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 5 students' scores on the exam after completing the course: 10,9,20,13,12 Using these data, construct a 99% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 2 of 4: Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.

Answer 1

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=12.8`

The sample deviation calculated
`s=4.324 \approx 4.3`

`12.8-4.604(4.324)/(√(5))=3.896`

`12.8+ 4.604(4.324)/(√(5))=21.704`

So on this case the 99% confidence interval would be given by (3.896;21.704)

Explanation:

Data: 10,9,20,13,12

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".

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

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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

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

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=12.8`

The sample deviation calculated
`s=4.324`

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

`df=n-1=5-1=4`

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,4)".And we see that
`t_(\alpha/2)=4.604`

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

`12.8-4.604(4.324)/(√(5))=3.896`

`12.8+ 4.604(4.324)/(√(5))=21.704`

So on this case the 99% confidence interval would be given by (3.896;21.704)