I am a clinical researcher and orthopedic surgeon and one of my PhD student is starting on the last part of her thesis. Her work is centered around the potential benefits of interprofessional teaching at an orthopedic ward. The study is a retrospective cohort study on register data for hip fracture patients. Our hypothesis is that an interprofessional ward (with medical students and nurse students guided by real doctors and nurses) does not put patient at higher risk for adverse events than a control ward. As the primary outcome variable we have a proxy variable for adverse events; readmission rate at 3 months. The number of patients treated at the interprofessional ward is about 1:4 compared to the control ward.

My question: Assuming a 3-month readmission rate of about 20% (about normal for this patient group) in the control ward, how large a sample size do I need when I aim at detecting a 5% difference in readmission rates? This is for 80% power and $p \le 0.05$.

When performing this calculation as a superiority calculation for proportions in for instance SamplePower 3, I wind up at approximately 500 and 2000 in the interprofessional and control ward, respectively. Is this calculation correct, or should I do a non-inferiority analysis for proportions instead where the minimum clinical difference I am interested in is 5%? I cant find any such non-inferiority calculators where the number of subjects in the two groups are of different numbers, as in this study.

  • 1
    $\begingroup$ This is nearly impossible to say without understanding how the student intends to carry out the analysis. Will this be carried out with some sort of mixed effects model? Generalized Estimating Equations? Will propensity scores be used? $\endgroup$ Commented Jan 18, 2019 at 20:05
  • $\begingroup$ What's interesting is that you are testing to see if your null hypothesis is "true". Hypothesis tests are generated such that we seek evidence to reject, not support, a null hypothesis. Unfortunately if you're looking to detect a difference of about 5% then you're looking at quite a high sample size. Curious, what analysis are you planning? I feel like this study could be a great candidate for a survival analysis. $\endgroup$
    – Tommyixi
    Commented Jan 24, 2020 at 20:20

1 Answer 1


This is my calculation and it could be wrong because I am studying statistics by myself. Any improvement or correction is welcome. Please don't mind my awkward English. Thank you.

Let's say readmission rates for the interprofessional ward and the control ward are $p_1$ and $p_2$ respectively and think about the hypothesis for our test.

$H_0$ : $p_1=p_2$ (Their readmission rates are same)

$H_1$ : $p_1<p_2$

We can reject the null hypothesis if the estimation of $p_2-p_1$ is big enough.

The estimation $\hat \Delta=\hat p_2-\hat p_1$ has the normal distribution with mean $\Delta=p_2-p_1$ and the variance $\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}$ approximately.

Under the null hypothesis, the mean $\Delta_0=0$ and so $D=\frac{\hat \Delta}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}}$ has the standard normal distribution.

If we define our test like this : reject $H_0$ when $D>z_{0.05}$, the significance of the test is 0.05.


When the difference of the rates is 5%, the power would be * $\begin{align}P(D>z_{0.05})& =P\left(\frac{\hat \Delta-5\%}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}} > z_{0.05}-\frac{5\%}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}} \right)\\& =1-\Phi\left(z_{0.05}-\frac{5\%}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}}\right)\\ &=\Phi\left(-z_{0.05}+\frac{5\%}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}}\right) \end{align}$.

Because $\frac{\hat \Delta-5\%}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}}$ has the standard normal distribution under this 5% difference hypothesis.

And this power should be bigger than 80% as you want.

$-z_{0.05}+\frac{5\%}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}}\geq z_{0.2}$

$\frac{0.05}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}} \geq z_{0.05}+z_{0.2}$

And say the total number of patients is $n$, then $n_1=\frac{4}{5}n$, $n_2=\frac{1}{5}n$.

$n \geq \left(\frac{z_{0.05}+z_{0.2}}{0.05}\right)^2 \left({1.25 \hat p_1(1-\hat p_1)+5\hat p_2(1-\hat p_2)}\right)$

After putting in $p_1=0.2$ and $p_2=0.25$, we get $n\geq 2814$.


*Actually in this case we don't have to put hats on the radical $p_1$ and $p_2$.


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