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 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:

\[
\boxed{7.6\%}
\]