Certainly! Let's go through the step-by-step solution for determining how many households must be surveyed to achieve the desired confidence level and margin of error.
### Given Values:
1. Previous proportion of households using email, [tex]\(\hat{p} = 0.76\)[/tex]
2. [tex]\(z^\)[/tex]-score for a 95% confidence level, [tex]\(z^ = 1.96\)[/tex]
3. Margin of error, [tex]\(E = 0.02\)[/tex] (which is 2%)
### Formula:
The formula to find the required sample size [tex]\(n\)[/tex] is:
[tex]\[ n = \hat{p} (1 - \hat{p}) \left( \frac{z^*}{E} \right)^2 \][/tex]
### Step-by-Step Solution:
1. Calculate the term inside the square:
[tex]\[
\left( \frac{z^*}{E} \right)^2 = \left( \frac{1.96}{0.02} \right)^2
\][/tex]
[tex]\[
= (98)^2
\][/tex]
[tex]\[
= 9604
\][/tex]
2. Calculate [tex]\(\hat{p} (1 - \hat{p})\)[/tex]:
[tex]\[
\hat{p} (1 - \hat{p}) = 0.76 \times (1 - 0.76)
\][/tex]
[tex]\[
= 0.76 \times 0.24
\][/tex]
[tex]\[
= 0.1824
\][/tex]
3. Combine the results to find [tex]\(n\)[/tex]:
[tex]\[
n = 0.1824 \times 9604
\][/tex]
[tex]\[
= 1751.7696
\][/tex]
### Conclusion:
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 is approximately 1752 households (since we generally round up to ensure the sample size is sufficient).
So, the final answer is:
1752 households
