Sampling Distribution of x-bar
Introduced| English |
|---|
| sample mean/ˈsæmpl miːn/ |
| standard error/ˈstændəd ˈerə/ |
The distribution of x-bar
- A sample mean 样本均值 $\bar{x}$ estimates the true population mean $\mu$.
- Its sampling distribution is centered at $\mu$: $\mu_{\bar{x}} = \mu$ (so $\bar{x}$ is unbiased).
- Each sample gives a different $\bar{x}$; the spread shrinks with sample size.
- This is the workhorse statistic for inference about means.
The standard deviation
- The standard deviation of $\bar{x}$ is:
-
$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$
- The population SD divided by $\sqrt{n}$ — often called the standard error.
- Quadruple $n$ and the spread only halves (because of the square root).
Normality via the CLT
- Is $\bar{x}$ normal? Two ways to justify it:
- If the population is normal, $\bar{x}$ is exactly normal for any $n$.
- Otherwise, the Central Limit Theorem gives approximate normality when $n \ge 30$.
- Always state which justification your problem relies on.
Conditions and probabilities
- Check the 10% condition ($n \le 0.10N$) when sampling without replacement.
- With normality justified, standardize: $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$.
- Then find probabilities for $\bar{x}$ from the standard normal.
- This answers "how likely is a sample mean this far from $\mu$?"
Use $\sigma/\sqrt{n}$, not $\sigma$, for the spread of $\bar{x}$ — the sample mean varies far less than individual data values. A frequent error is dividing by $n$ instead of $\sqrt{n}$. And justify normality explicitly: either a normal population or the CLT with a large $n$ — don't assume $\bar{x}$ is normal for a small sample from a skewed population.
Population mean $\mu = 100$, SD $\sigma = 20$; take $n = 25$.
- Center: $\mu_{\bar{x}} = 100$. SD: $\sigma_{\bar{x}} = \dfrac{20}{\sqrt{25}} = \dfrac{20}{5} = 4$.
- $P(\bar{x} > 108)$: $z = \dfrac{108 - 100}{4} = 2$ → about $2.5\%$.
- So a sample mean above $108$ is fairly unlikely.
The sample mean $\bar{x}$ has $\mu_{\bar{x}} = \mu$ and $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$. Justify a normal model either by a normal population or by the Central Limit Theorem ($n\ge 30$); check the 10% condition when sampling without replacement, then find probabilities via $z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$.
The sampling distribution of x-bar
Centered at μ, spread σ/√n (the standard error).
Population σ = 20, n = 25. Find the SD of x-bar, σ/√n.
20/√25 = 20/5 = 4.
With μ=100 and SD of x-bar = 4, find the z-score of x-bar = 108.
z = (108 − 100)/4 = 2.
The standard deviation of the sample mean is...
Divide σ by √n — not by n.
If the population is already normal, x-bar is normal for any sample size n.
A normal population gives an exactly normal x-bar at any n.
To make the sampling distribution of x-bar have half the spread, you should...
Spread ∝ 1/√n, so 4× n halves the SD.