A) \[\overline{X}<{{\overline{X}}_{1}}\]
B) \[\overline{X}>{{\overline{X}}_{2}}\]
C) \[\overline{X}=\frac{{{\overline{X}}_{1}}+{{\overline{X}}_{2}}}{2}\]
D) \[{{\overline{X}}_{1}}<\overline{X}<{{\overline{X}}_{2}}\]
Correct Answer: D
Solution :
Let \[{{n}_{1}}\] and \[{{n}_{2}}\] be the number of observations, in two groups having means \[{{\overline{X}}_{1}}\] and \[{{\overline{X}}_{2}}\] respectivley. Then, \[\overline{X}=\frac{{{n}_{1}}{{\overline{X}}_{1}}+{{n}_{2}}{{\overline{X}}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] Now, \[\overline{X}=\overline{{{X}_{1}}}=\frac{{{n}_{1}}\overline{{{X}_{1}}}+{{n}_{2}}\overline{{{X}_{2}}}}{{{n}_{1}}+{{n}_{2}}}-\overline{{{X}_{1}}}\] \[=\frac{{{n}_{2}}({{\overline{X}}_{2}}-{{\overline{X}}_{1}})}{{{n}_{1}}+{{n}_{2}}}>0\] \[\Rightarrow \] \[\overline{X}>{{\overline{X}}_{1}}\] ?(i) and \[\overline{X}-{{X}_{2}}\,=\frac{n({{\overline{X}}_{1}}-{{\overline{X}}_{2}})}{{{n}_{1}}+{{n}_{2}}}>0\] \[\Rightarrow \] \[\overline{X}<{{\overline{X}}_{2}}\] \[\therefore \] From Eqs. (i) and (ii), we get\[{{\overline{X}}_{1}}<\overline{X}<{{\overline{X}}_{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
