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

Test statistic, If population S.D is unknown $t=\frac{({\stackrel{\u2015}{x}}_{1}-{\stackrel{\u2015}{x}}_{2})-({\mu}_{1}-{\mu}_{2})}{S\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}\sim t{n}_{1}+{n}_{2}-2$

Under Ho If population S.D is known $Z=\frac{({\stackrel{\u2015}{x}}_{1}-{\stackrel{\u2015}{x}}_{2})-({\mu}_{1}-{\mu}_{2})}{S\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}\sim N(o,1),$

Under Ho where, ${\stackrel{\u2015}{x}}_{1}=$ sample mean from ${1}^{s}t$ group. ${\stackrel{\u2015}{x}}_{2}=$ sample mean from ${2}^{n}d$ group. ${S}_{1}^{2}=$ sample mean from ${1}^{s}t$ group. ${S}_{2}^{2}=$ sample mean from ${2}^{n}d$ group. ${n}_{1}=$ sample mean from ${1}^{s}t$ group. ${n}_{2}=$ sample mean from ${2}^{n}d$ group. and ${S}^{z}=\frac{{n}_{1}{S}_{1}^{2}+{n}_{2}{S}_{2}^{2}}{{n}_{1}+{n}_{2}-2}$ (assuming equal variances).