Let's go through the problem step-by-step to find the margin of error for the survey results.
1. Identify the given data:
- Total number of students surveyed, [tex]\( n \)[/tex], is 80.
- Number of students willing to pay extra, [tex]\( x \)[/tex], is 26.
- Desired confidence level [tex]\( z^* \)[/tex]-score is 1.645 for a 90% confidence interval.
2. Calculate the sample proportion [tex]\( \hat{p} \)[/tex]:
[tex]\[
\hat{p} = \frac{x}{n} = \frac{26}{80} = 0.325
\][/tex]
3. Write down the formula for the margin of error:
[tex]\[
E = z^* \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}
\][/tex]
4. Substitute the known values into the margin of error formula:
[tex]\[
E = 1.645 \sqrt{\frac{0.325 \cdot (1 - 0.325)}{80}}
\][/tex]
5. Perform the calculation inside the square root first:
[tex]\[
\hat{p} (1 - \hat{p}) = 0.325 \cdot 0.675 = 0.219375
\][/tex]
[tex]\[
\frac{0.219375}{80} = 0.0027421875
\][/tex]
6. Take the square root of the result:
[tex]\[
\sqrt{0.0027421875} \approx 0.05236
\][/tex]
7. Multiply by the [tex]\( z^* \)[/tex]-score:
[tex]\[
E = 1.645 \cdot 0.05236 \approx 0.08614
\][/tex]
8. Convert the margin of error to a percentage:
[tex]\[
E \times 100 \approx 8.614
\][/tex]
So, the margin of error for the survey is approximately 8.614%. Comparing with the given options, we find that 8% is the closest available choice. Therefore, the margin of error is:
[tex]\[
\boxed{8 \%}
\][/tex]
