Why is the formula for standard error of the sample mean and standard error of the sample proportion different?

Trying to understand this intuitively and haven't been able to find any explanations for this.

If the formula for the standard error of the sample proportion is: $$\sigma_\hat p = \frac{\sigma_X}{n}=\frac{\sqrt{n\!\cdot\!p(1-p)}}{n}=\sqrt{\frac{p(1-p)}{n}}$$

where $$X$$ is the sum of the results of $$n$$ independent trials of picking a $$1$$ or an $$0$$,

why is the formula for the standard error of the sample mean,

$$\sigma_\bar x=\frac{\sigma}{\sqrt{n}}$$

?

In both formulas, the sample size, $$n$$, is deflating the standard error, but why is one deflating by $$n$$ and the other one by $$\sqrt{n}$$ ?

• You are not comparing the same things. Consider a dataset consisting only of zeros and ones and apply both formulas: you will see they are equivalent.
– whuber
Commented Mar 30, 2020 at 17:03

You have various different standard deviations and standard errors there. Trying to unpick them, for given $$p$$:
• An individual trial has variance $$p(1-p)$$ and standard deviation $$\sqrt{p(1-p)}$$
• The sum of $$n$$ independent trials has variance $$np(1-p)$$ and standard deviation $$\sqrt{np(1-p)}$$
• The mean of $$n$$ independent trials has variance $$\frac1n p(1-p)$$ and standard deviation $$\sqrt{\frac1n p(1-p)}$$; this last expression is sometimes called the standard error of the sample mean or of the sample proportion
So the standard deviation of the mean is indeed $$\frac1n$$ of the standard deviation of the sum, and is $$\sqrt{\frac1n}$$ of the standard deviation of an individual trial, as you say in your question. You are comparing different things, as whuber has commented.