$\text{(i) The null hypothesis H$_0$ : μ$_1$= μ$_2$.
Alternative hypothesis H$_a$ : μ$_1$≠ μ$_2$.} \ $
$\text{(ii) Calculation of test statistic: Since the sample size is small , we use t – distribution}$
$\text{S$_p$ = $\sqrt{ \frac{∑(X_i- \bar{X})^2 +∑(Y_i- \bar{Y})^2 }{(n_1+n_2-2)}}$ = $\sqrt{ \frac{(26.94+18.73)}{(9+7-2)}}$ = 1.81}$
$\text{S.E.= S$_p$ $\sqrt{(\frac{1}{n_1} + \frac{1}{n_2 })}$ = 1.81 $\sqrt{( \frac{1}{9}+ \frac{1}{7})}$ = 0.91 }$
$\text{t = $\frac{( \bar{X_1} - \bar{X_2})}{(S.E.)}$ = $\frac{(196.42-198.82)}{(0.91)}$ = - 2.64}$
$\text{(iii) Level of significance : α =0.05}$
$\text{(iv) Critical value : the value of t$_α$ at 5% level of significance for ν= 9+7 - 2 =14 degrees of freedom is 2.145}$
$\text{(v) Decision : since the calculated value of |t| =2.64 is greater than the table value t$_α$=2.145, the null hypothesis is rejected.}$
$\text{∴ The samples cannot be considered to have been drawn from the same population.}$