2 1 2 1 2 1 1 1 n n S X X T p Tests on the Difference in Means of Two Normal

2 1 2 1 2 1 1 1 n n s x x t p tests on the difference

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2 1 2 1 2 1 1 1 ) ( n n S X X T p
Tests on the Difference in Means of Two Normal Distributions, Variances Unknown and Equal Null hypothesis: H 0 : 1   2   0 Test statistic: (10-14) Alternative Hypothesis P-Value Rejection Criterion for Fixed-Level Tests H 1 : 1   2   0 Probability above t 0 and probability below  t 0 or H 1 : 1   2   0 Probability above t 0 H 1 : 1   2   0 Probability below t 0 2 1 0 2 1 0 1 1 n n S X X T p 2 , 2 / 0 2 1 n n t t 2 , 2 / 0 2 1 n n t t 2 , 0 2 1 n n t t 2 , 0 2 1 n n t t
EXAMPLE 10-5 Yield from a Catalyst Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in Table 10-1. Is there any difference between the mean yields? Use   0.05, and assume equal variances. 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 s 1 2.39 s 2 2.98 255 . 92 1 x 733 . 92 2 x
EXAMPLE 10-5 Yield from a Catalyst - Continued The seven-step hypothesis-testing procedure is as follows: 1. Parameter of interest: The parameters of interest are 1 and 2 , the mean process yield using catalysts 1 and 2, respectively. 2. Null hypothesis: H 0 : 1   2 3. Alternative hypothesis: H 1 : 1   2 4. Test statistic: The test statistic is 5. Reject H 0 if: Reject H 0 if the P -value is less than 0.05. 2 1 2 1 0 1 1 0 n n s x x t p
EXAMPLE 10-5 Yield from a Catalyst - Continued 6. Computations: From Table 10-1 we have , s 1 = 2.39, n 1 = 8, , s 2 = 2.98, and n 2 = 8. Therefore and 7. Conclusions: From Appendix Table V we can find t 0.40,14 0.258 and t 0.25,14 0.692. 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. Interpretation: At 5% 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. 255 . 92 1 x 2 92.733 x 70 . 2 30 . 7 30 . 7 2 8 8 ) 98 . 2 ( 7 ) 39 . 2 ( ) 7 ( 2 ) 1 ( ) 1 ( 2 2 2 1 2 2 2 2 1 1 2 p p s n n s n s n s 35 . 0 8 1 8 1 70 . 2 733 . 92 255 . 92 1 1 70 . 2 2 1 2 1 0 n n x x t
10-4: Paired t -Test Null hypothesis: H 0 : D   0 Test statistic: Alternative Hypothesis P-Value Rejection Criterion for Fixed-Level Tests H 1 : D 0 Probability above t 0 and probability below  t 0 H 1 : D   0 Probability above t 0 H 1 : D   0 Probability below t 0 n S D T D / 0 0 (10-24) 1 , 2 0 1 , 2 0 n n t t or t t 1 , 0 n t t 1 , 0 n t t
Example 10-11 Shear Strength of Steel Girder An article in the Journal of Strain Analysis [1983, Vol. 18(2)] reports a comparison of several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied to nine specific girders, are shown in the table below.

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