To determine the number of households that must be surveyed to be 95% confident that the current estimated population proportion is within a 2% margin of error, we will follow the steps below:
1. Identify the given information:
- Proportion of households using email five years ago, [tex]\(\hat{p}\)[/tex] = 0.76.
- Z-score for 95% confidence, [tex]\(z^*\)[/tex] = 1.96.
- Margin of error, [tex]\(E\)[/tex] = 0.02.
2. Use the formula for sample size:
[tex]\[
n = \hat{p}(1 - \hat{p}) \left( \frac{z^*}{E} \right)^2
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
n = 0.76 \times (1 - 0.76) \left( \frac{1.96}{0.02} \right)^2
\][/tex]
4. Calculate the components step-by-step:
- First, calculate [tex]\(\hat{p} (1 - \hat{p})\)[/tex]:
[tex]\[
0.76 \times (1 - 0.76) = 0.76 \times 0.24 = 0.1824
\][/tex]
- Next, calculate [tex]\(\left( \frac{z^*}{E} \right)\)[/tex]:
[tex]\[
\left( \frac{1.96}{0.02} \right) = 98
\][/tex]
- Then, square the result of [tex]\(\left( \frac{z^*}{E} \right)\)[/tex]:
[tex]\[
98^2 = 9604
\][/tex]
- Finally, multiply the results together to find [tex]\(n\)[/tex]:
[tex]\[
n = 0.1824 \times 9604 = 1751.7696
\][/tex]
Thus, the sociologist must survey 1752 households (rounding 1751.7696 to the nearest whole number) to be 95% confident that the current estimated population proportion of households using email is within a 2% margin of error.
