Student-Newman-Keuls - Post Hoc Test

Student-Newman-Keuls Test (SNK)

The step of working on the comparison between the mean treatments is similar to the use of the Duncan Test.  The difference lies only in the comparative value used.

Calculation steps:

1. Sort the mean value of the treatment (usually ascending order)
2. Calculate the shortest significant region for the region of various intermediate values using the following formula:
   $$\begin{matrix}W_p=w_{\alpha,p,\nu}s_{\bar{Y}}\\W_p=w_{\alpha,p,\nu}\sqrt{\frac{MSE}{r}}\end{matrix}$$
   Where:
   • MSE     =  Mean Squared Error
   • R          =  replication
   • $w_{\alpha,p,\nu}$    = the significant region value of the student
   • p = distance (2, 3, .. t); v  = degree of freedom;  $\alpha$ = significant level

3. Test criteria:
   • Compare the absolute value of the second difference in means that we will see the difference with the shortest significant region value (W_p) with the following test criteria:
     $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>Wp\ \mathbf{reject}\ H_0,\ two\ means\ are\ significantly\ different\\\le W p\ \mathbf{accept}\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$

Examples of Using the SNK Test

As an illustration, we reuse Red Clover's experimental data.  The full stages are as follows: 

1. Step 1: Calculate the shortest significant region value (W_p):
   • Determine the value of the MSE and its degree of freedom obtained from the Analysis of variance Table.
     • MSE = 11.7887
     • ν = df = 24
   • Determine its critical value from the student's significant region table based on the error-degree of freedom and the number of treatments to be compared. 
   • There are three parameters needed to determine the value of w_p, namely the significant level (α), p = the number of treatments to be compared, and the error-degree of freedom (df).  In this example, p = 2, 3, 4, 5, 6, the value of df = 24 (see df error in its Analysis of variance table) and α = 0.05. Next, specify the value of w_{0.05(p, 24)}.
   • To find the value of w_{0.05(p, 24)} we can see it in the same table as the table for the Tukey test (studentized range distribution table), i.e., the wp value at the significant level α = 0.05 with p = 1, 2, ... , 6 and degree of freedom (v)= 24.  Consider the following image to specify the w-table.
     
   • From the table we get the values of the values of _w,p,a which are 2.92; 3.53; 3.90; 4.17; and 4.37 _n
   • Calculate the shortest significant region (W_p):
     
   • Test criteria:
     • Compare the absolute value of the second difference in means that we will see the difference with the shortest significant region value (W_p) with the following test criteria:
       $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>Wp\ \mathbf{reject}\ H_0,\ two\ means\ are\ significantly\ different\\\le W p\ \mathbf{accept}\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$
2. Step 2: Sort the treatment mean value from small to large or vice versa. In this example, the mean treatment is sorted from small to large

First Method

Create a Matrix (Crosstabulation) the mean difference between all combinations of treatment pairs.  Since it is a handlebar, it is enough to create a lower triangle matrix table as presented in the Table below.

Matrix Table of mean differences of treatment

Comparator Values Table (Wp)

 Information: 

