Mathematics: A certain test preparation course is designed to help students improve their scores on the MCAT 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 7 students scores on the exam after completing the course: 37,12,12,17,13,32,23 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

X - 45=30Step-by-step explanation:...

A certain test preparation course is designed to

sza2016

03.02.2022

The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Step-by-step explanation:

The net change in 7 students' scores on the exam after completing the course are:

S = {37 ,12 ,12 ,17 ,13 ,32 ,23}

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{7}\times 146=20.857\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}}=\sqrt{\frac{1}{7}\times 622.8571}=10.189$

As the population standard deviation is not known, a t-interval will be formed.

Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:

$t_{\alpha/2, (n-1)}=t_{0.20/2, (7-1)}=t_{0.10,6}=1.415$

*Use a t-table.

Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times\frac{s}{\sqrt{n}}$

$=20.857\pm 1.415\times\frac{10.189}{\sqrt{7}}\\\\ =20.857\pm 5.4493\\\\=(15.4077, 26.3063)\\\\\approx (15.4,26.3)$

Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

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