Question

Find the 90% confidence interval for the variance and standard deviation of the ages of seniors at oak park college if a random sample of 24 students has a standard deviation of 2.3 years. assume the variable is normally distributed.

Answer 1

Given that the standard deviation, s = 2.3 and that there are 24 students, thus the degree of freedom is 24 - 1 = 23.

For 90% confidence interval, $\alpha =0.1$ and $\frac{ \alpha }{2} =0.05$

From the chi-square table, the value of $\chi^2_{1- \frac{ \alpha }{2} }=\chi^2_{0.95}=35.172$ and the value of $\chi^2_{ \frac{ \alpha }{2} }=\chi^2_{0.05}=13.091$.

The 90% confidence interval for $\sigma^2$ is given by:

$\frac{(n-1)s^2}{\chi^2_{1- \frac{ \alpha }{2} }} \ \textless \ \sigma^2\ \textless \ \frac{(n-1)s^2}{\chi^2_{ \frac{ \alpha }{2} }} \\ \\ = \frac{(24-1)(2.3)^2}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{(24-1)(2.3)^2}{13.091} \\ \\ = \frac{23(5.29)}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{23(5.29)}{13.091} = \frac{121.67}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{121.67}{13.091} \\ \\ =3.46\ \textless \ \sigma^2\ \textless \ 9.29$

And the 90% confidence interval for $\sigma$ is given by:

$\sqrt{\frac{(n-1)s^2}{\chi^2_{1- \frac{ \alpha }{2} }}} \ \textless \ \sigma\ \textless \ \sqrt{\frac{(n-1)s^2}{\chi^2_{ \frac{ \alpha }{2} }}} \\ \\ = \sqrt{3.46} \ \textless \ \sigma\ \textless \ \sqrt{9.29} =1.86\ \textless \ \sigma\ \textless \ 3.05$