Question

High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with 92% confidence?

Answer 1

Answer:

n=1915

Step-by-step explanation:

1) 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

$\sigma=50$ represent the population standard deviation  (assumed)

n represent the sample size (variable of interest)  

ME=2 represent the margin of error desired

Confidence =0.92 or 92%

$\alpha=0.08$ represent the significance level

2) 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)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$    (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$   (b)

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04,0,1)", and we got $z_{\alpha/2}=1.75$, replacing into formula (b) we got:

$n=(\frac{1.75(50)}{2})^2 =1914.06 \approx 1915$

So the answer for this case would be n=1915 rounded up to the nearest integer