Answers 755Chi- Number
Stepa Effects Squarec df Sig. of Iterations
0 Generating Classb
Deleted Effect 1
1 Generating Classb
Deleted Effect 1
2
3
2 Generating Classb
Deleted Effect 1
2
3 Generating Classb
aAt each step, the effect with the largest significance level for the Likelihood Ratio Change is deleted, provided the
significance level is larger than .050.
bStatistics are displayed for the best model at each step after step 0.
cFor ‘Deleted Effect’, this is the change in the Chi-Square after the effect is deleted from the model.
Race*Gender*Intercourse
Race*Gender*Intercourse
Race*Gender,Race*Intercourse,
Gender*Intercourse
Race*Gender
Race*Intercourse
Gender*Intercourse
Race*Intercourse,
Gender*Intercourse
Race*Gender
Race*Intercourse
Gender*Intercourse
Race*Intercourse,
Gender*Intercourse.000 0
.065 1 .798 2
.065 1 .7981.686
30.359
8.475 1 .194 2
1 .000 2
1 .004 21.752 2 .417
28.977 1
7.093 1 .000 2
1.752 2 .008 2Backward Elimination Statistics17.15 We could not include multiple behaviors in the same de-
sign because the observations would not be independent.
Each person would contribute data on each behavior.
17.17 Death penalty data
The optimal model includes DefRace*VictimRace and
VictimRace*DeathPen.17.19The answers depend on the software packages the stu-
dent uses.
Chapter 18
18.1 (a) WS 5 23; W.025 5 27.
(b) Reject H 0 and conclude that older children include
more inferences in their summaries.
18.3 z 52 3.15; reject H 0.
18.5 (a) T 5 8.5; T.025 5 8; do not reject H 0.
(b) We cannot conclude that we have evidence support-
ing the hypothesis that there is a reliable increase in
hypothesis generation and testing over time. (Here is
a case in which alternative methods of breaking ties
could lead to different conclusions.)
18.7 I would randomly assign the order within each pair of
Before and After scores, and for each set of assign-
ments I would calculate a statistic. (That statistic
could be the mean of the difference scores, or a ttest
on the difference scores.) I would then calculate the
number of times I came out with a result as extreme
as the one I actually obtained, and that, divided by the
number of resamples, would give me the probability
under the null.
18.9 z 52 2.20; p(z$ 2.20) 5 .0278. Again reject H 0 ,
which agrees with our earlier conclusion.
18.11 The scatter plot shows that the difference between the
pairs is heavily dependent upon the score for the first born.
18.13 The Wilcoxon Matched-pairs signed-ranks test tests the
null hypothesis that paired scores were drawn from iden-
tical populations or from symmetric populations with the
same mean (and median). The corresponding t test tests
the null hypothesis that the paired scores were drawn from
populations with the same mean and assumes normality.
18.15 Rejection of the H 0 by a ttest is a more specific state-
ment than rejection using the appropriate distribution
free test because, by making assumptions about normal-
ity and homogeneity of variance, the ttest refers specif-
ically to population means.
18.17 H 5 6.757; reject H 0.
18.19 Take the data for all Nsubjects and shuffle them to ran-
dom order. Then take the first n 1 observations and assign
them to Treatment 1, the next n 2 observations and assign
them to Treatment 2, and so on. Then calculate an Fsta-
tistic on that set of resampled data and record the F.
Repeat this a large number of times (e.g.,1000) and look
at the sampling distribution of F. The proportion of Fval-
ues that are equal to, or greater than, the Fobtained on the
original data will give you the probability under the null.6Results for Exercise 17.3