I've been asked to analyze some data but I'm quite new in statistic. I want to compare the means of a continuous parameter in two independent population A and B: A can be seen as the treatment group whereas B can be seen as the control group. I have two samples, for population A and for population B, whose dimensions are 79 and 15 respectively. I would like to do a hypothesis test with null hypothesis $H_0: \mu_A=\mu_B$ (I haven't seen the data yet, but I expect to reject $H_0$). Obviously I will use the sample mean $\overline{X}_A$ as estimator for $\mu_A$ and $\overline{X}_B$ as estimator for $\mu_B$, $H_0$ can be written as $H_0: \mu_A-\mu_B=0$, so the estimator I have to use for $\mu_A-\mu_B$ is $\overline{X}_A-\overline{X}_B$
The variance of $\overline{X}_A$ and of $\overline{X}_B$ are unkown, so I have to use the sample standard deviations $s_A$ and $s_B$, hence I extimate the standard deviation of $\overline{X}_A-\overline{X}_B$ with $\sqrt{s_A^2+s_B^2}$. I use a t-test with $$t=\frac{\overline{X}_A-\overline{X}_B}{\sqrt{s_A^2+s_B^2}}.$$ The question is: why do I use a t-statistic? Is it correct? I read that, when the standard deviation is unkown, the t-statistic is appropriate provided that $\overline{X}_A-\overline{X}_B$ is normally distributed... But I have no idea if $\overline{X}_A-\overline{X}_B$ is normally distributed.
Is it ok the proportion between the dimensions of sample A and B? We have 79 vs 15 so 84% vs 16%. What are the problems I can run into if the proportion is not appropriate?
For example, given that using a t-test as above is correct, if I fix the probability of a I type error of 5%, can I be sure that it is a real 5% or is it bigger? What about the power of the test? I expect to reject $H_0$, so I hope my test is powerful enough in order to do that.
If this partition (84|16) is not good, is increasing the dimension of the control group the only way to make a more significant statistic?
Any suggestion or good reference will be more than welcome.