Discuss the basic strategy for comparing the means of two populations based on independent simple random samples.

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:

${\mu }_{\overline{x_1}-\overline{x_2}}={\mu }_1-{\mu }_2$

${\sigma }_{\overline{x_1}-\overline{x_2}}=\sqrt{{\left(\frac{{\sigma }_1}{n_1}\right)}^2+{\left(\frac{{\sigma }_2}{n_2}\right)}^2}$

And $\overline{x_1}-\overline{x_2}$is normally distributed

Test statistics:

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

$z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{{\left(\frac{{\sigma }_1}{n_1}\right)}^2+{\left(\frac{{\sigma }_2}{n_2}\right)}^2}}$

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

$t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{s_p\sqrt{\left(\frac{1}{n_1}\right)+\left(\frac{{\sigma }_2}{n_2}\right)}}~t_{\left(n_1+n_2-2\right)}$

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.

$t=\frac{\left(\overline{x_1}-\overline{\left.x_2\right)}-\left({\mu }_1-{\mu }_2\right)\right.}{\sqrt{\left(\frac{{s_1}^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)}}~t_{\left(\Delta \right)}$

Where

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