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 2014

$n_1 = 32$ and $n_2 = 36$ (\gt 30 it is large sample) $$\overline x_1=72, \overline x_2=70,S_1=8, S_2=8$$

Step 1:

Null Hypothesis $(H_0): µ_1 = µ_2$ (performance of boys and girls are equal)

Alternative Hypothesis $(H_a): µ_1 = µ_2$ (i.e boys perform better than girls) [o0ne tailed test]

Step 2:

Los = 1% [Two tailed test]

Los = 2% [one tailed test]

Critical value $(z_x)= 2.33$

Step 3:

Since samples are large

$S.E=\sqrt{\dfrac {s_1^2}{n_1}+\dfrac {s_2^2}{n_2}}\\ =\sqrt{\dfrac {8^2}{32}+\dfrac {6^2}{36}} \\ =1.732$

Step 4: Test statistic

$z_{cal}=\dfrac {\overline x_1 -\overline x_2}{S.E}\\ =\dfrac {72-70}{1.732} \\ =1.1547$

Step 5: Decision

Since $|z_{cal}| \lt Z_x, H_0$ is accepted.

Boys do not perform better than the girls.