Given -

$$\overline{x_1}=67.5\ ,\ \overline{x_2}=68 n{}_{1}=1000 n{}_{2}=2000$$

Step 1 :-

Null Hypothesis (H${}_{0}$):$\mu$${}_{1}$=$\mu$${}_{2}$ Alternative Hypothesis (H${}_{\mathrm{\infty}}$)=$\mu$${}_{1}$$\mathrm{\neq}$$\mu$${}_{2}$ Step 2 :- Test Statistic:- $$\overline{x_1}-\overline{x_2}=67.5-68.0=-0.5$$

Since S.D. of the population is known

$$S.E. S=\sqrt{\frac{{\sigma }^2_1}{n_1}+\frac{{\sigma }^2_2}{n_2}}$$ $$=\sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$ $$=\sqrt{\frac{1}{1000}+\frac{1}{2000}}$$ $$=0.097$$ $$\mathrm{\therefore } z=\frac{\overline{x_1}-\overline{x_2}}{s}=\frac{-0.5}{0.097}=-5.15$$

$$\mathrm{\therefore } |z| = 5.15$$

Step 3 :- Level of significance $\mathrm{\propto }$=0.27%

Step 4 :-

Critical Value

The value of Z${}_{\mathrm{\propto }}$ at 0.27% level of significance from table is 3

Step 5 :- Decision

Since the computed value of |z|=5.15 is greater than critical value z${}_{\mathrm{\propto }}$=3 the hypothesis is rejected.

$\mathrm{\therefore }$ The sample can not be regarded as drawn from same population.