Question

A certain test preparation course is designed to help students improve their scores on the USMLE 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 5 students' scores on the exam after completing the course: 14,11,18,9,19 Using these data, construct a 99% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 3 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

The critical value is
`T = 4.604`.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

Explanation:

The first step to solve this question is finding the sample mean and sample standard deviation:

14,11,18,9,19

Sample mean is sum of all values divided by the number of values. Thus:

`\overline{x} = (14+11+18+9+19)/(5) = 14.2`

The sample standard deviation is the square root of the division of the sum of the subtractions squared of each value and the mean, and the number of values. Thus:

`s = \sqrt{((14-14.2)^2+(11-14.2)^2+(18-14.2)^2+(9-14.2)^2+(19-14.2)^2)/(5)} = 3.868`

Confidence interval:

We have the standard deviation for the sample, and so we use the t-distribution to build the confidence interval.

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 = 5 - 1 = 4

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 4.604.

The critical value is
`T = 4.604`.

The margin of error is:

`M = T(s)/(√(n)) = 4.604(3.868)/(√(5)) = 7.964`

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 14.2 - 7.964 = 6.236.

The upper end of the interval is the sample mean added to M. So it is 14.2 + 7.964 = 22.164.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).