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.

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:

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:

The degrees of freedom for this sample size are:

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:

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

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

i think the answer is b.

step-by-step explanation:

not 100% sure though.

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x