Given n = 17, s² = 18.56, and that the population is normally distributed, the 80% confidence interval for the population variance is ________. A)11.4372 $\le$ $\sigma$ ² $\le$ 36.3848 B)23.5418 $\le$ $\sigma$ ² $\le$ 9.31223 C)12.6141 $\le$ $\sigma$ ² $\le$ 31.8892 D)11.2929 $\le$ $\sigma$ ² $\le$ 37.2989 E)14.2929 $\le$$\sigma$ ² $\le$ 39.2989

Question

Quizplus · Accepted Answer

To find the confidence interval for the population variance in a normally distributed population, we use the chi-square distribution. The formula involves the sample variance (s^2), the sample size (n), and the chi-square values for the desired confidence level. The correct interval is calculated using the chi-square distribution values for the degrees of freedom (n-1) and the confidence level (80% in this case), which leads to the interval given in option C.

Given N = 17, S2 = 18 ≤\le≤ σ\sigmaσ 2 ≤\le≤ 363848 B)23

Given N = 17, S² = 18 $\le$ $\sigma$ ² $\le$ 363848
B)23