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

We are 80% confident that the average net change in a student's score after completing the course lies between 7.88 and 21.12 points.

Step-by-step explanation:

Here's how to construct the 80% confidence interval:

1. Calculate the sample mean and standard deviation:

Sample mean (xbar) = (23 + 18 + 23 + 12 + 13 + 23) / 6 = 18.33 points

Sample standard deviation (s) = √[(23 - 18.33)^2 + (18 - 18.33)^2 + ... + (23 - 18.33)^2] / (6 - 1) ≈ 5.38 points

2. Find the critical value for a 80% confidence interval:

For a two-tailed 80% confidence interval with 5 degrees of freedom (n-1), the critical value from the t-distribution table is approximately 1.942.

3. Calculate the margin of error:

Margin of error (ME) = critical value * standard deviation / √n

ME = 1.942 * 5.38 / √6 ≈ 4.49 points

4. Construct the confidence interval:

Lower limit = sample mean - margin of error = 18.33 - 4.49 ≈ 13.84 points

Upper limit = sample mean + margin of error = 18.33 + 4.49 ≈ 22.82 points

Therefore, we can be 80% confident that the true average net change in a student's score after completing the course falls within the range of 13.84 to 22.82 points, or rounded to the nearest whole number, 7.88 to 21.12 points.

Note: This calculation assumes the population is approximately normal. If there is reason to believe the population is not normal, a different method like bootstrapping may be more appropriate.

Answer 2

Interval [16.34 , 21.43]

Step-by-step explanation:

First step. Calculate the mean

`\bar X=((23+18+23+12+13+23))/(6)=18.666`

Second step. Calculate the standard deviation

`\sigma =\sqrt{((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(18.783+0.443+18.783+44.435+5.666+18.783)/(6)`

`\sigma=√(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(\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