18
Chapter 1
infinite sample size.
BexpL from Table 1.1
z~/2 = 1.960 N = 24
Substituting
0.731934B 2.28635 _ The solution from SAS for 24 data points from Eq. (1.53)
2.69271 _< B _< 3.55471
for infinite sample size O use Eq. (1.38)
0.8741670 < O < 1.143950
2813.91 < O < 3682.33
The solution from SAS for 24 data points from Eq. (1.53)
(1.58)
(1.59)
(1.60)
EXAMPLE 1.4. Aluminum casting, 26 tensile strength data, from
Problem 1.1 casting A is used to find the best fit of a Gaussian curve or
1462.96 < 0 < 4974.96
(1.61)
The intercept ~ for 24 points has an average or mean and what appears as a
standard deviation. The only option is to bound ~ is by use of an ~A equation
similar to Eq. (1.54) for the infinite sample size
~A =~±kA S~
YA = 36,497 q- 3.25(1,634)
(1.62)
31,187 _< 7A --< 41,808 psi
The SAS solution for 24 points gives from Eq. (1.53) yields
34,863 < 7 -< 38,131 psi
(1.63)
Now for the infinite Weibull distribution Eq. (1.14)
/3 is obtained from Eq. (1.58)
O is Eq. (1.60) while Eq. (1.61) has more spread
~ is from Eq. (1.62)