Question

n a survey of 110 shoppers at a grocery store, 23 said they use coupons regularly. Given a 95% level of confidence with a z-score of 1.96, what is the margin of error, rounded to the nearest tenth of a percent?(1 point)

Answer 1

To calculate the margin of error in a proportion at a 95% confidence level, we can use the formula:

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

Where:
- \( z \) is the z-score for the desired confidence level (for 95%, \( z = 1.96 \)),
- \( p \) is the sample proportion,
- \( n \) is the sample size.

1. **Calculate the sample proportion \( p \)**:
\[
p = \frac{x}{n} = \frac{23}{110} \approx 0.2091
\]

2. **Calculate \( 1 - p \)**:
\[
1 - p = 1 - 0.2091 \approx 0.7909
\]

3. **Calculate the margin of error**:
\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.2091 \times 0.7909}{110}}
\]

1. Calculate the value inside the square root:
\[
0.2091 \times 0.7909 \approx 0.1653
\]

2. Divide by \( n \):
\[
\frac{0.1653}{110} \approx 0.001503
\]

3. Take the square root:
\[
\sqrt{0.001503} \approx 0.03877
\]

4. Finally, multiply by the z-score:
\[
\text{Margin of Error} = 1.96 \times 0.03877 \approx 0.0760
\]

4. **Convert to percentage**:
\[
\text{Margin of Error (in \%)} \approx 0.0760 \times 100 \approx 7.60\%
\]

5. **Round to the nearest tenth of a percent**:
\[
\text{Rounded Margin of Error} \approx 7.6\%
\]

Thus, the margin of error, rounded to the nearest tenth of a percent, is **7.6%**.