Question

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 1 2 3 2 1 1 1 2 4 NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
(1) State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.)
(2) What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)
(3) What is the p-value? (Round your answer to four decimal places.)
(4) Alpha (Enter an exact number as an integer, fraction, or decimal.)
α =

Answer 1

Answer:

(a) The test statistic would follow a t distribution with n - 1 = 10 - 1 = 9 degrees of freedom.

(b) The test statistic is -2.2361.

(c) The p-value of the test is 0.026.

(d) α = 0.05.

Step-by-step explanation:

In this case a hypothesis test is to be performed to determine whether the mean difference in the husband's versus the wife's satisfaction level is negative.

(a)

The data provided is paired data.

So a paired t-test would be used.

The test statistic would follow a t distribution with n - 1 = 10 - 1 = 9 degrees of freedom.

The hypothesis can be defined as follows:

$\text{H}_{0}:\ \mu_{d}=0\\\\\text{H}_{a}:\ \mu_{d}<0$

(b)

Compute the test statistic as follows:

$t=\frac {\bar d-\mu_{d}}{\sigma_{d}/\sqrt{n}}$

  $=\frac{-0.50-0}{0.7071/\sqrt{10}}\\\\=-2.2360894\\\\\approx -2.2361$

The test statistic is -2.2361.

(c)

Compute the p-value as follows:

$p-value=P(t_{n-1}<t)\\\\=P(t_{9}<-2.2361)\\\\=0.026$

The p-value of the test is 0.026.

(d)

It is provided that the significance level of the test is, α = 0.05.