A drug maker wants to design a study in which two medications are compared. The first group has seen a improvement in 40%. Researchers want to see if the newer drug will improve this by 10% so they design a study. Assuming a study with power of 80% and a two-sided significance level of 5%. How many participants are required to detect a difference of 10% between the two groups.

My attempt and understanding

From the information gathered above there are two critical values $Z_{1-\alpha/2} = 1.96$ and $Z_{\beta}=0.84$ given the significance level of $\alpha = 0.05$. From the proportions, $p_{0} = 0.4$ and the proposed new drug should have the second group $p_{1} = 0.5$. The difference $\theta = p_{1}-p_{0} = 0.1$. The hypotheses

\begin{align} H_{0}:\theta &= 0 \\ H_{1}: \theta &\neq 0 \end{align}

However, I am unsure what formula has been used to calculate this. I have run command in R which has given me $\approx 388$ per group.

> power.prop.test(p1 = 0.4, p2 = 0.5, alternative = "two.sided",
+                 sig.level = 0.05,
+                 power = 0.80)

     Two-sample comparison of proportions power calculation 

              n = 387.3385
             p1 = 0.4
             p2 = 0.5
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number in *each* group

However, I am unsure what formula has been used to calculate the sample size. From my understanding this has been a comparison of two binomial proportions. If I could get any assistance in understand how this was calculated, it would be greatly appreciated!

  • $\begingroup$ Minor nitpick in the phrasing of question, there isn't one sample size that lets you detect "a difference of 10%", as that depends on the actual values that differ - this is specifically the number of patients needed to detect the difference between 40% and 50%. Finding the 10% difference between 40% and 50% requires more than five times as many subjects as finding the 10% difference between 0% and 10%! $\endgroup$ Apr 14, 2023 at 17:41

2 Answers 2


A formula based on a large sample test for the equality of two independent proportions is:

\begin{align} n_1 &= \kappa n_2 \\ n_2 &= \frac{(z_{\alpha/2} + z_{\beta})^2}{(p_1 - p_2)^2}\left[\frac{p_1(1 - p_1)}{\kappa} + p_2(1 - p_2)\right] \end{align}

Where $p_1$ and $p_2$ are the proportions, $\kappa$ is the allocation ratio and $\alpha$ and $\beta$ are the significance level and 1-Power.


Chow S-C, Shao J, Wang H, Lokhnygina Y (2018): Sample size calculations in clinical research. 3rd ed. CRC Press.


It isn't precisely from a formula. According to the code inside power.prop.test, the power for any value of n is calculated as

pnorm((sqrt(n) * abs(p1 - p2) - (qnorm(sig.level/2,lower.tail = FALSE) * 
   sqrt((p1 + p2) * (1 - (p1 + p2)/2))))/
   sqrt(p1 * (1 - p1) + p2 * (1 - p2)))

or in traditional notation $$\Phi\left(\frac{\sqrt{n}|p_1-p_2|-z_{\alpha/2}\sqrt{(p_1+p_2)(1-(p_1+p_2)/2)}}{\sqrt{p_1(1-p_2)+p_2(1-p_2)}} \right) $$ and numerical root-finding methods are used to get the value of n that makes this equal to 80%.

That's not necessarily how you'd do the computations by hand.

  • $\begingroup$ That is interesting. How would one go about calculating sample sizes by hand given the context? $\endgroup$ Apr 14, 2023 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.