Means of two random samples of size 9 & 7 are 196.42 & 198.82 respectively. Sum of squares of deviation from means is 26.94 & 18.73. Can sample be considered to have been drawn from same population?

4.1kviews

written 8.7 years ago by

teamques10 ★ 69k

• modified 8.7 years ago

n ${}_{1}$ =9 \& n ${}_{2}$ =7( $\lt$ 30 , so it is small sample)

$\overline{x_1}=196.42\ ;\ \overline{x_2}=198.82\ ;\ \sum{{(x_{1i}-\overline{x_2})}^2=26.94\ ;\ {(x_{2i}-\overline{x_2})}^2=18.73}$

Step 1 :-

Null Hypothesis (H ${}_{0}$ ): $\mu$ ${}_{1}$ = $\mu$ ${}_{2}$ (i.e. Samples are drawn from the same population) Alternative Hypothesis (H ${}_{\mathrm{\infty}}$ )= $\mu$ ${}_{1}$ $\mathrm{\neq}$ $\mu$ ${}_{2}$ (i.e. Samples are not drawn from the same population) (Two tailed test) Step 2 :- LOS=5% (Two tailed test) Degree of freedom = n ${}_{1}$ +n ${}_{2}$ -2=9+7-2=14 $\mathrm{\therefore }$ Critical value (t ${}_{\mathrm{\infty}}$ )=2.145 Step 3:-

$Since\ Sample\ is\ small,\ sp=\sqrt{\frac{\sum{{\left(x_1-\overline{x_1}\right)}^2+\sum{{\left(x_{2t}-\overline{x_2}\right)}^2}}}{n_1+n_2-2}}=\sqrt{\frac{26.94+18.73}{9+7-2}}$

$=1.8061$

$S.E.=sp\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

$\mathrm{S.E.=1.8061\ }\sqrt{\frac{1}{9}+\frac{1}{7}}$

$=0.9102$

Step 4 :- Test Statistic

$t_{cal} =\frac{\overline{x_{1} }-\overline{x_{2} }}{S.E.} =\frac{196.42-198.82}{0.9102} =-2.6368$

Step 5 :- Decision

Since |t ${}_{cal}$ | $\gt$ t ${}_{\mathrm{\infty}}$ , H ${}_{0}$ is rejected $\mathrm{\therefore }$ The samples cannot be considered to have to have been drawn from the same population.

ADD COMMENT EDIT