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
`( \alpha )/(2) =0.05`

From the chi-square table, the value of
`\chi^2_{1- ( \alpha )/(2) }=\chi^2_(0.95)=35.172` and the value of
`\chi^2_{ ( \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- ( \alpha )/(2) }} \ \textless \ \sigma^2\ \textless \ \frac{(n-1)s^2}{\chi^2_{ ( \alpha )/(2) }} \\ \\ = ((24-1)(2.3)^2)/(35.172) \ \textless \ \sigma^2\ \textless \ ((24-1)(2.3)^2)/(13.091) \\ \\ = (23(5.29))/(35.172) \ \textless \ \sigma^2\ \textless \ (23(5.29))/(13.091) = (121.67)/(35.172) \ \textless \ \sigma^2\ \textless \ (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- ( \alpha )/(2) }}} \ \textless \ \sigma\ \textless \ \sqrt{\frac{(n-1)s^2}{\chi^2_{ ( \alpha )/(2) }}} \\ \\ = √(3.46) \ \textless \ \sigma\ \textless \ √(9.29) =1.86\ \textless \ \sigma\ \textless \ 3.05`