Stat 312 Homework Problems - Week 10

The hypotheses are h01 2 0 versus ha 0 that is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 Section 9.2 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. Hand Chiseled H0: 2 - 1 = 0 versus Ha: 2 - 1 ≠ 0 The test statistic is . The individual standard errors are The degrees of freedom are n 12 12 mean 19.20 23.13 s 1.58 4.01 and , so = 14 degrees of freedom will be used. The p-value is 2P(t14...
View Full Document

Ask a homework question - tutors are online