What is the 90% confidence interval for the variance of exam scores for 28 algebra students, if the standard deviation of their last exam was 12.7?

Question

Quizplus · Accepted Answer

We know that the sample size (n) is 28 and the standard deviation (s) is 12.7. 
Using a chi-squared distribution table, we can find the critical values for a 90% confidence interval with 27 degrees of freedom (df = n-1). The critical values are 14.611 and 40.118. 
The confidence interval formula for variance is: 
[(n-1)*s^2]/chi-squared critical value 
Substituting in our values, we get: 
[(28-1)*12.7^2]/40.118 = 138.007 
Therefore, the 90% confidence interval for the variance of exam scores is (11ebb61d_fc12_f6a5_92a9_9713a1239305_TB8376_11) D: (82.786, 234.469). We can use the chi-square distribution to calculate the confidence interval for the variance. Since we want a 90% confidence interval, and there are 28 students, we will use a chi-square distribution with 27 degrees of freedom (one less than the sample size). The formula for the confidence interval is:

[ (n-1)s^2 / X^2(a/2,n-1), (n-1)s^2 / X^2(1-a/2,n-1) ]

where n is the sample size, s is the sample standard deviation, X^2(a/2,n-1) is the chi-square value at the lower tail of the distribution with area a/2 to its right and n-1 degrees of freedom, and X^2(1-a/2,n-1) is the chi-square value at the upper tail with area 1-a/2 to its right and n-1 degrees of freedom.

Plugging in the values, we get:

[ (28-1)(12.7)^2 / X^2(0.05/2,27), (28-1)(12.7)^2 / X^2(1-0.05/2,27) ]

Using a chi-square table or calculator, we can find that X^2(0.025,27) = 42.98 and X^2(0.975,27) = 15.07.

Plugging those values in, we get:

[ (27)(12.7)^2 / 42.98, (27)(12.7)^2 / 15.07 ]

= [336.4, 950.4]

Therefore, the 90% confidence interval for the variance is [336.4, 950.4], and the best choice is D.

What Is the 90% Confidence Interval for the Variance of Exam