To calculate the margin of error (MOE) for the given proportions at a 95% confidence level, we use the formula:

$MOE = Z_{\alpha/2} \times \sqrt{\frac{p(1 - p)}{n}}$

Where:

• $Z_{\alpha/2}$ is the critical value from the standard normal distribution for a 95% confidence level, which is approximately 1.96.
• $p$ is the sample proportion of students who answered "yes."
• $n$ is the total sample size.

Step 1: Find the sample proportion $p$

The total number of students surveyed is: $n = 14,255 + 115,335 = 129,590$ The proportion of students who answered "yes" is: $p = \frac{14,255}{129,590} \approx 0.1100$

Step 2: Apply the formula

Now, we substitute $p$, $n$, and $Z_{\alpha/2}$ into the MOE formula: $MOE = 1.96 \times \sqrt{\frac{0.1100(1 - 0.1100)}{129,590}}$ Let’s calculate this.

The margin of error (MOE) for the sample proportions at a 95% confidence level is approximately 0.0017, or 0.17%.

Would you like further details or have any questions?

Here are five related questions:

1. How does the confidence level affect the margin of error?
2. What happens to the MOE if the sample size increases?
3. Can the margin of error be negative?
4. How would the MOE change if the confidence level were 99% instead of 95%?
5. What is the difference between margin of error and standard error?

Tip: Larger sample sizes generally reduce the margin of error, making estimates more precise.