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Kids and Leisure Young children require a lot of time and 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 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

- Identify the problem

The problem asks us to construct and interpret a 90% confidence interval for the mean difference in daily leisure time between adults with no children under the age of 18 and adults with children under the age of 18.

02

- Define the variables

Let \( \mu_1 \) be the mean daily leisure time for adults with no children under 18, and \( \mu_2 \) be the mean daily leisure time for adults with children under 18. We need to estimate the difference in means, \( \mu_1 - \mu_2 \).

03

- Gather the sample data

From the problem statement: \( \bar{x}_1 = 5.62 \) hours, \( s_1 = 2.43 \) hours, \( n_1 = 40 \) (adults with no children) and \( \bar{x}_2 = 4.10 \) hours, \( s_2 = 1.82 \) hours, \( n_2 = 40 \) (adults with children).

04

- Calculate the standard error

The standard error (SE) for the difference in means is calculated using the formula \[ SE = \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } = \sqrt{ \frac{2.43^2}{40} + \frac{1.82^2}{40} } \] This results in \[ SE = \sqrt{ \frac{5.9049}{40} + \frac{3.3124}{40} } = \sqrt{ 0.1476 + 0.0828 } = \sqrt{0.2304} \approx 0.480 \]

05

- Find the critical value

For a 90% confidence interval and a large sample size, we use the standard normal distribution (Z-distribution). The critical value (Z*) for a 90% confidence interval is approximately 1.645.

06

- Calculate the margin of error

The margin of error (ME) is given by \[ ME = Z^* \times SE = 1.645 \times 0.480 \approx 0.7896 \]

07

- Determine the confidence interval

The 90% confidence interval for the difference in means is given by \[ (\bar{x}_1 - \bar{x}_2) \pm ME = (5.62 - 4.10) \pm 0.7896 \] This results in \[ 1.52 \pm 0.7896 \] or the interval \[ (0.7304, 2.3096) \]

08

- Interpret the confidence interval

We are 90% confident that the mean difference in daily leisure time between adults with no children under the age of 18 and adults with children under the age of 18 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

To understand the concept of a mean difference, let’s look at our example. Here, we are comparing the daily leisure time between two groups of adults: those with children under 18 and those without children under 18. The mean difference is simply the difference between the average (mean) leisure time of these two groups.
In our example, the mean daily leisure time for adults without children is 5.62 hours, while for those with children, it is 4.10 hours. So, the mean difference is: \( 5.62 - 4.10 = 1.52 \)
This calculation provides an understanding of how much more (or less) one group may be enjoying their leisure time compared to the other.

Standard Error

The standard error (SE) measures the accuracy with which a sample represents a population. When comparing two sample means, the standard error helps us estimate the variability in the difference between the two sample means.
Using the formula for the standard error of the mean difference:
\[ SE = \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } \]
we plug in our values: \[ SE = \sqrt{ \frac{2.43^2}{40} + \frac{1.82^2}{40} } = \sqrt{ \frac{5.9049}{40} + \frac{3.3124}{40} } = \sqrt{ 0.1476 + 0.0828 } \]
This results in:
\( SE = \sqrt{0.2304} \approx 0.480 \)
This value tells us how much we expect the mean difference to vary from sample to sample due to random chance.

Critical Value

In statistics, the critical value is a threshold that defines a certain probability level in the context of confidence intervals or hypothesis testing. For a 90% confidence interval, we look at the Z-distribution (standard normal distribution) to find the critical value.
The critical value (Z*) for a 90% confidence interval is typically around 1.645. This value ensures that 90% of the data falls within this threshold inside the normal distribution.
So, we use:
\( Z^* = 1.645 \)
This critical value is then used to calculate the margin of error, helping to establish the range within which we expect the true mean difference to lie.

Margin of Error

The margin of error (ME) quantifies the range of uncertainty around the sample mean difference. It incorporates both the level of confidence we want (determined by the critical value) and the standard error.
To find the margin of error, we use the formula:
\( ME = Z^* \times SE \)
For our example, the margin of error is:
\( ME = 1.645 \times 0.480 \approx 0.7896 \)
This value indicates the range above and below the sample mean difference within which we believe the true mean difference lies. In other words, the true mean difference in daily leisure time between adults with no children under 18 and adults with children under 18 is likely somewhere between 0.7896 hours more and 0.7896 hours less than the observed mean difference of 1.52 hours.