# Suppose the U.S. president wants to estimate the proportion of the population that supports his...

## Question:

Suppose the U.S. president wants to estimate the proportion of the population that supports 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 advisers found a similar survey from two years ago that reported that 54% of people supported health care revisions.

How large of a sample is required? (Round up answer to the next whole number.)

How large of a sample would be necessary if no estimate were available for the proportion supporting current policy? (Round up answer to the next whole number.)

## Confidence interval

Point estimated for a population proportion is the sample proportion which is the ratio of the number of succes and a total number of trials in the distribution or total outcomes. The value of the sample proportion always lies between 0 to 1. The formula for point estimation is given as:

{eq}\begin{align*} \hat p = \dfrac{X}{n}\\ {\rm{where \ X \ is \ number \ of \ success \ and \ n \ is \ sample \ size}} \end{align*}{/eq}

Given information

• Margin of error: 0.03
• Sample proportion: 0.54

The critical value is obtained from the standard normal table at level of significance (0.02)

The critical value is 2.326

The sample size is calculated as follows:

{eq}\begin{align*} n &= \dfrac{{{{\left( {{Z_{\alpha /2}}} \right)}^2} \times p\left( {1 - p} \right)}}{{{\varepsilon ^2}}}{\rm{ \ where \ }}\varepsilon {\rm{ \ is \ margin \ of \ error}}\\ &= \dfrac{{{{2.326}^2} \times 0.54\left( {1 - 0.54} \right)}}{{{{0.03}^2}}}\\ &= 1493.23 \simeq 1494 \end{align*}{/eq}

Therefore, required sample size is 1494

If no estimate were available for the proportion supporting current policy, therefore we let sample proportion is 0.5

The sample size is calculated as follows:

{eq}\begin{align*} n &= \dfrac{{{{\left( {{Z_{\alpha /2}}} \right)}^2} \times p\left( {1 - p} \right)}}{{{\varepsilon ^2}}}{\rm{ \ where \ }}\varepsilon {\rm{ \ is \ margin \ of \ error}}\\ &= \dfrac{{{{2.326}^2} \times 0.5\left( {1 - 0.5} \right)}}{{{{0.03}^2}}}\\ &= 1502.85 \simeq 1503 \end{align*}{/eq}

Therefore, required sample size is 1503 