Questions: A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 7% margin of error at a 90% confidence level, what size of sample is needed? Give your answer in whole people.

Solution Steps

We need to determine the required sample size \( n \) for a political candidate's poll, ensuring a margin of error of \( 7\% \) at a \( 90\% \) confidence level. The formula for the margin of error in a proportion is given by:

\[ \text{Margin of Error} = Z \times \sqrt{\frac{p(1-p)}{n}} \]

For a \( 90\% \) confidence level, the Z-score \( Z \) is approximately \( 1.645 \). We will use \( p = 0.5 \) to maximize variability. The margin of error is given as \( 0.07 \).

To find the sample size \( n \), we rearrange the formula:

\[ n = \frac{Z^2 \times p(1-p)}{\text{Margin of Error}^2} \]

Substituting the known values into the formula:

\[ n = \frac{(1.645)^2 \times 0.5 \times (1 - 0.5)}{(0.07)^2} \]

Calculating the components:

1. \( Z^2 = (1.645)^2 \approx 2.706025 \)
2. \( p(1-p) = 0.5 \times 0.5 = 0.25 \)
3. \( \text{Margin of Error}^2 = (0.07)^2 = 0.0049 \)

Now substituting these values back into the equation:

\[ n = \frac{2.706025 \times 0.25}{0.0049} \approx \frac{0.67650625}{0.0049} \approx 137.5 \]

Since the sample size must be a whole number, we round up to the nearest whole number:

\[ n = 138 \]

Final Answer