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. ($1.03, $2.74)
c. ($1.06, $7.51)
d. ($0.84, $2.24)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 10\). So \(df=10-1 = 9\).

Step2: Find the critical values

For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\). The lower - tail critical value \(\chi_{1-\frac{\alpha}{2},df}^2=\chi_{0.975,9}^2 = 2.700\) and the upper - tail critical value \(\chi_{\frac{\alpha}{2},df}^2=\chi_{0.025,9}^2=19.023\) (from the chi - square distribution table).

Step3: 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 \(s = 1.50\), \(n = 10\), \((n - 1)s^{2}=(10 - 1)\times(1.50)^{2}=9\times2.25 = 20.25\).
Substitute the values: \(\frac{20.25}{19.023}\leq\sigma^{2}\leq\frac{20.25}{2.700}\).
\(\frac{20.25}{19.023}\approx1.06\) and \(\frac{20.25}{2.700}=7.5\).

Step4: Calculate the confidence interval for the population standard deviation \(\sigma\)

Take the square root of each part of the variance interval. \(\sqrt{1.06}\approx1.03\) and \(\sqrt{7.5}\approx2.74\)

Answer:

B. \((\$1.03,\$2.74)\)