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

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$  

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

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

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 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:  

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

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

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$  

And rounded up we have that n=421