The 99% confidence interval estimate for a population variance when a sample standard deviation of 12 is obtained from a sample of 10 items is
A)4.58 to 62.25
B)46.53 to 422.17
C)54.94 to 747.01
D)62.04 to 562.89

Question

Quizplus · Accepted Answer

To obtain the confidence interval for population variance, we use the chi-square distribution. The formula for the confidence interval is:
[(n-1)s^2)/X^2(U/2), (n-1)s^2)/X^2(1-(U/2))]
where n is the sample size, s is the sample standard deviation, X^2 is the chi-square value with n-1 degrees of freedom and U is the confidence level.
Substituting the given values, we get:
[(10-1)(12^2))/X^2(0.995), (10-1)(12^2))/X^2(0.005)]
= [1295.64/X^2(0.995), 82.08/X^2(0.005)]
From chi-square distribution table, X^2(0.995) = 23.209 and X^2(0.005) = 3.940.
Substituting these values, we get:
[1295.64/23.209, 82.08/3.940]
= [55.928, 20.837]
Thus, the 99% confidence interval for population variance is (55.928, 20.837), which is closest to choice C: 54.94 to 747.01.

The 99% Confidence Interval Estimate for a Population Variance When