# Confidence Interval on the Difference in Means, Variances Unknown and not Assumed Equal

How can I prove that $$T=\dfrac{\overline{X}_1-\overline{X}_2-(\mu_1-\mu_2)}{\sqrt{\dfrac{S_1^2}{n_1}+\dfrac{S_2^2}{n_2}}}$$ has a $t$ distribution?

Assumptions:

1. $X_{11},\ldots,X_{1n_{1}}$ is a random sample from population 1.
2. $X_{21},\ldots,X_{2n_{2}}$ is a random sample from population 2.
3. The two populations represented by $X1$ and $X2$ are independent.
4. $\sigma_{1}$, $\sigma_{2}$ unknown, and $\sigma_1 \neq \sigma_2$.
5. Both populations are normal.

What I know:

• $Z=\dfrac{\overline{X}_1-\overline{X}_2-(\mu_1-\mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}} \sim N(0,1)$.
• $U=\dfrac{(n_1-1)S_1^2}{\sigma_1^2}+ \dfrac{(n_2-1)S_2^2}{\sigma_2^2} \sim \chi^{2}_{n_1+n_2-2}$
• If $Z \sim N(0,1)$, $U \sim \chi_{n_1+n_2-2}^{2}$ and $Z$ and $U$ are independent, then

$$\dfrac{Z}{\sqrt{\dfrac{U}{n_1+n_2-2}}} \sim t_{n_1+n_2-2}$$

My attempt: I have formed the quotient

$$\dfrac{\dfrac{\overline{X_{1}}-\overline{X_{2}}-(\mu_{1}-\mu_{2})}{\sqrt{\dfrac{\sigma_{1}^{2}}{n_{1}}+\dfrac{\sigma_{2}^{2}}{n_{2}}}}}{\sqrt{\left[\dfrac{(n_{1}-1)S_{1}^{2}}{\sigma_{1}^{2}}+ \dfrac{(n_{2}-1)S_{1}^{2}}{\sigma_{2}^{2}}\right]\dfrac{1}{n_{1}+n_{2}-2}}} \sim t_{n_1+n_2-2}$$

The problem with which I found is that unknown variances can not be eliminated as in the case $\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}$

• Where you wrote $X_{11},\ldots,X_{1n}$ and $X_{21},\ldots,X_{2n},$ did you intend $X_{11},\ldots,X_{1n_1}$ and $X_{21},\ldots,X_{2n_2} \text{ ?} \qquad$ – Michael Hardy May 25 '18 at 19:13
• @MichaelHardy I already corrected $n_1$ and $n_2$ – Jacob S. May 25 '18 at 19:45

• 5.4.21 Discuss the problem of finding a confidence interval for the difference $\mu_{1}-\mu_{2}$ between the two means of two normal distributions if the variances $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$ are known but not necessarily equal. – Jacob S. May 25 '18 at 19:56