Question

a study was conducted to estimate μ, the mean commute distance that all employed u.s. adults travel to work. suppose a random sample of 49 employed u.s. adults gives a mean commute distance of 22 miles and that from prior studies, the population standard deviation is assumed to be σ = 8.4 miles. we are 95% confident that the mean commute distance to work of all employed u.s. adults falls between which of the following intervals? a. 18.4 to 25.6 b. 5.2 to 38.8 c. 20.8 to 23.2 d. 19.6 to 24.4

Explanation:

Step1: Identify the formula for confidence interval

For a large - sample (n ≥ 30) confidence interval of the population mean when the population standard deviation $\sigma$ is known, the formula is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Here, $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and n is the sample size.

Step2: Determine the values of parameters

We are given that $\bar{x}=22$, $\sigma = 8.4$, $n = 49$, and for a 95% confidence interval, $\alpha=1 - 0.95=0.05$, so $\alpha/2=0.025$. The $z -$score $z_{\alpha/2}=z_{0.025}=1.96$. Also, $\sqrt{n}=\sqrt{49}=7$, and $\frac{\sigma}{\sqrt{n}}=\frac{8.4}{7}=1.2$.

Step3: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times1.2 = 2.352$.

Step4: Calculate the confidence interval

The lower limit of the confidence interval is $\bar{x}-E=22 - 2.352=19.648\approx19.6$. The upper limit of the confidence interval is $\bar{x}+E=22 + 2.352=24.352\approx24.4$.

Answer:

D. 19.6 to 24.4