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?
The margin of error is ___%

Answer 1

To calculate the margin of error (ME) for a proportion, you can use the following formula:

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

Where:
- \( z \) is the z-score corresponding to the desired confidence level (1.96 for 95% confidence level)
- \( p \) is the sample proportion
- \( n \) is the sample size

First, we need to find the sample proportion \( p \). Given that 23 out of 110 shoppers use coupons regularly, we have:

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

Next, we calculate \( 1 - p \):

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

Now we can plug these values into the margin of error formula:

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

Calculating the value inside the square root:

\[
0.2091 \times 0.7909 \approx 0.1655
\]

\[
\frac{0.1655}{110} \approx 0.001505
\]

Now take the square root:

\[
\sqrt{0.001505} \approx 0.0388
\]

Finally, calculate the margin of error:

\[
ME = 1.96 \times 0.0388 \approx 0.0762
\]

To express the margin of error as a percentage, we multiply by 100:

\[
ME \approx 0.0762 \times 100 \approx 7.62\%
\]

Rounding to the nearest tenth of a percent:

\[
\text{Margin of Error} \approx 7.6\%
\]

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