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.

Answer 1

To calculate the margin of error (ME) for the proportion of adults who eat cereal for breakfast at least twice a week, we can use the formula for the margin of error for proportions at a 95% confidence level:

\[
\text{ME} = z \times \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \( z \) is the z-score for a 95% confidence level (which is approximately 1.96),
- \( p \) is the sample proportion (0.46 in this case),
- \( n \) is the sample size (3,236).

Now we can perform the calculations:

1. Calculate \( p(1 - p) \):
\[
p(1 - p) = 0.46(1 - 0.46) = 0.46 \times 0.54 = 0.2484
\]

2. Calculate \( \frac{p(1 - p)}{n} \):
\[
\frac{p(1 - p)}{n} = \frac{0.2484}{3236} \approx 0.0000769
\]

3. Take the square root:
\[
\sqrt{\frac{p(1 - p)}{n}} \approx \sqrt{0.0000769} \approx 0.00877
\]

4. Calculate the Margin of Error:
\[
\text{ME} = 1.96 \times 0.00877 \approx 0.0172 \text{ or } 1.72\%
\]

Now rounding that to the nearest hundredth gives:

\[
\text{The margin of error is } 1.72\%.
\]

### Interpretation:

The survey reports with 95% confidence that between:
\[
46\% - 1.72\% \text{ and } 46\% + 1.72\%
\]

Calculating these bounds:

- Lower bound: \( 46\% - 1.72\% = 44.28\% \)
- Upper bound: \( 46\% + 1.72\% = 47.72\% \)

Thus, the final statement is:

**The margin of error is 1.72%. The survey reports with 95% confidence that between 44.28% and 47.72% of adults eat cereal for breakfast at least twice a week.**