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:23,18,23,12,13,23Using 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.

Answer 1

Answer:

Interval [16.34 , 21.43]

Step-by-step explanation:

First step. Calculate the mean

$\bar X=\frac{(23+18+23+12+13+23)}{6}=18.666$

Second step. Calculate the standard deviation

$\sigma =\sqrt{\frac{(23-18.666)^2+(18-18.666)^2+(23-18.666)^2+(12-18.666)^2+(13-18.666)^2+(23-18.666)^2}{6}}$

$\sigma=\sqrt\frac{18.783+0.443+18.783+44.435+5.666+18.783}{6}$

$\sigma=\sqrt{17.815}=4.22$

As the number of data is less than 30, we must use the t-table to find the interval of confidence.

We have 6 observations, our level of confidence DF is then 6-1=5 and we want our area A to be 80% (0.08).  

We must then choose t = 1.476 (see attachment)

Now, we use the formula that gives us the end points of the required interval

$\bar X \pm t\frac{\sigma}{\sqrt n}$

where n is the number of observations.

The extremes of the interval are then, rounded to the nearest hundreth, 16.34 and 21.43