Question

select the correct answer.
surveys conducted in american high schools concluded that 90% of the students in a sample of 400 students had more than one active email
account. what would the margin of error be for the population proportion?
a. ±1.5%
b. ±3%
c. ±4.5%
d. ±9%

Explanation:

Step1: Recall the formula for margin of error for proportion

The formula for margin of error \(E\) for a proportion is \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a simple case (assuming a 95% confidence level, \(z\approx2\), common in basic problems when not specified otherwise), \(p = 0.9\) (90% as a decimal), and \(n=400\).

Step2: Substitute values into the formula

First, calculate \(\sqrt{\frac{p(1 - p)}{n}}\):

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Then, since \(z = 2\) (approximate for 95% confidence in basic set - ups), \(E=2\times0.015=0.03\) or \(3\%\)

Answer:

B. ±3%