A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.33 hours, with a standard deviation of 2.46 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.39 hours, with a standard deviation of 1.53 hours. Construct and interpret a $95 \%$ confidence interval for the mean difference in leisure time between adults with no children and adults. with children $\left(\mu_{1}-\mu_{2}\right)$. Let $\mu_{1}$ represent the mean leisure hours of adults with no children under the age of 18 and $\mu_{2}$ represent the mean leisure hours of adults with children under the age of 18. The $95 \%$ confidence interval for $\left(\mu_{1}-\mu_{2}\right)$ is the range from hours to hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? A. There is $95 \%$ confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours. B. There is $95 \%$ confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. C. There is a $95 \%$ probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. D. There is a $95 \%$ probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.
Solution
Step 1 ：We are given two samples, one of adults with no children under the age of 18 and another of adults with children under the age of 18. We are asked to construct a 95% confidence interval for the difference in mean leisure time between these two groups.
Step 2 ：The formula for a confidence interval for the difference between two means is: \[\bar{x}_1 - \bar{x}_2 \pm z \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, \(n_1\) and \(n_2\) are the sample sizes, and \(z\) is the z-score corresponding to the desired level of confidence.
Step 3 ：In this case, we have \(\bar{x}_1 = 5.33\), \(s_1 = 2.46\), \(n_1 = 40\), \(\bar{x}_2 = 4.39\), \(s_2 = 1.53\), \(n_2 = 40\), and \(z = 1.96\) for a 95% confidence level.
Step 4 ：We can plug these values into the formula to calculate the confidence interval.
Step 5 ：The $95 \%$ confidence interval for the difference in mean leisure time between adults with no children and adults with children is \(\boxed{(0.04, 1.84)}\) hours. This means we are $95 \%$ confident that the true difference in mean leisure time between the two groups lies within this interval. Since the interval does not contain zero, we can conclude that there is a significant difference in the number of leisure hours between adults with no children and adults with children. Therefore, the correct interpretation is option A.
