Apologies if this is a little confusing. If anything needs clarifying, I'll edit the question as best I can. Thanks.

If we take a sample of, say, 100 people, we could measure their heights, and find the mean and standard deviation. We can calculate an estimate of the standard error from this single sample to find where the population mean is likely to lie.

If we repeated this sampling and measured further samples of 100 people from the population, then finding the means for each sample would give a sampling distributions of means. The mean of this sampling distribution estimates the population mean, while the standard deviation of this sampling distribution also estimates the standard error.

This all seems fairly clear.

Now, consider a large number of trials, e.g. 20000 trials of 100 coin flips, where the number of heads in each trial is recorded and the fraction of heads obtained in each trial found. To me, it seems less obvious as to what constitutes a sample here.

Would each one of these 20000 trials be considered as a single sample, so that we have 20000 samples?

If that's right, then when we average the fractions of heads found over all 20000 trials, then — just as for the height example — the mean is an estimate of the population mean, and the standard deviation would be an estimate of the standard error.

The alternative I'm thinking of would be to treat each set of 100 coin flips as a single data point, which would give a single large sample of 20000 measurements.

Treating this data as 20000 individual samples does seem like it should be correct since it's analogous to the height example. However, it doesn't seem immediately intuitive to me. Maybe this is because each trial results in a single fraction value, instead of dealing with the 100 individual flips of each trial. So, thinking of the result of each trial as a individual sample mean seems less obvious here, as the trial effectively yields a single value. On the other hand, I can see that the fraction of heads is effectively a mean value, where we code heads as 1, tails as 0 and then divide by the total number of flips in a trial.

What confuses me further is if we then decide to work with the raw number of heads found in each trial instead of the fraction of heads. The total number of heads in a trial is a single value, rather than a mean. However, my thinking is that the same would be true: the mean of the 20000 raw numbers of heads would be an estimate of the population mean, and that the standard deviation would estimate the standard error of the mean. I'm not sure if or why this would be the case though.

Edit: Really, I think what I'm trying to understand about this:

  1. If I take repeated trials of a fixed number of flips and compile the fraction of heads, do these give a distribution of sample means? I think from Michael Chernick's answer that this is correct.

  2. Next, if we look at the distribution of the raw scores of heads obtained in the repeated trials of a fixed number of flips, can this be again considered as a distribution of sample means? This is exactly the same as in 1, except the measured values are multiplied by the sample size. Because of this, I assume the answer is yes, but it just seems a little dissonant. When you reframe the data like that, you're using a total value, rather than the mean number of heads (the mean being equivalent to the fraction of heads).

  • $\begingroup$ I am not sure I understand what you are asking, but one possible help is that 100 people are, presumably, a sample from some defined population of people. What is the population of coin flips? If you can't define a population, I don't see how you can define a sample from it. $\endgroup$
    – Peter Flom
    Sep 29, 2012 at 19:53
  • $\begingroup$ It's not really the coin experiment itself that it's important; I'm just trying to make sure that I've got ideas of standard deviations and error down. It doesn't need to be coin flips, I could equally pose the question in terms of anythig with a binary outcome, e.g. yes/no survey of samples of a finite population. $\endgroup$ Sep 29, 2012 at 22:41

3 Answers 3


The answer to both of your questions (at the end of your post) is yes--they just represent sample means of different random variables. In the first case you are sampling a Bernoulli($p$) random variable with a sample size of $N$, and taking the mean. In the second case you are sampling a Binomial($N$,$p$) random variable with a sampling size of one and taking the mean. In both cases it's fair to talk about the distribution of the sample mean, though the second case is kind of trivial (because the sample size is 1).


What confuses me about your question is what is you objective? Are you trying to estimate a single proportion or determine the sampling distribution for a proportion estimate when the sample size is 100?

1. What to do if you want to estimate a single proportion.

If your objective is to get a point or interval estimate for a single proportion, then since each sample fo size 100 is independently chosen from a single population you can pool the data to get a best estimate for teh parameter.

2. If you want to know the sampling distribution empirically from independent samples of size 100

Compute the 20000 estimates and use that that discrete distribution as an approximation to the sampling distribution.


In the way you have defined it, you would have 20000 samples.

There is a discrete sample space describing all possible outcomes of 100 coin flips.

Each time you flip 100 coins, that's an observation (though this is perhaps semantically counter-intuitive).

All of the observations constitute the sample.


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