Question
In a survey of 3,236 adults, 46% said that they eat cereal for breakfast at least twice a week. Determine the margin of error for this study. Round it to the nearest hundredth. Interpret its meaning in context. (2 points) The margin of error is □ %. The survey reports with 95% confidence that between □°% and □ % of adults eat cereal for breakfast at least twice a week.
Round or nearest whole number
Round or nearest whole number
Answers
GPT-4o mini
To calculate the margin of error for the survey at a 95% confidence level, we can use the formula for the margin of error (ME) in estimating proportions:
\[
ME = z \times \sqrt{\frac{p(1-p)}{n}}
\]
where:
- \( z \) is the z-score that corresponds to the desired confidence level,
- \( p \) is the sample proportion,
- \( n \) is the sample size.
1. **Determine the values:**
- Given \( p = 0.46 \) (46%),
- Sample size \( n = 3236 \),
- For a 95% confidence level, the z-score \( z \) is approximately 1.96.
2. **Calculate the margin of error:**
First, calculate the standard error (SE):
\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46(1-0.46)}{3236}} = \sqrt{\frac{0.46 \times 0.54}{3236}} = \sqrt{\frac{0.2484}{3236}} \approx \sqrt{0.0000768} \approx 0.00876
\]
Now, compute the margin of error:
\[
ME = 1.96 \times 0.00876 \approx 0.01717
\]
Converting this to a percentage:
\[
ME \approx 0.01717 \times 100 \approx 1.72\%
\]
3. **Rounding:**
The margin of error is approximately **1.72%** when rounded to the nearest hundredth.
4. **Interpretation:**
The survey reports with 95% confidence that between \( 46% - 1.72% = 44.28% \) and \( 46% + 1.72% = 47.72% \) of adults eat cereal for breakfast at least twice a week.
**Final Answers:**
- The margin of error is **1.72%**.
- The survey reports with 95% confidence that between **44%** and **48%** of adults eat cereal for breakfast at least twice a week (rounded to the nearest whole number).
\[
ME = z \times \sqrt{\frac{p(1-p)}{n}}
\]
where:
- \( z \) is the z-score that corresponds to the desired confidence level,
- \( p \) is the sample proportion,
- \( n \) is the sample size.
1. **Determine the values:**
- Given \( p = 0.46 \) (46%),
- Sample size \( n = 3236 \),
- For a 95% confidence level, the z-score \( z \) is approximately 1.96.
2. **Calculate the margin of error:**
First, calculate the standard error (SE):
\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46(1-0.46)}{3236}} = \sqrt{\frac{0.46 \times 0.54}{3236}} = \sqrt{\frac{0.2484}{3236}} \approx \sqrt{0.0000768} \approx 0.00876
\]
Now, compute the margin of error:
\[
ME = 1.96 \times 0.00876 \approx 0.01717
\]
Converting this to a percentage:
\[
ME \approx 0.01717 \times 100 \approx 1.72\%
\]
3. **Rounding:**
The margin of error is approximately **1.72%** when rounded to the nearest hundredth.
4. **Interpretation:**
The survey reports with 95% confidence that between \( 46% - 1.72% = 44.28% \) and \( 46% + 1.72% = 47.72% \) of adults eat cereal for breakfast at least twice a week.
**Final Answers:**
- The margin of error is **1.72%**.
- The survey reports with 95% confidence that between **44%** and **48%** of adults eat cereal for breakfast at least twice a week (rounded to the nearest whole number).