Suppose that we need to compare means of two sample spaces, say P, Q. Let $\mu$ and $\nu$ be the expectation value of $P$ and $Q$, respectively. Consider $m$ independent random samples from $P$ and $n$ independent random samples from $Q$. Suppose that $m$ and $n$ are large enough to simplify $m-1 \approx m$ and $n-1 \approx n$ for simplicity. From the standard statistical argument, the $Z$-estimator can be obtained $$ Z = \frac{\left(\overline{X}-\overline{Y} \right)-\left(\mu-\nu \right)}{S_p \sqrt{\frac{1}{m} + \frac{1}{n}}} $$ which could be simplified to $$ Z = \frac{D-\left(\mu-\nu \right)}{\sqrt{\frac{S_1^2}{n}+\frac{S_2^2}{m}}} $$ under our assumption that $m$ and $n$ are very large where $S_1^2$ and $S_2^2$ are sample variances and $S_p^2$ be the pooled variance and $D=\overline{X}-\overline{Y}$.
Now consider the following simple but wrong process. We may regard $\overline{X}$, $\overline{Y}$, $D$ as normal random variables. Since $m$ and $n$ are large enough, $S_1^2$ and $S_2^2$ are good estimators for the parameters say $\sigma_1^2$ and $\sigma_2^2$. Since the variance of $D$ is given by $\frac{\sigma_1^2}{m}+\frac{\sigma_2^2}{n}$, I concluded that $$ Z' = \frac{D-(\mu-\nu)}{\sqrt{\frac{S_1^2}{m}+\frac{S_2^2}{n}}}$$ would be a good $Z$-estimator. However $Z$ and $Z'$ looks very differently especially when $m$ and $n$ are significantly different. How could I explain the difference and conclude that $Z'$ does not follows the standard normal distribution even if our assumption holds?