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What sample size should be obtained if he wants to be within 4 percentage points with 96​% confidence if he uses an estimate of 52​% obtained from a​ poll?

Respuesta :

jushmk
e = Z* Sqrt {pq/N} e = 4% = 0.04; Z = 2.054 at 96% confidence, p= 52% = 0.052, q= 1-p = 1-0.52 = 0.48, N = Sample size required

Therefore,
N = (e/Z)^2*pq = (2.054/0.04)^2*0.52*0.48 = 658.15 ≈ 659 respondents

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