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: 1 / 4 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. < μ <

Explanation:

Step1: Find the z - value

For a 90% confidence interval, the significance level \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\).
The z - value \(z_{\alpha/2}\) corresponding to a right - tail area of \(0.05\) is \(z_{0.05}\approx1.645\) (from the standard normal distribution table).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\).
Given \(n = 50\), \(\sigma=3.8\), and \(z_{\alpha/2}=1.645\).
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{3.8}{\sqrt{50}}\approx\frac{3.8}{7.071}\approx0.537\).
Then \(E=1.645\times0.537\approx0.890\).

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu <\bar{x} + E\).
Given \(\bar{x}=29\).
\(\bar{x}-E=29 - 0.890=28.11\approx28.1\)
\(\bar{x}+E=29+0.890 = 29.89\approx29.9\)

Answer:

$$28.1 < \mu < 29.9$$