This is a question I am struggling with. I would appreciate it if I can get some help on this proof and an outline what you would normally do if variance is not available.

image problem

  • 2
    $\begingroup$ You might be overthinking this, or perhaps you are not distinguishing the population variance from the variance of a sample. Usually we don't ever know $\sigma$. But when you have a sample, you can estimate $\sigma$. $\endgroup$ – whuber Apr 5 '17 at 22:12
  • $\begingroup$ this needs self-study tag. Please provide it! $\endgroup$ – kjetil b halvorsen Apr 15 '17 at 22:11

First part

The standard error of $\overline X$ is another word for the standard deviation of $\overline X$. Now notice how

$$\text{Var}(\overline X) = \text{Var}\left( \frac{1}{n} \sum_{i=1}^n X_i\right) = \frac{1}{n^2} \sum_i \text{Var}(X_i) = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n}$$

Then $\text{SE}(\overline X) = \sqrt{\text{Var}(\overline X)} = \dfrac{\sigma}{\sqrt{n}}$.

Second part

If the variance is unkown one would estimate $\sigma^2$ with the sample variance $$s^2 = \dfrac{1}{n-1}\sum_{i=1}^n(X_i-\overline X)^2$$ Dividing by $\frac{1}{n-1}$ avoides bias.


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