Save \& Exit Certify Lesson: 8.1 Estimating Population Means (... LACARA OWENS Question 3 of 10, Step 1 of 1 $2 / 10$ Correct 1 A survey of 73 randomly selected homeowners finds that they spend a mean of $\$ 72$ per month on home maintenance. Construct a $95 \%$ confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is $\$ 14$ per month. Round to the nearest cent. Answer Tables Keypad Keyboard Shortcuts submit Answer
Solution
Step 1 ：Given values are: sample mean = \$72, sample size = 73, population standard deviation = \$14, and z-score = 1.96 for a 95% confidence interval.
Step 2 ：Calculate the margin of error using the formula: \(z \times \frac{\sigma}{\sqrt{n}}\), where \(z\) is the z-score, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
Step 3 ：Substitute the given values into the formula to get the margin of error: \(1.96 \times \frac{14}{\sqrt{73}} \approx 3.21\).
Step 4 ：Calculate the confidence interval using the formula: \(\bar{x} \pm \text{margin of error}\), where \(\bar{x}\) is the sample mean.
Step 5 ：Substitute the given values into the formula to get the confidence interval: \(72 \pm 3.21\).
Step 6 ：The 95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners is approximately \(\boxed{(\$68.79, \$75.21)}\).
