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

At a constant it will take one hour to travel 32 mi.at that speed it will take 1/2 hr to travel the remaining 16 mi.it will take 1+1/2 hr to travel the 48mi @ 32mph.so 1 hr = 60 min& 1/2 hr = 30 min1+1/2 hr = 90 min.there are 60s in each 90 × 60 = 3600 seconds.the real question seriously is he why are they driving so slow? ! lol.

Answer from: Quest

Here for a good time of free question

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