Question

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 6 students' scores on the exam after completing the course: 8,6,12,15,42,23 Using these data, construct a 80% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 4 of 4 : Construct the 80% confidence interval. Round your answer to one decimal place.

Answer 1

(9.6, 25.7) is a 80% confidence interval for the average net change in a student's score after completing the course.

Explanation:

We have n = 6,
`\bar{x} =  17.6667` and s = 13.3367. The confidence interval is given by

`\bar{x}\pm t_(\alpha/2)((s)/(√(n)))` where
`t_(\alpha/2)` is the
`\alpha/2`th quantile of the t distribution with n-1=5 degrees of freedom. As we want the 80% confidence interval, we have that
`\alpha = 0.2` and the confidence interval is
`17.6667\pm t_(0.1)((13.3367)/(√(6)))` where
`t_(0.1)` is the 10th quantile of the t distribution with 5 df, i.e.,
`t_(0.1) = -1.4759`. Then, we have
`17.6667\pm (1.4759)((13.3367)/(√(6)))` and the 80% confidence interval is given by (9.6, 25.7)