Question

3) In a study of student loan subsidies, I surveyed 100 students. In this sample, students will owe a mean of $25,000 at the time of graduation with a standard deviation of $2,000.

(a) Develop a 96% confidence interval for the population mean.

(b) Develop a 96% confidence interval for the population standard deviation.

Answer 1

Here, n = 100

Sample mean = 25000

Sample standard deviation (s)= 2000

Degree of freedom ()= n-1= 99

Level of significance () = 0.04

a) Since, mean and standard deviation of the population both are unknown, mean will follow t distribution.

Hence, Confidence interval for mean is given by -

Where, is the critical value of the t distribution for two tailed test at level of significance and degrees of freedom. However, since the sample size is large (>30), critical value of t will be same as z. Therefore, this critical value = = 2.05 ( This can be obtained from z table by looking at z value corresponding to the probability 0.94)

= [ 25000 - 410, 25000 + 410]

= [ 24590, 25410 ]

b) Confidence interval for variance is given by-

Where,   and   are critical value of chi square at respective probabilities( 0.02 and 0.98) and degrees of freedom (99).

= 129.996

= 72.288

This can be obtained from chi square table by noting the chi square value corresponding to given probability and degrees of freedom.

So, CI(2) = [3046247.577, 5478087.649]

So, confidence interval for standard deviation will be obtained by taking square root of confidence limits of variance,

CI() = [1745.35, 2340.53]