9.4 - Example 9-2: Ratios in a population-based study (relative risks, relative rates or prevalence ratios)

Example 9-2 Section

Suppose the rate of disease in an unexposed population is 10/100 person-years. You hypothesize an exposure has a relative risk of 2.0. How many persons must you enroll assuming half are exposed and half are unexposed to detect this increased risk? \(\alpha=0.05\) and 90% power.

Here are the hypotheses:

Null hypothesis

\(H_0\colon \text{Incidence}_{(Unexposed)} \le \text{Incidence}_{(Exposed)}\)

Alternative hypothesis

\(H_A\colon \text{Incidence}_{(Unexposed)} \le \text{Incidence}_{(Exposed)}=\lambda\)

Where:

\(\lambda \gt 0\)
\(\text{Incidence}_{(Exposed)}=p(\text{Disease|Exposed})\)
\(\text{Incidence}_{(Unexposed)}=p(\text{Disease|Not Exposed})\)

and the resulting formula:

\(n=\dfrac{r+1}{r(\lambda -1)^{2}\pi^{2} }\left [ z_{\alpha }\sqrt{(r+1)p_{c}(1-p_{c})}+z_{\beta }\sqrt{\lambda \pi (1-\lambda \pi)+r\pi(1-\pi )} \right ]^{2}\)

where \(\pi=\pi_2\) is the proportion in the reference group and \(p_c\) is the common proportion over the two groups, which is estimated as:

\(p_{c}=\dfrac{\pi (r\lambda +1)}{r+1}\)

When r = 1 (equal-sized groups), the formula above reduces to:

\(p_{c}=\dfrac{\pi (\lambda +1)}{2}=\dfrac{\pi_{1}+\pi_{2} }{2}\)

Let's take a look at tabulated results:

Click the button below to find sample size for detecting RR of 2 under conditions above.

Try It! Section

What happens to the necessary sample size as:
  1. Incidence rate increase \((\pi)\)?
  2. Relative risk decreases \((\lambda)\)?
  3. How would you use this table to determine sample size for 'protective' effects (i.e., nutritional components or medical procedures which prevent a negative outcome), as opposed to an increased risk?
  4. What is the minimal detectable relative risk if you had funds for 1000 subjects?
  1. n decreases
  2. Largest n is closest to l
  3. Protective effects would be those with \(\lambda \lt 1\)
  4. With a background rate of 10/100 and 1000 subjects, a relative risk of about 1.65 could be detected.