The CAPABILITY Procedure |
This example compares the use of three-parameter and two-parameter Weibull Q-Q plots for the failure times in months for 48 integrated circuits. The times are assumed to follow a Weibull distribution.
data Failures; input Time @@; label Time='Time in Months'; datalines; 29.42 32.14 30.58 27.50 26.08 29.06 25.10 31.34 29.14 33.96 30.64 27.32 29.86 26.28 29.68 33.76 29.32 30.82 27.26 27.92 30.92 24.64 32.90 35.46 30.28 28.36 25.86 31.36 25.26 36.32 28.58 28.88 26.72 27.42 29.02 27.54 31.60 33.46 26.78 27.82 29.18 27.94 27.66 26.42 31.00 26.64 31.44 32.52 ; run;
[See CAPQQ3 in the SAS/QC Sample Library]If no assumption is made about the parameters of this distribution, you can use the WEIBULL option to request a three-parameter Weibull plot. As in the previous example, you can visually estimate the shape parameter by requesting plots for different values of and choosing the value of that linearizes the point pattern. Alternatively, you can request a maximum likelihood estimate for , as illustrated in the following statements produce Weibull plots for 1, 2 and 3:
symbol v=plus; title 'Three-Parameter Weibull Q-Q Plot for Failure Times'; proc capability data=Failures noprint; qqplot Time / weibull(c=est theta=est sigma=est) square href=0.5 1 1.5 2 vref=25 27.5 30 32.5 35; run;
Note:When using the WEIBULL option, you must either specify a list of values for the Weibull shape parameter with the C= option, or you must specify C=EST.
Output 5.23.1 displays the plot for the estimated value . The reference line corresponds to the estimated values for the threshold and scale parameters of (=24.19 and =5.83, respectively.
[See CAPQQ3 in the SAS/QC Sample Library]Now, suppose it is known that the circuit lifetime is at least 24 months. The following statements use the threshold value to produce the two-parameter Weibull Q-Q plot shown in Output 5.23.2:
symbol v=plus; title 'Two-Parameter Weibull Q-Q Plot for Failure Times'; proc capability data=Failures noprint; qqplot Time / weibull2(theta=24 c=est sigma=est) square href= -4 to 1 vref= 0 to 2.5 by 0.5; run;
The reference line is based on maximum likelihood estimates =2.08 and =6.05. These estimates agree with those of the previous example.
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