$$\overline {x_1}=980,\overline {x_2}=1012 : \sigma_1=75,\sigma_2=80,$$

Test the hypothesis that $µ_1 = µ_2$ at $5\%$ level of significance.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 04

Year : DEC 2015

When population standard deviation $\overline {v_1}$ and $\overline {v_2}$ are known we can assume $\overline {x_1}- \overline {x_2}$ be normal with mean zero and $S.E=\sqrt{\dfrac {\overline {v_1}^2}{n_1}}+ \dfrac {\overline {v_2}^2}{n_2}$ and hence use z – distribution.

1) Null hypothesis $H_0 = µ_1 = µ_2$

Alternate hypothesis $µ_1 \neq µ_2$

2) Calculation of test statistic:

$$SE=\sqrt{\dfrac {\sigma_1^2}{n_1}+\dfrac {\sigma_2^2}{n_2}}=\sqrt{\dfrac {75^2}{15}+\dfrac {80^2}8}=\sqrt{375+800}=\sqrt{1175}=34.28$$

$\therefore z=\dfrac {\overline x_1-\overline x_2}{S.E}=\dfrac {980-1012}{34.28}=-0.93\\ \therefore |z|=0.93$

3) Level of significance $\alpha= 0.05$

4) Critical value: The table value of z at $α = 0.05$ is $Z_α = 1.96$

5) Decision: Since the computed value $|z|= 0.93$ is less than the table value $1.96$ the hypothesis is accepted.

The population means are equal $µ_1 = µ_2.$