Applied Statistics and Probability for Engineers

340 CHAPTER 10 STATISTICAL INFERENCE FOR TWO SAMPLES

3. H 1 :  1  2


4. 0.05


5. The test statistic is

6. Reject H 0 if t 0

 t0.025,142.145 or if t 0 t0.025,142.145.

7. Computations: From Table 10-1 we have 92.255, s 1 2.39, n 1 8, 92.733,
   s 2 2.98, and n 2 8. Therefore

and

8. Conclusions: Since2.145t 0 0.352.145, 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 t0.40,140.258 and t0.25,140.692. Therefore, since 0.2580.350.692, we

conclude that lower and upper bounds on the P-value are 0.50P0.80. Therefore, since

the P-value exceeds 0.05, the null hypothesis cannot be rejected.

t 0 

x 1 x 2

2.70 ̨

B

1

n 1

1

n 2



92.25592.733

2.70

B

1

8

1

8

0.35

sp 2 7.302.70

s^2 p

1 n 1  12 s^211 n 2  12 s^22

n 1 n 2  2



1721 2.39 2271 2.98 22

88  2

7.30

x 1 x 2

t 0 

x 1 x 2  0

sp

B

1

n 1

1

n 2



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

92.255 92.733

s 1 2.39 s 2 2.98

x 1 x 2

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