBAR: an estimate of MU above.
e(i): = Y(i)  MU(HAT) ! A deviation, the difference = Y(i)  Y(i)HAT ! between an observed inten = Y(i)  YBAR ! sity setting, Y(i), and what we would estimate for the ith intensity setting, Y(i)HAT. Here, Y(i)HAT = MU(HAT) = YBAR.
b. PARAMETER ESTIMATE   MU MU(HAT) = YBAR = (SUM(i=1,20)(Y(i)))/n = 1820/20 = 91
SIGMA**2 S**2 = (SUM(i=1,20)((Y(i)  YBAR)**2))/(n  1) = ((SUM(i=1,20)(Y(i)**2))(SUM(Y(i))**2)/20)/(n1) = [(96**2 + ... + 93**2)  ((1820)**2/20)! / 19 = 39.789 with 19 df
CONFIDENCE LIMITS (Variance Estimated):
General form for limits: parameter estimate +/ (t)*(estimated standard error of parameter estimate)
In this case, limits for MU = MU(HAT) +/ t(ALPHA=.05, df=19) * (S(MU(HAT))) = YBAR +/ (1.729) * (S(YBAR)) = 91 +/ (1.729) * (1.410) = from 88.56 to 93.44
i.e., 90% of the time that we draw a random sample of 20 opera tors and calculate an interval in this way, we will get an in terval that contains the true mean, MU. (This assumes that the model used is appropriate.)
c. CONFIDENCE LIMITS (Variance Known):
General form for limits: parameter estimate +/ (Z)* (known standard error of parameter estimate)
In this case, limits for MU = MU(HAT) +/ (Z) * (SIGMA(MU(HAT))) = YBAR +/ (1.64) * (SIGMA(YBAR)) = 91 +/ (1.64) * (1.34) = from 88.8 to 93.2
d. i Y(i) Y(i)HAT=YBAR e(i)     1 96 91 +5 2 83 91 8 3 97 91 +6 4 93 91 +2 5 99 91 +8 6 95 91 +4 7 97 91 +6 8 91 91 0 9 100 91 +9 10 92 91 +1 11 88 91 3 12 89 91 2 13 85 91 6 14 94 91 +3 15 90 91 1 16 92 91 +1 17 91 91 0 18 78 91 13 19 77 91 14 20 93 91 +2
SUM(i=1,20)(e(i)**2) = (5)**2 + ... + (2)**2 = 756
[SUM(i=1,20)(e(i)**2)!/(n  P) = 756/19 = 39.789 with 19 df
since Y(i)HAT = YBAR.
103.
Molybedenum rods are produced by a production line setup. It is desired to check whether the process is in control. Let X = length of such a rod. Assume X is approximately normally distributed with mean = MU and variance = SIGMA**2, where MU and SIGMA**2 are unknown.
Take N = 400 sample rods, with sample average length XBAR = 2 inches, and SUM((XXBAR)**2) = 399.
The correct confidence interval for MU at ALPHA = 8% is closest to:
a. 2.2 +/ (1.75/20) b. 2 +/ (1.41)SQRT(399/400) c. 2.2 +/ (2.06/20) d. 2 +/ (1.75/20) e. 2 +/ (1.67)SQRT(399/400)
Answer: d. 2 +/ (1.75/20)
S**2 = 399/(4001) = 1 S(XBAR) = SQRT(S**2/n) = SQRT(1/400) = 1/20
C.I. = XBAR +/ Z(ALPHA=.08/2) * S(XBAR) = 2 +/ (1.75) * (1/20)
104.
John has done an experiment on gallons of water per second that flow in a sewer main in the city. He makes 16 measurements of this flow and finds that their average is 100 and their variance is 9. Find a 98% confidence interval for the mean flow.
Answer: C.I. = XBAR +/ [t(df=15,ALPHA=.01)*S/SQRT(n)! C.I. = 100 +/ [2.602*(3/SQRT(16))! C.I. = 100 +/ 1.95 = from 98.05 to 101.95
105.
The calculated nitrogen content of pure benzanilide is 7.10%. Five repeat analyses of "representative" samples yielded values of 7.11%, 7.08%, 7.06%, 7.06%, and 7.04%. Using an ALPHA level of size 5%, can we conclude that the experimental mean differs from the expected value? Assume that the measured values are approximately normally distributed.
Answer: H(O): MU = 7.10 H(A): MU =/= 7.10
YBAR = 7.07
S(Y) = 0.0265
t = (YBAR  MU)/S(YBAR) = (7.07  7.10)/(0.0265/SQRT(5)) = 2.53
t(critical, ALPHA=.05, df=4) = +/ 2.776
Since the calculated value of t is not in the critical region, continue H(O) that the nitrogen content has a true value of 7.10%, i.e., the 0.03% difference is ascribable to random error.
or
YBAR +/ t*(S(Y)/SQRT(n)) YBAR +/ 2.776*(0.0265)/(SQRT(5)) P(7.037 <= MU <= 7.103) = 0.95
Continue H(O) that the nitrogen content has a true value of 7.10% at 95% level since 7.10 lies within the 95% confidence interval.
106.
Wire cable is being manufactured by two processes. We need to determine if the processes are having different effects on the mean breaking strength of the cable. Randomly selected samples from each process were submitted to the lab for testing as though they were regular samples. Coded values of the load required to break the cables (tension) are given below:
Process No. 1: 9, 4, 10, 7, 9, 10
Process No. 2: 14, 9, 13, 12, 13, 8, 10
Determine if there is any difference in the mean breaking strength for the two processes at the 95% probability level.
Answer: HO: MU(1)  MU(2) = 0
n(1) = 6 n(2) = 7 MU(1) = 8.167 MU(2) = 11.285 S(1)**2 = 4.47 S(2)**2 = 4.49
F = 4.49/4.47 = 1.004 F(ALPHA = .05, df = 5,6) = 4.39
Since F(calculated) < F(critical), we can assume at the 95% probability level that there is no difference in the standard deviations. It is, therefore, acceptable to pool the variances.
S(P)**2 = [(n(1)  1)(S(1)**2) + (n(2)  1)(S(2)**2)!/(n(1) + n(2)  2) = (5*4.47 + 6*4.49)/11 = 4.48
S(1BAR  2BAR) = SQRT(4.48/6 + 4.48/7) = 1.177
t = [(8.167  11.285)  0!/1.177 = 2.66
t(ALPHA = .05, df = 11) = 2.201
Since t(calculated) > t(critical), reject H(0). Therefore, the two methods are different.
Using a confidence interval: X(2)BAR  X(1)BAR +/ 2.201 * 1.77 3.118 +/ 2.591
C.I. = P(.527 <= MU(2)  MU(1) <= 5.71) = .95
107.
Past production units of a certain jet engine model showed the mean military thrust to be 7600 pounds. The first ten production units manufactured after a model change yielded military thrusts of 7620, 7680, 7570, 7700, 7650, 7720, 7600, 7540, 7670, and 7630. Is there sufficient evidence (use ALPHA = 0.05) that the model change resulted in a higher average military thrust?
Answer: Using ALPHA = .05 and a onetailed ttest we test: H(0): MU <= 7600 H(A): MU > 7600
Finding: YBAR = 7638 S(Y) = SQRT((583,420,000  ((76,380)**2/10))/9) = 57.3
t = (YBAR  MU)/(S(Y)/SQRT(N)) t = (7638  7600)/(57.3/SQRT(10)) t = 2.097
t(critical) = 1.83
Since t(calculated) is larger than t(critical) for a onesided test at ALPHA = .05, reject the null hypothesis. At the 95% confidence level, the sample evidence indicates a detectable increase.
108.
Past experience shows that, if a certain machine is adjusted properly, 5 percent of the items turned out by the machine are defective. Each day the first 25 items produced by the machine are inspected for defects. If three or fewer defects are found, production is continued without interruption. 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 the alternative p > .05, p being the probability that the machine turns out a defective item. In test terminology, the engineer is asked to make adjustments only when the hypothesis is rejected.
Interpret the quality control procedure described above as a test of the 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.
109.
