# Solved:Suppose the U.S. president wants an estimate of the proportion of the population who support his current policy toward revisions in the health

#### ByDr. Raju Chaudhari

Aug 23, 2020

Suppose the U.S. president wants an estimate of the proportion of the population who support his current policy toward revisions in the health care system. The president wants the estimate to be within 0.03 of the true proportion. Assume a 98% level of confidence. The president's political advisors estimated the proportion supporting the current policy to be 0.54. (Use z Distribution Table.)

a. How large of a sample is required? (Round the z-values to 2 decimal places. Round up your answer to the next whole number.)
b. How large of a sample would be necessary if no estimate were available for the proportion supporting current policy?

#### Solution

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

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

a. Given that $p = 0.54$.

Given that margin of error $E =0.03$. The confidence coefficient is $0.98$. Assume that the proportion is $p =0.54$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.33$.

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.54(1-0.54)\bigg(\frac{2.33}{0.03}\bigg)^2\\ &=1498.3764\\ &\approx 1499. \end{aligned}
Thus, the sample of size $n=1499$ will ensure that the $98$% confidence interval for the proportion will have a margin of error $0.03$.

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

Given that margin of error $E =0.03$. The confidence coefficient is $0.98$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.33$.

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{2.33}{0.03}\bigg)^2\\ &=1508.0278\\ &\approx 1509. \end{aligned}

Thus, the sample of size $n=1509$ will ensure that the $98$% confidence interval for the proportion will have a margin of error $0.03$.