For independent samples $X_1,\cdots,X_n $from $\textit{Bernoulli }(p_1)$ and $Y_1,\cdots,Y_m$ from $\textit{Bernoulli }(p_2)$,where $n$ and $m$ are large. Then,the Central Limit Theorem tell us $\frac{\bar X_n-p_1}{\sqrt{\frac{p_1(1-p_1)}{n}}}$ converges in distribution to $\textit{Normal }(0,1)$ and $\frac{\bar Y_m-p_2}{\sqrt{\frac{p_2(1-p_2)}{m}}}$ converges in distribution to $\textit{Normal } (0,1)$.Can we get $\frac{(\bar X_n-\bar Y_m)-(p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n}+\frac{p_2(1-p_2)}{m}}} $ converges in distribution to $\textit{Normal }(0,1)$, when $n$ and $m$ are large enough?
I cannot find some conclusions said that if $X_n$ converges in distribution to $X$ and $Y_n$ converges in distribution to $Y$,then $X_n\pm Y_{n}$ converges in distribution to $X\pm Y.$
In hypothesis test concerning the difference between two population proportions, we consider two binomial populations with parameters $n$, $p_1$ and $m$, $p_2$, respectively, where $n$ and $m$ are large. Then, we are usually interested in testing hypotheses such as $$H_0: p_1 − p_2 \leq 0 \quad\textrm{versus}\quad H_1: p_1 − p_2 > 0 \quad (a)$$$$H_0: p_1 − p_2 \geq 0 \quad\textrm{ versus}\quad H_1: p_1 − p_2 < 0 \quad (b)$$$$H_0: p_1 − p_2 = 0 \quad\textrm{ versus}\quad H_1: p_1 − p_2 \ne 0 \quad (c)$$ Why we assume that the test statistic $T= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n}+\frac{p_2(1-p_2)}{m}}}$,where $\hat{p}_1=\frac{1}{n}\sum_{i=1}^{n}x_{i}$and $\hat{p}_2=\frac{1}{m}\sum_{k=1}^{m}y_{k}$, is approximately distributed by the standard normal $\textit{Normal }(0, 1)$?