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 4 students' scores on the exam after completing the course: 12,7,13,11 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 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

Answer:

The 80% confidence interval for the average net change is (8.596, 12.904).

Critical value t=1.638.

Step-by-step explanation:

First, we calculate the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{4}(12+7+13+11)\\\\\\M=\dfrac{43}{4}\\\\\\M=10.75\\\\\\s=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2\\\\\\s=\dfrac{1}{3}((12-10.75)^2+(7-10.75)^2+(13-10.75)^2+(11-10.75)^2)\\\\\\s=\dfrac{20.75}{3}\\\\\\s=6.92$

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=10.75.

The sample size is N=4.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.63}{\sqrt{4}}=\dfrac{2.63}{2}=1.315$

The degrees of freedom for this sample size are:

$df=n-1=4-1=3$

The t-value for a 80% confidence interval and 3 degrees of freedom is t=1.638.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.638 \cdot 1.315=2.154$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 10.75-2.154=8.596\\\\UL=M+t \cdot s_M = 10.75+2.154=12.904$

The 80% confidence interval for the average net change is (8.596, 12.904).