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Re: Significant Digits for the Mean
Posted:
Jun 14, 1996 6:26 AM
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In article <1996Jun11.134058@mint.hnrc.tufts.edu>, jerry@mint.hnrc.tufts.edu (Jerry Dallal) writes:
|>In article <4pjolu$40e7@b.stat.purdue.edu>, hrubin@b.stat.purdue.edu (Herman Rubin) writes:
|>> In the case of integer data, the number of "significant" digits in
|>> the mean depends heavily on the sample size; with an extremely large
|>> data set, the number grows, but not that fast. One way to see that
|>> this must be so is in reporting an estimate of a probability from
|>> Bernoulli trials. The dictum in 1. above would limit this to one
|>> decimal place, which is clearly absurd.
|>
|> Consider an urn containing 10 balls, an unknown proportion
|> of which are red. Suppose a ball is drawn from the urn
|> repeatedly, with replacement.
|>
|> Let Xi = 1, i-th ball is red
|> 0, otherwise
|> (a Bernoulli trial!)
|>
|> Now, as the sample size increases, the sample mean can be
|> expressed with real meaning to more significant digits. Yet,
|> the larger the sample size, the more sense it makes to
|> estimate the population mean by reporting the sample mean to
|> only 1 significant digit! That is, as the sample size
|> increases, the sample mean expressed to only 1 significant
|> digit is superior to the "full precision" sample mean as an
|> estimate of the proportion of red balls in the urn by most
|> commonly used measures of accuracy.
Let the urn contain 2 balls, one red and one blue. Consider the stocastic process that draws
with replacement the red ball with probability p and the blue with probability 1-p. Let
Xi be as defined. The mean is then p, cleary not necessarily an integer.
The described process is of course the binomial with 1 trial. With n trials the mean
is np.
Consequently, there is no general rule about the number of decimal places for the mean.
-Jarle
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