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 3 students' scores on the exam after completing the course:
15,20,18
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.

Answer:

The 90% confidence interval for the average net change in a student's score after completing the course is (13.42, 21.92).

Step-by-step explanation:

Average change in the sample:

[tex]M_{s} = \frac{15+20+18}{3} = 17.67[/tex]

Standard deviation of the sample:

[tex]s = \sqrt{\frac{(15-17.67)^2+(20-17.67)^2+(18-17.67)^2}{2}} = 2.52[/tex]

We have the standard deviation for the sample, so the t distribution is used.

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 = 3 - 1 = 2

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 2 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 2.92

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.92\frac{2.52}{\sqrt{3}} = 4.25[/tex]

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 17.67 - 4.25 = 13.42

The upper end of the interval is the sample mean added to M. So it is 17.67 + 4.25 = 21.92

The 90% confidence interval for the average net change in a student's score after completing the course is (13.42, 21.92).

Can someone please answer this

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