Data Reduction
7
EXAMPLE 1.1. In order to illustrate the concepts for estimating
and a or ~ for small samples, N < 30, select two test stress values 40,000 psi
and 45,000 psi. These are from Eq. E.I in the range of ultimates for 6061-T6
aluminum, therefore, the final answers can be compared to MILHDBK 5F
[1.18] for 0.010-0.249 sheet
A Basis - ~rut = 42 Ksi ayt = 36 Ksi
(1.25)
B Basis - aut = 43 Ksi ~yt = 38 Ksi
(1.26)
The mean 2 and standard deviation, s, for test samples of two will be calculated
w-range = 45,000 - 40,000 --- 5000 psi
(1.27)
Zxi 1
2-means - ~- - 2 (45,000 + 40,000) = 42,500 psi
(1.28)
s-STD Deviation =
[.(45x103-42.5x 103)2+(40× 103 - 42.5 x103)211/2
2- 1
(1.29)
= 3,536 psi (this for N = 1 will not work out)
Another estimate, range = 6s, and s is 833 psi. This value will be used and
checked against the final ¢ or ~ from [1.8]. Now to find # from Eq. (1.21)
for 95% confidence
S
S
SC -- 10.975 ~-~_ _ < /g _ < 2 + t0.975 N ----~i(1.30)
d.f.=2-1=l
/0.975 = /0.025 = 12.706 (Table E.2.)
42,500-12.706~-~ ~ 31,916psi _< g < 53,084psi
range infinite sample size for 95% confidence from a sample size of two. Next
estimate o- or ~ from the data using Eq. (1.23) for 95% confidence
-- < ~ <-(1.31)
X0,025
X0.975
from Table E.3. for d.f.=N-1 =2-1 = 1
x0.0252 = 0.0009
x0.025 = 0.0313
X0.9752 = 5.02
X0.975 = 2.2405