A public relations officer of William Paterson University wants to estimate the mean IQ of the university students. If she wants to be 99% confident that her sample mean is off by no more than 3 points, how many students she has to test in order to come up with a valid estimation?

Assuming this complete problem: "A public relations officer of William Paterson University wants to estimate the mean IQ of the university students. If she wants to be 99% confident that her sample mean is off by no more than 3 points, how many students she has to test in order to come up with a valid estimation?. A recent study shows that IQ of New Jersey students has standard deviation of 15 points"

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".

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)

And on this case we have that [tex]ME =\pm 3[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.58(15)}{3})^2 =166.41 \approx 167[/tex]

So the answer for this case would be n=167 rounded up to the nearest integer