$\text{We are given n$_1$ = 8, n$_2$=10 ; $\bar{X_1}$ =118.4 , $\bar{X_2}$ =121.0 ; s$_1$=12.17 , s$_2$ =12.88.}$
$\text{(i) The null hypothesis H$_0$ : μ$_1$= μ$_2$. Alternative hypothesis H$_1$: μ$_1$≠ μ$_2$}$
$\text{(ii) Calculation of test statistic: Since the sample size is small , we use t – distribution}$
$\text{S$_p$ =$\sqrt{\frac{n_1 (s_1)^2+n_2 (s_2)^2}{(n_1+n_2-2)}}$ = $\sqrt{\frac{(8(12.17)^2+10(12.88)^2)}{(8+10-2)}}$ = 13.33}$
$\text{S.E.= S$_p$ $\sqrt{ \frac{1}{n_1} + \frac{1}{n_2 }}$ = 13.33 $\sqrt{ \frac{1}{8}+\frac{1}{10}}$ =6.32}$
$\text{t = $\frac{ \bar{X_1}- \bar{X_2} }{S.E.}$ = $\frac{118.4-121.0}{6.32}$ = - 0.41}$
$\text{(iii) Level of significance : α =0.05}$
$\text{(iv) Critical value : the value of t_α at 5% level of significance for ν= 8+10 -2 =16 degrees of freedom is 2.12}$
$\text{(v) Decision : since the calculated value of |t| =0.41 is less than the table value t$_α$=2.12 , the null hypothesis is accepted i.e. μ$_1$= μ$_2$}$