Question

overweight men for a random sample of 50 overweight men, the mean of the number of pounds that they were overweight was 29. the standard deviation of the population is 3.8 pounds. part 1 of 4 (a) the best point estimate of the mean is 29 pounds. part 2 of 4 (b) find the 90% 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. 28.1 < μ < 29.9 part: 2 / 4 part 3 of 4 (c) 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.

Explanation:

Step1: Determine the formula

The formula for the confidence interval when the population standard deviation \(\sigma\) is known is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)

Step2: Identify the values

Given \(\bar{x} = 29\), \(\sigma=3.8\), \(n = 50\)

For a \(95\%\) confidence interval, \(\alpha=1 - 0.95=0.05\), so \(\alpha/2=0.025\)

From the standard normal distribution table, \(z_{0.025}=1.96\)

Step3: Calculate the margin of error \(E\)

\(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{3.8}{\sqrt{50}}\)

First, \(\sqrt{50}\approx7.071\)

\(\frac{3.8}{7.071}\approx0.537\)

\(E = 1.96\times0.537\approx1.052\)

Step4: Calculate the confidence interval

The lower limit is \(\bar{x}-E=29 - 1.052=27.948\approx27.9\)

The upper limit is \(\bar{x}+E=29+1.052 = 30.052\approx30.1\)

So the \(95\%\) confidence interval is \(27.9<\mu <30.1\)

Answer:

\(27.9<\mu <30.1\)