Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be 80% confident that you estimate is within 2.5% of the true population proportion. How large of a sample size is required?

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{1.28})^2}=655.36$  

And rounded up we have that n=656

Step-by-step explanation:

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 80% of confidence, our significance level would be given by $\alpha=1-0.80=0.20$ and $\alpha/2 =0.10$. And the critical value would be given by:

$z_{\alpha/2}=\pm 1.28$

Solution to the problem

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)  

Since we don't have prior info for the proportion of interest we can use $\hat p=0.5$ as estimator. And on this case we have that $ME =\pm 0.025$ 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)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{1.28})^2}=655.36$  

And rounded up we have that n=656