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.
Accepted Solution
A:
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, [tex] \alpha =0.1[/tex] and [tex] \frac{ \alpha }{2} =0.05[/tex]
From the chi-square table, the value of [tex]\chi^2_{1- \frac{ \alpha }{2} }=\chi^2_{0.95}=35.172[/tex] and the value of [tex]\chi^2_{ \frac{ \alpha }{2} }=\chi^2_{0.05}=13.091[/tex].
The 90% confidence interval for [tex]\sigma^2[/tex] is given by: