We are interested in testing the following hypothesis:

\(\begin{array}{l}
\mathrm{H}_{0}\colon \pi=\pi_{0} \\
\mathrm{H}_{1}\colon \pi=\pi_{1}=\pi_{0}+d
\end{array}\)

Where \(\pi\) is the true proportion, \(\pi_0\) is some specified value for the proportion we wish to test (30% in our example), and \(\pi_1\) (which differs from \(\pi_0\) by an amount d (d= 2% in our example)) is the alternative value.  
The formula needed to calculate the sample size is:

\(\displaystyle{n=\frac{1}{d^{2}}\left[z_{\alpha} \sqrt{\pi_{0}\left(1-\pi_{0}\right)}+z_{\beta} \sqrt{\pi_{1}\left(1-\pi_{1}\right)}\right]^{2}}\)

Where

• \(\pi_0\) = null hypothesized proportion
• d = estimated change in proportion

Note that we can replace \(z_a\) by \(z_{\alpha / 2}\) for a two-sided test.
The z terms can be found from a standard normal distribution table, and common values are shown below:

(Chapter 8.5, p 305, Woodward book)

From the formula, we can calculate that n= 4,417.

The table below can also be used to estimate the necessary sample size.  Note that the formula and the Woodward table define d as a positive change. Since we are testing a decrease (30% down to 28%), we need to assume \(pi_{0}\) is 70%, and that it will go up to 72%.  We can think of it as testing the “non-smoking” prevalence.  
Note that Woodward offers additional tables in his textbook which can be used for different power, and for a two-sided versus a one-sided test.  

Sample Size Statement:  A total sample size of n=4,417 is needed to detect a change in smoking prevalence of 2% (30% down to 28%) using a one-group chi-squared test with one-sided alpha of 0.05 and 90% power.  
 

Table B.8. Sample size requirements for testing the value of a single proportion.

(Tables from Woodward, M. Epidemiology Study Design and Analysis. Boca Raton: Chapman and Hall:, 2013)

We are interested in testing the following hypothesis:

\(\begin{array}{l}
\mathrm{H}_{0}\colon \pi_{1}=\pi_{2} \\
\mathrm{H}_{1}\colon \pi_{1}-\pi_{2}=\delta
\end{array}\)

But it is usually more convenient to consider the ratio (i.e. relative risk = λ), so we can consider this hypothesis:

\(\begin{array}{l}
\mathrm{H}_{0}: \pi_{1}=\pi_{2} \\
\mathrm{H}_{1}: \pi_{1} / \pi_{2}=\lambda
\end{array}\)

The formulas needed to calculate the total sample size are:

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

and

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

where

\(\pi=\pi_{2}\) is the proportion in the reference group

\(\mathrm{r}=\mathrm{n}_{1} / \mathrm{n}_{2}\) (ratio of sample sizes in each group)

\(p_{0}=\) the common proportion over the two groups

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

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

From the formula, we can calculate that n=433 total, thus n=217 per group.

The table below can also be used to estimate the necessary sample size.  For the column with \(\pi\)=0.10, with \(\lambda\)=2.0, we see that n=448 total, with n=224 per group.  Approximately the same as from the formula.  

Sample Size statement:  A sample size of n=217 per group (total of 434) is needed to detect an increased risk of disease (relative risk=2.0) when the proportions are 10% in the unexposed and 20% in the exposed groups, using a two group chi-squared test with one-sided alpha of 0.05 and 90% power.  
 

Table B.9. Total sample size requirements (for the two groups combined) for testing the ratio
of two proportions (relative risk) with equal numbers in each group.

(Tables from Woodward, M. Epidemiology Study Design and Analysis. Boca Raton: Chapman and Hall, 2013)

Due to the design of unmatched case-control studies, where unequal sampling rates are used for the exposed and unexposed, we cannot estimate relative risks for case-control studies, and must instead estimate odds ratios.  The hypothesis we wish to test is slightly altered

\(\begin{array}{l}
\mathrm{H}_{0}^{*}\colon \pi_{1}^{*}=\pi_{2}^{*} \\
\mathrm{H}_{1}^{*}\colon \pi_{1}^{*} / \pi_{2}^{*}=\lambda^{*}
\end{array}\)

Where

\(\begin{array}{l}
\pi_{1}^{*}=p(\text { Exposed } \mid \text { Disease })=p(\text { Exposed } \mid \text { Case }) \\
\pi_{2}^{*}=p(\text { Exposed } \mid \text { No disease })=p(\text { Exposed } \mid \text { Control })
\end{array}\)

The formulas are similar to the formula for relative risk, but with additional parameters.

\(\begin{aligned}
n=\frac{(r+1)(1+(\lambda-1) P)^{2}}{r P^{2}(P-1)^{2}(\lambda-1)^{2}}[ & z_{\alpha} \sqrt{(r+1) p_{c}^{*}\left(1-p_{c}^{*}\right)} \\
& \left.+z_{\beta} \sqrt{\frac{\lambda P(1-P)}{[1+(\lambda-1) P]^{2}}+r P(1-P)}\right]^{2}
\end{aligned}\)

and

\(\displaystyle{p_{c}^{*}=\frac{P}{r+1}\left(\frac{r \lambda}{1+(\lambda-1) P}+1\right)}\)

Where

• \(\mathrm{P}=\) exposure prevalence
• \(\lambda=\) estimated relative risk
• r = ratio of cases to controls

From the formula, we can calculate that n=306 total, thus 153 cases and 153 controls.
The table below can also be used to estimate the necessary sample size.  For the column with P=0.30, with \(\lambda\)=2.0, we see that n=306 total, with n=153 cases, and n=153 controls.  

