Question

The time married men with children spend on child care averages 5.9 hours per week. You belong to a professional group on family practices that would like to do its own study to determine if the time married men in your area spend on child care per week differs from the reported mean of 5.9 hours per week. A sample of 40 married couples will be used with the data collected showing the hours per week the husband spends on child care. The sample data are contained in the table below.

7.7	10.6	10.4	10.4
3.8	9.1	11.5	2.5
4.6	4.0	10.6	11.0
9.6	6.6	0.9	10.6
7.4	6.3	6.3	8.6
3.5	7.7	5.9	8.1
2.8	11.5	8.1	2.1
7.0	4.1	6.6	7.3
5.5	8.0	5.7	3.5
5.4	0.9	4.5	4.1

a. What are the hypotheses if your group would like to determine if the population mean number of hours married men are spending in child care differs from the mean reported in your area?

H₀ : μ _______ 5.9 (less than, greater than, equal to, less than or equal to, greater than or equal to, not equal to)

Ha : μ _______ 5.9 (less than, greater than, equal to, less than or equal to, greater than or equal to, not equal to)

What is the sample mean?

What is the test statistic?

the p-value is ______. (lower than .01, between .01 and .025, between .025 and .05, between .05 and .10, between .10 and .20, greater than .20)

c. Select your own level of significance. What is your conclusion?

Do we reject the hypothesis? do we/don't we have enough evidence?

Answer 1

Given the time married men with children spend on child care averages = 5.9 hours per week. the sample size taken to conduct the hypothesis is n = 40.

Since the population standard deviation is unknown we will conduct a t-distribution hypothesis test for the given sample.

but before this, we need to find the mean and sample standard deviation of the sample which is calculated as:

Mean = (0.9 + 0.9 + 2.1 + 2.5 + 2.8 + 3.5 + 3.5 + 3.8 + 4 + 4.1 + 4.1 + 4.5 + 4.6 + 5.4 + 5.5 + 5.7 + 5.9 + 6.3 + 6.3 + 6.6 + 6.6 + 7 + 7.3 + 7.4 + 7.7 + 7.7 + 8 + 8.1 + 8.1 + 8.6 + 9.1 + 9.6 + 10.4 + 10.4 + 10.6 + 10.6 + 10.6 + 11 + 11.5 + 11.5)/40
= 264.8/40
Mean = 6.62

and sample standard deviation as:

s = 2.99

a) Based on the claim the hypotheses are:

Test Statistic:

The test statistic is calculated as:

P-value:

The P-value for t-distribution can be computed using excel formula for t-distribution using T-score caluclated above and at degree of freedom which is n-1 = 40-1=39.

the formula used for P-value caluclation is =T.DIST.2T(1.523, 39), thsi results in P-value = 0.1358

So, P-value is between .10 and .20

c) Since we do not have any level of significance we assumne the level of significance as 0.05 to test the hypothesis but the P-value is greater than 0.5 we do not reject the null hypothesis and conclude that there is insufficient evidence at any reasonable level of significance to support the claim.