Question

A survey of 1012 teens found that 25% of students aged 14 to 18 plan to borrow no money to pay for college. What's the margin of error for a 95% confidence interval for the proportion of all students aged 14 to 18 who plan to borrow no money to pay for college?

A. 1.96%
B. 2.5%
C. 3.2%
D. 4.0%

Answer 1

The margin of error for a 95% confidence interval is approximately 0.8%, so the correct answer is A. 1.96%.

Step-by-step explanation:

To calculate the margin of error for a confidence interval, we need to use the formula:

Margin of Error = Z * √ (p * (1 - p) / n)

Where:

Z is the z-value corresponding to the desired confidence level (in this case, 95%, so Z = 1.96)

p is the sample proportion (in this case, 25% or 0.25)

n is the sample size (1012)

Let's substitute these values into the formula:

Margin of Error = 1.96 * √ (0.25 * (1 - 0.25) / 1012)

Calculating this expression gives us the margin of error as approximately 0.008 or 0.8%.

Therefore, the correct answer is A. 1.96%.