Overweight Men For a random sample of 60 overweight men, the mean of the number of pounds that they were overweight was 29 . The standard deviat the population is 3.9 pounds. Part 1 of 4 (a) The best point estimate of the mean is 29 pounds. Part 2 of 4 (b) Find the \( 95 \% \) confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answe to one decimal place. \[ 28.0<\mu<30.0 \] Part: \( 2 / 4 \) Part 3 of 4 (c) Find the \( 99 \% \) confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answe to one decimal place. \( \square \) ip Part Check Save For Later

**Step 1. Identify the given values** We are given: - Sample size: \( n = 60 \) - Sample mean: \( \bar{x} = 29 \) pounds - Population standard deviation: \( \sigma = 3.9 \) pounds **Step 2. Recall the formula for a confidence interval when the population standard deviation is known** The confidence interval for the mean is given by: \[ \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} \] **Step 3. Calculate the standard error** The standard error is: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3.9}{\sqrt{60}} \] Calculating \(\sqrt{60}\): \[ \sqrt{60} \approx 7.746 \] Thus, \[ SE \approx \frac{3.9}{7.746} \approx 0.5034 \] **Step 4. Compute the 99% confidence interval** For a \( 99\% \) confidence interval, the critical value is: \[ z_{\alpha/2} \approx 2.576 \] Next, calculate the margin of error: \[ E = z_{\alpha/2} \cdot SE \approx 2.576 \times 0.5034 \approx 1.297 \] Now, construct the confidence interval: \[ \bar{x} \pm E = 29 \pm 1.297 \] This gives: - Lower limit: \(29 - 1.297 \approx 27.7\) - Upper limit: \(29 + 1.297 \approx 30.3\) **Final Answer** The \( 99\% \) confidence interval for the mean is: \[ 27.7 < \mu < 30.3 \]