Consider the following random sample from a normal population: 14,10,13,16,12,18,15,and 11.What is the 95% confidence interval for the population variance?

Question

Quizplus · Accepted Answer

To find the confidence interval for population variance, we need to use the chi-square distribution. For a 95% confidence interval with degrees of freedom (n-1), the chi-square values for the lower and upper bounds are 2.18 and 14.07, respectively.
Now,
Sample Variance = S^2 = ∑(Xi - X̅)^2 / (n-1)
= (14-13)^2+(10-13)^2+(13-13)^2+(16-13)^2+(12-13)^2+(18-13)^2+(15-13)^2+(11-13)^2/7
= 5.71
Lower Bound = (n-1)*S^2 / Chi-square upper (n-1) ,α/2 = (7*5.71) / 14.07 = 2.83
Upper Bound = (n-1)*S^2 / Chi-square lower (n-1) ,α/2 = (7*5.71) / 2.18 = 18.72
Hence, the 95% confidence interval for population variance is 3.11 to 29.51.

Consider the Following Random Sample from a Normal Population: 14,10,13,16,12,18,15,and