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) The t-distribution is most suitable for this problem.

2) Test statistic = 2.356

3) p-value = 0.0214

4) Alpha = 5% = 0.05

5) The p-value is greater than the significance level at which the test was performed, meaning that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the husbands are happier in dual couple marriages.

Step-by-step Explanation:

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

To 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, we first take the difference in the respomses of wives and husbands

x = (wife's score) - (husband's score)

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

Difference | 0 | 1 | 1 | 0 | 2 | 1 | 0 | 0 | 0 | 0

To use the hypothesis test method, we have to make sure that the distribution is a random sample of the population and it is normally distributed.

The question already cleared these two for us that this sample size is randomly selected from the population and each variable is independent from the other.

The question also already explained that the distribution is assumed to be normally distributed.

1) The distribution to use for this test is the t-distribution. This is because the sample size isn't very large and we have no information about the population mean and standard deviation.

For any hypothesis testing, we must first define the null and alternative hypothesis

Since we want to investigate whether the husbands are happier, that the mean difference is negative, that is less than 0,

The null hypothesis, which normally counters the claim to be investigated, would be that there isnt evidence to conclude that the husbands are happier in dual couple marriages. That is, the mean difference in happiness isn't less than 0, that it is equal to or greater than 0.

And the alternative hypothesis, which usually confirms the claim to be tested, is that there is significant evidence to conclude that the husbands are happier in dual couple marriages. That is, the mean difference in happiness is less than 0.

Mathematically, if μ is the mean difference in happiness of wives and husbands,

The null hypothesis is represented as

H₀: μ ≥ 0

The alternative hypothesis is represented as

Hₐ: μ < 0

2) To obtain the test statistic, we need the mean and standard deviation first.

Mean = (sum of variables)/(number of variables) = (5/10) = 0.5

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = 6(0 - 0.5)² + 3(1 - 0.5)² + (2 - 0.5)² = 1.5 + 0.75 + 2.25 = 4.5

σ = √(4.5/10) = 0.671

we compute the t-test statistic

t = (x - μ)/σₓ

x = sample mean difference = 0.50

μ = 0

σₓ = standard error of the sample mean = (σ/√n)

where n = Sample size = 10,

σ = Sample standard deviation = 0.671

σₓ = (0.671/√10) = 0.2122

t = (0.50 - 0) ÷ 0.2122

t = 2.356

3) checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 10 - 1 = 9

Significance level = 5% = 0.05

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 2.356, at 0.05 significance level, df = 9, with a one tailed condition) = 0.021441 = 0.0214

4) Alpha = significance level = 5% = 0.05

5) The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.0214

0.0214 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the husbands are happier in dual couple marriages.

Hope this Helps!!

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9987

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