7. Product and Process Comparisons
7.4. Comparisons based on data from more than two processes
7.4.7. How can we make multiple comparisons?

## Scheffe's method

Scheffe's method tests all possible contrasts at the same time Scheffé's method applies to the set of estimates of all possible contrasts among the factor level means, not just the pairwise differences considered by Tukey's method.
Definition of contrast An arbitrary contrast is defined by $$C = \sum_{i=1}^r c_i \, \mu_i \, ,$$

where $$\sum_{i=1}^r c_i = 0 \, .$$

Infinite number of contrasts Technically there is an infinite number of contrasts. The simultaneous confidence coefficient is exactly $$1 - \alpha$$, whether the factor level sample sizes are equal or unequal.
Estimate and variance for $$C$$ As was described earlier, we estimate $$C$$ by: $$\hat{C} = \sum_{i=1}^r c_i \, \bar{Y}_{i \scriptsize{\, \bullet}}$$ for which the estimated variance is: $$s_{\hat{C}}^2 = \hat{\sigma}_e^2 \, \sum_{i=1}^r \frac{c_i^2}{n_i} \, .$$
Simultaneous confidence interval It can be shown that the probability is $$1 - \alpha$$ that all confidence limits of the type $$\hat{C} \pm \sqrt{(r-1) F_{\alpha; \, r-1, \, N-r}} \,\, \cdot s_{\hat{C}}$$ are correct simultaneously.
Scheffe method example
Contrasts to estimate We wish to estimate, in our previous experiment, the following contrasts $$\begin{eqnarray} C_1 & = & \frac{\mu_1 + \mu_2}{2} - \frac{\mu_3 + \mu_4}{2} \\ & & \\ C_2 & = & \frac{\mu_1 + \mu_3}{2} - \frac{\mu_2 + \mu_4}{2} \, , \\ \end{eqnarray}$$ and construct 95 percent confidence intervals for them.
Compute the point estimates of the individual contrasts The point estimates are: $$\begin{eqnarray} \hat{C}_1 & = & \frac{\bar{Y}_1 + \bar{Y}_2}{2} - \frac{\bar{Y}_3 + \bar{Y}_4}{2} = -0.5 \\ & & \\ \hat{C}_2 & = & \frac{\bar{Y}_1 + \bar{Y}_3}{2} - \frac{\bar{Y}_2 + \bar{Y}_4}{2} = 0.34 \, .\\ \end{eqnarray}$$
Compute the point estimate and variance of $$C$$ Applying the formulas above we obtain in both cases: $$\sum_{i=1}^4 \frac{c_i^2}{n_i} = \frac{4(1/2)^2}{5} = 0.2$$ and $$s_{\hat{C}}^2 = \hat{\sigma}_\epsilon^2 \, \sum_{i=1}^4 \frac{c_i^2}{4} = 1.331(0.2) = 0.2661 \, ,$$

where $$\sigma_\epsilon^2$$ = 1.331 was computed in our previous example. The standard error is 0.5158 (square root of 0.2661).

Scheffe confidence interval For a confidence coefficient of 95 percent and degrees of freedom in the numerator of $$r$$ - 1 = 4 - 1 = 3, and in the denominator of 20 - 4 = 16, we have: $$\sqrt{(r-1) F_{\alpha; \, r-1, \, N-r}} = \sqrt{3 \, F_{0.05; \, 3, \, 16}} = 3.12 \, .$$

The confidence limits for $$C_1$$ are -0.5 ± 3.12(0.5158) = -0.5 ± 1.608, and for $$C_2$$ they are 0.34 ± 1.608.

The desired simultaneous 95 percent confidence intervals are $$-2.108 \le C_1 \le 1.108$$ $$-1.268 \le C_2 \le 1.948 \, .$$

Comparison to confidence interval for a single contrast Recall that when we constructed a confidence interval for a single contrast, we found the 95 percent confidence interval: $$-1.594 \le C \le 0.594 \, .$$ As expected, the Scheffé confidence interval procedure that generates simultaneous intervals for all contrasts is considerabley wider.
Comparison of Scheffé's Method with Tukey's Method
Tukey preferred when only pairwise comparisons are of interest If only pairwise comparisons are to be made, the Tukey method will result in a narrower confidence limit, which is preferable.

Consider for example the comparison between $$\mu_3$$ and $$\mu_1$$.

Tukey:    1.13 $$< \mu_3 - \mu_1 <$$ 5.31
Scheffé:  0.95 $$< \mu_3 - \mu_1 <$$ 5.49

which gives Tukey's method the edge.

The normalized contrast, using sums, for the Scheffé method is 4.413, which is close to the maximum contrast.

Scheffe preferred when many contrasts are of interest In the general case when many or all contrasts might be of interest, the Scheffé method tends to give narrower confidence limits and is therefore the preferred method.