So if, as in an example they used, the mean difference of 231 comparisons was 0.2 mm with s = 3 mm and the limits of 95% – 5.8 to + 6.1 mm, the default error would be 332231 = 0.34, and the 95% confidence interval is ± 1.96 × 0.34 = 0.67, so that the 95% confidence intervals are: 5.8 ± 0.67 = − 5.13 to − 6.47 and for the ceiling = 6.1 ± 0.67 = 5.43 to 6.77. Ludbrook (2010) followed. Instead of 95% compliance limits, he prefers to use the concept of 95% tolerance with 95% confidence. Ludbrook modified the difference from mean diagram to display the mean difference (as in the usual Bland Altman diagram) and two pairs of boundary lines; The inner pair is the upper and lower limit of 95% confidence for the population, and the outer torque is the 95% tolerance with 95% confidence. For the internal limits, he proposed to use a slightly more precise formula: Xdiff ̄±t0.05N-1sdiff1+1N. For the example previously used, the 95% population confidence limits are 0.2+ ̄1.96×31+1231=0.2+ ̄5.89=6.09to-5.69. These are slightly different from those calculated from the simpler formula. For the tolerance limits, Ludbrook used the mean difference ± ksdiff, k being taken from the tolerance tables referred to in Chapter 7. For N = 231 is k ~ 2.131, and tolerances are 0.2 ± 2.131 × 3 = 6.56 to – 6.16. If the differences increase in proportion to the size of the measurement, Ludbrook recommends, instead of taking logarithms, to perform the regression of the differences from the Model I means, and then calculate the hyperbolic limit values of 95% for the points; These are essentially the tolerance limits. Exercise 15.5. Figure 15.3 shows the preoperative versus postoperative plasma silicazium levels of DB5 breast implants.
Interpret the results of the Bland Altman diagram. These differ slightly from those calculated from the simpler formula….