Question

6 Parameter Estimation 11. Thirty-five randomly selected students took the calculus final. If the sample mean was 82 and standard deviation was 4.6construct score of all students Provide estimates and circle one of the multiple choice answers. (6 points) a 99% confidence interval for the mean ,

Answer 1

Mean $\mu$ = 82

Standard deviation $\sigma$ = 4.6

Standard deviation of sample mean = $\sigma / \sqrt{n}$ = $4.6/ \sqrt{35}$ = 0.7775419

As, we do not know the population standard deviation, we will use t statistic to calculate the 99% confidence interval.

Degree of freedom = n-1 = 35 - 1 = 34

Critical t value for df = 34 and 99% confidence interval. is  2.728

99% confidence interval of the sample mean is,

(82 - 2.728 * 0.7775419, 82 + 2.728 * 0.7775419)

(79.87887, 84.12113)