Question

In a survey of 110 shoppers at a grocery store 23 said they use coupons regularly give 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

Answer 1

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

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

Where:
- \(z\) is the z-score (1.96 for a 95% confidence level),
- \(p\) is the sample proportion of shoppers who use coupons regularly,
- \(n\) is the total number of shoppers surveyed.

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

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

Now we can compute \(1 - p\):

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

Next, we can substitute these values into the margin of error formula:

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

Calculating the values under the square root:

\[
0.2091 \times 0.7909 \approx 0.1654
\]
\[
\frac{0.1654}{110} \approx 0.0015036
\]
\[
\sqrt{0.0015036} \approx 0.0388
\]

Now substitute this back into the margin of error formula:

\[
ME = 1.96 \times 0.0388 \approx 0.0761
\]

To convert the margin of error into a percentage:

\[
ME \approx 0.0761 \times 100 \approx 7.61\%
\]

Rounded to the nearest tenth of a percent, the margin of error is:

\[
\boxed{7.6\%}
\]