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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
nonSufferers 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 nonsufferers of Reynaud’s syndrome exceeds that of
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sufferers by more than one cal/cm /min.
The test statistic is
. At the = .01 level of significance, the critical value for this
onesided test on the left is z.99 = 2.33. Since 2.90 < 2.33, we reject the null hypothesis and conclude that the mean
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heat output of Reynaud’s syndrome suffers is more than one cal/cm /min than the mean for nonsuffers.
b. The pvalue = 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.
 Spring '04
 Chung
 Standard Error

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