Question

Dylan wants to determine a 90 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must he have to get a margin of error less than 0.03

Answer 1

`n=(0.5(1-0.5))/(((0.03)/(1.64))^2)=747.11`

And rounded up we have that n=748

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 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 0.03` 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)

The best estimatr for the proportionis 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))/(((0.03)/(1.64))^2)=747.11`

And rounded up we have that n=748