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 15.3 pounds. assume the sample is from a normally distributed population.
a. (125.474, 582.21)
b. (2.864, 6.169)
c. (11.763, 22.333)
d. (11.202, 24.129)

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, \(\alpha=1 - 0.95 = 0.05\), so \(\frac{\alpha}{2}=0.025\) and \(1-\frac{\alpha}{2}=0.975\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{0.025,14}^{2}=26.119\) and \(\chi_{R}^{2}=\chi_{0.975,14}^{2}=5.629\).

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

The formula for the confidence interval of the population variance is \(\frac{(n - 1)s^{2}}{\chi_{U}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\).
Given \(n = 15\), \(s = 15.3\), then \(s^{2}=15.3^{2}=234.09\) and \(n - 1=14\).
Substitute the values: \(\frac{14\times234.09}{26.119}\leq\sigma^{2}\leq\frac{14\times234.09}{5.629}\).
\(\frac{3277.26}{26.119}\leq\sigma^{2}\leq\frac{3277.26}{5.629}\).
\(125.474\leq\sigma^{2}\leq582.21\).

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

Take the square - root of each part of the variance interval.
\(\sqrt{125.474}\leq\sigma\leq\sqrt{582.21}\).
\(11.202\leq\sigma\leq24.129\).

Answer:

D. \((11.202,24.129)\)