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

Answer:

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

Step-by-step explanation:

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

$\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ 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 $17.6667\pm t_{0.1}(\frac{13.3367}{\sqrt{6}})$ 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 $17.6667\pm (1.4759)(\frac{13.3367}{\sqrt{6}})$ and the 80% confidence interval is given by (9.6, 25.7)