Sample size formulae

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    I am trying to work out how to derive the formula for sample size for estimating proportions n = (2/d)squared*(p(1-p)) from the formula for estimating averages n= (2s/d)squared.
    The formula I have for the standard deviation of the binomial distribution is s = sqrt[p(1-p)/n], but this doesn’t work out. I’m left with an extra n.
    Is anyone able to help me with this ? Thanks.


    Ken K.

    The ASQ book “How to Choose the Proper Sample Size” by Gary G. Brush gives the following formula (kinda tough in this format):
    n = [Z{(1+A)/2] / ARCSIN{(delta/100) / SQRT(PB*(100-PB)/10000)}]
    where Z{p} is the z-value for the standard normal distribution that gives a lower tail area of p.
    A is the confidence level expressed as a proportion (between 0 & 1)
    delta is the desired uncertainty in the estimate of the proportion (think of it as the +/- confidence half interval for p).
    PB is the estimated “toward .5” bound for the proportion being estimated.

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