Question

select the correct answer.
a random sample of 100 undergraduate students at a university found that 78 of them had used the university library’s website to find resources for a class. what is the margin of error for the true proportion of all undergraduates who had used the library’s website to find resources for a class?
a. 0.04
b. 0.08
c. 0.1
d. 0.12

Explanation:

Step1: Recall Margin of Error Formula

For a proportion, the margin of error (ME) for a 95% confidence interval (common, though not stated, assumed here) is \( ME = z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion, \( n \) is the sample size, and \( z^* \) for 95% CI is approximately 1.96 (or sometimes 2 for simplicity).

Step2: Calculate Sample Proportion

\( \hat{p} = \frac{78}{100} = 0.78 \), so \( 1 - \hat{p} = 0.22 \), \( n = 100 \).

Step3: Plug into Formula (Using \( z^* = 2 \) for approximation)

\( ME = 2 \sqrt{\frac{0.78 \times 0.22}{100}} \)
First, calculate inside the square root: \( \frac{0.78 \times 0.22}{100} = \frac{0.1716}{100} = 0.001716 \)
Then, square root: \( \sqrt{0.001716} \approx 0.0414 \)
Multiply by 2: \( 2 \times 0.0414 \approx 0.0828 \approx 0.08 \) (or with \( z^* = 1.96 \), \( 1.96 \times 0.0414 \approx 0.0811 \approx 0.08 \))

Answer:

B. 0.08