Question

A certain test preparation course is designed to help students improve their scores on the USMLE 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 3 students' scores on the exam after completing the course: 15,20,18 Using these data, construct a 90% 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

`17.67-2.92(2.517)/(√(3))=13.427`

`17.67+2.92(2.517)/(√(3))=21.913`

So on this case the 90% confidence interval would be given by (13.427;21.913)

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".

`\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=17.67`

The sample deviation calculated
`s=2.517`

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

`df=n-1=3-1=2`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,2)".And we see that
`t_(\alpha/2)=2.92`

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

`17.67-2.92(2.517)/(√(3))=13.427`

`17.67+2.92(2.517)/(√(3))=21.913`

So on this case the 90% confidence interval would be given by (13.427;21.913)