91Ó°ÊÓ

The basic approach for comparing the means of two populations using independent simple random samples will be discussed.

Let's pretend that $x$is a regularly distributed variable in both groups.

Then, given independent samples of size $n_1$and $n_2$from the two populations, do the following:

${\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}={\mathrm{μ}}_1-{\mathrm{μ}}_2$

${\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{{\left(\frac{{\mathrm{σ}}_1}{n_1}\right)}^2+{\left(\frac{{\mathrm{σ}}_2}{n_2}\right)}^2}$

And $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$is normally distributed

Test statistics:

When the population standard deviations are known, compare two population means.

$z=\frac{\left(\stackrel{¯}{x_1}-\stackrel{¯}{x_2}\right)-\left({\mathrm{μ}}_1-{\mathrm{μ}}_2\right)}{\sqrt{{\left(\frac{{\mathrm{σ}}_1}{n_1}\right)}^2+{\left(\frac{{\mathrm{σ}}_2}{n_2}\right)}^2}}$

When the population standard deviations are unknown and presumed to be identical, compare two population means.

$t=\frac{\left(\stackrel{¯}{x_1}-\stackrel{¯}{x_2}\right)-\left({\mathrm{μ}}_1-{\mathrm{μ}}_2\right)}{s_p\sqrt{\left(\frac{1}{n_1}\right)+\left(\frac{{\mathrm{σ}}_2}{n_2}\right)}}~t_{\left(n_1+n_2-2\right)}$

Where $s_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}}$

When the population standard deviations are unknown and assumed to be unequal, compare two population means.

$t=\frac{\left(\stackrel{¯}{x_1}-\stackrel{¯}{\left.x_2\right)}-\left({\mathrm{μ}}_1-{\mathrm{μ}}_2\right)\right.}{\sqrt{\left(\frac{{s_1}^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)}}~t_{\left(\mathrm{Δ}\right)}$

Where

$\mathrm{Δ}=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{{\left(\frac{s_1^2}{n_1}\right)}^2/\left(n_1-1\right)+{\left(\frac{s_2^2}{n_2}\right)}^2/\left(n_1-1\right)}$