Question

A professor wishes to estimate the variance of student test scores. A random sample of 18...

A professor wishes to estimate the variance of student test scores. A random sample of 18 scores had a sample standard deviation of 10.4. Find a 98% confidence interval for the population variance and interpret it. The lower confidence limit is _________ and the upper confidence limit is ________ . What assumption was necessary to calculate this interval estimate? Answer to 2 decimal places (PLEASE MAKE THE ANSWER ACCURATE)

Homework Answers

Answer #1

Given that, sample size (n) = 18 and

sample standard deviation (s) = 10.4

confidence level = 0.98

=> significance level = 1 - 98 = 0.02

Degrees of freedom = 18 - 1 = 17

Lower critical value is,

Upper critical value is,

The 98% confidence interval for population variance is,

Therefore, The lower confidence limit is 55.04 and the upper confidence limit is 286.94.

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