# Suppose you take independent random samples from populations with means mu1 and mu2 and standard deviations sigma1 and sigma2. Furthermore, assume eit

Suppose you take independent random samples from populations with means $\mu 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mu 2$ and standard deviations $\sigma 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sigma 2$. Furthermore, assume either that (i) both populations have normal distributions, or (ii) the sample sizes () are large. If are the random sample means, then how does the quantity
$\frac{\left(\stackrel{―}{{x}_{1}}-\stackrel{―}{{x}_{2}}\right)-\left({\mu }_{1}-{\mu }_{2}\right)}{\sqrt{\frac{{\sigma }_{1}^{2}}{{n}_{1}}+\frac{{\sigma }_{2}^{2}}{{n}_{2}}}}$
Give the name of the distribution and any parameters needed to describe it.

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Step 1
Given:
Two independent random samples from populations with
means-${\mu }_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mu }_{2}$
standard deviations - ${\sigma }_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\sigma }_{2}$
Assumptions:
i.Both populations have normal distributions
ii.The sample sizes(${n}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{2}$) are large
${X}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{X}_{2}$ are random samples then the quantity
$t\left(say\right)=\frac{\left(\stackrel{―}{{x}_{1}}-\stackrel{―}{{x}_{2}}\right)-\left({\mu }_{1}-{\mu }_{2}\right)}{\sqrt{\frac{{\sigma }_{1}^{2}}{{n}_{1}}+\frac{{\sigma }_{2}^{2}}{{n}_{2}}}}$
this quantity $t\left(say\right)\sim N\left(0,1\right)⇒$ quantity t has standard normal distribution with mean zero and variance 1 the parameters are ${\mu }_{1}-{\mu }_{2}$ is the difference between the population means and ${\sigma }_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\sigma }_{2}$ are the population standard deviations.