Sample Size statement:  A total sample size of n=306 (153 cases and 153 controls) is needed to detect an OR of 2.0, assuming the prevalence of exposure is 30%, with one-sided alpha of 0.05 and 90% power.  
 

Table B.10. Total sample size requirements (for the two groups combined) for unmatched case–control studies with equal numbers of cases and controls.

(Tables from Woodward, M. Epidemiology Study Design and Analysis. Boca Raton: Chapman and Hall, 2013)

In matched case/control study designs, useful data come from only the discordant pairs of subjects. Useful information does not come from the concordant pairs of subjects. Matching of cases and controls on a confounding factor (e.g., age, sex) may increase the efficiency of a case-control study, especially when the moderator's minimal number of controls is rejected.

The sample size for matched study designs may be greater or less than the sample size required for similar unmatched designs because only the pairs discordant on exposure are included in the analysis. The proportion of discordant pairs must be estimated to derive sample size and power. The power of matched case/control study design for a given sample size may be larger or smaller than the power of an unmatched design.

The hypothesis to be tested is essentially that the number of discordant pairs that have an exposed case is 50% compared to the alternative that it is different from 50%.

The formulas for sample size calculation for matched case-control study are:

\(\displaystyle{d_{p}=\frac{\left[z_{\alpha}(\lambda+1)+2 z_{\beta} \sqrt{\lambda}\right]^{2}}{(\lambda-1)^{2}}}\)    and  \(\displaystyle{n=2 d_{p} / \pi_{d}}\)

Where

• \(\mathrm{Dp}=\) number of discordant pairs needed
• n = total number of matched pairs
• \(\lambda =\) estimated relative risk
• \(\boldsymbol{\pi}_{\mathrm{d}}=\) probability of a discordant pair

From the formula, we can calculate that \(d_{p}\) = 73.19, and then n=292.7, so rounding up to the next nearest even number, the study needs 294 individuals - that is, 147 pairs.
In this scenario, conducting a matched case-control study provides a saving of 12 compared with the unmatched version.

Sample Size statement:  A total sample size of n=294 (147 matched case-control pairs) is needed to detect an OR of 2.0, assuming the prevalence of exposure is 30%, with one-sided alpha of 0.05 and 90% power.  
 

We are interested in testing the following hypothesis:

\(\mathrm{H}_{0}\colon \mu=\mu_{0}\)

\(\mathrm{H}_{1}\colon \mu=\mu_{1}\)

The formula needed to calculate the sample size is:

\(\displaystyle{n=\frac{\left(z_{\alpha}+z_{\beta}\right)^{2} \sigma^{2}}{\left(\mu_{1}-\mu_{0}\right)^{2}}}\)

Where

• \(\mu_{0}\)= null hypothesized value
• \(\mu_{1}\)= alternative hypothesized value
• \(\sigma\) = standard deviation

From the formula, we can calculate that n=67.1, so rounding to the next whole number would be n=68 .

To use the table below, we can calculate S= (6.0 – 5.5)/1.4 = 0.3571. This exact value does not appear in Table B.7. In these situations, we can get a rough idea of sample size by taking the nearest figure for S. In the example, the nearest tabulated figure is 0.35, which has n = 70 (for one-sided 5% significance and 90% power). This is only slightly above the true value of 67 for S = 0.3571. However, this process can lead to considerable error when S is small, so it is preferable to use the formula. 

Sample Size Statement:  A total sample size of n=68 is needed to detect a 0.5 mmol/l increase in mean cholesterol compared to an historical value of 5.5 mmol/l using a one-group t- test with one-sided alpha of 0.05 and 90% power, and assuming a standard deviation of 1.4.  
 

Table B.7. Sample size requirements for testing the value of a single mean or the difference between two means.

from Woodward, M. Epidemiology Study Design and Analysis. Boca Raton: Chapman and Hall, 2013, p.770

We are interested in testing the following hypothesis:

\(\begin{array}{l}
\mathrm{H}_{0}\colon \mu_{1}=\mu_{2} \\
\mathrm{H}_{1}\colon \mu_{1}-\mu_{2}=\delta,
\end{array}\)

The formula needed to calculate the sample size is:

\(\displaystyle{n=\frac{(r+1)^{2}\left(z_{\alpha}+z_{\beta}\right)^{2} \sigma^{2}}{\delta^{2} r}}\)

Where...

• \(\mu_{1}\)= hypothesized mean in group 1
• \(\mu_{2}\)= hypothesized mean in group 2
• \(\delta\)= difference in means (null hypothesis \(\delta = 0\), alternative hypothesis \(\delta \ne 0\))
• \(\sigma\) = standard deviation
• \(r = \dfrac{n_1}{n_2}\)

From the formula, we can calculate that n=137, but after rounding to the next highest even number, n=138, with 69 per group.

To use the table below, we can calculate S= (6.0-5.3)/1.4 = 0.5.  For a one-sided alpha of 0.05, we need to use the column for alpha =5%, and 90% power. Reading down to S=0.5, we see n=35. Back to the table header directions, we see that for a test of the difference between two means, we need to multiply the value by 4. Thus, 35*4 = 140 is the total sample size needed.

Sample Size Statement:  A total sample size of n=138 (69 per group) is needed to detect a 0.7 mmol/l difference in mean cholesterol using a two-group t- test with one-sided alpha of 0.05 and 90% power, and assuming a common standard deviation of 1.4.  
 

Table B.7. Sample size requirements for testing the value of a single mean or the difference between two means.

(from Woodward, M. Epidemiology Study Design and Analysis. Boca Raton: Chapman and Hall, 2013, p.770)