The daily yield of a chemical manufactured in a chemical plant, recorded for n = 49 days, produced a mean and standard deviation equal 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 = 3 XBAR(crit) = MU(M) + Z(crit)S(M) = 880 + ((1.65)*3) = 875.05
Since 870 < 875.05, we reject H(0) and conclude that MU < 880.
BETA is the probability of committing a type II error. Using the above decision rule and given H(A), it is the probability that XBAR is 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
110.
An economist is interested in the possible influence of "Miracle Wheat" on the average yield of wheat in a district. To do so he fits a linear regression of average yield per year against year after introduction of "Miracle Wheat" for a ten year period. The fitted trend line is
YHAT(j) = 80 + 1.5*X(j) (Y(j): Average yield in j year after introduction) (X(j): j year after introduction).
a. What is the estimated average yield for the fourth year after introduction? b. Do you want to use this trend line to estimate yield for, say, 20 years after introduction? Why? What would your estimate be?
Answer: a. 80 + 1.5*4 = 86 b. No. I would not want to extrapolate that far. If I did, my estimate would be 110, but some other factors probably come into play with 20 years.
111.
A management analyst is studying production in an electronic component assembly factory. Workers individually assemble components into final products. Each worker is given 100 sets of components to assemble each day. Employees clock out at the time they finish assembling the 100 sets into final products. The analyst has average hourly production rates for each individual worker. Which mean should be used to calculate the overall average production per labor hour?
a. arithmetic mean b. geometric mean c. 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.
112.
A management analyst is studying production in an electronic component assembly factory. Workers individually assemble components into final products. Workers assemble as many units as they can in an eight hour day. The analyst has average hourly production rates for each individual worker. To calculate the factory's overall average hourly production per worker, which mean should be used?
a. arithmetic mean b. geometric mean c. 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.
113.
A computer programmer reports that the average time required to run a particular program is 11.67 minutes, and that the variance is 8.55 with 5 df. In the Appendix of his report, he lists the following values for time to run the program:
12, 17, 9, 13, 11, 8.
a. What model for time to run the program was implicit in what he re ported?
b. What does this report (or the model) say about factors that might affect run times?
Answer: a. Y(I) = MU + EPSILON(I) Where:
Y(I): time to run job I MU: population mean run time EPSILON(I): random error term associated with job I, usually assumed to be normally distributed with a mean of zero and a variance of SIGMA**2.
b. The model that he uses in his report says that all important factors that affect runtime have been held constant.
114.
Suppose that you have at your disposal the information below for each of 30 drivers. Propose a model (including a very brief indication of symbols used to represent independent variables) to explain how miles per gallon vary from driver to driver on the basis of the factors measured.
Information: 1. miles driven per day 2. weight of car 3. number of cylinders in car 4. average speed 5. miles per gallon 6. number of passengers
Answer: Y(j) = b(0) + b(1)*X(1) + b(2)*X(2) + b(3)*X(3) + b(4)*X(4) + b(5)*X(6) + e(j)
where the dependent variable is variable 5  miles per gallon and the independent variables are X(1)  miles driven per day X(2)  weight of car X(3)  number of cylinders in car X(4)  average speed X(6)  number of passengers
115.
A hospital is considering use of a new device for measuring patient temperatures. For each of 50 patients it is proposed that there be one time when the patient's temperature is taken with both a standard thermometer and the new device. The order of using devices is to be randomly determined for each patient. The model proposed to describe the temperatures recorded is:
Y(i,j) = MU + TAU(i) + RHO(j) + EPSILON(i,j)i=1 or 2, j=1, 2, ..., 50.
a. What is Y(i,j)? b. What is TAU(i)? c. What is RHO(j)? d. What design is proposed?
Answer: a. Y(i,j) is the response measured for treatment i at level j of the blocking factor (Person j). b. TAU(i) is the effect of treatment i, treatment 1: standard thermo meter, treatment 2: new thermometer. c. RHO(j) is the effect of block j, (Person j). d. The Randomized Complete Block design has been proposed.
116.
In the attached Table 1, results for the routine measurement of nickel in a steel standard are reported. This determination was made daily over a long period of time to establish a quality control program.
In Table 2, the data have been plotted as a tally sheet of individual values. Clearly, a grouped tally sheet would be more effective 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 Standard
Date % Ni Date % Ni Date % Ni
Mar. 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 4.98 Apr. 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 5.03
Table 2. Frequency Table and Tally Sheet for the Data in Table 1
Ni 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 Cum Midpoints 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 ___ 100
b) 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 == 0.11
c) 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 = 5.03
e) S(YBAR) = S(Y)/SQRT(n) = 0.108/SQRT(100) = 0.0108 == 0.011
f) 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.
117.
A coffee dispensing machine provides servings that have a population mean 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.), the distribution of differences will have a mean of ______ and a standard deviation of ______.
Answer: a. MU = 0 b. SIGMA = SQRT(.09/1 + .09/1) = .424
118.
The administrator of a loan program for small farmers (five foot and under) institutes a new objective scale by which his field investigators are asked to rate small farms on their profit potential. He suspects that two of his investigators are applying the standard quite differently, which offends his sense of order. To check on them, he asks both of them to rate 12 randomly chosen farms. The results:
FARM # 1 2 3 4 5 6 7 8 9 10 11 12
A RATING 90 80 75 80 60 80 55 40 80 65 70 60 B RATING 65 50 50 65 40 50 55 45 55 55 45 45
a) Use an appropriate statistical test to see whether this is strong enough evidence to reject the hypothesis that they rate in the same way.
b) Make a scatter diagram of the same data. Fit a straight line to the set of points by eye. Estimate the equation of this line using your graph.
c) How could the information in (a) and (b) TAKEN TOGETHER be useful to the administrator?
Answer: a) The appropriate test in this case appears to be the paired (re lated samples) ttest.
H(O): MU(Y)  MU(X) = 0 H(A): MU(Y)  MU(X) =/= 0
Calculations:
Y ^ X ^ D = Y  X ^ d = (D  DBAR) ^ d**2  90 ^ 65 ^ 25 ^ 7.08 ^ 50.17 80 ^ 50 ^ 30 ^ 12.08 ^ 146.01 75 ^ 50 ^ 25 ^ 7.08 ^ 50.17 80 ^ 65 ^ 15 ^  2.92 ^ 8.51 60 ^ 40 ^ 20 ^ 2.08 ^ 4.34 80 ^ 50 ^ 30 ^ 12.08 ^ 146.01 55 ^ 55 ^ 0 ^ 17.92 ^ 321.01 40 ^ 45 ^  5 ^ 22.92 ^ 525.17 80 ^ 55 ^ 25 ^ 7.08 ^ 50.17 65 ^ 55 ^ 10 ^  7.92 ^ 62.67 70 ^ 45 ^ 25 ^ 7.08 ^ 50.17 60 ^ 45 ^ 15 ^  2.92 ^ 8.51  835 ^ 620 ^ 215 ^ 0.00 ^ 1422.91
YBAR = 69.58 XBAR = 51.67 DBAR = 17.92
S(D) = SQRT(1422.91/11) = 11.37 S(DBAR) = S(D)/SQRT(12) = 3.28
t(calc) = 17.92/3.28 = 5.457
t(crit, ALPHA=.05, df=11, twotailed) = +/ 2.201
Since t(calc) > +t(crit), reject H(O). Thus, the evidence is strong enough that the hypothesis that they rate in the same way can be rejected.
b) ^ ^ Y ^ ^ 90 + * B ^ 85 + ^ 80 + 2 * ^ 75 + * NOTE: for a line pos ^ sibly fit by eye draw 70 + * a line through points ARATING ^ A and B. Also note 65 + A * that 2 indicates there ^ are two data points 60 + * * located at this posi ^ tion. 55 + * ^ 50 + ^ 45 + ^ 40 + * ^ +++++++++> X 40 45 50 55 60 65 70 75 80 BRATING
The equation associated with the above line fitted by eye is: YHAT = 20 + (1.0*X)
The estimated equation found by the least squares method is: YHAT = 14.69 + (1.063*X)
c) The information in part (a) does imply that the investigators do apply the standards quite differently. However, using the infor mation in part (b), the administrator can estimate one of the ratings given the other.
