If the formula for a hypothesis test is

$$ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma^2_{x_1}}{n_{x_1}} + \frac{\sigma^2_{x_2}}{n_{x_2}}}} $$

And the variance for a binomial variable is given by the equation

$$ \sigma^2 = np(1-p) $$

Then why is the sum of the variances given by

$$ \sigma_{x_1 - x_2} = \sqrt{\frac{p_1(1-p_1)}{n_{x_1}} + \frac{p_2(1-p_2)}{n_{x_2}}} $$

Rather than

$$ \sigma_{x_1 - x_2} = \sqrt{p_1(1-p_1)+ p_2(1-p_2)} $$

  • 1
    $\begingroup$ When you divide a random variable by $n$ you divide its variance by $n^2$ $\endgroup$
    – Glen_b
    Oct 3, 2016 at 4:16

1 Answer 1


The correct formulas are $$ \text{st.dev}\left(\frac{x_1}{n_{x_1}}+\frac{x_2}{n_{x_2}}\right)=\sqrt{\frac{p_1(1-p_1)}{n_{x_1}}+\frac{p_2(1-p_2)}{n_{x_2}}} $$ and $$ \text{st.dev}\left(x_1+x_2\right)=\sqrt{p_1(1-p_1)+p_2(1-p_2)}. $$ Let me derive the first formula: $$ \text{st.dev}\left(\frac{x_1}{n_{x_1}}+\frac{x_2}{n_{x_2}}\right)=\sqrt{\text{var}\left(\frac{x_1}{n_{x_1}}\right)+\text{var}\left(\frac{x_2}{n_{x_2}}\right)} $$ and because $$ \text{var}\left(\frac{x}{n_{x}}\right)=\frac{1}{n^2}\text{var}\left(x\right)=\frac{p(1-p)}{n} $$ you get the first formula.

  • $\begingroup$ I don't follow what you did. For a binomial random variable the Var(x) = np(1-p). Where does var(x1/nx1) come from? $\endgroup$
    – hot_whisky
    Oct 2, 2016 at 21:12
  • $\begingroup$ @hot_whiskey I am not sure what you are asking; perhaps you didn't see the step var(X/n-Y/n)=var(X/n)+var(Y/n)? Or perhaps you did not recognize that your 'correct' formula(your 3rd) is wrong if sigma_(x1-x2) is notation for the standard deviation of x1-x2 $\endgroup$
    – svendvn
    Oct 2, 2016 at 21:27
  • $\begingroup$ Sorry, I don't mean to be obtuse. I don't understand the st.dev() notation, and I don't understand why the first formula you give doesn't have an n in front of p1 and p2, where everywhere I have read gives the variance equation as σ2=np(1−p). onlinecourses.science.psu.edu/stat414/node/71. Then, in the third equation what are you taking the variance of? $\endgroup$
    – hot_whisky
    Oct 2, 2016 at 21:59

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