Suppose that a random sample of $n$ observations is taken from some population with mean $\mu_X$ and variance $\sigma_X^2$. Each random variable $X_1,X_2,…,X_n$ of the random sample will have the same distribution (iid). If such is the case, then the sample mean, $\bar{X}$, will have mean and variance as given below:

$$\begin{align*} \mu_\bar{X}&=\mathbb{E}(\bar{X})=\mu\\ \sigma_\bar{X}^2&=\mathbb{V}(\bar{X})=\frac{\sigma_X^2}{n} \end{align*} $$

Note the following points:

these results hold true regardless of the underlying probability distribution of $X$.
only the expected value and variance of the sample mean $\bar{X}$ can be calculated - not the distribution of $\bar{X}$.

The expected value of the sample mean is:

$$\begin{align*} \mathbb{E}(\bar{X})&=\mathbb{E} \Big(\frac{1}{n}\sum^{n}_{i=1}X_i\Big)\\ &=\frac{1}{n}\sum^{n}_{i=1}\mathbb{E}(X_i)\\ &=\frac{1}{n}\sum^{n}_{i=1}\mu_X\\ &=\frac{1}{n}(n\mu_X)\\ &=\mu_X\\ \end{align*}$$

The variance of the sample mean is:

$$\begin{align*} \mathbb{V}(\bar{X})&=\mathbb{V}\Big(\frac{1}{n}\sum^{n}_{i=1}X_i\Big)\\ &=\frac{1}{n^2}\sum^{n}_{i=1}\mathbb{V}(X_i)\\ &=\frac{1}{n^2}\sum^{n}_{i=1}\sigma^2_X\\ &=\frac{1}{n^2}(n\sigma^2_X)\\ &=\frac{\sigma^2_X}{n}\ \end{align*}$$

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Definition.

Standard error

The standard error is defined as the standard deviation of an estimator or statistic.

Example.

Standard error of the sample mean

What is the standard error of the sample mean?

When we try to estimate the population mean using the sample mean, we know that the sample mean won't exactly equal the population mean. In order to know how good the estimate is, we use a statistic called the standard error of the sample mean.

A high value for the standard error means that there is high uncertainty in our estimate, whereas a low standard error means that we can be certain our estimate isn't off by much.

Recall from theorem 5 that the variance of the sample mean $\bar{X}$ is:

$$\begin{align*} \sigma_\bar{X}^2&=\mathbb{V}(\bar{X})=\frac{\sigma_X^2}{n} \end{align*} $$

The standard error of the sample mean $\bar{X}$ is the standard deviation of $\bar{X}$:

$$\begin{align*} \sigma_\bar{X}&=\sqrt{\mathbb{V}(\bar{X})}=\frac{\sigma_X}{\sqrt{n}} \end{align*} $$

As we can see, the standard error decreases as the sample size $n$ increases. This should be intuitive because the more data we use to compute the sample mean, the more precise the sample mean is in estimating the population mean. As the sample size tends to infinity, the standard error approaches zero.

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Published by Isshin Inada

Edited by 0 others

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