119.
We want to compare two machines for production line speed of beer bottle manufacturing. At the end of each of n(1) = 9 days, the number of bottles produced by machine(1) yield XBAR(1) = 19, S(1)**2 = 4. For machine(2), we have n(2) = 6 days and XBAR(2) = 17, S(2)**2 = 9. Assume independence of samples, normality, and equality of unknown variances.
To test H(0): MU(1) = MU(2) vs. H(1): MU(1) =/= MU(2) at ALPHA=.05, we need a table value equal to:
a. 1.771 b. 1.960 c. 2.160 d. 2.131 e. 2.145
Answer: c. 2.160
t(df=n(1)+n(2)2=13, ALPHA=.05, Twotailed) = 2.160
120.
Four replicate analyses of each of two ends of a special metal rod were made. All eight analyses were made in random order. Results for copper analyses on end A were: 4.02, 4.04, 4.08, and 4.05. On end B, they were: 4.08, 4.06, 4.12, and 4.10. At the 95% probability level, can we reject the hypothesis of no difference in copper content for the two ends? At the 99% level?
Answer: Mean for end A = (4.02 + 4.04 + 4.08 + 4.05)/4 = 4.0475 Variance for end A = .000625 Mean for end B = (4.08 + 4.06 + 4.12 + 4.1)/4 = 4.09 Variance for end B = .000667
S(P)**2 = [(n(1)  1)(S(1)**2) + (n(2)  1)(S(2)**2)!/(n(1) + n(2)  2) = (3*.000625 + 3*.000667)/6 = .000646
S(ABAR  BBAR) = SQRT(.000646/4 + .000646/4) = .0180
HO: MU(A)  MU(B) = 0
t = (.0425  0)/.0180 = 2.36
t(ALPHA=.05, df=6) = 2.447 Do not reject the null hypothesis. t(ALPHA=.01, df=6) = 3.707 Do not reject the null hypothesis.
121.
A biologist is working with dangerous chemical residues found in wild skunks. Of particular interest is the possible relationship between the % Mercury accumulation in the liver and the % Telurium accumulation in the lungs. He purchases seven "chemically clean" skunks, and subjects them to a diet containing Telurium and Mercury. The amounts absorbed by an animal will, of course, vary from animal to animal. The results were:
% Mercury (X) % Telurium (Y)   3 3 5 4 2 2 4 5 6 2 1 0 7 1
Possible useful summaries:
SUM(X) = 28.00000 SUM(Y) = 17.00000 SUM(X**2) = 140.00000 SUM(Y**2) = 29.00000 SUM([XXBAR!**2) = 28.00000 SUM([YYBAR!**2) = 17.71429 SUM(X*Y) = 72.00000 SUM([XXBAR!*[YYBAR!) = 4.00000
At the 5% level of significance, do you think there is a linear rela tionship between % Mercury accumulation and the % Telurium accumulation?
Answer: Using Pearson's Product Moment correlation coefficient as an indicator of the strength of a linear relaionship:
r = [SUM([XXBAR!*[YYBAR!)!/ [SQRT([SUM([YYBAR!**2)!*[SUM([XXBAR!**2)!)! = [4.000!/[SQRT([17.71429!*[28.00000!)! = 0.1796
To see if this is suspiciously large we may refer to special tables or use the approximate ttest.
H(O): Correlaion is zero. H(A): Correlation is other than zero.
i.e. H(O): RHO = 0 H(A): RHO =/= 0
t(calc.) = [r!/[SQRT([1(r**2)!/[n2!)! = [0.1796!/[SQRT([1(0.1796**2)!/[5!)! = 0.4082 with 5 df
t(crit., df=5, ALPHA=.05, twotailed) = +/ 2.571
Since t(crit.) < t(calc.) < +t(crit.) continue (do not reject) H(O). It seems there is no relationship between Mercury and Telurium accumulations.
However, a plot of the data reveals:
* = DATA ^ ^ ^ ^ 5 ^ * ^ ^ ^ ^ 4 ^ * ^ ^ ^ ^ % of 3 ^ * Telerium ^ ^ ^ ^ 2 ^ * * ^ ^ ^ ^ 1 ^ * ^ ^ ^ ^ 0 ^ * 1 2 3 4 5 6 7 8 % of Mercury
It is obvious that Mercury and Telurium are related by a curve having a maximum Y when X=4. Why didn't the ttest reveal this? It is be cause the test only looks at linear correlation. These two variables are correlated but not in a linear manner.
122.
We want to know which of two types of filters should be used over an oscilloscope to help the operator pick out the image on the cathode ray tube. A test was designed in which the strength of a signal could be varied from zero up to the point where the operator first detects the image. At this point, the intensity setting was read. The lower the reading when the image was first detected, the better the filter is. Because people vary in their ability to detect the image, twenty opera tors were selected and each one made one reading for each filter. From the results which are tabulated below, test the null hypothesis of no detectable difference in the filters. If they do differ at some ALPHA level of less than .10, tell which is best.
Operator F1 F2 Operator F1 F2 Operator F1 F2
1 96 92 8 91 90 15 90 89 2 83 84 9 100 93 16 92 90 3 97 92 10 92 90 17 91 90 4 93 90 11 88 88 18 78 80 5 99 93 12 89 89 19 77 80 6 95 91 13 85 86 20 93 90 7 97 92 14 94 91
Answer: DBAR = SUM(D)/N = 40/20 =2
F1 F2 D=F1F2 DDBAR (DDBAR)**2
96 92 4 2 4 83 84 1 3 9 97 92 5 3 9 93 90 3 1 1 99 93 6 4 16 95 91 4 2 4 97 92 5 3 9 91 90 1 1 1 100 93 7 5 25 92 90 2 0 0 88 88 0 2 4 89 89 0 2 4 85 86 1 3 9 94 91 3 1 1 90 89 1 1 1 92 90 2 0 0 91 90 1 1 1 78 80 2 4 16 77 80 3 5 25 93 90 3 1 1   40 140
S(D) = SQRT[SUM((DDBAR)**2)/(N1)! = SQRT(140/19) = 2.7145
Using a paired t test:
H(O): MU(F1  F2) = 0 H(A): MU(F1  F2) =/= 0
t(calculated) = (DBAR  MU(F1F2))/(S(D)/SQRT(N)) = (2  0)/(2.7145/SQRT(20)) = 3.30
t(ALPHA = .01, df=19) = 2.86 and t(ALPHA = .001, df=19) = 3.88
At ALPHA = .01, t(calculated) > t(critical), so you would reject H(O). At ALPHA = .001, t(calculated) < t(critical), so you would continue H(O).
Since F1BAR = 91 and F2BAR = 89, F2 should be considered the best.
123.
Two methods were used in a study of the latent heat of fusion of ice. Both method A (an electrical method) and method B (a method of mixtures) were conducted with the specimens cooled to 0.72 degrees C. The data represent the change in total heat from 0.72 degrees C to water at 0 degrees C, in calories per gram of mass.
METHOD A METHOD B 79.98 80.02 80.04 79.94 80.02 79.98 80.04 79.97 80.03 79.97 80.03 80.03 80.04 79.95 79.97 79.97 80.05 80.03 80.02 80.00 80.02
Is there any difference in the 2 methods at the 5% probability level?
Answer: H(0): MU(A) = MU(B) H(A): MU(A) =/= MU(B)
YBAR(A) = 80.02 YBAR(B) = 79.98 S(A)**2 = 0.000574 S(B)**2 = 0.000984 n = 13 n = 8
F = 984/574 = 1.71 F(ALPHA=.05, df=7,12) = 2.92
Since F(calculated) < F(critical), we can assume at the .95 level that there is no difference in the standard deviations. Therefore, it is acceptable to pool.
S(P) = SQRT[((.12*.000574) + (7*.000984))/(12+7)! = 0.0269
t = [((80.0279.98)0)/(0.0269*SQRT(1/13+1/8))! = .04/(0.0269*.45) = 3.30
t(ALPHA=.05, df=19, twotailed) = 2.093
Since t(calculated) exceeds t(critical), reject H(0), i.e. the 2 methods are different.
