# Solved: A researcher wishes to estimate the percentage of adults who support abolishing the penny

#### ByDr. Raju Chaudhari

Feb 27, 2021

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 33 percentage points with 99% confidence if

a. he uses a previous estimate of 38%?
b. he does not use any prior estimates?

#### Solution

The formula to estimate the sample size required to estimate the proportion is

 $$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$
where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

a. Given that $p = 0.38$.

Given that margin of error $E =0.33$. The confidence coefficient is $0.99$. Assume that the proportion is $p =0.38$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.58$.

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.38(1-0.38)\bigg(\frac{2.58}{0.33}\bigg)^2\\ &=14.4008\\ &\approx 15. \end{aligned}

Thus, the sample of size $n=15$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.33$.

b. Assume that $p$ is unknown. Let $p =0.5$.

Given that margin of error $E =0.33$. The confidence coefficient is $0.99$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.58$.

The minimum sample size required to estimate the proportion is
 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{2.58}{0.33}\bigg)^2\\ &=15.281\\ &\approx 16. \end{aligned}

Thus, the sample of size $n=16$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.33$.