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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)?

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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).

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  • $\begingroup$ 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? $\endgroup$ – puckT Sep 28 '18 at 0:12
  • $\begingroup$ 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? $\endgroup$ – Sebastian Sep 28 '18 at 7:02
  • $\begingroup$ Sorry for this error I corrected it $\endgroup$ – Sebastian Sep 28 '18 at 7:09
  • $\begingroup$ 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? $\endgroup$ – puckT Sep 28 '18 at 15:19

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