124.
Propose and justify your proposal for a relation, if any, of the following variables on steam use. Steam Use Production Wind Days Worked Days Below 32o Temperature 1. 10.98 .61 7.4 20 22 35.3 2. 11.13 .64 8.0 20 25 29.7 3. 12.51 .78 7.4 23 17 30.8 4. 8.40 .49 7.5 20 22 58.8 5. 9.27 .84 5.5 21 0 61.4 6. 8.73 .74 8.9 22 0 71.3 7. 6.36 .42 4.1 11 0 74.4 8. 8.50 .87 4.1 23 0 76.7 9. 7.82 .75 4.1 21 0 70.7 10. 9.14 .76 4.5 20 0 57.5 Values are on a monthly basis for a manufacturing firm. Wind and temperature entries are monthly means. Days worked and Days Below 32 degrees are number of days in a month.
Answer: The model proposed here is: Y(Steam use) = B(0) + B(1) * X2(prod) + B(2) * X6(temp) + E This description based on temperature (X6) and production (X2) accounts for around 96% of the variation in steam use. The fitted equation is: YHAT = 11.31 + 4.56*X2  .09*X(6). The tests of individual b values and F test of regression mean square are significant. The residual patterns seem to be acceptable. Since Days below 32 degrees and Days worked are highly correlated with Temperature and Production (respectively) we suspect that they will not add much to the regression.
Model with Temperature and Production df SS M.sq. Regression 2 27.90 13.95 Error 7 1.21 .17 R**2 = .958
Model with all five variables: df SS M.sq. Regression 5 28.42 5.68 Error 4 .69 .17 R**2 = .976
125.
Average June August Yield Minimum Temperature June Rainfall Y X(1) X(2)
13.1 50.4 3.1 14.1 51.0 5.0 15.7 49.1 6.7 14.3 51.2 5.2 15.2 48.1 6.9 16.7 48.0 7.8 13.8 51.0 5.6 12.4 49.6 4.0 11.5 53.1 3.7 15.3 48.2 6.5 14.4 52.2 4.8 13.3 50.5 4.3 12.5 54.2 1.9 12.7 50.1 5.6 16.5 49.9 6.8
The above data was studied with the aid of a computer. It is data typical of actual corn yield information recorded in Oklahoma. The correlations were as follows:
yield vs. temperature : .657 yield vs. rainfall : .846 temperature vs. rainfall: .796
Now, although the correlation between yield and temperature is strong and negative, the least squares equation given by the computer print out was:
Y = 7.79 + .0379X(1) + .847X(2).
Is the sign of the temperature variable X(1) consistent with the nega tive correlation coefficient? Explain.
Answer: The printout is correct. The correlation need not have the same sign as the coefficient of the variable in the least squares fit. In the presence of X(2) the effect of X(1) need not be the same as the effect of X(2) alone. This makes sense in this experiment. Corn needs mois ture, and rain is usually accompanied by cool weather, but the best com bination for corn yield is warm and wet weather.
126.
Listed below is some fictitious data concerning 1. Amount of oil needed to fill the tank for the heating plant of a building. 2. Days elapsed since last oil delivery. 3. Average outside temperature (Fahrenheit) during the time since last delivery
Oil Days since last delivery Average Temperature (F) 60 16 10.1 41 17 22 50 16 15.4 29 13 18 81 19 7.3 74 26 28 57 25 34
a. Write and fit 2 models 1. One relating oil consumption to average temperature alone. 2. One relating oil consumption to both temp. and days since last delivery. b. Compare the estimated regression coefficients for temperature from these two models. Why do you think there is this difference (of lack of difference)? c. Compare the estimated intercepts for the two models. Does either of these values seem reasonable? Why?
Answer: a. 1. Y = b(0) + b(1)*X(3) + e Y = 61.216  .27084*X(3) 2. Y = b(0) + b(2)*X(2) + b(3)*X(3) + e Y = .42296 + 5.0193*X(2) 2.0289*X(3)
b. Regression coefficients for temp. Model 1: .27084 Model 2: 2.0289 The coefficients for average temperature are different because the variable "Days since last delivery" provide important information for exploring variation in oil use. When average temperature is fitted without using information about time since last delivery all times are treated as if the same and the coefficient for temperature is calculated as if times were the same. When variation in time since last delivery is taken into account, the coefficient for temperature no longer is calculated as if there were no variation in time since last delivery.
c. Intercepts Model 1: 61.2 Model 2: .42296 RSquare Model 1: .02 Model 2: .99
I would go with Model 2 because Model 1 does not account for much of the variation, and the intercept for Model 2 does make more sense. If it has been zero days since the last delivery, they should not need much oil, even if it has been very cold (0 degrees) that day.
127.
It is known that longthin titanium curtain rods lengthen with in creasing temperature. A sample of n = 20 identical titanium rods are selected. Each is subjected to a particular uniform tempera ture X for a specified time. Let Y denote the change in length. The readings are (X(1), Y(1)), ..., (X(20), Y(20)), with data XBAR = 2 (in hundreds of degrees fahrenheit), YBAR = 3 (in milliinches), SUM((X  XBAR)**2) = 10, SUM((Y  YBAR)**2) = 40, and SUM[(X  XBAR) (Y  YBAR)! = 16.
The least squares regression line of the form Y = a + bX has values a = __________and b = __________respectively.
(a) 4/5, 7/5 (d) 7/4, 1/2 (b) 3/4, 3/2 (e) 1/5, 13/5 (c) 8/5, 1/5
Answer: (c) 8/5, 1/5
a = YBAR  b(XBAR)
b = SUM(i = 1, n)[(X(i)  XBAR)(Y(i)  YBAR)!/ SUM(i = 1, n)[(X(i)  XBAR)**2!
b = 16/10 = 8/5
a = 3  8/5*2 = 1/5
128.
It is known that long, thin titanium curtain rods lengthen with in creasing temperature. A sample of n = 20 identical titanium rods is selected. Each is subjected to a particular uniform temperature X for a specified time. Let Y denote the change in length. The readings are (X(1), Y(1)), ..., (X(20), Y(20)), with data XBAR = 2 (in hundreds of degrees F), YBAR = 3 (in Milliinches), SUM((X  XBAR)**2) = 10, SUM ((Y  YBAR)**2) = 40, and SUM[(X  XBAR)(Y  YBAR)! = 16.
When the temperature is set at 400 degrees (i.e., X = 4), then the predicted value of the lengthening of the rod is closest to ______ milliinches.
(a) 24/5 (d) 3/5 (b) 17/5 (e) 31/5 (c) 5
Answer: (e) 31/5
b = SUM(i = 1, n)[(X(i)  XBAR)(Y(i)  YBAR)!/ SUM(i = 1, n)[(X(i)  XBAR)**2!
b = 16/10 = 8/5
a = YBAR  b(XBAR)
a = 3  8/5*2 = 1/5
Hence, YHAT = a + bX = 1/5 + 8/5X
Now X = 4
Hence, YHAT = 1/5 + 8/5*4 = 31/5
129.
The following data were obtained in a study of road width and the number of accidents occurring per hundred million vehicle miles.
Width Number of Accidents 73 42 50 83 62 58 30 93 25 90
The Department of Transportation wishes to use width to predict number of accidents. Determine an equation which will enable them to do this. Can the department significantly improve its prediction of number of accidents by using the data on width, over what the prediction would be not using the width data? (HINT: do a hypothesis test.) (Get the appropriate formulas set up, data inserted, then approximate.)
Answer: X = Width Y = Number of Accidents
SUM(X) = 240 SUM(Y) = 366 SUM(X)**2 = 13198 SUM(Y)**2 = 28766 SUM(X*Y) = 15852
b = (15852  (240*366)/5)/(13198  (240**2)/5) = (1716)/1678 = 1.02
a = 366/5  ((1.02)*240/5) = 122.16
Regression equation: YHAT = 122.16  1.02*X
Source df SS MSQ F  Regression 1 1754.86 1754.86 23.93 Deviations 3 219.94 73.31 Total 4 1974.8
Critical Region: F > 10.1
Reject H(O) and conclude that width data can significantly improve the prediction of accidents.
