Homework 7.4 Taylisha 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 deviation of the population is 3.9 pounds. Part 1 of 4 (a) The best point estimate of the mean is \( \square \) 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 answers to one decimal place. \[ 28.0<\mu<30.0 \] 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 answers to one decimal place. \[ 27.7<\mu<30.3 \] Part: 3 / 4 Part 4 of 4 (d) Which interval is larger? Why? The \( \square \) \% \% confidence interval is larger. An interval with a \( \square \) greater range of values than the \( \square \) \% confidence interval will to contain the true population mean. Try again Skip Part Recheck Save For Later Submit Assignn :icGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center Acce

**(a)** The best point estimate of the mean is the sample mean, which is \[ 29 \text{ pounds}. \] **(b)** For a \(95\%\) confidence interval when the population standard deviation is known, we use the formula \[ \bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \] where \(\bar{x} = 29\), \(\sigma = 3.9\), and \(n = 60\). 1. Find the standard error: \[ \text{SE} = \frac{3.9}{\sqrt{60}} \approx \frac{3.9}{7.746} \approx 0.503. \] 2. For a \(95\%\) confidence interval, \(z_{\alpha/2} = 1.96\). Compute the margin of error: \[ \text{Margin} = 1.96 \times 0.503 \approx 0.986. \] 3. The confidence interval is: \[ 29 - 0.986 < \mu < 29 + 0.986, \] which simplifies to \[ 28.0 < \mu < 30.0. \] **(c)** For a \(99\%\) confidence interval, the formula remains the same, but the critical value changes. 1. The standard error is still: \[ \text{SE} \approx 0.503. \] 2. For a \(99\%\) confidence interval, \(z_{\alpha/2} \approx 2.576\). Compute the margin of error: \[ \text{Margin} = 2.576 \times 0.503 \approx 1.296. \] 3. The confidence interval is: \[ 29 - 1.296 < \mu < 29 + 1.296, \] which simplifies to \[ 27.7 < \mu < 30.3. \] **(d)** The \(99\%\) confidence interval is larger because a higher confidence level requires a wider interval to ensure that it captures the true population mean. Thus, the interval with a \(99\%\) confidence level has a greater range compared to the \(95\%\) confidence interval.