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Kids and Leisure Young children require a lot of time. This time commitment cuts into a parent's leisure time. A sociologist wanted to estimate the difference in the amount of daily leisure time (in hours) of adults who do not have children under the age of 18 years and the amount of daily leisure time (in hours) of adults who have children under the age of 18 years. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.62 hours with a standard deviation of 2.43 hours. A random sample of 40 adults with children under the age of 18 years results in a mean daily leisure time of 4.10 hours with a standard deviation of 1.82 hours. Construct and interpret a \(90 \%\) confidence interval for the mean difference in leisure time between adults with no children and adults with children. (Source: American Time Use Survey)

Step by step solution

01

Define the Parameters

Let \(\bar{X_1}\) represent the mean daily leisure time for adults without children under 18, \(\bar{X_2}\) represent the mean daily leisure time for adults with children under 18. Let \(s_1\) and \(s_2\) be the standard deviations, and \(n_1\) and \(n_2\) be the sample sizes.

02

State the Given Information

From the problem statement: \(\bar{X_1} = 5.62\) hours, \(s_1 = 2.43\) hours, \(n_1 = 40\); \(\bar{X_2} = 4.10\) hours, \(s_2 = 1.82\) hours, \(n_2 = 40\).

03

Calculate the Mean Difference

The mean difference \(\bar{D}\) is given by \(\bar{X_1} - \bar{X_2} = 5.62 - 4.10 = 1.52\) hours.

04

Calculate the Standard Error

The standard error (SE) of the difference can be calculated using the formula: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{2.43^2}{40} + \frac{1.82^2}{40}} = \sqrt{\frac{5.9049}{40} + \frac{3.3124}{40}} = \sqrt{0.1476225 + 0.08281} = \sqrt{0.2304325} = 0.480\] hours.

05

Find the Critical Value

For a 90% confidence interval and given sample sizes, use the t-distribution with degrees of freedom \(df = n_1 + n_2 - 2 = 40 + 40 - 2 = 78\). The critical value \(t_{\alpha/2, df}\) for 90% confidence level and 78 degrees of freedom can be approximated as 1.645.

06

Calculate the Margin of Error

The margin of error (ME) is calculated as: \[ME = t_{\alpha/2} \times SE = 1.645 \times 0.480 = 0.7896\] hours.

07

Construct the Confidence Interval

The 90% confidence interval for the mean difference is given by: \[CI = (\bar{D} - ME, \bar{D} + ME) = (1.52 - 0.7896, 1.52 + 0.7896) = (0.7304, 2.3096)\] hours.

08

Interpret the Results

There is a 90% confidence that the true mean difference in daily leisure time between adults with no children and adults with children is between 0.7304 hours and 2.3096 hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Difference

The mean difference is an essential concept when comparing two groups. In this exercise, we're looking at the difference in daily leisure time between adults with no children under 18 and those with children under 18. To find the mean difference, simply subtract the average leisure time of one group from the other. Here, \(\bar{X_1} - \bar{X_2} = 5.62 - 4.10 = 1.52\) hours. This 1.52 hours represents our \( \bar{D} \), the mean difference in daily leisure time between the two groups.

Standard Error

The standard error (SE) quantifies the variability of the mean difference. It helps us understand how much the sample mean difference might vary from the true population mean difference. In this exercise, we calculate the standard error using the formula: \[ SE = \sqrt{\frac{ s_1^2 }{ n_1 } + \frac{ s_2^2 }{ n_2 }} = \sqrt{\frac{2.43^2}{40} + \frac{1.82^2}{40}} = \sqrt{\frac{5.9049}{40} + \frac{3.3124}{40}} = \sqrt{0.1476225 + 0.08281} = \sqrt{0.2304325} = 0.480 \] hours. This tells us there is a 0.480-hour error margin in our mean difference due to sampling variability.

Degrees of Freedom

Degrees of freedom (df) are crucial in determining the critical value from the t-distribution. It is related to the sample size of each group and defines the shape of the T-distribution we use. For two independent samples, degrees of freedom can typically be calculated as: \[ df = n_1 + n_2 - 2 \] In our problem, \( n_1 \) and \( n_2 \) are both 40, so the degrees of freedom equals \( 40 + 40 - 2 = 78 \). This degree of freedom is used to find the critical t-value.

Critical Value

The critical value represents a cutoff point from the t-distribution corresponding to the desired confidence level. For a 90% confidence interval and 78 degrees of freedom, we can approximate the critical value (\( t_{\alpha/2, df} \)) as 1.645. This value determines how far our sample mean can deviate from the true population mean before we consider it too far off.

Margin of Error

The margin of error (ME) combines the standard error and the critical value to tell us how far the true mean difference could be from our sample mean difference with a certain level of confidence. It's calculated by: \[ ME = t_{\alpha/2} \times SE = 1.645 \times 0.480 = 0.7896 \] hours. So, the possible error range around our mean difference is 0.7896 hours, giving us the final confidence interval of \( [0.7304, 2.3096] \) hours for the true mean difference. This tells us that we can be 90% confident the true mean difference in daily leisure time lies within this interval.