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A campus survey wants to estimate, with 95% confidence and a margin of error no greater than 5%, the true proportion of students who voted in the 2012 presidential election. With no prior estimate, using p'=q'=0.5 and z(0.025)=1.96, how many students must be interviewed?
A196 students
B385 students
C1,960 students
D50 students
Answer & Solution
Correct answer: B. 385 students
1. n = z^2*p'q'/EBP^2 = (1.96)^2(0.5)(0.5)/(0.05)^2.
2. (1.96)^2 = 3.8416, and (3.8416)(0.25) = 0.9604.
3. 0.9604/0.0025 ≈ 384.16, which rounds up to 385 students.
4. Option C, 1,960, mistakes the z-value itself for part of the final sample-size calculation rather than squaring and dividing it correctly.
_Source: OpenStax Introductory Statistics (CC BY 4.0), Ch 8 "Confidence Intervals", section 8.3 A Population Proportion_
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