Page 396 - Applied Statistics And Probability For Engineers

c10.qxd 5/16/02 1:30 PM Page 340 RK UL 6 RK UL 6:Desktop Folder:TEMP WORK:MONTGOMERY:REVISES UPLO D CH114 FIN L:Quark Files:

340 CHAPTER 10 STATISTICAL INFERENCE FOR TWO SAMPLES

Table 10-1 Catalyst Yield Data, Example 10-5
Observation
Number Catalyst 1 Catalyst 2
1 91.50 89.19
2 94.18 90.95
3 92.18 90.46
4 95.39 93.21
5 91.79 97.19
6 89.07 97.04
7 94.72 91.07
8 89.21 92.75
x 1 92.255 x 2 92.733
s 1 2.39 s 2 2.98

3. H : 2
1
1
4. 0.05
5. The test statistic is

x x 0
1
2
t
0
1 1
s n n
p
B 1 2
6. Reject H if t
t 0.025,14 2.145 or if t t 0.025,14 2.145.
0
0
0
7. Computations: From Table 10-1 we have x 1 92.255, s 2.39, n 8, x 2 92.733,
1
1
s 2.98, and n 8. Therefore
2
2
2 2 2 2
1n 12s 1 1n 12s 2 17212.392 712.982
2
1
2
s p 7.30
n 1 n 2 2 8 8 2
s 27.30 2.70
p
and
x x 2 92.255 92.733
1
t 0.35
0
1 1 1 1
2.70
2.70˛ n n
B 1 2 B8 8
8. Conclusions: Since 2.145 t 0 0.35 2.145, the null hypothesis cannot be
rejected. That is, at the 0.05 level of significance, we do not have strong evidence to
conclude that catalyst 2 results in a mean yield that differs from the mean yield when
catalyst 1 is used.
A P-value could also be used for decision making in this example. From Appendix Table IV
we find that t 0.40,14 0.258 and t 0.25,14 0.692. Therefore, since 0.258 0.35 0.692, we
conclude that lower and upper bounds on the P-value are 0.50 P 0.80. Therefore, since
the P-value exceeds 0.05, the null hypothesis cannot be rejected.