To calculate the margin of error for the study, we can use the formula for the margin of error (ME) in a proportion at a 95% confidence level, which is given by:
\[ ME = z \times \sqrt{\frac{p(1-p)}{n}} \]
where:
- \( z \) is the z-score corresponding to the desired confidence level (for 95%, \( z \approx 1.96 \)),
- \( p \) is the sample proportion (0.46 in this case),
- \( n \) is the sample size (3,236 in this case).
- Calculate the standard error (SE):
\[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46 \times (1 - 0.46)}{3236}} = \sqrt{\frac{0.46 \times 0.54}{3236}} = \sqrt{\frac{0.2484}{3236}} \approx \sqrt{0.0000769} \approx 0.00876 \]
- Calculate the margin of error (ME):
\[ ME = 1.96 \times SE \approx 1.96 \times 0.00876 \approx 0.0172 \]
To percentage format, we multiply by 100:
\[ ME \approx 0.0172 \times 100 \approx 1.72% \]
Rounding to the nearest whole number, the margin of error is approximately 2%.
- Interpret the results:
The survey reports with 95% confidence that between \( 46% - 2% = 44% \) and \( 46% + 2% = 48% \) of adults eat cereal for breakfast at least twice a week.
Thus, to summarize:
The margin of error is 2%. The survey reports with 95% confidence that between 44% and 48% of adults eat cereal for breakfast at least twice a week.