# How can I use confidence intervals for the population mean µ?

Aug 4, 2015

$m \pm t s$

Where $t$ is the $t$-score associated with the confidence interval you require.
[ If your sample size is greater than 30 then the limits are given by
$\mu$ = $\overline{x} \pm \left(z \times S E\right)$]

#### Explanation:

Calculate the sample mean ($m$) and sample population ($s$) using the standard formulas.

$m = \frac{1}{N} \sum \left({x}_{n}\right)$

s=sqrt(1/(N-1)sum(x_n-m)^2

If you assume a normally distributed population of i.i.d. (independent identically distributed variables with finite variance) with sufficient number for the central limit theorem to apply (say $N > 35$) then this mean will be distributed as a $t$-distribution with $\mathrm{df} = N - 1$.

The confidence interval is then:

$m \pm t s$

Where $t$ is the $t$-score associated with the confidence interval you require.

If you know the population standard deviation and do not need to estimate it ($\sigma$), then replace $s$ with $\sigma$ and use a Z score from the normal distribution rather than a $t$-score since your estimate will be normally distributed rather than $t$ distributed (using the above assumptions about the data).

[$\overline{x}$ = sample Mean
z = critical value
SE is standard Error
SE = $\frac{\sigma}{\sqrt{n}}$ Where n is sample size.

Upper limit of the population --$\mu$ = $\overline{x} + \left(z \times S E\right)$
Lower limit of the population - $\mu$ = $\overline{x} - \left(z \times S E\right)$

If your sample size is less than 30 use the 't' value]