Question

School vouchers A small pilot study estimated that $44\%$of all American

adults agree that parents should be given vouchers that are good for education at any public or private school of their choice.

a. How large a random sample is required to obtain a margin of error of at most $0.03$with $99\%$confidence? Answer this question using the pilot study’s result as the guessed value for $\hat{p}$

b. Answer the question in part (a) again, but this time use the conservative guess$\hat{p}=0.5$. By how much do the two sample sizes differ?

Short answer

a. $n=1816$

b. Increase in sample size is$26$.

Step-by-step solution

Given Information

It is given that $E=0.03$

$\hat{p}=44\%=0.44$

Calculation of n

Sample size is $\hat{p}\text{known:}n=\frac{{\left[z_{a/2}\right]}^2\hat{p}\hat{q}}{E^2}$

$=\frac{{\left[z_{a/2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

and $\hat{p}\text{unknown:}n=\frac{{\left[z_{a/2}\right]}^{2}0.25}{E^2}$

Now for $1-\alpha =0.99$, using table $z_{a/2}=2.575$

Hence, $n=\frac{{\left[z_{\alpha /2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$=\frac{2.{575}^2\times 0.44(1-0.44)}{0.{03}^2}$

$n=1816$

Increase in Sample Size

As$\hat{p}$is known, hence

$n=\frac{{\left[z_{a/2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$=\frac{2.{575}^2\times 0.50(1-0.50)}{0.{03}^2}$

$n=1842$

Difference between two sizes is $1842-1816=26$