GIve a correct answer for this question: when we are comparing two population means from two groups (suppose they are called Group 1 and Group 2), which of the following statistics do we use to make a confidence interval or perform a hypothesis test for a difference in means?

GIve a correct answer for this question: when we are comparing two population means from two groups (suppose they are called Group 1 and Group 2), which of the following statistics do we use to make a confidence interval or perform a hypothesis test for a difference in means?
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Aamina Herring

Let ${\mu }_{1}$ be the population means of group 1. ${\mu }_{2}$ be the population means of group 2. To test

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

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

Under Ho where, ${\stackrel{―}{x}}_{1}=$ sample mean from ${1}^{s}t$ group. ${\stackrel{―}{x}}_{2}=$ 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).

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