Question

A certain test preparation course is designed to help students improve their scores on the MCAT 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: 19, 15, 9, 17, 7, 16. Using these data, construct a 95% confidence interval for the average net change in a students's score after completing the course. Assume the population is approx. normal.Step 1. calculate the sample mean for the given sample data (round answer to 1 decimal place)Step 2. Calculate the sample standard deviation for the given sample data ( round answer to 1 decimal place)Step 3. Find the critical value that should be used in constructing the confidence interval. ( round answer to 3 decimal placed)Step 4. construct the 95% confidence interval (round answer to 1 decimal place)

Answer 1

1)
`\bar X =(\sum_(i=1)^n X_i)/(n)=13.8`

2)
`s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}=4.8`

3)
`t_(\alpha/2)=2.571`

4) The 95% confidence interval would be given by (8.8;18.8)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and sample deviation with the following formulas:

Step 1. calculate the sample mean for the given sample data (round answer to 1 decimal place)

`\bar X =(\sum_(i=1)^n X_i)/(n)=13.8`

Step 2. Calculate the sample standard deviation for the given sample data ( round answer to 1 decimal place)

`s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}=4.8`

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n=6 represent the sample size

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

Step 3. Find the critical value that should be used in constructing the confidence interval. ( round answer to 3 decimal places)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=6-1=5`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,5)".And we see that
`t_(\alpha/2)=2.571`

Step 4. construct the 95% confidence interval (round answer to 1 decimal place)

Now we have everything in order to replace into formula (1):

`13.8-2.571(4.8)/(√(6))=8.8`

`13.8+2.571(4.8)/(√(6))=18.8`

So on this case the 95% confidence interval would be given by (8.8;18.8)