To calculate the margin of error for a proportion, you can use the formula for the margin of error (ME) at a certain confidence level. Commonly, a 95% confidence level is used, which corresponds to a z-score of approximately 1.96.
The formula for the margin of error for a proportion is:
\[ ME = z \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \( z \) is the z-score for the desired level of confidence (1.96 for 95% confidence),
- \( p \) is the sample proportion (0.46 in this case),
- \( n \) is the sample size (3,236 adults).
Step 1: Calculate \( p(1-p) \)
\[ p(1-p) = 0.46 \times (1 - 0.46) = 0.46 \times 0.54 = 0.2484 \]
Step 2: Calculate the standard error (SE)
\[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2484}{3236}} \approx \sqrt{0.00007679} \approx 0.00875 \]
Step 3: Calculate the margin of error (ME)
\[ ME = 1.96 \times SE \approx 1.96 \times 0.00875 \approx 0.01715 \]
Step 4: Convert margin of error to a percentage
\[ ME \approx 0.01715 \times 100 \approx 1.72% \]
Step 5: Rounding
The margin of error is approximately 2% when rounded to the nearest whole number.
Interpretation in context
The survey reports with 95% confidence that between \( (46% - 2%) \) and \( (46% + 2%) \) of adults eat cereal for breakfast at least twice a week. Therefore, the confidence interval is:
\[ (44%, 48%) \]
Thus, the final interpretation is:
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.