# Suppose you know σ and you want an 85 percent confidence level. What value would you use to multiply the standard error of the mean by?

Apr 29, 2018

If the sample size is $\le 40$, use t_(alpha//2," " n–1).
If the sample size is $> 40$, use ${z}_{\alpha / 2} \text{ } \approx 1.44$.

(alpha = 100% - 85%=0.15)

#### Explanation:

$t$ and $z$ as mentioned above are multipliers that determine how many standard errors wide our margin of error will be for a desired confidence level.

When estimating the population mean $\mu$ with a small sample size $\left(n \le 40\right) ,$ the statistic $\frac{\overline{X} - \mu}{\sigma}$ is assumed to come from a $t$ distribution with $n - 1$ degrees of freedom (d.f.). This is when we use t_(alpha//2," " n–1).

The family of $t$ distributions looks similar to the standard normal $z$ distribution, but with heavier tails. As the sample size $n$ increases, the t_(n–1) distribution approaches the $z$ distribution.

When the sample size is sufficiently large (i.e. $> 40$), the $t$ distribution (with $n - 1$ d.f.) and the $z$ distribution are approximately equivalent, so it is widely accepted to use the $z$ distribution when $n > 40$.

For an 85% C.I., we have $\alpha = 0.15$. Assuming your sample size is large enough, you would multiply the standard error $\frac{\sigma}{\sqrt{n}}$ for the mean $\mu$ by

${z}_{\alpha / 2} = {z}_{0.15 / 2} = {z}_{0.075} \approx 1.44$

This value of 1.44 can be found by reverse-lookup in a $z$-table, or with Statistics software.