Question

assume the age of night school students is normally distributed. A simple random sample of 24 night school students had an average age of 27.3 years. use this information and a sample standard deviation of 5.6 to find 90% confidence interval estimate for the population variance

Answer 1

Let
n = number of data
s = standard deviation (sample)
S = standard deviation (population)

The working equations is

$\frac{(n-1) s^{2} }{ x^{2}_{right} } \ \textless \ S^{2} \ \textless \ \frac{(n-1) s^{2} }{ x^{2}_{left} }$

To find $x^{2}_{right}$, : (1 - 0.90)/2 = 0.05

To find $x^{2}_{left}$, : 1 - 0.05 = 0.95

Degrees of freedom = n-1 = 24 - 1 = 23

This is shown in the figure attached. Since there is no row for df=23, we interpolate. Thus,

 $x^{2}_{left} = 13.093$ 

 $x^{2}_{right} = 35.17$ 


Substitute all values,

$\frac{(24-1) 5.6^{2} }{ 35.17} \ \textless \ S^{2} \ \textless \ \frac{(24-1) 5.6^{2} }{ 13.093} }$

Thus the answer is,

$20.51\ \textless \ S^{2} \ \textless \ 55.09$