A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.23 hours, with a standard deviation of 2.38 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.34 hours. with a standard deviation of 1.67 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (My - #2)
Let u1 represent the mean leisure hours of adults with no children under the age of 18 and f2 represent the mean leisure hpurs of adults with children under the age of 18.
The 95% confidence interval for is the range from hours to hours
(Round to two decimal places as needed.)
Answer
\(\boxed{\text{Final Answer: The 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children is approximately from -0.01 hours to 1.79 hours. This means we are 95% confident that the true mean difference in leisure time between the two groups lies within this interval.}}\)
Step 1 ：Let \(u1\) represent the mean leisure hours of adults with no children under the age of 18 and \(f2\) represent the mean leisure hours of adults with children under the age of 18.
Step 2 ：We are asked to construct a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children. The formula for a confidence interval is: \[\bar{x} \pm z \frac{s}{\sqrt{n}}\] where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level (for a 95% confidence level, \(z = 1.96\)), \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 3 ：We can calculate the mean difference and the standard error of the difference, and then use these to construct the confidence interval.
Step 4 ：Given that the mean leisure time for adults with no children is 5.23 hours, the standard deviation is 2.38 hours, and the sample size is 40. For adults with children, the mean leisure time is 4.34 hours, the standard deviation is 1.67 hours, and the sample size is also 40.
Step 5 ：The mean difference in leisure time between the two groups is 0.89 hours and the standard error of the difference is approximately 0.46 hours.
Step 6 ：Using the formula for the confidence interval, we find that the 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children is approximately from -0.01 hours to 1.79 hours.
Step 7 ：\(\boxed{\text{Final Answer: The 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children is approximately from -0.01 hours to 1.79 hours. This means we are 95% confident that the true mean difference in leisure time between the two groups lies within this interval.}}\)
