First slide

Introduction to statistics

Question

tf ${\bar{X}}_{1} and {\bar{X}}_{2}$ , are the means of two distributions such that ${\bar{X}}_{1} < {\bar{X}}_{2} and \bar{X}$ is the mean of the combined distribution, then

Moderate

A

$\bar{X} < {\bar{X}}_{1}$
B

$\bar{X} > {\bar{X}}_{2}$
C

$\bar{X} = \frac{{\bar{X}}_{1} + {\bar{X}}_{2}}{2}$
D

${\bar{X}}_{1} < \bar{X} < {\bar{X}}_{2}$

Solution

Let n₁, and n₂ be the number of observations in two groups having means ${\bar{X}}_{1} and {\bar{X}}_{2}$ , respectively. Then,
$\begin{array}{l} \bar{X} = \frac{n_{1} {\bar{X}}_{1} + n_{2} {\bar{X}}_{2}}{n_{1} + n_{2}} \\ Now, \bar{X} - {\bar{X}}_{1} = \frac{n_{1} {\bar{X}}_{1} + n_{2} {\bar{X}}_{2}}{n_{1} + n_{2}} - {\bar{X}}_{1} \\ = \frac{n_{2} ({\bar{X}}_{2} - {\bar{X}}_{1})}{n_{1} + n_{2}} > 0 [∵ {\bar{X}}_{2} > {\bar{X}}_{1}] \end{array}$
${\begin{array}{lr} \Rightarrow \bar{X} > {\bar{X}}_{1} - - - (i) \\ and \bar{X} - {\bar{X}}_{2} = \frac{n_{1} ({\bar{X}}_{1} - {\bar{X}}_{2})}{n_{1} + n_{2}} < 0 \\ \Rightarrow \bar{X} < {\bar{X}}_{2} - - - - (ii) \end{array}}_{}$
From Eqs. (i) and (ii), we get
${\bar{X}}_{1} < \bar{X} < {\bar{X}}_{2}$

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