Test at 1% level of significance whether the boys perform better than the girls.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2015

1) Null Hypothesis $H_0=µ_1= µ_2$

Alternative hypothesis is $H_a : µ_1 \neq µ_2$

2) Calculation of statistic $:\overline x_1-\overline x_2 =72-70=2 \\ S.E. =\sqrt{\dfrac {S_1^2}{n_1}+\dfrac {S_2^2}{n_2}} =\sqrt{\dfrac {64}{32} +\sqrt{36}{36}} =\sqrt 3\\ z= \dfrac {\overline x_1-\overline x_2}{S.E}=\dfrac 2{\sqrt 3}= 1.15\\ \therefore |z|=1.15$

3) Level of significance $α=1\%$

4) Critical value: The value of $Z α$ at $1\%$ level of significance from the table is $2.58$

5) Decision: Since the computed value of $|z|=1.15$ is less than the critical value Z $α=2.56$ the hypothesis is accepted.

$\therefore $ Boys do not perform better than girls