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)
The margin of error is %

Answer 1

To calculate the margin of error for this survey, we will follow these steps:

1. **Identify the sample proportion (p)**: In this case, 23 out of 110 shoppers use coupons regularly. Thus, the sample proportion \( p \) is:

\[
p = \frac{23}{110} \approx 0.2091
\]

2. **Calculate the sample size (n)**: The sample size \( n \) is 110.

3. **Calculate the standard error (SE)** using the formula for standard error of a proportion:

\[
SE = \sqrt{\frac{p(1 - p)}{n}}
\]

Substituting in the values:

\[
SE = \sqrt{\frac{0.2091(1 - 0.2091)}{110}} \approx \sqrt{\frac{0.2091 \times 0.7909}{110}} \approx \sqrt{\frac{0.1651}{110}} \approx \sqrt{0.001501} \approx 0.0387
\]

4. **Calculate the margin of error (ME)** using the formula:

\[
ME = z \times SE
\]

Where \( z \) is the z-score for a 95% confidence level (1.96):

\[
ME = 1.96 \times 0.0387 \approx 0.0759
\]

5. **Convert the margin of error to a percentage**:

\[
ME \text{ (in percentage)} = 0.0759 \times 100 \approx 7.59
\]

6. **Round to the nearest tenth of a percent**:

\[
7.59 \% \approx 7.6 \%
\]

Thus, the margin of error is **7.6%**.