To determine the number of households Marcus needs to survey, we'll use the formula for calculating sample size for a population proportion:
[tex]\[
n = \hat{p}(1 - \hat{p}) \left(\frac{z^*}{E}\right)^2
\][/tex]
Where:
- [tex]\( \hat{p} \)[/tex] is the estimated proportion of households with a landline.
- [tex]\( z^ \)[/tex] is the z-score corresponding to the desired confidence level (99% confidence level in this case).
- [tex]\( E \)[/tex] is the margin of error.
Here are the given values:
- Confidence level: 99%, which corresponds to a [tex]\( z^ \)[/tex]-score of 2.58.
- Margin of error ([tex]\( E \)[/tex]): 6%, which is 0.06 as a decimal.
- Since Marcus does not have any previous information, we use [tex]\( \hat{p} = 0.5 \)[/tex].
Now, substituting the given values into the formula:
[tex]\[
n = 0.5 \times (1 - 0.5) \times \left(\frac{2.58}{0.06}\right)^2
\][/tex]
First, calculate the fraction inside the square:
[tex]\[
\frac{2.58}{0.06} = 43
\][/tex]
Now square the result:
[tex]\[
43^2 = 1849
\][/tex]
Then, multiply by 0.5 and 0.5 (which is 0.25):
[tex]\[
n = 0.5 \times 0.5 \times 1849 = 0.25 \times 1849 = 462.25
\][/tex]
Since we can't survey a fraction of a household, we round to the nearest whole number:
[tex]\[
n \approx 462
\][/tex]
Therefore, Marcus needs to survey approximately 462 households to meet his requirements with 99% confidence and a 6% margin of error.
Given the provided options, the correct answer is:
- 463 households
