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

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1\), where \(n = 16\), so \(df=16-1 = 15\).
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.995,15}^{2}=4.601\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,15}^{2}=32.801\).
The sample standard deviation \(s = 1.8\).

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

The formula for the confidence interval of the population variance is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\).
Substitute the values: \(\frac{(16 - 1)\times(1.8)^{2}}{32.801}\leq\sigma^{2}\leq\frac{(16 - 1)\times(1.8)^{2}}{4.601}\).
First, \((16 - 1)\times(1.8)^{2}=15\times3.24 = 48.6\).
Then, \(\frac{48.6}{32.801}\approx1.4817\) and \(\frac{48.6}{4.601}\approx10.563\).

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

Take the square - root of the interval for \(\sigma^{2}\).
\(\sqrt{1.4817}\approx1.217\) and \(\sqrt{10.563}\approx3.25\).

Answer:

D. \((1.217,3.25)\)