20
Chapter 1
A class C K= 5.8
~c = 46,507 - 5.8(2158) = 33,991 psi
The/~ or .~ and ~ = s can be corrected from N = 26 to N infinite using Eqs.
(1.30) and (1.31). The d.f. =26-1 =25, from Table E.2 the t value for 0.025
is 2.060 then x 2 from Table E.3 the 0.025 value is 40.65 with square root of
6.376 and the 0.975 value of 13.12 or X =3.622.
Using Eq. (1.30)
46,507 - 2 060
2158
" 25 -<#1 -<46,507+
2. 060 2158
25
(1.68)
46,329 < ~I -< 46, 685 psi
Using Eq. (1.31)
2158~/-~
2158,v/~
-- <
-6.376 - - 3.622
(1.69)
1,726 < a~ _< 3,038 psi
The Weibull parameter Eq. (1.65), for 26 data points will be converted from
26 points to an infinite sample size. The process starts with Eq. (1.58) and
z ~/2 = 1.960 Table 1.1 and N = 26 points
/3 = B = 1.5609
0 = 3766.922
y = 43,108
yields
0.740950 B - 1.15655 (1.70)
The solution from SAS for 26 data points from Eq. (1.64)
1.1299 < B < 1.99925
(1.71)
The 69 conversion from 26 data points to infinite sample size follows Eq.
(1.60) with
z ~/2 = 1.960 and N = 26 with 0 = 3766.922
0.7721520 -< 6) -< 1.29508
(1.72)
2908.64 -< 6) _< 4878.47
The solution for 26 data points from Eq. (1.64)
3027.92 < 0 < 4505.92
(1.73)
The ~ conversion to infinite sample size follows Eq. (1.62) with KA----3.15