The time married men with children spend on child care averages 6.4 hours per week (Time, March 12, 2012). 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 6.4 hours per week. A sample of 40 married couples are surveyed to collect the hours per week the husband spends on child care. The sample mean is 7.0 hours per week, and the sample standard deviation is 2.4276 hours per week. Use Excel to answer the following questions. 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 by Time in your area? b. What is the test statistics of the population mean? c. Using a .05 level of significance, what are the critical values? d. What is your conclusion?

d. There is not enough evidence to support the claim that the time married men in your area spend on child care per week differs significantly from the reported mean of 6.4 hours per week.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the time married men in your area spend on child care per week differs significantly from the reported mean of 6.4 hours per week.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=6.4\\\\H_a:\mu\neq 6.4[/tex]

The significance level is 0.05.

The sample has a size n=40.

The sample mean is M=7.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.4276.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2.4276}{\sqrt{40}}=0.384[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7-6.4}{0.384}=\dfrac{0.6}{0.384}=1.563[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=40-1=39[/tex]

This test is a two-tailed test, with 39 degrees of freedom and t=1.563, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=2\cdot P(t>1.563)=0.126[/tex]

As the P-value (0.126) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the time married men in your area spend on child care per week differs significantly from the reported mean of 6.4 hours per week.

If the critical value approach is used, for 39 degrees of freedom and a significance level of 0.05, the critical values are tc=±2.021.

As the test statistic falls within the acceptance region (larger than -2.021 and smaller than 2.021), the null hypothesis failed to be rejected.