130.
Find the regression line of the stopping distance Y on the speed X of cars based on the following data:
X = speed (mph) 20 30 40 50 Y = stopping distance (ft) 50 90 150 210
Answer: Y = 64 + (5.4*X)
131.
The following data are the result of a thermodynamic experiment:
X surface area 2 1 3 5 Y heat loss 0 6 9 10
a. Find the least squares line to fit these data, and make a sketch of the points and the line.
b. Estimate the loss of heat for a surface area of 4.0.
c. Would you feel safe in using this line to estimate heat loss for a surface area of 10? Explain.
Answer: a. Computations:
X ^ Y ^ X**2 ^ Y**2 ^ X*Y ^^^^ 2 ^ 0 ^ 4 ^ 0 ^ 0 1 ^ 6 ^ 1 ^ 36 ^  6 3 ^ 9 ^ 9 ^ 81 ^ 27 5 ^ 10 ^ 25 ^ 100 ^ 50 ^^^^ 5 ^ 25 ^ 39 ^ 217 ^ 71
XBAR = 5/4 = 1.25
YBAR = 25/4 = 6.25
bHAT = [(4*71)(5*25)!/[(4*39)(5**2)! = 1.214
aHAT = 6.25  (1.214*1.25) = 4.733
^ Y ^ ^ 10 +^^^^* (Note: draw a ^ ^ ^ ^ ^ ^ line through ^ ^ ^ ^ ^ ^ points A and 9 +^^*^^ B for an ^ ^ ^ ^ ^ ^ approximation ^ ^ ^ ^ ^ ^ to the least 8 +^^^^^ squares line.) ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ Heat Loss 7 +^^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ * 6 +B^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 5 +^^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 4 +^^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 3 +^^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 2 +^^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ A 1 +^^^^^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ <+*+++++++> X 3 2 1 0 1 2 3 4 5 Surface Area
b. YHAT = 4.733 + (X*1.214) = 4.733 + (4.0*1.214) = 9.589
c. No. Since the observed values for surface area range from 2 to 5, I would feel very apprehensive about extrapolating to a surface area of 10.
132.
An engineer is interested in the flow rate of a river (volume/min.) at a downstream location, D. He has a poor set of records for this location, but an extensive set of records for an upstream location U. He would like to find a way to estimate flow rates at D corres ponding to various flow rates at U.
a) If we use a regression model for a straight line, what are the usual symbols and assumptions that apply to measurements taken at U and D?
b) If there were no streams of any consequence entering the river between U and D, how would our choice of model be influenced and what parameters are to be estimated for a straight line re lation?
c) If a number of major streams entered the river between U and D, how would our choice of model be influenced and what parameters are to be estimated for a straight line relation?
d) Suppose that there are several streams entering the river between U and D. Estimate the slope of a straight line linking these measurements.
Flow Rate at U Flow Rate at D  
1 3 2 7 3 9 4 9 5 12
Answer: a) The measurements at U are values for the independent variable usually represented by X and are assumed to be measured with negligible error. The measurements at D are values for the de pendent variable usually represented by Y such that each Y(i) is assumed to be normally and independently distributed with mean = ALPHA + BETA*X(i) and variance = SIGMA**2.
b) We might force the straight line through the origin (X = 0, Y = 0). Parameters to be estimated would be BETA and SIGMA**2.
c) We should then be reluctant to claim that there were any values of X where we did not have to estimate the corresponding value of Y. We would use a model for a straight line through (XBAR, YBAR) and estimate ALPHA, BETA and SIGMA**2.
d) For a straight line through (X, Y),
b = [SUM(X(j)*Y(j))!/[SUM(X(j)**2)!
SUM(X(j)*Y(j)) = 20 SUM(X(j)**2) = 10
BETA(HAT) = b = 2
133.
A few months ago, Road & Track magazine compared the performance of about 25 sports cars with respect to attainable top speed and fuel eco nomy. Regressions were run to investigate how both top speed and fuel economy were affected by the horsepower capability of the engine. The findings are summarized below.
Where: M(i) = miles per gallon of the ith car. S(i) = top speed in miles per hour of the ith car. H(i) = horsepower rating of the ith car, measured as the actual number of horses. All cars tested had between 50 and 300 horsepower engines.
Equation (1): MHAT(i) = 30  .05*H(i) r**2 = .55 Equation (2): SHAT(i) = 60 + .20*H(i) r**2 = .72
(1) Interpret the regression coefficients (slope and intercept) of Equation (1) precisely.
(2) What do the results suggest about the relative usefulness of horse power rating in predicting fuel economy on one hand and top speed on the other hand?
(3) Why would you hesitate to use the regression results for predicting the performance characteristics of cars which have less than a 50 horsepower engine?
(4) Based on the results of equations (1) and (2), (as well as on in tuition), one could say that there exists a "tradeoff" between top speed and fuel economy; i.e., in order to generate an improvement in one, you must sacrifice some of the other. Compute the magni tude of the tradeoff; the number of miles per hours we would pre dict would be sacrificed for each additional mile per gallon of fuel economy.
Answer: (1) Slope = .05 indicating m.p.g. decreases with an inrease in horse power.
Intercept = 30 indicating all cars got less than 30 m.p.g. and, in fact, since minimum horsepower was 50, all cars got less than 27.5 m.p.g.
(2) Horsepower is more useful in predicting speed, (i.e. r**2 is larger).
(3) No such cars were in the sample, therefore one cannot safely extra polate unless one has reason to believe (through other similar experiments) that the relationship follows a similar linear pattern below 50 h.p.
(4) 20 horsepower per 1 mile change in economy, and 20 horsepwer would increase speed by 4 m.p.h. Therefore, an increase of 1 m.p.g. in fuel economy should be accompanied by a decrease of 4 m.p.h. in the speed of the car.
134.
Once upon a time an investigator was concerned about fuel consumption and speed of travel of automobiles. He measured miles per gallon, mpg, for static tests at 25 miles per hour (mph) and 55 miles per hour for many brands of cars.
His report stated that a regression line had been fitted for each brand and that a straight line describes the relation between mpg and mph perfectly since r**2 = 100 for every brand.
Do you subscribe to the claim that a straight line perfectly describes the relation of mpg to speeds between 25 and 55 mph? Why?
Answer: No, I don't, because he has fit regression lines with only two observations, so naturally his r**2 equals 100. You need at least three observations to test for(or to observe) departures from a straight line. Two points determine one and only one line.
135.
Sometimes a fitted regression equation will do a good job of explaining how response varies with the independent variables measured and fail miserably to agree with theory or previous observations. For example, an equation relating yield for a crop to rate of Nitrogen (N) applica tion might fit well and indicate that increased N reduces yield. How can there be such an inconsistency?
Answer: A regression equation that fits observed responses well is concerned with summarizing the data set at hand. If the message conveyed by that equation doesn't fit with theory or previous experience, it may well be that the data set at hand involves variable settings different from those envisioned by theory or encountered in previous experience. Such conflicts should not be dismissed quickly. They probably indicate that response is being observed under "new" conditions that warrant detailed comparison with those that have been considered previously (These "new" conditions may be due to variation in factors other than those ordina rily thought to be important. e.g., If we obtained data relating the volume of an ideal gas to pressure, we would obtain consistent results as long as all other factors were kept constant. But, the results that applied under constant conditions would fail if we allowed, say, tempe rature to vary. These results would not fit previous experiences or theory. But, if temperature had been measured, these results could be reconciled with previous experience by further analysis).
136.
It is known that longthin titanium curtain rods lengthen with in creasing temperature. A sample of n = 20 identical titanium rods are selected. Each is subjected to a particular uniform tempera ture X for a specified time. Let Y denote the change in length corresponding to X.
