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


`((n-1) s^(2) )/( x^(2)_(right) ) \ \textless \ S^(2) \ \textless \ ((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,


`image`

Thus the answer is,


`20.51\ \textless \ S^(2) \ \textless \ 55.09`