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Insurance companies want to know what proportion of drivers always buckle up, aiming for 95% confidence within EBP = 0.03 with no prior estimate of p'. Using p'=q'=0.5 and z(0.025)=1.96, what is the minimum number of drivers to survey?
A1,960 drivers
B300 drivers
C9 drivers
D1,068 drivers
Answer & Solution
Correct answer: D. 1,068 drivers
1. n = z^2*p'q'/EBP^2 = (1.96)^2(0.5)(0.5)/(0.03)^2.
2. (1.96)^2 = 3.8416, and (3.8416)(0.25) = 0.9604.
3. 0.9604/0.0009 ≈ 1067.1, which rounds up to 1,068 drivers.
4. Option B, 300, is far too small to reach a margin of error as tight as 0.03 at 95% confidence.
_Source: OpenStax Introductory Statistics (CC BY 4.0), Ch 8 "Confidence Intervals", section 8.3 A Population Proportion_
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