A medical study is to be designed to estimate the population proportion p having a certain disease. How many people should be examined if the researcher wishes to be 95% certain that the error of estimation is below 0.1 when
a. there is no prior knowledge about the value of p?
b. the proportion is known to be about 0.2?

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

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Part a

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

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

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

And on this case we have that [tex]ME =0.1[/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{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

Since we don't have prior estimation for p we can use 0.5 for the value of [tex]\hat p[/tex]. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04[/tex]

And rounded up we have that n=97

Part b

For this case we have a prior estimation for [tex]\hat p =0.2[/tex] and if we replace in formula (b) we got

[tex]n=\frac{0.2(1-0.2)}{(\frac{0.1}{1.96})^2}=61.47[/tex]

And rounded up we have that n=62