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Unformatted text preview: th the new alloy. The
hypotheses are H0:1 - 2 = 0 versus Ha: - > 0 (that is, the alloy beam has smaller deflection). At = .01 the critical
value for the one-sided test on the right is 2.33. Since 1 = 2 = .05, . Both sample sizes will be equal. If the true value of - is .04, then
. Setting .05 2 gives , so n = 7.02 = 49.4, or n = 50. 15. The hypotheses are H0:1 - 2 = 0 versus Ha: - > 0. Under the alternate hypothesis
a. is defined as ’. . Anything that makes
the type II error probability smaller if < 0. Increases in either m or n will cause smaller will make to increase. . Since the value of ’ is greater than 0, 0 -’ < 0, so to achieve a b.
lower probability, n will have to increase. Steel
22. The data is summarized in the table. Assuming that slant shear is normally
W ire Brushed
distributed in both populations, we can use a t test for the following hypotheses.
H0: 2 - 1 = 0
Ha: 2 - 1 ≠ 0
The test statistic is
. The individual standard errors are
The degrees of freedom are n
4.01 and , so = 14 degrees of freedom will be used. The p-value is 2P(t14...
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- Spring '04
- Standard Error