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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
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
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
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
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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.
- Spring '04
- Standard Error