Question

A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 3 percentage points with 95​% confidence if

​(a) he uses a previous estimate of 32​%?

​(b) he does not use any prior​ estimates?

Answer 1

a)
`n=(0.32(1-0.32))/(((0.03)/(1.96))^2)=928.81`

And rounded up we have that n=929

b)
`n=(0.5(1-0.5))/(((0.03)/(1.96))^2)=1067.11`

And rounded up we have that n=1068

Explanation:

Part a

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)

For a confidence of 95% we have that the significance is
`\alpha=0.05` and the critical value would be:

`z = 1.96`

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

`n=(0.32(1-0.32))/(((0.03)/(1.96))^2)=928.81`

And rounded up we have that n=929

Part b

For this case since we don't have prior info we can use as estimator for the true proportion the value
`\hat p=0.5` and replacing we got:

`n=(0.5(1-0.5))/(((0.03)/(1.96))^2)=1067.11`

And rounded up we have that n=1068