Questions: A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate wants a 6% margin of error at a 99% confidence level, what sample size is needed? Use technology to find the z-score and do not round between steps. Make sure you use the correct rounding rule for samples size, and give your answer in whole people.

Solution Steps

To determine the required sample size for a political candidate's poll with a \(99\%\) confidence level, we first need to find the z-score corresponding to this confidence level. The area in each tail for a \(99\%\) confidence level is given by:

\[ \alpha = 1 - 0.99 = 0.01 \]

Thus, the area in each tail is:

\[ \frac{\alpha}{2} = \frac{0.01}{2} = 0.005 \]

Using the cumulative distribution function \( \Phi \), we find:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.995) - \Phi(-\infty) = 0.8401 \]

The z-score for a \(99\%\) confidence level is:

\[ Z_{end} = 0.995 \]

Next, we use the formula for the margin of error in a proportion to calculate the required sample size \(n\):

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

Rearranging the formula to solve for \(n\):

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

Assuming the worst-case scenario for \(p\) (the estimated proportion of the population) is \(p = 0.5\), we substitute the values:

• \(z = 0.995\)
• \(\text{Margin of Error} = 0.06\)
• \(p(1-p) = 0.5 \times (1 - 0.5) = 0.25\)

Now substituting these values into the formula:

\[ n = \left( \frac{0.995}{0.06} \right)^2 \times 0.25 \]

Calculating \(n\):

\[ n = \left( 16.5833 \right)^2 \times 0.25 \approx 69.0001 \]

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

\[ n = 69 \]

Final Answer