Ahuman resources manager at a large company wants to estimate the proportion of employees that would be interested in reimbursment for college courses. if she wishes to be 95% confident that her estimate is within 4% of the true proportion, how many employees would need to be sampled?

And rounded up we have that n=601  

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".  

represent the real population proportion of interest

represent the estimated proportion for the sample

n is the sample size required (variable of interest)

represent the critical value for the margin of error

The population proportion have the following distribution  

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 95% 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 the margin of error is given (equivalent to 4%) and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

(b)

We can assume that the estimated proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:  

And rounded up we have that n=601  

its c or d   take a lucky guess my dude