Question

a study was conducted to estimate μ, the mean number of weekly hours that u.s. adults use computers at home. suppose a random sample of 81 u.s. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. we are 95% confident that the mean number of weekly hours that u.s. adults use computers at home falls between which of the following intervals? a. 7.3 and 9.7 b. 6.5 and 10.5 c. 7.7 and 9.3 d. 8.1 and 8.9 e. 8.4 and 8.6

Explanation:

Step1: Identify the formula for confidence - interval

For a 95% confidence interval when the population standard - deviation $\sigma$ is known, the formula is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ 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 $\bar{x}$, $\sigma$, $n$, and $z_{\alpha/2}$

We are given that $\bar{x} = 8.5$, $\sigma=3.6$, $n = 81$. For a 95% confidence interval, $\alpha=1 - 0.95=0.05$, so $\alpha/2=0.025$. The $z$ - score $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate the margin of error $E$

$E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{3.6}{\sqrt{81}}=1.96\times\frac{3.6}{9}=1.96\times0.4 = 0.784\approx0.8$.

Step4: Calculate the confidence interval

The lower limit is $\bar{x}-E=8.5 - 0.8 = 7.7$.
The upper limit is $\bar{x}+E=8.5 + 0.8 = 9.3$.

Answer:

C. 7.7 and 9.3