A) \( \Large \overline{x}\ <\ \overline{x}_{1} \) |
B) \( \Large \overline{x}\ >\ \overline{x}_{2} \) |
C) \( \Large \overline{x} = \frac{\overline{x}_{1} \times \overline{x}_{2}}{2} \) |
D) \( \Large \overline{x}_{1}\ <\ \overline{x}\ <\ \overline{x}_{2} \) |
D) \( \Large \overline{x}_{1}\ <\ \overline{x}\ <\ \overline{x}_{2} \) |
Let \( \Large n_{1}\ and\ n_{2} \) be the number of observations in two groups having means \( \Large \overline{x}_{1}\ and\ \overline{x}_{2} \), respectively.
then, \( \Large \overline{x}=\frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}} \)
Now, \( \Large \overline{x}-\overline{x}_{1} = \frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}-\overline{x}_{1}=n_{2}\frac{ \left(\overline{x}_{2}+\overline{x}_{1}\right) }{n_{1}+n_{2}} > 0 \)
\( \Large \left(\because\ \overline{x}_{2} > \overline{x}_{1}\right) \) .. (i)
=> \( \Large \overline{x} > \overline{x}_{1} \)
and \( \Large \overline{x} > \overline{x}_{2}=\frac{n_{1} \left(\overline{x}_{1}-\overline{x}_{2}\right) }{n_{1}+n_{2}} \)
\( \Large \left(\because\ \overline{x}_{2} > \overline{x}_{1}\right) \) .. (ii)
From relation (i) and (ii), we get
\( \Large \overline{x}_{1} < \overline{x} < \overline{x}_{2} \)