I have a jar with white and black balls. Total number of balls in the jar is 100000. I want to estimate the proportion of white balls. My constraint is that the sample size for estimation should be low, lets assume 500 balls. I am debating between two approaches.

  1. Draw a single sample of 500 balls, $\hat{p}$ = number of white balls divided by 500
  2. Draw 10 samples of 50 balls each. Calculate the proportion of white balls in each sample, i. e., $[r_1, r_2, ..., r_{10}]$. Estimated $\hat{p}$ = average of $[r_1, r_2, ..., r_{10}]$.

Which method should I use so that I am less susceptible to sampling error?

  • 1
    $\begingroup$ Both methods will give exactly the same answer, I'm afraid. $\endgroup$
    – Pat
    Oct 17, 2013 at 10:07
  • $\begingroup$ With this sort of question it is important to consider whether balls are replaced after drawing. Would you a) replace each ball before drawing the next one, or b) for approach 2, replace the balls after each sample of 50, or c) not replace at all? $\endgroup$ Oct 17, 2013 at 10:11
  • $\begingroup$ I ran an experiment with both approaches. In approach 2, each sample of 50 is drawn without replacement. But I draw the sample of the next 50 with replacement. Like Pat mentions I found that both estimations yield similar results. My follow up question is will it make more sense to draw the sample of 50 with replacement? $\endgroup$
    – ryk
    Oct 17, 2013 at 10:17
  • $\begingroup$ Ohh, I hadn't thought about sampling with/without replacement. I just assumed the former. Disregard my earlier comment. $\endgroup$
    – Pat
    Oct 17, 2013 at 10:24
  • $\begingroup$ @ryk I edited your question for consistent notation and terminology, please check if it is still what you wanted to ask (or roll back with my apologies). $\endgroup$
    – Momo
    Oct 17, 2013 at 10:39

1 Answer 1


If there is no replacement of balls after drawing, then approaches 1 and 2 are equivalent. With approach 2, you can find the average of the proportions of white balls in each of the 10 samples, or find the total number of white balls in the 10 samples combined and express this as a proportion of 500. Both calculations will give the same result (given that, as is the case here, the samples are all of the same size).

If however for approach 2 the balls are replaced after each sample of 50, then the sampling error will be slightly higher than with approach 1. One way to see this is to consider the extreme case in which rather than 100000 there are only 500 balls in the jar. In that case approach 1, which would then be a 100% sample, would be guaranteed to estimate the true proportion correctly. But approach 2 would still be subject to sampling error because each sample of 50 would be only a 10% sample. With a much larger number of balls this effect is still present, albeit greatly diminished.


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