Question

The prices of a random sample of

2525

new motorcycles have a sample standard deviation of

$ 3749$3749.

Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance

sigma squaredσ2

and the population standard deviation

sigmaσ.

Use a

99 %

level of confidence. Interpret the results.

Answer 1

sample size =n = 25

Degree of freedom = n -1 = 25 - 1= 24

sample standard deviation = s = $ 3749

Here we have to find the 99% confidence interval for population variance and population standard deviation

Here for population variance

Lower Limit = (n-1)s²/a/2

[ _a/2 = CHIINV(0.005, 24) = 45.5586 ]

Lower Limit = (25 - 1) * 3749 * 3749/45.5586 = 7404091

Upper Limit = (n - 1)s²/_1-a/2

[ _1-a/2 = CHIINV(0.995, 24) = 9.8862 ]

Upper Limit = (25 - 1) * 3749 * 3749/9.8862 = 34120176

so here for population variance

7404091 < < 34120176

for standard deviation , 99%confidence interval is

sqrt(7404091) < < sqrt(34120176)

2721.05 < < 5841.25

Lower Limit = 2721.05

Upper Limit = 5841.25