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 ;
See CAPQQ3 in the SAS/QC Sample LibraryIf 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 c by requesting plots for different values of c and choosing the value of c that linearizes the point pattern. Alternatively, you can request a maximum likelihood estimate for c, as illustrated in the following statements produce Weibull plots for c = 1, 2 and 3:
ods graphics off; 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 c with the C= option, or you must specify C=EST.
Output 5.23.1 displays the plot for the estimated value c = 1.99. 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 LibraryNow, 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.