GIve a correct answer for this question: when we are comparing two population means from...

Let ${\displaystyle {\mu }_1}$ be the population means of group 1. ${\displaystyle {\mu }_2}$ be the population means of group 2. To test ${\displaystyle H_0:{\mu }_1={\mu }_{2}i.e.({\mu }_1-{\mu }_2)=0vs{H}_1:{\mu }_1\ne q{\mu }_{2}i.e.({\mu }_1-{\mu }_2)\ne q0}$

Test statistic, If population S.D is unknown ${\displaystyle t=\frac{({\bar{x}}_1-{\bar{x}}_2)-({\mu }_1-{\mu }_2)}{S\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\sim t{n}_1+n_2-2}$

Under Ho If population S.D is known ${\displaystyle Z=\frac{({\bar{x}}_1-{\bar{x}}_2)-({\mu }_1-{\mu }_2)}{S\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\sim N(o,1),}$

Under Ho where, ${\displaystyle {\bar{x}}_1=}$ sample mean from ${\displaystyle 1^{s}t}$ group. ${\displaystyle {\bar{x}}_2=}$ sample mean from ${\displaystyle 2^{n}d}$ group. ${\displaystyle S_1^2=}$ sample mean from ${\displaystyle 1^{s}t}$ group. ${\displaystyle S_2^2=}$ sample mean from ${\displaystyle 2^{n}d}$ group. ${\displaystyle n_1=}$ sample mean from ${\displaystyle 1^{s}t}$ group. ${\displaystyle n_2=}$ sample mean from ${\displaystyle 2^{n}d}$ group. and $S^z=\frac{n_1{S}_1^2+n_2{S}_2^2}{n_1+n_2-2}$ (assuming equal variances).