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)