Question

The amount of time the husband and the wife spend on house work is measured for 15 women and their 15 husbands. For the wives the mean was 7 hours/week and for the husbands the mean was 4.5 hours/week. The standard deviation of the differences in time spent on house work was 2.85. What is the value of the test statistic for testing the difference in mean time spent on housework between husbands and wives

Answer 1

The null hypothesis is that there is no difference in the mean time spent on housework between husbands and wives, and the alternative hypothesis is that there is a difference.

Step-by-step explanation:

The hypothesis test can be conducted using a t-test for comparing means. The null hypothesis, H0, would state that there is no difference in the mean length of time spent on housework between husbands and wives. The alternative hypothesis, Ha, would state that there is a difference in the mean length of time spent on housework between husbands and wives.

To determine the value of the test statistic, the formula for calculating the t-score can be used:

t = (x1 - x2) / (s / sqrt(n))

where:

• x1 is the mean time spent on housework by the wives
• x2 is the mean time spent on housework by the husbands
• s is the standard deviation of the differences in time spent on housework
• n is the sample size (which is 15 in this case)

Plugging in the given values, the test statistic can be calculated.

Answer 2

The value of the test statistic is z = -0.877.

Step-by-step explanation:

Testing the difference in mean time spent on housework between husbands and wives.

At the null hypothesis, we test if there is no difference, that is, the subtraction of the means is 0:

`H_0: \mu_H - \mu_W = 0`

At the alternate hypothesis, we test if there is a difference, that is, the subtraction of the means is different from 0.

`H_1: \mu_H - \mu_W \\eq 0`

The test statistic is:

`z = (X - \mu)/(s)`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that
`\mu = 0`

For the wives the mean was 7 hours/week and for the husbands the mean was 4.5 hours/week. The standard deviation of the differences in time spent on house work was 2.85.

This means that
`X = 4.5 - 7 = -2.5, s = 2.85`

What is the value of the test statistic for testing the difference in mean time spent on housework between husbands and wives?

`z = (X - \mu)/(s)`

`z = (-2.5 - 0)/(2.85)`

`z = -0.877`

The value of the test statistic is z = -0.877.