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Resolve por aproximação o tal do empate técnico estatístico da margem de erro
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# Qual é a probabilidade de A estar a frente de B dada a diferença entre eles e o erro amostral? | |
# Abaixo, a diferença é de .505 - .495 = 1% | |
A = .505 | |
B = .495 | |
n = c(2500, 1111, 625, 400) | |
# Resolve a margem de erro por N | |
(moe = sqrt((A*(1-A)/n)+(B*(1-B)/n)+(2*((A*B)/n)) ) ) | |
[1] 0.02 0.03 0.04 0.05 | |
# Resolve a probabilidade de A - B > 0 por aproximacao normal | |
round((probs = pnorm(A - B, mean=0, sd=moe)) , 2) | |
[1] 0.69 0.63 0.60 0.58 |
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