# Standard error for the sample distribution of a random binomial variable

This is a followup question to Standard error for the mean of a sample of binomial random variables

The previous answer mentions that the standard error of the sampling distribution is sqrt(kpq/n), where k is the number of trials in each binomial experiment, p is the probability of success, q is (1-p) and n is the number of experiments in generating the sampling distribution.

For the normal approximation of the binomial confidence interval, the standard error is sqrt(pq/n). Does that mean the normal approximation is achieving this approximation by reducing the binomial experiments to bernoulli trials (k=1)?

If I had a case where each binomial experiment had 25 trials and 100 experiments were run (k=25, n=100), would the standard error used for the binomial confidence interval be sqrt(pq/4)?

## 1 Answer

welcome to stack exchange. First of all note that you can use MathJax for mathematical notation on this site: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

To your question:

1. The standard error for the normal approximation is $$\sqrt{\frac{pq}{n}}$$ if you draw once from a $$Bi(n,p)$$ and average over these n draws, i.e. divide by $$n$$, or this is - as you say - equivalent to drawing $$n$$ times from $$Bi(1,p)$$ and considering their mean. So yes, if I understand you correctly, we apply the normal approximation to the $$Bi(n,p)$$ (if n sufficiently large), because it is the sum of $$n$$ Bernoulli trials and we use the normal approximation if we consider a large enough sum of independent Random Variables.

2. This is basically explained in the link that you mention: The Variance for a $$Bi(25,p)$$ is $$25\cdot p(1-p)$$ (as the variance of a Bernoulli trial is $$p(1-p)$$ and $$Var(\sum_{i=1}^nX_i)= nVar(X_i)$$, if $$X_i$$ are iid). In order to get the variance of the sum of $$100$$ of these Random Variables you obtain $$100\cdot25\cdot p(1-p)$$ (using the same rule as before). If you consider the mean and not the sum of these 100 experiments you divide by $$100^2$$, as ( $$Var(X/n)=\frac{1}{n^2}Var(X)$$). Taking the square root of the variance of the mean of these 100 experiments gets you $$\sqrt{\frac{25\cdot p(1-p)}{100}}=\sqrt{\frac{p\cdot(1-p)}{4}}$$, which is the standard error for binomial distribution, i.e. $$[25\cdot \hat{p}-\sqrt{\frac{p(1-p)}{4}}, 25\cdot \hat{p}+\sqrt{\frac{p(1-p)}{4}}]$$ contains the value $$25\cdot p$$ around 68% of the time, where $$\hat{p}=\frac{1}{400}\sum_{i=1}^{2500}x_i$$, i.e. the mean of all the 100 binomial trials (or 2500 Bernoulli trials).

• Thank you for your reply. Everything you've explained is clear, except I don't get the very last part of the 400 bernoulli trials. Do you mean that 100 samples of binomial experiments of 25 trials can be considered 2500 Bernoulli trials? – puckT Sep 28 '18 at 0:12
• Yes, one draw from a $Bi(25,p)$ is equivalent to considering the sum of 25 draws from a Bernoulli. Imagine that you want to estimate the probability of heads of a coin. Does it matter whether you toss the coin 100 times for 25 times or once 2500 times? – Sebastian Sep 28 '18 at 7:02
• Sorry for this error I corrected it – Sebastian Sep 28 '18 at 7:09
• That makes sense. In that case, should $\hat{p} = \frac{1}{2500} \sum_{i=1}^{2500} x_i$? Also, in the confidence interval you showed, there is both $\hat{p}$ and p (in the standard error). Would we also use $\hat{p}$ for p in the case where we are estimating the true p? – puckT Sep 28 '18 at 15:19