How do I do a power calculation where the effect size and values of $\alpha$ and $\beta$ are known? I am planning a study where the endpoint of the current practice is 40% successful. The null hypothesis will be rejected if the intervention produces a 60% success rate (treatment effect of interest).  We are happy with a significance criterion of 0.05 and a $\beta$ 0.8.  The study will be to intervention or to current treatment, randomised and blinded.
What is the sample size needed? 
 A: I'm assuming you're taking $H_0: p_1 = p_2 = 0.4$, $H_a: p_1 < p_2$, the specific alternative $p_2 - p_1 > 0.6 - 0.4 = 0.2$, and $n_1 = n_2$.  Under these assumptions, the test statistic is $$z = \frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}.$$ 
With $\alpha = 0.05$, you're saying you want $P(z \leq z^*) = 0.05$ when $p_1 = p_2 = 0.4$, which gives $z^* = -1.645$.  With $\beta = 0.8$, you're saying you want $P(z \leq z') = 0.8$ when $p_2 - p_1 = 0.2$, which gives $z' = 0.84$.
Based on the assumptions on $z^*$ and $z'$ we have
$$z^* = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\frac{0.4(0.6)}{n} + \frac{0.4(0.6)}{n}}} = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\frac{0.48}{n}}}$$ and
$$z' = \frac{(\hat{p}_1 - \hat{p}_2) - (-0.2)}{\sqrt{\frac{0.4(0.6)}{n} + \frac{0.6(0.4)}{n}}} = \frac{\hat{p}_1 - \hat{p}_2 + 0.2}{\sqrt{\frac{0.48}{n}}}.$$
Thus to find $n$ we solve $$-1.645 + \frac{0.2}{\sqrt{\frac{0.48}{n}}} = 0.84,$$
which (rounding up) yields $n = n_1 = n_2 = 75$.  
Incidentally, there are online tools that will do this calculation for you, such as this one.  They give the sample size as $n_1 = n_2 = 77$.  I'm not sure the reason for the discrepancy; maybe they are using a continuity correction or using slightly different values from the normal distribution or taking the conservative $p_1 = p_2 = 0.5$ in the $z$ score calculation.
