1
$\begingroup$

I'm reading Hypothesis Testing: The Basics, there is such an experiment:

So, we have a coin. Our null hypothesis is that this coin is fair. We flip it 100 times and it comes up heads 51 times. Do we know whether the coin is biased or not?

The author mentioned Central Limit Theorem and said the random variable is the proportion of heads in our sample of 100 coin flips. In our case, it is equal to 0.51.

But by the central limit theorem we also know that p approximates a normal distribution. This means we can estimate the standard deviation of p as $$\sigma=\sqrt{\frac{p(1-p)}{N}}$$

Why the standard deviation is calculated using this equation? According to Wikipedia, the std of a binomial distribution is $\sqrt{np(1-p)}$. What am I missing?

$\endgroup$
  • $\begingroup$ The binomial is a count of successes; you're not looking at the count. $\endgroup$ – Glen_b -Reinstate Monica Jan 18 '19 at 9:47
1
$\begingroup$

If numerous random samples or repetitions of the same size $n$ are taken, the distribution of possible values of $\hat{p}$ is approximately a normal curve distribution with mean $p$ and $SD=\sqrt{\frac{p(1-p)}{N}}$. Have a look here. And, for the distinction from binomial distribution please have a look at this thread.

EDIT: To the explanation asked in the comment. $\hat{p} = X/n$, where $X$ is a binomial distribution with parameters $n$ and $p$. $n$ is the number of trials and $p$ is the probability of success. We are interested in the fraction of successes.

$E(\hat{p}) = E(X/n) = np/n = p$ and $Var(\hat{p}) = Var(X/n) = Var(X)/n^2 = np(1-p)/n^2 = p(1-p)/n$. So, $SD=\sqrt{\frac{p(1-p)}{N}}$

$\endgroup$
  • $\begingroup$ Thank you for the answer. Could you please explain a bit more on why there is $\frac{1}{N}$ in the SD calculation? $\endgroup$ – Tyler傲来国主 Jan 21 '19 at 5:38
  • $\begingroup$ Edited the answer to answer this query. Thank you! $\endgroup$ – naive Jan 21 '19 at 5:53
1
$\begingroup$

The variance for the Binomial distribution is in fact equal to $np(1-p)$, however you are estimating the variance for the proportion $p$, which is a Bernoulli trial, which has variance $p(1-p)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.