Question

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)

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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