Question

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

Short answer

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

2)$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{s_{\mathrm{p}}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}~t_{\left(n_1+n_2-2\right)}$

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

Step-by-step solution

Step 1:Given information

Step 2:Concepts required 

The basic strategy for comparing the means of two populations based on independent simple random samples is given in the following steps:

1. Independently and randomly take a sample from Population 1 and a sample from Population 2

2. Compute the mean $\overline{x_1}$of the sample from Population 1 and the mean $\overline{x_2}$of the sample from Population 2

3. Reject the null hypothesis if the sample means$\overline{x_1}$ and $\overline{x_2}$differ by too much; otherwise, do not reject the null hypothesis.

Step 3:ExplaInation

Suppose that x is a normally distributed variable on each of two populations. Then, for independent samples of sizes $n_1$and $n_2$from the two populations,

$\text{-}{\mu }_{{\overline{x}}_1-{\overline{x}}_2}={\mu }_1-{\mu }_2$

width="219" height="37" role="math">σx¯1-x¯2=σ12/n1+σ22/n2,and

${\overline{x}}_1-{\overline{x}}_2\text{is normally distributed.}$

Test Statistic:

i. For comparing two population means when the population standard deviations are known.

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

ii. For comparing two population means when the population standard deviations are unknown and assumed to be equal.

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{s_{\mathrm{p}}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}~t_{\left(n_1+n_2-2\right)}$

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

iii. For comparing two population means when the population standard deviations are unknown and assumed to be unequal.

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

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