The basic formula for a confidence interval for a difference in population means from two independent samples is
(difference in sample means) ± (multiplier)(pooled estimate of sd)(1/n1+1/n1)
Scheffe's method uses this same format. The differences are that we will use all samples to estiamte the common sd and we will use a multiplier from the F distribution instead of the t distribution.
For a 95% confidence interval for the pairwise differences in mean cuckoo bird length (in mm) for the hedge sparrow, robin, and wren host populations, we calculate the multiplier to be
multiplier = sqrt( (g-1) F(1-alpha) ) = sqrt( 5*2.29 ) = 3.38.
Notice that this number is larger than the t distribution value with 114 degrees of freedom, 1.98.
For all the pairwise confidence intervals, the estimated sd is 0.91 (see previous notes). The sample sizes are different for each comparison. The three pairwise confidence intervals are:
Sparrow - Robin: (23.12 - 22.58) ± 3.38 * 0.91 * sqrt(1/14 + 1/16) or 0.54 ± 1.13.
Sparrow - Wren: (23.12 - 21.13) ± 3.38 * 0.91 * sqrt(1/14 + 1/15) or 1.99 ± 1.14.
Robin - Wren: (22.58 - 21.13) ± 3.38 * 0.91 * sqrt(1/16 + 1/15) or 1.45 ± 1.11.
We can be at least 95% confident that all three confidence intervals are correct. We are at least 95% confident that the mean egg size of cuckoo bird eggs laid in wren nests are smaller than those in robin or hedge sparrow nests. We do not have sufficient evidence to concluse that the means for robins and hedge sparrows are different.
We can repeat the previous example. In this case, for three comparisons and simultaneous 95% confidence, we would want the individual confidence levels to be 1 - 0.05/3 = 98 1/3%. This value is not in our table, but we can use S-PLUS to find that the approapriate t value is 2.43. The three pairwise confidence intervals are:
Sparrow - Robin: (23.12 - 22.58) ± 2.43 * 0.91 * sqrt(1/14 + 1/16) or 0.54 ± 0.81.
Sparrow - Wren: (23.12 - 21.13) ± 2.43 * 0.91 * sqrt(1/14 + 1/15) or 1.99 ± 0.82.
Robin - Wren: (22.58 - 21.13) ± 2.43 * 0.91 * sqrt(1/16 + 1/15) or 1.45 ± 0.79.
Notice that the margins of error are smaller for Bonferroni's method than Scheffe's. This will be true when the number of comparisons is small, but not true if the number of comparisons is large enough.
Bret Larget, email@example.com