In article <1996Jun11.email@example.com>, firstname.lastname@example.org (Jerry Dallal) writes: |>In article <email@example.com>, firstname.lastname@example.org (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.