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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.

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Try It!
Section* *

- Incidence rate increase \((\pi)\)?
- Relative risk decreases \((\lambda)\)?
- 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?
- What is the minimal detectable relative risk if you had funds for 1000 subjects?

*n*decreases- Largest
*n*is closest to*l* - Protective effects would be those with \(\lambda \lt 1\)
- With a background rate of 10/100 and 1000 subjects, a relative risk of about 1.65 could be detected.