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)