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

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

The working equations is



To find , : (1 - 0.90)/2 = 0.05

To find , : 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,

  

  

Substitute all values,



Thus the answer is,