Question

a random sample of 20 women have a mean height of 62.5 inches and a standard deviation of 2.2 inches. construct a 98% confidence interval for the population variance, \\( \sigma^{2} \\). assume the sample is from a normally distributed population.

a. (1.594, 3.471)
b. (1.155, 5.476)
c. (2.541, 12.048)
d. (2.675, 12.682)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1\), where \(n = 20\), so \(df=20-1 = 19\).
For a \(98\%\) confidence interval, \(\alpha=1 - 0.98=0.02\), and \(\frac{\alpha}{2}=0.01\).
From the chi - square distribution table, \(\chi_{1-\frac{\alpha}{2},df}^2=\chi_{0.99,19}^2 = 7.633\) and \(\chi_{\frac{\alpha}{2},df}^2=\chi_{0.01,19}^2=36.191\).
The sample variance \(s^{2}=(2.2)^{2}=4.84\).

Step2: Calculate the confidence interval for the population variance

The formula for the confidence interval for the population variance \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},df}^2}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},df}^2}\).
Substitute the values:
\(\frac{(20 - 1)\times4.84}{36.191}<\sigma^{2}<\frac{(20 - 1)\times4.84}{7.633}\)
First, calculate \(\frac{19\times4.84}{36.191}=\frac{91.96}{36.191}\approx2.541\)
Second, calculate \(\frac{19\times4.84}{7.633}=\frac{91.96}{7.633}\approx12.048\)

Answer:

C. \((2.541,12.048)\)