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.
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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.
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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:
\[ ME \approx 0.01717 \times 100 \approx 1.72% \]
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Rounding: The margin of error is approximately 1.72% when rounded to the nearest hundredth.
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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).