Chapter 10: Q. 98 (page 603)

98. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement "I'm pleased with the way we divide the responsibilities for childcare." The ratings went from one (strongly agree) to five (strongly disagree). Table 10.26 contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband's versus the wife's satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife).
Wife's Score
2
2
3
3
4
2
1
1
2
4
Husband's
Score
2
2
1
3
2
1
1
1
2
4

Short Answer

Expert verified

(a) The null hypothesis: $H_{0} : μ_{d} = 0$
(b) The alternate hypothesis: $H_{a} : μ_{1} < 0$
(c) The random variable $X_{d}$ represents the average difference in satisfaction levels between husband and wife.
(d) The distribution is normal.
(e) The test statistics: $- 1.86$ .
(f) The $p$ -value is $0.0478$ .
(g)(i) $α = 0.05$
(ii) Decision: reject the null hypothesis.
(iii) Reason for Decision: $α > p$ -value .
(iv) As a result: Since $α$ and $p$ are so near, this is a poor test.
Therefore, there is insufficient information to establish that the mean difference is $(-)$ negative.

Step by step solution

Given information

To conduct a hypothesis test to see if the mean difference in the husband's versus the wife's satisfaction level is negative .

Explanation

(a) The null hypothesis is indicated as:
$H_{0} : μ_{d} = 0$
(b) Alternate hypothesis is indicated as:
$H_{a} : μ_{1} < 0$
(c) The random variable $X_{d}$ represents the average difference in satisfaction levels between husband and wife.
(d) The distribution is normal.
(e) To determine the test static as:
To access the stat list editor, click STAT and then 1.

Then ENTER all values the OUTPUT will be:

Hence, the test statistics is $- 1.86$ .

Explanation

(f) The $p$ -value from the output is determined as $0.0478$ .

(g) Obtain a clear picture of the situation using the information from the last task.
The horizontal axis should be clearly labelled and scaled, and the region(s) corresponding to the $p$ -value should be shaded on the graph.

(i) $α = 0.05$
(ii) Decision: reject the null hypothesis.
(iii) Reason for Decision: $α > p$ -value .
(iv) As a result: Since $α$ and $p$ are so near, this is a poor test.
Therefore, there is insufficient information to establish that the mean difference is $(-)$ negative.