Question

A human 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?

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25$  

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".  

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

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 $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:  

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$  

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

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)  

And on this case we have that the margin of error is given$ME =\pm 0.04$ (equivalent to 4%) and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (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:  

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25$  

And rounded up we have that n=601