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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?
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| 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 $$ \sum_{i=1}^r c_i = 0 \, . $$ |
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| 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). |
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| 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 $$ -2.108 \le C_1 \le 1.108 $$ $$ -1.268 \le C_2 \le 1.948 \, . $$ |
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| 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 \(\mu_1\).
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. |
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| 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. | ||
