Scheffe - Post Hoc Test

Scheffe Test

Scheffe's test is compatible with the analysis of variance test, where it never states significant contrast if test F is not significant.  Scheffe's  test was developed by Henry Scheffe (1959) which was used for unnecessarily orthogonal comparisons.  This test controls the MEER for each contrast including on the pairwise comparison.  The testing procedure allows various types of comparison so that it is less sensitive in finding significant differences compared to other comparison procedures.

For paired tests with the same test, the difference in critical values is like the previous test.

$$\begin{matrix}LSD=t_{\alpha/2,dfe}.\times\sqrt{\frac{2MSE}{r}}\ \\\\SCD=\sqrt{f_t\ \ F_{\alpha(ft,fe)}}\times\sqrt{\frac{2MSE}{r}}\ \\if\ number\ of\ replicationsdifferent:\\SCD=\sqrt{f_t\ \ F_{\alpha(ft,fe)}}\times\sqrt{MSE(\frac{1}{r_i}+\frac{1}{r_j})}\ \\\end{matrix}$$

Scheffe’s test is primarily used for comparisons among unnecessarily orthogonal mean groups.  If we are more interested in the comparison of all the mean pairs of treatments, this method is less sensitive (Tukey is better).  To compare among the mean groups, we must first create a comparison (contrast, Q) defined as follows:

$ Q=\sum{c_i{\bar{Y}}_{i.}}$ with  $\sum{c_i=0}$  (or  $\sum{r_ic_i=0}$  if the replications are not the same)

Reject the Null (H0) hypothesis if the value of | Q| > the critical value of FS.  The general form for the FS test is as follows:

$ Critical\ value\ FS\ =\sqrt{f_t\ \ F_{\alpha(ft,fe)}}\times\sqrt{MSE\sum\frac{c_i^2}{r_i}}$

where ft = degree of treatment-free, fe = error-degree of freedom, Fα = Value of F at significant level α with treatment-degree of freedom = fr and error-degree of freedom = fe, MSE = Mean Square of Error; c = contrast coefficient, r = replication.

Suppose if we want to compare the mean pair, then the values ci = 1 and -1.  If we want to compare the control vs. the other three means, then the contrast coefficient is : +3; -1; -1; -1. 

Scheffe Test Calculation Procedure for paired comparison:

As an illustration of the use of the Scheffe test, we reuse experiments on the Nitrogen content of Red Clover plants.  In this experiment, we assumed the compound as the control treatment, so we compared all the treatments with the combined.

1. Calculate SCD value:
   • From the table of analysis of variance, we obtain the value: ft = 5; fe = 24; MSE = 11.79; r = 5.
     $\begin{matrix}SCD=\sqrt{f_t\ \ F_{\alpha(ft,fe)}}\times\sqrt{\frac{2MSE}{r}}\\=\sqrt{f_t\ \ F_{0.05(5;\ 24)}}\times\sqrt{\frac{2MSE}{r}}\\=\ \sqrt{5(2.62)}\times\sqrt{\frac{2(11.79)}{5}}\\=7.9\ \ mg\\\end{matrix}$
2. Test criteria:
   • Compare the absolute value of the difference between the two means that we will see the difference with the SCD value with the following test criteria:
     $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>SCD\ \mathbf{reject}\ H_0,\ two\ means\ are\ significantly\ different\\\le SCD\ \mathbf{accept}\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$
   • State the difference between the mean treatments when the difference is greater than the SCD value, 7.9 mg. 
3. The results are as follows:

Information: 

