If bar{x}_1 and bar{x}_2 are the means of two | vedclass

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(D) 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,the combined mean is g

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$\bar{x}_1 < \bar{x} < \bar{x}_2$

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(D) 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,the combined mean is given by $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$. Now,consider $\bar{x} - \bar{x}_1 = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} - \bar{x}_1 = \frac{n_2(\bar{x}_2 - \bar{x}_1)}{n_1 + n_2}$. Since $\bar{x}_2 > \bar{x}_1$ and $n_1, n_2 > 0$,we have $\bar{x} - \bar{x}_1 > 0$,which implies $\bar{x} > \bar{x}_1$ $(i)$. Similarly,$\bar{x} - \bar{x}_2 = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} - \bar{x}_2 = \frac{n_1(\bar{x}_1 - \bar{x}_2)}{n_1 + n_2}$. Since $\bar{x}_1 Combining $(i)$ and (ii),we get $\bar{x}_1 < \bar{x} < \bar{x}_2$.

If $\bar{x}_1$ 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

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