You have been hired by a college foundation to conduct a survey of graduates. a) if you want to estimate the percentage of graduates who made a donation to the college after graduation, how many graduates must you survey if you want 99% confidence that your percentage has a margin of error of 5.5 percentage point?

And rounded up we have that n=551

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".  

The population proportion have the following distribution

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by and . And the critical value would be given by:

The margin of error for the proportion interval is given by this formula:  

   (a)  

And on this case we have that (Or 5.5% points) and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

Since we don't have a prior estimation for we can assume 0.5 as a point of estimate. And replacing into equation (b) the values from part a we got:

And rounded up we have that n=551

the solution is h=10 because you add 9 to each side of the equation.