Question

a random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 1.8 inches. construct a 99% confidence interval for the population standard deviation, σ. assume the sample is from a normally distributed population.

a. (1.217, 3.25)
b. (1.261, 3.049)
c. (0.907, 2.422)
d. (1.257, 3.357)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 16\). So \(df=16-1 = 15\).

Step2: Find the critical values

For a \(99\%\) confidence interval (\(\alpha=1 - 0.99=0.01\)), and \(\frac{\alpha}{2}=0.005\).
Using the \(\chi^{2}\) - distribution table or a calculator, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.995,15}^{2}=4.601\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,15}^{2}=32.801\).

Step3: Calculate the confidence interval for \(\sigma\)

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{R}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{L}^{2}}}\).
Given \(s = 1.8\), \(n = 16\).
First, \(\sqrt{\frac{(16 - 1)\times1.8^{2}}{32.801}}=\sqrt{\frac{15\times3.24}{32.801}}=\sqrt{\frac{48.6}{32.801}}\approx1.22\)
Second, \(\sqrt{\frac{(16 - 1)\times1.8^{2}}{4.601}}=\sqrt{\frac{15\times3.24}{4.601}}=\sqrt{\frac{48.6}{4.601}}\approx3.25\)

Answer:

A. \((1.217,3.25)\)