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.