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The average of marks scored by 32 boys is 72 with standard deviation 8 while that of 36 girls is70 with standard deviation 6.Test at 1% level of significance whether the boys perform better than girls
$\text{(i) The null hypothesis H$_0$: μ$_1$= μ$_2$. Alternative hypothesis H$_1$:μ$_1$≠ μ$_2$}$ (ii) Calculation of test statistic: S.E = $\sqrt{\frac{(s_1)^2}{n_1} +\frac{(s_2)^2}{n_2}}$ = $\sqrt{\frac{64}{32}+\frac{36}{36}}$ = 1.732; Z = $\frac{(X_1-X_2)}{(S.E)}$ = $\frac{(72-70)}{1.732}$ = 1.15 $\text{(iii) Level of significance : α =0.01}$ $\text{(iv) Critical value : the value of Z$_α$at 1% level of significance = 2.58}$ $\text{(v) Decision : since the calculated value of |Z| =1.15 is less than the table value Z$_α$=1.15. Therefore, the null hypothesis is accepted.}$