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%
The margin of error is
%
Its not 7.6%
Answers
GPT-4o mini
To calculate the margin of error for a proportion, we can use the following formula:
\[
\text{Margin of Error} = Z \times \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( Z \) is the z-score (1.96 for 95% confidence),
- \( p \) is the sample proportion,
- \( n \) is the sample size.
1. **Calculate the sample proportion \( p \)**:
\[
p = \frac{\text{number of shoppers using coupons}}{\text{total shoppers}} = \frac{23}{110} \approx 0.2091
\]
2. **Calculate \( 1 - p \)**:
\[
1 - p = 1 - 0.2091 \approx 0.7909
\]
3. **Plug in the values into the margin of error formula**:
\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.2091 \times 0.7909}{110}}
\]
4. **Calculate the standard error**:
\[
\text{Standard Error} = \sqrt{\frac{0.2091 \times 0.7909}{110}} \approx \sqrt{\frac{0.1653}{110}} \approx \sqrt{0.001503} \approx 0.0388
\]
5. **Calculate the margin of error**:
\[
\text{Margin of Error} = 1.96 \times 0.0388 \approx 0.076
\]
6. **Convert to percentage**:
\[
\text{Margin of Error} \approx 0.076 \times 100 \approx 7.6\%
\]
Given that you mentioned the margin of error is not 7.6%, let’s double-check the calculations.
After recalculating:
\[
\text{Margin of Error} = 1.96 \times 0.0388 \approx 0.076
\]
So:
\[
\text{Margin of Error} = 7.6\%
\]
If you believe this is incorrect, please verify if all input values are accurate or if there's any other aspect of the question that might change the inputs. However, based on standard calculations, the margin of error for the data provided remains approximately **7.6%**.
\[
\text{Margin of Error} = Z \times \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( Z \) is the z-score (1.96 for 95% confidence),
- \( p \) is the sample proportion,
- \( n \) is the sample size.
1. **Calculate the sample proportion \( p \)**:
\[
p = \frac{\text{number of shoppers using coupons}}{\text{total shoppers}} = \frac{23}{110} \approx 0.2091
\]
2. **Calculate \( 1 - p \)**:
\[
1 - p = 1 - 0.2091 \approx 0.7909
\]
3. **Plug in the values into the margin of error formula**:
\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.2091 \times 0.7909}{110}}
\]
4. **Calculate the standard error**:
\[
\text{Standard Error} = \sqrt{\frac{0.2091 \times 0.7909}{110}} \approx \sqrt{\frac{0.1653}{110}} \approx \sqrt{0.001503} \approx 0.0388
\]
5. **Calculate the margin of error**:
\[
\text{Margin of Error} = 1.96 \times 0.0388 \approx 0.076
\]
6. **Convert to percentage**:
\[
\text{Margin of Error} \approx 0.076 \times 100 \approx 7.6\%
\]
Given that you mentioned the margin of error is not 7.6%, let’s double-check the calculations.
After recalculating:
\[
\text{Margin of Error} = 1.96 \times 0.0388 \approx 0.076
\]
So:
\[
\text{Margin of Error} = 7.6\%
\]
If you believe this is incorrect, please verify if all input values are accurate or if there's any other aspect of the question that might change the inputs. However, based on standard calculations, the margin of error for the data provided remains approximately **7.6%**.