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So here's the exercise: enter image description here

So the solution of the exercise is as follow: enter image description here

is this not a mistake?? The exercise says that the two groups are normally distributed. It doesn't say anything about them having the same variance. All the working out done in the question is the typical one that we use when we have two normal samples with different means and same variance and we want to check whether they have the same mean. But this is clearly not that situation.

Can we still use the same procedure? And if it is correct, why is it?

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Excellent question. This is an old problem called Behrens-Fisher Problem. If the variances for the two normal populations are unequal, no known exact test exists. Asymptotic tests, however, are known. It is an open problem to find a test statistic for all $n$ having suitable null distribution if the variances differ.

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    Can we not just proceed as follows: We have $Y_{1i} \sim N(\mu_1,\sigma_1^2)$ and $Y_{2i} \sim N(\mu_2,\sigma_2^2)$ hence $\hat{\mu_1}=\bar{Y_1}=\frac{1}{n_1}\sum_{I=1}^{n_1}(Y_{1i}-\bar{Y_1})^2$ and similarly $\hat{\mu_2}=\bar{Y_2}=\frac{1}{n_2}\sum_{I=1}^{n_2}(Y_{2i}-\bar{Y_2})^2$ so that $$\hat{\mu_1}-\hat{\mu_2} \sim N(\mu_1-\mu_2,\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2})$$ and then use the CLT ? For example then $Z = \frac{\hat{\mu_1}-\hat{\mu_2}-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \sim N(0,1)$ – 2017-02-19
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    then we just need to find an estimate for the denominator – 2017-02-19
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    That's what I am saying: you will estimate $\sigma_i^2$ by $s_i^2$ which is fine as long as you use large enough $n$. If you want an exact test, this is not possible. Of course, as $n\to\infty$ one can do many things. – 2017-02-19
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    Oh okay right, now I understand what you mean. – 2017-02-19
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    do you think that in the solution of the exercise it was assumed that the variances were equal? – 2017-02-19
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    Yes, they have used an exact test, and they assumed equal variance. Otherwise, as I mentioned, nobody as of now knows how to proceed. You may think about it and if you can come up with a solution, you will have to your credit a very influential paper !! :D – 2017-02-19
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    I'll post it here as an open source then, give me a couple of minutes to think about it ;) – 2017-02-19
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    I kind of managed to find something involving loads of gammas but got stuck ahah – 2017-02-19
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    It is good to get stuck; only then you will realize whether a problem is genuinely worth your time. Note you need chi square in the denominator for a standard distribution to work. Only gamma won't really help. – 2017-02-19
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    true. I was trying to get an expression similar to that in the denominator playing around with chi-squared, but then I always end up messing up the scale parameter, so that it wasn't a chisquared anymore – 2017-02-19