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 (MOE) for a proportion, you can use the formula:

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

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

In this case:
- The sample size \( n = 110 \),
- The number of shoppers using coupons regularly \( x = 23 \).

First, we calculate the sample proportion \( p \):

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

Now, calculate \( 1-p \):

\[
1-p = 1 - 0.2091 \approx 0.7909
\]

Next, we can substitute the values into the MOE formula:

\[
\text{MOE} = 1.96 \times \sqrt{\frac{0.2091 \times 0.7909}{110}}
\]

Calculating the component inside the square root:

\[
0.2091 \times 0.7909 \approx 0.1653
\]
\[
\frac{0.1653}{110} \approx 0.001503
\]
\[
\sqrt{0.001503} \approx 0.0388
\]

Now, substitute back into the MOE formula:

\[
\text{MOE} = 1.96 \times 0.0388 \approx 0.0761
\]

Convert the margin of error from a proportion to a percentage by multiplying by 100:

\[
\text{MOE} \approx 0.0761 \times 100 \approx 7.61\%
\]

Rounding to the nearest tenth of a percent:

\[
\text{MOE} \approx 7.6\%
\]

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