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 =

Step-by-step explanation:

Corresponding response ratings of wives and husbands form matched pairs.

The data for the test are the differences between the response ratings.

μd = the wives response ratings minus their husband's response ratings.

Wives. Husbands diff

2 2 0

2 1 1

3 2 1

3 3 0

4 2 1

2 1 1

1 1 0

1 1 0

2 2 0

4 4 0

Sample mean, xd

= (0 + 1 + 1 + 0 + 1 + 1 + + 0 + 0 + 0 + 0)/10 = 0.4

xd = 5.67

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (0 - 0.4)^2 + (1 - 0.4)^2 + (1 - 0.4)^2+ (0 - 0.4)^2 + (1 - 0.4)^2 + (1 - 0.4)^2 + (0 - 0.4)^2 + (0 - 0.4)^2 + (0 - 0.4)^2 + (0 - 0.4)^2 = 2.4

Standard deviation = √(2.4/10

sd = 0.49

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

1) The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 10 - 1 = 9

2) The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (0.4 - 0)/(0.49/√10)

t = 2.581

3) We would determine the probability value by using the t test calculator.

p = 0.0148

4) Assume alpha = 0.05

Since alpha, 0.05 > than the p value, 0.0148, then we would reject the null hypothesis.

Therefore, at 5% significance level, we can conclude that the mean difference in the husband's versus the wife's satisfaction level is negative

the answer is 28

step-by-step explanation:

5

step-by-step explanation: