Question

construct a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight of 165.2 pounds with a standard deviation of 13.4 pounds. assume the sample is from a normally distributed population.

a. (96.246, 446.587)
b. (9.81, 21.133)
c. (2.68, 5.773)
d. (10.302, 19.559)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1=15-1 = 14\). For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\).
From the chi - square distribution table, \(\chi_{1-\frac{\alpha}{2},df}^2=\chi_{0.975,14}^2 = 5.629\) and \(\chi_{\frac{\alpha}{2},df}^2=\chi_{0.025,14}^2=26.119\)

Step2: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval of the population variance is \(\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},df}^2}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},df}^2}\)
Given \(n = 15\), \(s = 13.4\), then \(s^{2}=(13.4)^{2}=179.56\)
\(\frac{(15 - 1)\times179.56}{26.119}\leq\sigma^{2}\leq\frac{(15 - 1)\times179.56}{5.629}\)
\(\frac{14\times179.56}{26.119}\leq\sigma^{2}\leq\frac{14\times179.56}{5.629}\)
\(\frac{2513.84}{26.119}\leq\sigma^{2}\leq\frac{2513.84}{5.629}\)
\(96.246\leq\sigma^{2}\leq446.587\)

Step3: Calculate the confidence interval for the population standard deviation \(\sigma\)

Take the square - root of each part of the variance interval.
\(\sqrt{96.246}\leq\sigma\leq\sqrt{446.587}\)
\(9.81\leq\sigma\leq21.133\)

Answer:

B. \((9.81, 21.133)\)