# Standard error for the mean of a sample of binomial random variables

Suppose I'm running an experiment that can have 2 outcomes, and I'm assuming that the underlying "true" distribution of the 2 outcomes is a binomial distribution with parameters $n$ and $p$: ${\rm Binomial}(n, p)$.

I can compute the standard error, $SE_X = \frac{\sigma_X}{\sqrt{n}}$, from the form of the variance of ${\rm Binomial}(n, p)$: $$\sigma^{2}_{X} = npq$$ where $q = 1-p$. So, $\sigma_X=\sqrt{npq}$. For the standard error I get: $SE_X=\sqrt{pq}$, but I've seen somewhere that $SE_X = \sqrt{\frac{pq}{n}}$. What did I do wrong?

It seems like you're using $n$ twice in two different ways - both as the sample size and as the number of bernoulli trials that comprise the Binomial random variable; to eliminate any ambiguity, I'm going to use $k$ to refer to the latter.

If you have $n$ independent samples from a ${\rm Binomial}(k,p)$ distribution, the variance of their sample mean is

$${\rm var} \left( \frac{1}{n} \sum_{i=1}^{n} X_{i} \right) = \frac{1}{n^2} \sum_{i=1}^{n} {\rm var}( X_{i} ) = \frac{ n {\rm var}(X_{i}) }{ n^2 } = \frac{ {\rm var}(X_{i})}{n} = \frac{ k pq }{n}$$

where $q=1-p$ and $\overline{X}$ is the same mean. This follows since

(1) ${\rm var}(cX) = c^2 {\rm var}(X)$, for any random variable, $X$, and any constant $c$.

The standard error of $\overline{X}$is the square root of the variance: $\sqrt{\frac{ k pq }{n}}$. Therefore,

• When $k = n$, you get the formula you pointed out: $\sqrt{pq}$

• When $k = 1$, and the Binomial variables are just bernoulli trials, you get the formula you've seen elsewhere: $\sqrt{\frac{pq }{n}}$

• When $X$ is a bernoulli random variable, then ${\rm var}(X) = pq$. When $X$ has a binomial random variable based on $n$ trials with success probability $p$, then ${\rm var}(X) = npq$ Jun 1 '12 at 16:48
• Thanks! You lifted my confusion. Sorry that it was so elementary, I'm still learning :-) Jun 1 '12 at 17:02
• So is it clear to Frank that we are using the fact that for any constant c Var(cX) =c$^2$Var(x)? Since the sample estimate of the proportion is X/n we have Var(X/n)=Var(X)/n$^2$ =npq/n$^2$ =pq/n and SEx is the square root of that. I think it is clearer for everyone if we spell out all the steps. Jun 1 '12 at 21:42
• @MichaelChernick, I've clarified the details you mentioned. Based on the problem description, I figured that Frank knew these facts but you're right that it would be more educational for future readers to include the details. Jun 1 '12 at 22:41
• Sol Lago - In this case k=1. If you flipped a coin 50 times and calculated the number of successes and then repeated the experiment 50 times, then k=n=50. A flip of a coin results in a 1 or 0. It is a Bernoulli r.v. May 10 '14 at 19:35

It's easy to get two binomial distributions confused:

• distribution of number of successes
• distribution of the proportion of successes

npq is the number of successes, while npq/n = pq is the ratio of successes. This results in different standard error formulas.

• This is by far the most succinct explanation.
– mrp
Dec 18 '20 at 19:28

We can look at this in the following way:

Suppose we are doing an experiment where we need to toss an unbiased coin $n$ times. The overall outcome of the experiment is $Y$ which is the summation of individual tosses (say, head as 1 and tail as 0). So, for this experiment, $Y = \sum_{i=1}^n X_i$, where $X_i$ are outcomes of individual tosses.

Here, the outcome of each toss, $X_i$, follows a Bernoulli distribution and the overall outcome $Y$ follows a binomial distribution.

The complete experiment can be thought as a single sample. Thus, if we repeat the experiment, we can get another value of $Y$, which will form another sample. All possible values of $Y$ will constitute the complete population.

Coming back to the single coin toss, which follows a Bernoulli distribution, the variance is given by $pq$, where $p$ is the probability of head (success) and $q = 1 – p$.

Now, if we look at Variance of $Y$, $V(Y) = V(\sum X_i) = \sum V(X_i)$. But, for all individual Bernoulli experiments, $V(X_i) = pq$. Since there are $n$ tosses or Bernoulli trials in the experiment, $V(Y) = \sum V(X_i) = npq$. This implies that $Y$ has variance $npq$.

Now, the sample proportion is given by $\hat p = \frac Y n$, which gives the 'proportion of success or heads'. Here, $n$ is a constant as we plan to take same no of coin tosses for all the experiments in the population.

So, $V(\frac Y n) = (\frac {1}{n^2})V(Y) = (\frac {1}{n^2})(npq) = pq/n$.

So, standard error for $\hat p$ (a sample statistic) is $\sqrt{pq/n}$

• You can use Latex typesetting by putting dollars around your math, e.g. $x$ gives $x$. Jun 28 '16 at 20:52
• Note that the step $V(\sum X_i)=\sum V(X_i)$ really deserves some justification! Jun 28 '16 at 20:53
• There is typo in the last deduction, V(Y/n) = (1/n^2)*V(Y) = (1/n^2)*npq = pq/n should be the correct deduction. Jun 29 '16 at 2:05
• Apologies, I introduced that when doing the typesetting. Hopefully sorted now. Jun 29 '16 at 2:45
• That's true if the $X_i$ are uncorrelated - to justify this, we use the fact that the trials are assumed to be independent. Jun 29 '16 at 10:18

I think there is also some confusion in the initial post between standard error and standard deviation. Standard deviation is the sqrt of the variance of a distribution; standard error is the standard deviation of the estimated mean of a sample from that distribution, i.e., the spread of the means you would observe if you did that sample infinitely many times. The former is an intrinsic property of the distribution; the latter is a measure of the quality of your estimate of a property (the mean) of the distribution. When you do an experiment of N Bernouilli trials to estimate the unknown probability of success, the uncertainty of your estimated p=k/N after seeing k successes is a standard error of the estimated proportion, sqrt(pq/N) where q=1-p. The true distribution is characterized by a parameter P, the true probability of success. If you did an infinite number of experiments with N trials each and looked at the distribution of successes, it would have mean K=P*N, variance NPQ and standard deviation sqrt(NPQ).