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 1 (strongly agree) to 5 (strongly disagree). The table below contains ten of the paired responses for husbands and wives. Conduct a hypothesis test at the 5% level 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

The dependent t-test (also known as the paired t-test or paired samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired t-test is used to determine if the mean difference in the husband's versus the wife's satisfaction level is negative .

The hypothesis for the test can be defined as follows:

H₀: The mean difference in the husband's versus the wife's satisfaction level is not negative, i.e. d ≥ 0.

Hₐ: The mean difference in the husband's versus the wife's satisfaction level is negative, i.e. d < 0.

The information computed from the provided data is:

[tex]n=10\\\bar d=0.50\\s_{d}=0.8498[/tex]

Compute the test statistic value as follows:

[tex]t=\frac{\bar d}{s_{d}/\sqrt{n}}=\frac{0.50}{0.8498/\sqrt{10}}=1.86[/tex]

The test statistic value is 1.86.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

The degrees of freedom is:

df = n - 1

= 10 - 1

= 9

Compute the p-value of the test as follows:

[tex]p-value=P(t_{9}<1.86)=0.048[/tex]

*Use a t-table.

The p-value of the test is 0.048.

p-value= 0.048 > α = 0.05

The null hypothesis was rejected at 5% level of significance.

Thus, it can be concluded that the mean difference in the husband's versus the wife's satisfaction level is negative.