Question 4 of 14, Step 1 of 1 $3 / 14$ Correct A survey of 76 randomly selected homeowners finds that they spend a mean of $\$ 68$ per month on home maintenance. Construct a $99 \%$ confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is $\$ 15$ per month. Round to the nearest cent. Answer Tables Keypad Keyboard Shortcuts

Step 1 ：The question is asking for a 99% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. We are given a sample mean ($\$ 68$), a population standard deviation ($\$ 15$), and a sample size (76).

Step 2 ：The formula for a confidence interval is: \[\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\] where: \(- \bar{x}\) is the sample mean, \(- Z\) is the Z-score, which we can find from a Z-table given the confidence level, \(- \sigma\) is the population standard deviation, \(- n\) is the sample size.

Step 3 ：For a 99% confidence interval, the Z-score is approximately 2.576.

Step 4 ：Let's plug in the values and calculate the confidence interval. The sample mean is 68, the population standard deviation is 15, the sample size is 76, and the Z-score is 2.576.

Step 5 ：The margin of error is calculated as \(Z \frac{\sigma}{\sqrt{n}}\), which is approximately 4.43.

Step 6 ：The confidence interval is then calculated as \(\bar{x} \pm\) margin of error, which gives us approximately (63.57, 72.43).

Step 7 ：This means that we are 99% confident that the true mean amount of money spent per month on home maintenance by all homeowners is between \$63.57 and \$72.43. This result makes sense given the provided sample mean and standard deviation.

Step 8 ：Final Answer: The 99% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners is approximately \(\boxed{(\$63.57, \$72.43)}\).