Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 90% confident that you esimate is within 2.5% of the true population proportion. How large of a sample size is required

Answer 1

`n=(0.5(1-0.5))/(((0.025)/(1.64))^2)=1075.84`

And rounded up we have that n=1076

Explanation:

Information given

`ME= 0.025` represent the desired margin of error

Solution

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 and . And the critical value would be given by:

`t_(\alpha/2)=-1.64, t_(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)

We can assume that the best estimate for the true proportion is
`\hat p=0.5`. 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=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

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

`n=(0.5(1-0.5))/(((0.025)/(1.64))^2)=1075.84`

And rounded up we have that n=1076