A) \[\bar{x}<{{\bar{x}}_{1}}\]
B) \[\bar{x}>{{\bar{x}}_{2}}\]
C) \[\bar{X}=\frac{{{{\bar{X}}}_{1}}+{{{\bar{X}}}_{2}}}{2}\]
D) \[{{\bar{x}}_{1}}<\bar{x}<{{\bar{x}}_{2}}\]
Correct Answer: D
Solution :
Let \[{{n}_{1}}\]and \[{{n}_{2}}\] be the number of observations in two groups having means \[{{\bar{x}}_{1}}\]and \[{{\bar{x}}_{2}}\] respectively. Then, \[\bar{x}=\frac{{{n}_{1}}{{{\bar{x}}}_{1}}+{{n}_{2}}{{{\bar{x}}}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] Now, \[\bar{x}-{{\bar{x}}_{1}}=\frac{{{n}_{1}}{{{\bar{x}}}_{1}}+{{n}_{2}}{{{\bar{x}}}_{2}}}{{{n}_{1}}+{{n}_{2}}}-{{\bar{x}}_{1}}\] \[=\frac{{{n}_{2}}({{{\bar{x}}}_{2}}-{{{\bar{x}}}_{1}})}{{{n}_{1}}+{{n}_{2}}}>0,\,\,\,\,\,[\because {{\bar{x}}_{2}}>{{\bar{x}}_{1}}]\] Þ \[\bar{x}\,\,\,>\,\,\,{{\bar{x}}_{1}}\] .....(i) and \[\bar{x}-{{\bar{x}}_{2}}=\frac{n({{{\bar{x}}}_{1}}-{{{\bar{x}}}_{2}})}{{{n}_{1}}+{{n}_{2}}}<0\], \[\left[ \because {{{\bar{x}}}_{2}}\,\,>\,{{{\bar{x}}}_{1}} \right]\] Þ \[\bar{x}\,\,<\,\,\,{{\bar{x}}_{2}}\] ......(ii) From (i) and (ii), \[{{\bar{x}}_{1}}<\bar{x}\,<\,\,{{\bar{x}}_{2}}\].
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