I have been doing the following problem and there is something I do not understand:
SCB's political party preference survey is conducted twice a year. Random selected people will answer the question "Which party would you vote for if it was a parliamentary election?". Below you will find the results of SCB's measurements in May 2018 and November 2017 for the "Alliansen", the "Rödgröna", and other parties("Övriga").
Estimate the change in support for the "Rödgröna" party and the standard error for this estimate.
So, in the solutions I have for the change of support, that $X~Bin(4632, p1)$ and $Y~Bin(4715, p2)$. Then $(p1)^*obs = 1945/4632 ≈ 0.4199$ $(p2)^*obs = 2126/4715 ≈ 0.4509$.
So, the change in support is:
Now, the part that I do not understand, it says: The standard error is an estimate of the standard deviation of the estimation. We estimate $p1-p2$ with $X/4632 - Y/4715$ that has the variance
Finally, the error:
My question: Why is $V(X)$ above divided by $4632^2$ and not just $4632$? I don't fully understand why is the standard error calculated this way? Any help would be appreciated.
You want to calculate $$Var\left(\frac{X}{n_1}+\frac{Y}{n_2}\right)$$
It is supposed that $X$ and $Y$ are independent. Thus we have $Var\left(\frac{X}{n_1}\right)+Var\left(\frac{Y}{n_2}\right)$
Now you can apply the rule $Var(cX)=c^2\cdot Var(X)$, where $c$ is an arbitrary constant. $\frac{1}{n_1}$ and $\frac{1}{n_2}$ are constants as well. Therefore we get
$$\frac{1}{n_1^2}\cdot Var(X)+\frac{1}{n_2^2}\cdot Var(Y)$$