Question

a random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.47 hours, with a standard deviation of 2.33 hours. a random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.06 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 ($\mu_1 - \mu_2$).
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 90% confidence interval for ($\mu_1 - \mu_2$) is the range from $\square$ hours to $\square$ hours. (round to two decimal places as needed.)

Explanation:

Step1: Determine the critical value

For a 90% confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\). Since \(n_1 = n_2=40\) (large - sample, we can use the standard normal distribution \(z\)). The critical value \(z_{\frac{\alpha}{2}}\) from the standard normal table is \(z_{0.05}\approx1.645\).

Step2: Calculate the difference in sample means

The sample mean for adults with no children \(\bar{x}_1 = 5.47\) and for adults with children \(\bar{x}_2 = 4.06\). The difference in sample means \(\bar{x}_1-\bar{x}_2=5.47 - 4.06=1.41\).

Step3: Calculate the standard error

The formula for the standard error \(SE=\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\), where \(s_1 = 2.33\), \(n_1 = 40\), \(s_2 = 1.82\), \(n_2 = 40\).

$$
LATEXBLOCK0
$$

Step4: Calculate the margin of error

The margin of error \(E = z_{\frac{\alpha}{2}}\times SE\). Substituting \(z_{\frac{\alpha}{2}} = 1.645\) and \(SE\approx0.47\), we get \(E=1.645\times0.47\approx0.77\).

Step5: Calculate the confidence interval

The confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1 - \bar{x}_2-E,\bar{x}_1 - \bar{x}_2 + E)\).
Substituting \(\bar{x}_1-\bar{x}_2 = 1.41\) and \(E = 0.77\), we get \((1.41-0.77,1.41 + 0.77)=(0.64,2.18)\).

Answer:

The 90% confidence interval for \((\mu_1-\mu_2)\) is the range from \(0.64\) hours to \(2.18\) hours.