The Analysis of Biological Data
2nd Edition
ISBN: 9781936221486
Author: Michael C. Whitlock, Dolph Schluter
Publisher: W. H. Freeman
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Question
Chapter 13, Problem 21AP
(a)
To determine
To construct: The figure of the difference in the stress level between the females
(b)
To determine
To explain: If the paired-test is appropriate for this data or not.
(c)
To determine
To explain: Whether the paired t-test would be appropriated after a log transformation.
(d)
To determine
To explain: if the sign-test is appropriate for this data or not.
(e)
To determine
To check: Whether the level of stress is same between the females using a sign test.
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In the journal Mental Retardation, an article reported the results of a peer tutoring program to help mildly mentally retarded children learn to read. In the experiment, the mildly retarded children were randomly divided into two groups: the experimental group received peer tutoring along with regular instruction, and the control group received regular instruction with no peer tutoring. There were n1 = n2 = 30 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 49.1. For the control group, the mean score on the same test was x2 = 353.8, with sample standard deviation s2 = 50.5. Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began.
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In a breeding experiment, chicken with white feathers, small comb were mated and the offspring categories white feathers, small comb (WS), white feathers. large comb (WL), dark feathers, small comb (DS) and dark feathers, large comb (DL) were expected to follow the ratio 9:3:3:1 the researcher observed that there were 100 WS, 32 WL, 40 DS and 20 DL offspring. In order to test if the observed frequencies follow the expected ratio, what should be the hull hypothesis?A. P(WS) = 100/192, P(WL) = 32/192, P(DS) = 40/192 , P(DL) = 20/192
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C.4
D.1
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A.7.81
B.3.84
C.6.81
D.5.99
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A.0.28
B.0.05
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The Analysis of Biological Data
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