To calculate the margin of error for a proportion at a 95% confidence level, you 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 (1.96 for 95% confidence),
- \( p \) is the sample proportion, and
- \( n \) is the sample size.
First, we need to calculate the sample proportion \( p \). In this case, 23 out of 110 shoppers use coupons regularly, so:
\[ p = \frac{23}{110} \approx 0.2091 \]
Now we calculate \( 1 - p \):
\[ 1 - p = 1 - 0.2091 \approx 0.7909 \]
Next, we can find the standard error:
\[ \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2091 \times 0.7909}{110}} \]
Calculating the numerator:
\[ 0.2091 \times 0.7909 \approx 0.1652 \]
Now dividing by \( n = 110 \):
\[ \frac{0.1652}{110} \approx 0.0015 \]
Now, we take the square root:
\[ \sqrt{0.0015} \approx 0.0387 \]
Finally, we multiply by the z-score to find the margin of error:
\[ \text{Margin of Error} = 1.96 \times 0.0387 \approx 0.0758 \]
To express the margin of error as a percentage, we multiply by 100:
\[ 0.0758 \times 100 \approx 7.58% \]
Rounding to the nearest tenth of a percent, the margin of error is:
\[ \boxed{7.6%} \]