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.
ROUND TO NEAREST WHOLE NUMBER FOR EAHC ANSWER
ROUND TO NEAREST WHOLE NUMBER FOR EAHC ANSWER
Answers
GPT-4o mini
To calculate the margin of error (MOE) for a proportion in a survey, we can use the formula:
\[
MOE = z \times \sqrt{\frac{p(1-p)}{n}}
\]
where:
- \( z \) is the z-score corresponding to the desired confidence level (for 95% confidence, \( z \approx 1.96 \)),
- \( p \) is the sample proportion (in this case, 0.46),
- \( n \) is the sample size (here, 3,236).
First, let's calculate the standard error (SE):
\[
p = 0.46
\]
\[
1 - p = 0.54
\]
\[
n = 3236
\]
Now we can calculate the standard error:
\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46 \times 0.54}{3236}}
\]
Calculating the numerator:
\[
0.46 \times 0.54 = 0.2484
\]
Now divide by \( n \):
\[
\frac{0.2484}{3236} = 0.00007681
\]
Taking the square root gives us the standard error:
\[
SE = \sqrt{0.00007681} \approx 0.00877
\]
Now we compute the margin of error:
\[
MOE = 1.96 \times 0.00877 \approx 0.0172
\]
Convert this into a percentage by multiplying by 100:
\[
MOE \approx 0.0172 \times 100 \approx 1.72\%
\]
Rounding to the nearest hundredth, we have:
\[
\text{Margin of error is } \approx 1.72\%
\]
Now we can find the confidence interval:
\[
\text{Lower bound} = p - MOE = 0.46 - 0.0172 = 0.4428 \approx 44\%
\]
\[
\text{Upper bound} = p + MOE = 0.46 + 0.0172 = 0.4772 \approx 48\%
\]
Thus, we can fill in the blanks:
- 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.
So, the complete answers are:
- 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.
\[
MOE = z \times \sqrt{\frac{p(1-p)}{n}}
\]
where:
- \( z \) is the z-score corresponding to the desired confidence level (for 95% confidence, \( z \approx 1.96 \)),
- \( p \) is the sample proportion (in this case, 0.46),
- \( n \) is the sample size (here, 3,236).
First, let's calculate the standard error (SE):
\[
p = 0.46
\]
\[
1 - p = 0.54
\]
\[
n = 3236
\]
Now we can calculate the standard error:
\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46 \times 0.54}{3236}}
\]
Calculating the numerator:
\[
0.46 \times 0.54 = 0.2484
\]
Now divide by \( n \):
\[
\frac{0.2484}{3236} = 0.00007681
\]
Taking the square root gives us the standard error:
\[
SE = \sqrt{0.00007681} \approx 0.00877
\]
Now we compute the margin of error:
\[
MOE = 1.96 \times 0.00877 \approx 0.0172
\]
Convert this into a percentage by multiplying by 100:
\[
MOE \approx 0.0172 \times 100 \approx 1.72\%
\]
Rounding to the nearest hundredth, we have:
\[
\text{Margin of error is } \approx 1.72\%
\]
Now we can find the confidence interval:
\[
\text{Lower bound} = p - MOE = 0.46 - 0.0172 = 0.4428 \approx 44\%
\]
\[
\text{Upper bound} = p + MOE = 0.46 + 0.0172 = 0.4772 \approx 48\%
\]
Thus, we can fill in the blanks:
- 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.
So, the complete answers are:
- 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.