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: 6,16,19,12,15,14.

Using these data, construct a 90% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval.

Answer 1

The critical value is
`T_c = 2.5706`.

The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).

Explanation:

Before building the confidence interval, we need to find the sample mean and the sample standard deviation:

Sample mean:

`\overline{x} = (6+16+19+12+15+14)/(6) = 13.67`

Sample standard deviation:

`s = \sqrt{((6-13.67)^2+(16-13.67)^2+(19-13.67)^2+(12-13.67)^2+(15-13.67)^2+(14-13.67)^2)/(5)} = 4.4121`

Confidence interval:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95`. So we have T = 2.5706, that is, the critical value is
`T_c = 2.5706`

The margin of error is:

`M = T(s)/(√(n)) = 2.5706(4.4121)/(√(6)) = 4.63`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 13.67 - 4.63 = 9.04.

The upper end of the interval is the sample mean added to M. So it is 13.67 + 4.63 = 18.30.

The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).