Keyshira Thompson
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Question 3, 9.3.9
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A simple random sample of size $n=40$ is drawn from a population. The sample mean is found to be $\bar{x}=121.7$ and the sample standard deviation is found to be $s=12.5$. Construct a $99 \%$ confidence interval for the population mean.

The lower bound is $\square$. (Round to two decimal places as needed.)

Step 1 ：We are given a simple random sample of size \(n=40\) drawn from a population. The sample mean is found to be \(\bar{x}=121.7\) and the sample standard deviation is found to be \(s=12.5\). We are asked to construct a \(99 \%\) confidence interval for the population mean.

Step 2 ：The formula for a confidence interval is \(\bar{x} \pm Z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score corresponding to the desired level of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：For a \(99\%\) confidence interval, the Z-score is \(Z = 2.576\).

Step 4 ：Substituting the given values into the formula, we get the lower bound of the confidence interval as \(\bar{x} - Z \frac{s}{\sqrt{n}}\).

Step 5 ：Substituting the given values, we get \(121.7 - 2.576 \frac{12.5}{\sqrt{40}}\).

Step 6 ：Solving the above expression, we get the lower bound of the confidence interval as approximately 116.61.

Step 7 ：Final Answer: The lower bound of the 99% confidence interval for the population mean is approximately \(\boxed{116.61}\).