The statistical evaluation of a sequence of measurements

Checking for a normal distribution ends with a measurement series whose points hug the probability-paper line. So take the random variable $X$ to be normally distributed, with the normal distribution of mean $\mu$ and standard deviation $\sigma$. Neither $\mu$ nor $\sigma$ is known. The empirical mean and empirical standard deviation $\bar{x}$ and $\Delta x$ are only estimates from one series. How far from them can the true values lie?

Pay for an unknown σ with t

Proceed in two steps.

1. Measure the quantity $X$ a total of $n$ times, giving $x_1,\dots ,x_n$.
2. Choose a small number $\alpha$ as the probability of being wrong, and look up $t_{\alpha ,m}$ in a table of Student's $t$-distribution with $m=n-1$ degrees of freedom.

Then the unknown mean satisfies

$$\bar{x}-t_{\alpha ,m}\frac{\Delta x}{\sqrt{n}}\le \mu \le \bar{x}+t_{\alpha ,m}\frac{\Delta x}{\sqrt{n}},$$

and this statement is wrong with probability $\alpha$.

Compare this with the interval for a known $\sigma$, which uses $z_{\alpha }$. Here $\sigma$ is unknown, so its estimate $\Delta x$ takes its place. The factor $t_{\alpha ,m}$ is larger than $z_{\alpha }$ to pay for that extra uncertainty, and the whole half-width shrinks like $1/\sqrt{n}$.

Example 1. The heights in the empirical measurement series give $n=8$, $\bar{x}=175$ and $\Delta x=4.8$. Choose $\alpha =0.01$. For $m=7$ the $t$-table gives $t_{\alpha ,m}=3.5$. Assuming height is normally distributed, with error probability $\alpha =0.01$,

$$169\le \mu \le 181.$$

Drag $n$ and $\Delta x$. The interval always stays centered on $\bar{x}$. Adding measurements narrows it twice over, because $t_{\alpha ,m}$ falls as $m$ grows and $\sqrt{n}$ grows in the denominator.

The χ² interval is lopsided

With error probability $\alpha$, the standard deviation satisfies

$$\frac{(n-1)(\Delta x{)}^2}{b}\le {\sigma }^2\le \frac{(n-1)(\Delta x{)}^2}{a}.$$

The numbers $a:={\chi }_{1-\alpha /2}^2$ and $b:={\chi }_{\alpha /2}^2$ are read from a table of the ${\chi }^2$-distribution with $m=n-1$ degrees of freedom.

Example 2. Take the same eight heights again. For $m=7$, the ${\chi }^2$-table gives $a=1.24$ and $b=20.3$. Taking square roots gives

$$2.8\le \sigma \le \mathrm{11.4.}$$

Drag $\Delta x$. Both ends scale with it, but they do not sit evenly around it: $\Delta x=4.8$ is about $2$ above the left end and more than $6$ below the right end. The ${\chi }^2$-values $a$ and $b$ are not mirror images of each other, so this interval is lopsided, unlike the one for $\mu$.

Eight measurements leave a wide band

Both estimates are rough: the mean is pinned down only to a band about $12$ units wide, and $\sigma$ only to a factor of four. Eight measurements are very few, so that width is expected. The interval for $\mu$ stays symmetric about $\bar{x}$; the interval for $\sigma$ does not.