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 \[\bar{X}=\frac{{{n}_{1}}{{{\bar{X}}}_{1}}+{{n}_{2}}{{{\bar{X}}}_{2}}}{{{n}_{1}}+{{n}_{2}}}-{{\bar{X}}_{2}}\] Now, \[\bar{X}-{{\bar{X}}_{1}}=\frac{{{n}_{1}}{{{\bar{X}}}_{1}}+{{n}_{2}}{{{\bar{X}}}_{2}}}{{{n}_{1}}+{{n}_{2}}}-{{\bar{X}}_{1}}\] \[=\frac{{{n}_{2}}({{{\bar{X}}}_{2}}-{{{\bar{X}}}_{1}})}{{{n}_{1}}+{{n}_{2}}}>0\] \[[\because \,\,{{\bar{X}}_{2}}>{{\bar{X}}_{1}}]\] \[\Rightarrow \] \[\bar{X}>{{\bar{X}}_{1}}\] ? (i) And \[\bar{X}-{{\bar{X}}_{2}}=\frac{n({{{\bar{X}}}_{1}}-{{{\bar{X}}}_{2}})}{{{n}_{1}}+{{n}_{2}}}<0\] \[[\because \,\,{{\overline{X}}_{2}}>{{\overline{X}}_{1}}]\] \[\Rightarrow \] \[\bar{X}<{{\bar{X}}_{2}}\] ... (ii) From Eqs. (i) and (ii),\[{{\bar{X}}_{1}}<\bar{X}<{{\bar{X}}_{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
