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 whole number. Interpret its meaning in context. (3 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.
Answers
GPT-4o mini
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.
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.