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
%
Its not 7.6%
23/110

Answer 1

To calculate the margin of error for the proportion of shoppers who use coupons regularly, we can use the formula for the margin of error (ME) at a 95% confidence level, which is given by:

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

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

First, we need to calculate the sample proportion \( p \):
\[
p = \frac{23}{110} \approx 0.2091
\]

Next, we calculate \( 1 - p \):
\[
1 - p \approx 1 - 0.2091 \approx 0.7909
\]

Now we can substitute these values into the formula:
\[
ME = 1.96 \cdot \sqrt{\frac{0.2091 \cdot 0.7909}{110}}
\]

Calculating the inside of the square root:
\[
\frac{0.2091 \cdot 0.7909}{110} \approx \frac{0.1654}{110} \approx 0.001503
\]

Now, take the square root:
\[
\sqrt{0.001503} \approx 0.0388
\]

Now calculate the margin of error:
\[
ME = 1.96 \cdot 0.0388 \approx 0.0761
\]

To convert this to a percentage:
\[
ME \approx 0.0761 \cdot 100 \approx 7.61\%
\]

Rounding to the nearest tenth of a percent:
\[
ME \approx 7.6\%
\]

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

Answer 2

Wrong it isnt 7.6 GPT-4o mini