Statistics Problem

Question

A principal randomly selects six students to take an aptitude test. Their scores are shown below. Determine a 99% confidence interval for the mean score of all students. Assume the population has a normal distribution

83.0 84.1 83.5 83.7 84.1 73.5

Answer 1

To compute the 99% confidence interval, we will use the critical t value in the computation of the margin of error. For a sample size of 6, the degrees of freedom is 5. In the t table you look for the tail probability of 0.005 and 5 degrees of freedom = 4.032.

We then need to compute the mean and standard deviation of our sample.

Mean = 81.98 and standard deviation = 4.18.

99% confidence interval =

81.98 +/- 4.032 * 4.18/sqrt(6)

81.98 +/- 4.032 * 1.71

(74.96,89.996)