Stat 312 Homework Problems - Week 10

The data is summarized in the table since the

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Unformatted text preview: ; –1 5. The data is summarized in the table. Since the population data is normal with known standard deviations, a “large sample” procedure is appropriate. The respective standard errors are and . Group Sufferers non-Sufferers n 10 10 .2 .4 mean .64 2.05 a. The hypotheses are H0: - = –1.0 (= 0) versus Ha: - < –1.0. The alternate hypothesis says that the mean heat output for non-sufferers of Reynaud’s syndrome exceeds that of 2 sufferers by more than one cal/cm /min. The test statistic is . At the = .01 level of significance, the critical value for this one-sided test on the left is z.99 = -2.33. Since -2.90 < -2.33, we reject the null hypothesis and conclude that the mean 2 heat output of Reynaud’s syndrome suffers is more than one cal/cm /min than the mean for non-suffers. b. The p-value = P(Z < -2.90) = .0019. c. In this problem, ’ represents the difference in means under the alternate hypothesis. That is, ’ - = –1.2. Now d. Using m = n, set So the goal will be accomplished if n = 66. 8. The data is summarized in the table. a. The hypotheses are H0: Y - X = 10 The test statistic is versus Wire Rod X=AISI 1064 Y=AIS...
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This note was uploaded on 09/17/2012 for the course STAT 312 taught by Professor Chung during the Spring '04 term at University of Wisconsin.

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