Question

Question In 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 (ME) for the given survey results, we can use the formula:

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

where:
- \( z \) is the z-score corresponding to the desired confidence level,
- \( p \) is the sample proportion (the proportion of shoppers using coupons regularly),
- \( n \) is the sample size.

In this case:

1. **Determine \( p \)**:
\[
p = \frac{23}{110} = 0.2091 \quad (\text{approximately})
\]

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

3. **Substitute into the formula**:
- The sample size \( n = 110 \).
- The z-score \( z = 1.96 \).

Now, calculate the standard error (SE):
\[
\text{SE} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.2091 \times 0.7909}{110}} = \sqrt{\frac{0.1652}{110}} = \sqrt{0.001502} \approx 0.0388
\]

4. **Calculate the margin of error (ME)**:
\[
\text{ME} = 1.96 \times 0.0388 \approx 0.0761
\]

5. **Convert to percentage**:
\[
\text{ME} \approx 0.0761 \times 100 \approx 7.61\%
\]

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