The readings are (X(1), Y(1)), ...(X(20), Y(20)), with data XBAR = 2 (in hundreds of degrees fahrenheit), YBAR = 3 (in milliinches), SUM(X  XBAR)**2 = 10, SUM(Y  YBAR)**2 = 40, SUM[(X  XBAR)(Y  YBAR)! = 16.
The sample correlation coefficient r =
(a) 4/5 (d) 1/4 (b) 3/4 (e) 1/25 (c) 1/5
Answer: (a) 4/5
Y = SUM((X  XBAR)(Y  YBAR))/ SQRT(SUM((X  XBAR)**2)SUM((Y  YBAR)**2))
= 16/SQRT(40*10) = 16/20 = 4/5
137.
Briefly discuss your evaluation of the following statement.
"Since the linear correlation coefficient (r) between IQ and earning potential is near 0 there is no relationship between the two."
Answer: The statement should probably be rewritten as:
"Since the linear correlation coefficient (r) between IQ and earning potential is near 0 there is no linear relationship between the two."
This would better emphasize the fact that no linear relationship may exist, however there remains the possibility that higher order rela tionships may exist.
138.
Thirty patients in a leprosorium were randomly selected to be treated for several months with one of the following:
A  an antibiotic B  a different antibiotic C  an inert drug used as a control
At the end of the test period, laboratory tests were conducted to provide a measure of abundance of leprosy bacilli in each patient.
Scores obtained were:
Patient 1 2 3 4 5 6 7 8 9 10 Drug A 6 0 2 8 11 4 13 1 8 0 Drug B 0 2 3 1 18 4 14 9 1 9 Drug C 13 10 18 5 23 12 5 16 1 20
a. Analyze and interpret the results of this trial (use ALPHA = .05). b. Write a model appropriate to this trial. Define all terms and estimate all parameters. c. To be consistent with the model you have specified, what order should have been followed in collecting samples from patients and in carrying out lab tests?
Answer: a. Using the program CARROT*** you get the following results:
Means for Antibiotics Treatment Mean (leprosy bacilli) Placebo 12.3 Drug A 5.3 Drug B 6.1
LSD (at ALPHA = .05) = 5.57113
LSD tests on means:
H(0): XBAR(1)  XBAR(2) = 0 H(A): XBAR(1)  XBAR(2) =/= 0
(XBAR(1)  XBAR(2)) +/ LSD Placebo and drug B
6.2 +/ 5.57 Interval is from +.63 to +11.77
The interval does not contain zero, therefore, we reject H(0).
Since both antibiotic treatments are significantly different from the placebo, one would conclude that they have a significant effect on reducing leprosy bacilli. However, antibiotics A and B are not significantly different from each other. The differences in their means could be attributed to chance variation alone.
b. Model: Y(I,J) = MU + TAU(I) + EPSILON(I,J)
Response = overall mean + treatment effect + error(assumed to have a mean of zero and variance = SIGMA**2)
Estimates: MU 7.9 TAU(1) drug A 2.6 TAU(2) drug B 1.8 TAU(3) Placebo 4.4 SIGMA**2 36.9
c. Suppose that the 30 patients had been assigned identification numbers 1, 2, ..., 30. Presumably, treatments were randomly assigned to patients on the basis of these numbers. To be con sistent with this set up, samples should have been collected and analyses run according to the identification numbers (i.e. sample collected first and analysis run first from patient 1 and so on). this amounts to randomly assigning order of sam ple collection and order of lab analysis as well as patients to treatments. (It's preferable to collecting samples and running analyses for all on Drug A, then all on Drug B, then all on placebo.)
139.
An imaginary investigation was conducted to compare three different locations at which body temperature can be obtained. Over a two month period, 20 occasions were randomly selected from a large set of times and beds that could reasonably be used in a ward. On those occasions, the temperature of the patient in the bed was taken at all three locations where the order of measurement was randomized independently with each patient. (This imaginary experiment must have been conducted by extraordinarily persuasive people.)
Imaginary results were:
Location: 1 2 3
Patient 1 98.8 98.9 98.6 2 98.2 98.2 98.0 3 99.0 99.1 99.1 4 100.6 100.6 100.5 5 99.8 99.9 99.6 6 101.3 101.5 101.2 7 100.4 100.4 100.2 8 101.6 101.7 101.6 9 98.5 98.5 98.4 10 98.1 98.2 97.9 11 98.8 99.0 98.7 12 100.9 100.9 100.7 13 98.4 98.6 98.3 14 99.0 99.1 98.8 15 101.4 101.4 101.3 16 99.8 99.9 99.6 17 99.2 99.3 99.2 18 98.7 98.9 98.6 19 99.2 99.3 99.2 20 101.5 101.6 101.5
a. Summarize the results of this investigation (use ALPHA = .05). b. What was the estimated variance (S**2)? What was the standard error of the difference between location means (S(dBAR))? c. Set 95% confidence limits for the temperature difference between location 1 and location 2.
Answer: a. The analysis needed for this trial is for an RCB design with patients corresponding to blocks. It is found that the three treatments (locations) all yield responses (temperatures) significantly different from one another.
The results were:
ANOVA
Source df SS M.Sq.
Total 60 595929.1280 Mean 1 595847.2000 Corrected total 59 81.9219 Location 2 0.4063 0.20313 Person 19 81.3906 4.28372 Experimental Error 38 0.1250 0.00329
Treatment means
Location 2 99.75 Location 1 99.66 Location 3 99.55
LSD (t*S(dBAR)) for these means at the .05 confidence level=.0366547
Location 2  Location 1 = .09 Location 2  Location 3 = .20 Location 1  Location 3 = .11
All of the differences between treatment means are larger than the least significant difference, therefore, I conclude that all the treatments yield responses significantly different from each other.
An F test of the blocking factor:
F(calc) = 4.28372/0.00329 = 1302 F(crit. df = 19.38; ALPHA = .05) == 1.85
indicates that the blocking factor was worthwhile.
b. The variance (S**2) is estimated by the mean square for experi mental error, which is 0.00329
The standard error of a difference between two means (S(dBAR)) is computed using the formula:
SQRT((2*S**2)/r)
Therefore, S(dBAR) == SQRT((2*.00329)/20) = .01813836
c. Confidence limits for the difference between Location 2 and Location 1:
(99.75  99.66) +/ t * S(dBAR) .09 +/ 2.021 * 0.018138 95% confidence interval is from .053 to .127.
140.
Treatments A, B, C and D are to be applied to a field pictured below. 3 blocks are to be used. The soil is known to improve as we go from east to west.
 ^ ^ ^ ^ East ^ ^ West ^ ^ ^ ^  >Soil gets better as we move in this direction>
1) Show in the field layout a typical randomized block design. (Indicate the 3 blocks.)
2) Write out a fixed effects model appropriate for the data resulting from such a design and experiment.
3) Write the sources of variation and the degrees of freedom for an ANOVA table associated with the experiment.
4) Write the expected mean square for treatments in the ANOVA table and explain how this EMS can be used to justify ANOVA F tests for the null hypothesis H(O): All treatment means are equal.
Answer: 1) Block 1 Block 2 Block 3  ^ D ^ B ^ A ^ ^ A ^ C ^ C ^ East ^ B ^ D ^ B ^ West ^ C ^ A ^ C ^ 
2) Let Y(ij) be the observation for the jth treatment in block i. Suppose Y(ij) = MU + BETA(i) + TAU(j) + EPSILON(ij), where MU, BETA(i), and TAU(j) are fixed, BETA(i) is the effect of the ith block, TAU(j) is the effect of the jth treatment, and EPSILON(ij) is the random error associated with observation Y(ij).
Suppose also that SUM(i=1,3)(BETA(i)=0), SUM(i=1,4)(TAU(j)=0), and EPSILON(ij) is distributed normally and independently with mean zero and homogeneous variance, SIGMA**2.
3) ANOVA table
Source of df SS MS EMS Variability
Total 12 Mean 1 Blocks 2 SSB Treatments 3 SST MST SIGMA**2+(3/3)SUM(j=1,4)(TAU(j)**2) Exp. Error 6 SSE MSE SIGMA**2
4) If H(O) is true then TAU(j) = 0 for all j and SUM(j=1,4)(TAU(j)**2)= 0. This means that MST and MSE are both estimating SIGMA**2. Hence an F ratio near 1 is compatible with H(0), and a large F ratio im plies rejection of H(0).