1. The number on the body of the Table is the value of the difference between the means of the treatment
2. Compare the mean difference with the SCD comparator = 7.9 mg. When it is smaller than the SCD value (ns), give it the same line to the right of it.  The granting of the same line is stopped when the mean difference value > the SCD value (*). Proceed to the next comparison of treatment.
   • For example, if we compare 3Dok13 with 3Dok4, compare the difference (1.38) with the value of SCD = 7.9. Since 1.38 ≤ 7.90 which indicates no difference, then we give the same line to the two means.
   • 3Dok13 vs Combined: Compare the difference (5.44) with SDC = 7.90. Because 5.44 ≤ 7.90 (ns); give the same line to the 3Dok4 column
   • 3Dok13 vs 3Dok7: 6.66 ≤ 7.90 (ns);
   • 3Dok13 vs 3Dok5: 10.72 > 7.90 (*); stop! The same line is not given again. Continue with the comparison of 3Dok4 vs others.  The procedure is the same as in the LSD test, Tukey HSD.
3. Ignore (discard) the red line, because the line is already represented by the previous line!
4. Finally, provide a letter code for the line (except for the line that is red).

Group Comparison Calculation Procedure:

As an illustration of the use of the Scheffe test, we reuse experiments on the Nitrogen content of Red Clover plants.

Example 1:

In this experiment, suppose we are going to compare the Combined with other treatments.

$ Q=\sum{c_i{\bar{Y}}_{i.}}$ with  $\sum{c_i=0}$  (in this experiment we use the same replication)

H0 = Q1 = 5M1 – M2 – M3 – M4 – M5 – M6 = 0

which means that the mean Nitrogen content of the Combined treatment is no different from the mean nitrogen content in other treatments. Note that ∑c_i = 0 : +5 – 1 – 1 – 1 – 1 – 1 = 0.

Q1 = 5(18.70) – 13.26 – 14.64 – 19.92  – 23.98  – 28.82

=-7.12

Next, we calculate the Fs value. Reject the Null hypothesis (H0) if the value of the | Q1| > the critical value of FS. 

$$\begin{matrix}Fs=\sqrt{f_t\ \ F_{\alpha(ft,fe)}}\times\sqrt{MSE\sum\frac{c_i^2}{r_i}}\\=\sqrt{f_t\ \ F_{0.05(5;\ 24)}}\times\sqrt{MSE\sum\frac{c_i^2}{r_i}}\\=\ \sqrt{5(2.62)}\times\sqrt{11.79\times\frac{\left[5^2+(-1)^2+(-1)^2+(-1)^2+(-1)^2+(-1)^2\right]}{5}}\\=\sqrt{13.10}\times\sqrt{70.74}\\=30.44\ \ mg\\\end{matrix}$$

Due to the value of | Q| ≤ the critical value of FS: |-7.12| ≤ 30.44, then Receive H0, which means there is no difference between the control (Combined) and other treatments.

Example 2:

In this experiment, suppose we are going to compare Combined vs 3Dok5 and 3Dok1.

H0 = Q2 = 2M1 – M2 – M3 = 0

which means that the mean Nitrogen content of the Combined treatment is not different from the mean nitrogen content in the 3Dok5 and 3Dok1 treatments. Note that  $\sum{c_i=0}$  : +2 – 1 – 1.

Q1 =2(18.70) – 23.98 – 28.82

=-15.40

Next, we calculate the Fs value. Reject the Null hypothesis (H0) if the value of the | Q2| > the critical value of FS.

$$\begin{matrix}Fs=\sqrt{f_t\ \ F_{\alpha(ft,fe)}}\times\sqrt{MSE\sum\frac{c_i^2}{r_i}}\\=\sqrt{f_t\ \ F_{0.05(5;\ 24)}}\times\sqrt{MSE\sum\frac{c_i^2}{r_i}}\\=\ \sqrt{5(2.62)}\times\sqrt{11.79\times\frac{\left[2^2+(-1)^2+(-1)^2\right]}{5}}\\=\sqrt{13.10}\times\sqrt{14.48}\\=13.6\ \ mg\\\end{matrix}$$

Due to the value of | Q| ≤ the critical value of FS: |-15.40| > 13.6, then Reject H0, which means Combined is different from 3Dok5 and 3Dok1.

Calculation using SmartstatXL Excel Add-In