This note describes how to choose the sample size to estimate the 95% limits of
aggreement. This approach to the agreement between two method of clinical
measurement is described in the paper
Statistical methods for assessing agreement between two methods of
clinical measurement, Bland JM, Altman DG. (1986). *Lancet*,
**i**, 307-310.

In our 1986 *Lancet* paper we gave a formula for the confidence interval
for the 95% limits of agreement.

The standard error of the 95% limit of agreement is approximately root( 3
s^{2}/*n*), where *s* is the standard deviation of the
differences between measurements by the two methods and *n* is the sample
size. The confidence interval is the estimate of the limit, *d* plus or
minus 1.96*s*, plus or minus 1.96 standard errors.

If you know how accurately you want to estimate the limits of agreement, you can use this to work out the sample size.

If you think of the 95% CI as +/- 1.96 root(3/*n*)*s*, you can see
that a sample of 12 gives a 95% CI approximately +/- *s*. This seems
pretty big. If we draw a little picture:

we can see that these confidence intervals are indeed wide. I usually
recommend 100 as a good sample size, which gives a 95% CI about +/-
0.34*s*, which looks something like this:

A sample of 200 subjects is even better, giving a 95% CI about +/-
0.24*s*. As with all estimation, to determine the appropriate sample size
the researcher must decide what accuracy is required.

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Last updated: 12 January, 2004.