141.
Suppose that you are to test the effects of these fertilizer rates on strength of cotton fiber. Rates (Potassium source in pounds/Acre) R1. 36 R2. 54 R3. 72 R4. 108 R5. 144
Field plots to be used for this test are arranged as below:
     1 2 3 4 5      6 7 8 9 10      11 12 13 14 15
(The numbers identify experimental units)
Randomly assign treatments within groups of 5 plots so that a randomized complete block design will be appropriate.
Answer: The program RCBPLN*** can be used to generate a random assignment of treatments to experimental units so that each treatment occurs once within each block. The following is an example:
1 2 3 4 5 Block 1 R3 R5 R2 R4 R1
6 7 8 9 10 2 R1 R4 R5 R3 R2
11 12 13 14 15 3 R4 R2 R3 R5 R1
15 Experimental units numbered 115 Any procedure which produces a random assignment of treatments within each block is acceptable. A random numbers table could be used where a method of determining I.D. numbers corresponding to the 5 treatments was performed for each of the 3 blocks.
142.
As a researcher working for the milk industry you have been asked to test four feed rations (named A, B, C, and D) and their effect on milk yield for cows. A total of 16 cows are available for testing purposes. These cows are not all alike, but do form four similar (homogeneous) groupings of four cows each.
a. Based on the information presented above, which of the more common experimental designs would you choose for this investigation? Explain. b. Write a model appropriate to the chosen design and define all terms. c. What constitutes an experimental unit? Explain. d. What are the treatments and how will you make comparisons among them? e. Produce a random assignment of treatments to experimental units for the specified design.
Answer: a. I would use a randomized complete block design because the cows form four similar groups of four cows, and I think that responses may vary among these groups. b. Y(I,J) = MU + TAU(I) + RHO(J) + EPSILON(I,J) Where Y is the response, (milk yield) MU is the overall mean TAU(I) are the treatment (ration) effects RHO(J) are the block (group of cows) effect EPSILON is the random element term with mean = 0 and variance = SIGMA**2 c. An experimental unit involves a cow being fed a particular ration over the test period. d. The treatments are the four feed rations, A, B, C, and D. I would feed these cows their assigned rations for the same length of time. During this time period, I will record milk yield for each of the cows at the end of the testing period. I will compute means for the treatments and a measure of the variance, and will then compute an LSD with which I could compare the means. e. I have obtained below one possible assignment of treatments to experimental units. Cows 1 through 4 are similar, 5 through 8 are similar, etc.
Blocks: 1 2 3 4 Treatments: 1 2 6 9 13 2 3 5 12 15 3 4 7 11 16 4 1 8 10 14
143.
An experiment involving two calculating machines [the Curta (hand operated) and the SR51 (electronicallyoperated)!, with the former designated as treatment C and the latter as treatment S, was conducted on 10 sets of 15 twodigit numbers. The yield data are seconds required to square and sum the 15 numbers. Since it was suspected that the time reqired to square and sum a set of numbers was shorter for the second operation on the same set of numbers than it was on the first, this source of variation was taken into account in designing the experiment. Each of the two treatments appeared five times in the first order of performing the calculation and five times in the second order, and both treatments appeared on each set of numbers; except for these restrictions the allocation of the treatments was random. The randomized plan for executing the experiment and the data obtained are given below (treatments are C and S):
 ^ ^ Set of numbers squared and summed ^Order^ ^ ^^ ^ ^Order^ 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7 ^ 8 ^ 9 ^ 10 ^Total^  ^ ^ ^ ^ ^ Time in seconds ^ ^ ^ ^ ^ ^ 1 ^ C ^ S ^ C ^ S ^ S ^ C ^ C ^ S ^ C ^ S ^ ^ ^ ^ 255^ 115 ^ 280 ^ 107 ^ 105 ^ 240^ 195 ^ 110 ^ 202 ^ 85 ^ 1694^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 2 ^ S ^ C ^ S ^ C ^ C ^ S ^ S ^ C ^ S ^ C ^ ^ ^ ^ 113^ 200 ^ 117 ^ 238 ^ 210 ^ 104^ 90 ^ 200 ^ 105 ^ 180 ^ 1557^  ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^Total^ 368^ 315 ^ 397 ^ 345 ^ 315 ^ 344^ 285 ^ 310 ^ 307 ^ 265 ^ 3251^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^Mean ^ 184^157.5^198.5^172.5^157.5^ 172^142.5^ 152 ^153.5^132.5 ^  ^ 
Overall mean: YBAR(...) = 162.55
Order means: YBAR(1..) = 169.4 and YBAR(2..) = 155.7
Treatment means: YBAR(.C.) = 220.0 and YBAR(.S.) = 105.1
Sum of squares of estimated random errors = SUM(eHAT(hij)**2) = 2219.50
a. The experimental unit is:
b. The experimental design is:
c. The number of degrees of freedom for error is:
d. Show how to obtain eHAT(1C1) in terms of the numbers above.
e. Show how to compute S(e)**2 = estimated variance of a single observation in terms of the numbers above.
f. For the above data, the estimated variance of a treatment mean equals:
g. For the above data, the estimated variance of a difference between two treatment means is:
h. The 95% confidence interval or interval estimate for the difference between the two treatment mean is from:
i. How do the computations in the preceding statement change in com puting the 80% confidence interval?
j. The sources of variation in the above experiment are:
k. What effects are orthogonal to each other in the above design?
l. The coefficient of variation for a single observation is:
Answer: a. The experimental unit is one set of 15 numbers.
b. The experimental design is a simple changeover (crossover).
c. The number of degrees of freedom for error is (21)(102) = 8.
d. eHAT(1C1) = 255  169.4  184.0  220.0 + 2(162.55) = 6.7
e. S(e)**2 = 2219.50/8 = 277.44
f. The estimated variance of a treatment mean = 2219.50/(8*10) = 2219.50/80 = 27.74
g. The estimated variance of a difference between two treatment means = (2*2219.50)/(8*10) = 2219.50/40 = 55.49
h. The 95% confidence interfal for the difference between the two treatment means from: 220.0  105.1  [2.31*SQRT(2219.5/40)! to 220.0  105.1 + [2.31*SQRT(2219.5/40)!; or from 97.69 to 132.11
i. The computations in the preceding statement change in that 2.31 is changed to 1.40.
j. The sources of variation are: overall mean, bias, order effects, set of numbers effects, calculator effects, and residual (random error).
k. Effects that are orthogonal to each other include:
mean and bias are not, but mean and bias, order, set, and calculator effects are orthogonal to each other.
l. The coefficient of variation = [SQRT(2219.5/8)!/[162.55! = .1025
144.
A test was conducted to compare the relative effectiveness of three waterproofing compounds, (A,B,C). A strip of cloth was subdivided into nine pieces   
Left Center Right _____ _____ _____ _____ _____ _____ _____ _____ _____
_____ _____ _____ _____ _____ _____ _____ _____ _____
Each piece was considered to be an experimental unit, but it was suspected that the pieces differed systematically from left to right in capacity to become waterproofed. Accordingly, the random 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); and II. 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 = 11.333
b. 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.
145.
A test has been conducted in which four tire brands have been tested using 12 experimental units where an experimental unit consisted of one tire position on one car. The random assignment of brands to experi mental units was restricted so that each brand was tested once on each car. Results (in amount of wear) were:
Front Right Front Left Rear Right Rear Left
Car 1 D, 7.17 A, 7.62 B, 8.14 C, 7.76 Car 2 B, 8.15 A, 8.00 D, 7.57 C, 7.73 Car 3 C, 7.74 B, 7.87 A, 7.93 D, 7.80
a. 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.
146.
Write out the sources of variation and the degrees of freedom for the following industrial experiment. Mention also the name of the design.
Three machines were used to produce parts made from four kinds of metal. Each machine made one part from each type of metal. The order with which the metals were assigned to the machines was established through a randomization procedure.
