Question

a student randomly selects 10 cds at a store. the mean is $13.75 with a standard deviation of $1.50. construct a 95% confidence interval for the population standard deviation, σ. assume the sample is from a normally distributed population.

a. ($0.99, $2.50)
b. ($1.06, $7.51)
c. ($1.03, $2.74)
d. ($0.84, $2.24)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1\), where \(n = 10\), so \(df=10-1 = 9\).
For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\), \(1-\frac{\alpha}{2}=0.975\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{0.025,9}^{2}=19.023\), \(\chi_{R}^{2}=\chi_{0.975,9}^{2}=2.700\).

Step2: Calculate the confidence interval for the population standard deviation

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{U}^{2}}}\lt\sigma\lt\sqrt{\frac{(n - 1)s^{2}}{\chi_{L}^{2}}}\).
Given \(s = 1.50\), \(n = 10\).
First, calculate the lower bound: \(\sqrt{\frac{(10 - 1)\times(1.50)^{2}}{19.023}}=\sqrt{\frac{9\times2.25}{19.023}}=\sqrt{\frac{20.25}{19.023}}\approx1.03\).
Then, calculate the upper bound: \(\sqrt{\frac{(10 - 1)\times(1.50)^{2}}{2.700}}=\sqrt{\frac{20.25}{2.700}}=\sqrt{7.5}\approx2.74\).

Answer:

C. ($1.03, $2.74)