A sample of 8 students of 16 years each shown up a mean systolic blood pressure of 118.4 mm of Hg with S.D. of 12.17 mm. While a sample of 10 students of 17 years each showed the mean systolic B.P.of 121.0 mm with S.D. of 12.88 during an investigation. The investigator feels that the systolic B.P. is related to age.Do you think that the data provides enough reasons to support investigators feeling at 5% LOS ?

$\text{We are given n$_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$}$