Mathematics, 25.04.2020 04:46 alondra6190

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.

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Answer from: DarinKing2005

$17.67-2.92\frac{2.517}{\sqrt{3}}=13.427$

$17.67+2.92\frac{2.517}{\sqrt{3}}=21.913$

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

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

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (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):

$17.67-2.92\frac{2.517}{\sqrt{3}}=13.427$

$17.67+2.92\frac{2.517}{\sqrt{3}}=21.913$

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

Answer from: Quest

idk i would have to be there to under stand it

step-by-step explanation:

idk i would have to be there to under stand it

Answer from: Quest

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

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