Question

The time married men with children spend on child care averages 6.0 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 6.0 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. 1.4 4.9 8.5 5.2 10.2 2.6 7.2 2.6 11.1 10.6 2.3 7.8 1.8 8.7 2.1 4.5 5.3 3.5 8.8 6.7 2.9 10.8 8.4 4.9 5.7 9.8 0.9 11.9 3.5 8.7 6.5 6.5 2.0 9.7 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? H : HA: # Ag b. What is the sample mean (to 1 decimal)? Calculate the value of the test statistic (to 2 decimals). What is the p-value? greater than .207 c. Select your own level of significance. What is your conclusion? Do not reject the null. We don't have enough evidence to disprove the reported time.

Answer 1

The null hypothesis (H0) would be that the population mean number of hours married men in your area spend on child care per week is equal to the reported mean of 6.0 hours per week.

The population mean number is equal to the reported mean according to null hypothesis. The alternative hypothesis (HA) would be that the population mean differs from the reported mean.

H0: μ = 6.0 hours/week

HA: μ ≠ 6.0 hours/week

b. To find the sample mean, we need to add up all of the values in the sample and divide by the number of values. The sample mean is 6.34 hours/week.

To calculate the test statistic, we can use the formula:

t = (x' - μ0) / (s / √n)

where x' is the sample mean, μ0 is the hypothesized mean (in this case, 6.0 hours/week), s is the sample standard deviation, and n is the sample size.

Plugging in the values, we get:

t = (6.34 - 6.0) / (3.23 / √40) = 0.34 / (0.53 / 6.32) = 0.34 / 0.08 = 4.25

The p-value is the probability of getting a result as extreme as the one we observed, given that the null hypothesis is true. In this case, we would need to use a t-distribution table or a computer program to find the p-value, since the sample size is small and the population standard deviation is unknown.

c. If we select a level of significance of 0.05, then we can use the p-value to make a decision about the null hypothesis. If the p-value is less than 0.05, we would reject the null hypothesis and conclude that the population mean differs from the reported mean. If the p-value is greater than 0.05, we do not have enough evidence to reject the null hypothesis and would conclude that the population mean is likely to be the same as the reported mean.

In this case, the p-value is greater than 0.207, so we do not have enough evidence to reject the null hypothesis. We would conclude that the population mean number of hours married men in your area spend on child care per week is likely to be the same as the reported mean of 6.0 hours per week.

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