Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

SECTION 17.3 | The Chi-Square Test for Independence 581
The column totals describe the overall distribution of alcohol-related problems. These totals
indicate that 160 out of 200 participants reported that they did not experience any alcoholrelated
problems. This proportion corresponds to 160/200, or 80% of the total sample.
Similarly, 40/200, or 20% reported that they did experience alcohol-related problems. The
null hypothesis (version 2) states that these proportions are the same for both groups of participants.
Therefore, we simply apply the proportions to each group to obtain the expected
frequencies. For the group of 80 teens who were not allowed to drink (top row), we obtain
80% of 80 = 64 expected to experience problems
20% of 80 = 16 expected not to experience problems
For the group of 120 teens who were allowed to drink (bottom row), we expect
80% of 120 = 96 expected to experience problems
20% of 120 = 24 expected not to experience problems
These expected frequencies are summarized in Table 17.9. Note that the expected frequencies
maintain the same row totals and column totals, and create an ideal frequency distribution
that perfectly represents the null hypothesis. Specifically, the proportions for the group
of 80 teens for whom alcohol was not allowed are the same as the proportions for the group
of 120 who were permitted to drink.
TABLE 17.9
The expected frequencies
( f e
values) if experience
with alcohol-related
problems is completely
independent of parents’
rules concerning underage
drinking.
Experience with
Alcohol-Related
Problems
No
Yes
Not Allowed to Drink 64 16 80
Allowed to Drink 96 24 120
160 40 n = 200
The chi-square statistic is now used to measure the discrepancy between the data (the
observed frequencies in Table 17.8) and the null hypothesis that was used to generate the
expected frequencies in Table 17.9.
x 2 5
s71 2 64d2
64
1
s9 2 16d2
16
1
s89 2 96d2
96
= 0.766 + 3.063 + 0.510 + 2.042
= 6.381
1
(31 2 24)2
24
STEP 4
Make a decision regarding the null hypothesis and the outcome of the study. The obtained
chi-square value exceeds the critical value (3.84). Therefore, the decision is to reject
the null hypothesis. In the literature, this would be reported as a significant result with
χ 2 (1, n = 200) = 6.381, p < .05. According to version 1 of H 0
, this means that we have
decided that there is a significant relationship between parents’ rules about alcohol and subsequent
problems. Expressed in terms of version 2 of H 0
, the data show a significant difference
in alcohol-related problems between teens whose parents allowed drinking and those
whose parents did not. To describe the details of the significant result, you must compare
the original data (Table 17.8) with the expected frequencies in Table 17.9. Looking at the
two tables, it should be clear that teens whose parents allowed drinking experienced more
alcohol-related problems than would be expected if the two variables were independent. ■