Answer: Source of Variation df  
Total 12 Mean 1 Metals 3 Machines 2 Residual 6 (Metal x Machine)
This is a randomized block experiment with metals playing the role of blocks.
147.
An investigator has at his disposal a garden in which there are 16 spaces for planting marigolds. The investigator is persuaded that a plant will respond equally well (produce the same number of flowers) in any one of these spaces. He wishes to compare 4 new marigold varieties. The design that matches his notion of the experimental material is:
a. Latin Square b. Randomized Block c. Completely Random
Answer: c., since his vision for uniform conditions is of equal response for all experimental units.
148.
The model proposed to describe the responses measured in an experiment is:
Y(i,j) = MU + TAU(i) + EPSILON(i,j) i=1, 2, 3, or 4
Where Y(i,j) is the number of flowers produced by a marigold plant j belonging to the variety i.
a. What is TAU(i)? b. What design corresponds to the model?
Answer: a. The treatment effect of variety i, which in this case represents the number of blossoms more or less than the overall mean produced by a plant belonging to variety i. b. Completely Random Design, since the model doesn't include any terms for blocking factors.
149.
Five laboratories were invited to participate in an experiment to test the chemical content of four materials known to vary over the range of interest. Each laboratory was given two samples of each material to analyze. The results were:
Laboratories  Material Specimens I II III IV V 
1 8,11 10, 8 7,10 9,12 10,13 2 14,19 11,15 13,11 10,13 17,19 3 20,16 21,18 21,20 22,25 24,22 4 19,13 11,12 17,15 19,17 9,11
Perform the appropriate calculations to determine if there is any systematic difference between laboratories.
Answer: ANOVA:
Source of Variation df SS MS F      Total 40 9696.00 Correction for mean 1 8761.00 Laboratories 4 36.65 9.16 1.97 Materials 3 628.20 Error 20 93.00 4.65 Interaction 12 176.55 14.71 3.16 *
F(critical, ALPHA=.05, df=12,20) = 2.28
In this case, the interaction is significant. This probably masks the systematic variation in laboratories, which turns out not to be significant.
To get a clearer picture, separate means were calculated for differ ent laboratories and material interactions.
ML (Material * Laboratory) Mean Response   11 9.5 12 9.0 13 8.5 14 10.5 15 11.5 21 16.5 22 13.0 23 12.0 24 11.5 25 18.0 31 18.0 32 19.5 33 20.5 34 23.5 35 23.0 41 16.0 42 11.5 43 16.0 44 18.0 45 10.0
Mean responses were plotted against materials for each laboratory. The graph indicated that the lab effects depend on the specific material being considered. No lab is consistently different from other labs for all materials. For example, Lab 5 gives highest mean response for material one and two, but lowest for material four. Similarly, Lab 4 gives highest mean response for material four only.
If available, consult file of graphs and diagrams that could not be computerized for appropriate graph.
150.
A completely randomized design was used for an experiment on light intensity in foot candle power units for three types of lights (M = mercury vapor, L = low pressure sodium vapor, and H = high pressure sodium vapor), in one large parking lot. Suppose the results ob tained were:
^ Treatment and Responses (Y(ij)) ^ ^ ^ ^ M L H ^ ^^^^ ^ Y(M1) = 12 ^ Y(L1) = 15 ^ Y(H1) = 20 ^ ^ Y(M2) = 10 ^ Y(L2) = 14 ^ Y(H2) = 12 ^ ^ Y(M3) = 11 ^ Y(L3) = 13 ^ Y(H3) = 8 ^ ^ Y(M4) = 9 ^ Y(L4) = 11 ^ Y(H4) = 7 ^ ^ Y(M5) = 8 ^ ^ Y(H5) = 23 ^ ^^^^ ^ ^ ^ ^ Overall mean Totals ^ Y(M.) = 50 ^ Y(L.) = 53 ^ Y(H.) = 70 ^ = 173/14 ^ ^ ^ ^ Means ^YBAR(M.)=10.0 ^YBAR(L.)=53/4 ^YBAR(H.)=14.0 ^ = YBAR(..) 
The estimated treatment effects in terms of the above data are computed as:
a. tHAT(M) = _______________. b. tHAT(L) = _______________. c. tHAT(H) = _______________.
The estimated random error effects in terms of the above data may be computed as:
d. eHAT(M2) = _______________. e. eHAT(L3) = _______________.
Answer: a. 10  173/14
b. 53/4  173/14
c. 14  173/14
d. 10  10
e. 13  53/4
151.
In the past a chemical fertilizer plant has produced an average of 1100 pounds of fertilizer per day. The record for the past year based on 256 operating days shows the following:
XBAR = 1060 lbs/day S = 320 lbs/day
where XBAR and S have the usual meaning. It is desired to test whether or not the average daily production has dropped significantly over the past year. Suppose that in this kind of operation, the traditionally acceptable level of significance has been .05. But the plant manager, in his report to his bosses, uses level of significance .01. Analyze the data at both levels after setting up appropriate hypotheses, and comment.
Answer: H(O): MU = 1100 H(A): MU < 1100
Since n = 256, use Z to approximate t.
S(XBAR) = 320/SQRT(256) = 320/16 = 20
Z(calculated) = (1060  1100)/20 = 40/20 = 2
Z(critical, ALPHA=.05, onetailed) = 1.645
Z(critical, ALPHA=.01, onetailed) = 2.33
Therefore, 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 his bosses 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.
152.
A test of the breaking strengths of two different types of cables was conducted using samples of n(1) = n(2) = 100 pieces of each type of cable.
CABLE I CABLE II  YBAR(1) = 1925 YBAR(2) = 1905 S(1) = 40 S(2) = 50
Do the data provide sufficient evidence to indicate a difference between the mean breaking strengths of the two cables? Use ALPHA = .10. Assume SIGMA(1)**2 = SIGMA(2)**2. The tabular value is 1.65.
Answer: Z = (1925  1905)/[SQRT(1600/100 + 2500/100)! = 3.1
Therefore the data indicates a difference.
153.
A standard method for determining the amount of active ingredient in propellants is known to have a standard deviation SIGMA = .8. Two new propellants, assumed to be homogeneous, were tested five times. The results of these tests are:
X(1): 63.2, 63.6, 62.7, 64.4, 63.1 X(2): 62.2, 64.8, 62.2, 60.2, 61.1
Test at the .05 level of significance whether there is a difference in the amount of active ingredient in the two propellants.
Answer: H(0): MU(1)  MU(2) = 0 H(1): MU(1)  MU(2) =/= 0
XBAR(1) = 63.4 XBAR(2) = 62.1
SIGMA(XBAR(1)XBAR(2) = SQRT([SIGMA(1)**2!/[n(1)!+[SIGMA(2)**2!/[n(2)! = SQRT(.64/5 + .64/5) = .506
Using a Ztest: Z = ((XBAR(1)  XBAR(2))  0)/SIGMA(XBAR(1)  XBAR(2)) = 1.3/.506 = 2.57
The critical twotailed value for Z with ALPHA = .05 is 1.96. Therefore reject H(0) at 5% significance level since 2.57 > 1.96.
154.
Suppose that you have been assigned to estimate the height of a group of corn plants arranged in 4 rows with 50 plants in each row. You may take measurements of 10 plants.
a. Outline a method for obtaining a random sample in such a situation. b. What advantages or disadvantages are in such a procedure?
Answer: a. Assign numbers to plants (1  200). Draw a random sample of size 10 using a random numbers table. Simplest procedure is to use sampling with replacement.
b. Advantage is that common formulas for mean and variance apply, but it's a nuisance to have to number plants and use random selection.
155.
In a random sample of flashlight batteries, the average useful life was 22 hours and the sample standard deviation was 2 hours. How large should the sample size be if you want the mean of your sample to be within 1 hour of MU 90 times out of 100 in repeated sampling?
a. 25 b. 11 c. 90 d. 35 e. both c & d. Since the calculated n is too small for the ntral limit theorem to apply, choose n >= 30.
Answer: b. 11
n = [[2**2![1.645**2!!/[1**2! = 10.8241 == 11
