Question

A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimate the proportion p who plan to get the vaccine.

How large does the sample size n need to be for X/n, the sample proportion of people who plan to get the vaccine, to have an 90% chance of lying within 0.04 of probability?

Answer 1

`n=(0.5(1-0.5))/(((0.04)/(1.64))^2)=420.25`

And rounded up we have that n=421

Explanation:

We know that the sample proportion have the following distribution:

`\hat p \sim N(p,\sqrt{(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 90% of confidence, our significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`z_(\alpha/2)=-1.64, z_(1-\alpha/2)=1.64`

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

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

And on this case we have that
`ME =\pm 0.04` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

We assume that a prior estimation for p would be
`\hat p =0.5` since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

`n=(0.5(1-0.5))/(((0.04)/(1.64))^2)=420.25`

And rounded up we have that n=421