Find the margin of error and​ 95% confidence interval for the survey result described. 
Surveying 1655 ​adults, a research center reported that​ "an overwhelming majority of Americans ​(76​%) support requiring all healthy schoolchildren to be vaccinated for​ measles, mumps and​ rubella."
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Part 1
Margin of errorequals
  
enter your response here​%
​(Type an integer or decimal rounded to one decimal place as​ needed.)

To find the margin of error (MOE) for the survey result, we can use the following formula for the margin of error at a 95% confidence level:

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

where:
- \( Z \) is the Z-score corresponding to the desired confidence level (for 95%, \( Z \approx 1.96 \))
- \( p \) is the sample proportion (76%, or 0.76)
- \( n \) is the sample size (1655)

### Step-by-Step Solution:

1. **Convert 76% to a decimal**: \( p = 0.76 \)
2. **Compute \( p(1 - p) \)**:
   \[
   p(1 - p) = 0.76 \times (1 - 0.76) = 0.76 \times 0.24 = 0.1824
   \]
3. **Divide by the sample size \( n \)**:
   \[
   \frac{0.1824}{1655} \approx 0.0001102
   \]
4. **Take the square root**:
   \[
   \sqrt{0.0001102} \approx 0.0105
   \]
5. **Multiply by the Z-score for a 95% confidence level** (1.96):
   \[
   \text{MOE} = 1.96 \times 0.0105 \approx 0.0206
   \]

6. **Convert

Find the margin of error and 95% confidence interval for the survey result described. Surveying 1655 adults, a research center reported that 'an overwhelming majority of Americans (76%) support requiring all healthy schoolchildren to be vaccinated for measles, mumps and rubella.'

Learn how to calculate the margin of error and 95% confidence interval for survey results, with a real-world example of a study on vaccination support. This problem covers key statistical concepts like sample proportions, Z-scores, and confidence intervals. Suitable for high school students (Grades 11-12).

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