Confidence interval

A 95 per cent confident for a population mean looks like.

$\mu ={\bar {x}}\pm 1.96{\frac {\sigma }{\sqrt {n}}}$

In practice we seldom know that $\sigma$ is known. Therefore most of the times an estimate of standard deviation in the sample $s$ (sometimes denoted as Failed to parse (unknown function "\Large"): {\displaystyle \Large\hat\sigma} ) is used:

$\mu ={\bar {x}}\pm z{\frac {s}{\sqrt {n}}}$

If we use: $e=z{\frac {s}{\sqrt {n}}}$

We can make a conclusion on determining the minimum sample size given a requested error level and confidence level:: $n=({\frac {zs}{e}})^{2}$

Adding up confidence intervals for the difference of means is given by: $(\mu _{1}-\mu _{2})=({\bar {x_{1}}}-{\bar {x_{2}}})\pm {\sqrt {({\frac {s_{1}^{2}}{n_{1}^{2}}}+{\frac {s_{2}^{2}}{n_{2}^{2}}})}}$

When the sample size comes closer to the population size, the error will become smaller. Using the finite population correction factor this can be taken into account:

${\sqrt {(1-{\frac {n}{N}})}}$

So corrected with the finite population correction factor a 95 per cent confident for a population mean looks like: $\mu ={\bar {x}}\pm 1.96{\sqrt {(1-{\frac {n}{N}})}}{\frac {s}{\sqrt {n}}}$

The T-distribution can be used instead of the Normal distribution when defining confidence intervals when sample size is small or the standard deviation is unknown.

Part of M7-Quantitative Decision Methods-S3 - Normal Distribution and confidence intervals