1. The number on the body of the Table is the value of the difference between the means of the treatment
2. Superscript (2); (3); ...; (6) = p, that is, the relative distance (rank) between the mean of one treatment and another treatment
3. Compare the mean difference with the comparison that corresponds to the rank (Wp). Give the code "ns" if the mean difference value is less or equal to the Wp value, and give the symbol "*"when the mean difference value is greater than the Wp value
4. On each column (starting from the 4th column) drag/give the same line starting from the number 0. Continue drawing the line in the same column at the mean difference value given the symbol "ns" and end when below it there is already a symbol "*".  Continue comparison the treatment in the next column.  The detailed stages are as follows:
   • In each column (in this example, columns 4 to 9) compare the difference in mean to the value of Wp, if it is smaller than the value of Wp (ns), give/draw the same line to the right of it.  The granting of the same line is stopped when the mean difference value > the Wp value (finds the "*" sign).
   • 3Dok13 vs other treatments (comparison in the 4th column)
     • For example, if we compare 3Dok13 with 3Dok4, compare the difference (1.38) with the corresponding Wp rating, Wp(2) = 4.50. Since 1.38 ≤ 4.50 which indicates no difference, then we give the same line to the two means.
     • 3Dok13 vs Combined: Compare the difference (5.44) with Wp(3) = 5.43. Because 5.44 > 5.43 (*); stop! The same line is not given again. Proceed to the next treatment comparison, which is the 3Dok4 vs other comparison.
   • 3Dok4 vs other treatments (comparison in the 5th column).
     • 3Dok4 vs Combined: Compare the difference (4.06) with Wp(2) = 4.50. Since 4.06 ≤ 4.50 (ns) so give the same line to column 3Dok4 (column 5);
     • 3Dok4 vs 3Dok7: 5.28 ≤ 5.43 (ns); so that the same line is given again
     • 3Dok4 vs 3Dok5: 9.34 > 5.43 (*): stop! the same line is not given again. Continue with the combined vs other comparison (6th column), etc... comes to the comparison in the 9th column.
5. Ignore (discard) the red line, because it is already represented by the previous line, which is the line contained in the Combined!
6. Finally, provide the letter code for the line.  The line that is notated is only the line that is black.  The first black line is given the letter "a", the second letter "b", the third letter "c", ignore the giving of the letter code to the red line (the line already represented by the other line, in this case the letter "c"), finally the 4th line of the letter "d".

Second Method

The full stages are as follows:

1. Step Forward #1:

• Give the letter "a" to the mean value of the first smallest treatment, which is 13.26.
• Compare the difference between 3Dok13 (13.26) and subsequent ranking treatment by using the appropriate Wp value.
• Give the same letter (in this case "a") if the difference is ≤ the corresponding Wp
• p = 2: 3Dok13 vs 3Dok4: |13.26 – 14.64|= 1.38. Compare this value with Wp(2) = 4.50.  Since 1.38 ≤ 4.50, which means no different, then 3Dok4 is given the letter "a" 
• p = 3: 3Dok13 vs Combined: |13.26 – 18.70| = 5.44. Compare this value with Wp(3) = 5.43.  Since 5.44 > 5.43, which means a significant difference, then the Combine is given a different letter, which is "b". 
• The comparison between 3Dok13 and the next treatment does not need to be done because it must be different.  Thus, at a step forward, subsequent comparison for the same group is stopped if the next treatment has been given a different notation.

2. Step Back:

• We've given the letter b to the Combined treatment, but we don't yet know if the combined is different from the other treatments located between 3Dok13 ("a") and The Combined ("b")!  In this example only one treatment, which is 3Dok4.  A more practical way is to check the step back, which is to compare the Combined with the previous treatment whose value is smaller than the Combined but greater than 3Dok13.
• p = 2: Combined vs. 3Dok4: |18.70 – 14.64| = 4.06. Compare this value with Wp(2) = 4.50.  Since 4.06 ≤ 4.50, which means no different, 3Dok4 besides being given the letter "a", is also given the letter "b" because it is not different from the combination. 

3. Step Forward #2:

• The starting point now starts from the smallest treatment that gets the letter "b", in this case on the 3Dok4 (14.64) treatment.
• Compare the difference between 3Dok4 (14.64) and the subsequent ranking treatment that has not been notated with letters i.e. 3Dok7 etc. using the appropriate Wp value.   Thus, p = 2:3Dok4 vs Combined: no need to do! because it had previously been given the notation "b"!
• Give the same letter (in this case "b") if the difference is ≤ the corresponding Wp
• p = 3: 3Dok4 vs 3Dok7: |14.64 – 19.92| = 5.28. Compare this value with Wp(3) = 5.43.  Since 5.28 ≤ 5.43, which means no significant difference, then 3Dok4 is given the letter "b".
• p = 4: 3Dok4 vs 3Dok5: |14.64 – 23.98| = 9.34. Compare this value with Wp(4) = 6.00.  Since 9.34 > 6.00, which means a significant difference, then 3Dok5 is given a different letter, i.e. "c".  The comparison between 3Dok4 and the next treatment does not need to be done because it must be different.

4. Step Back:

• We have already given the letter c to the 3Dok5 treatment, but we don't yet know if 3Dok5 is different from the previous treatment whose value lies between 3Dok5 and 3Dok4!  In this example Merge and 3Dok7. 
• p = 2: 3Dok5 vs 3Dok7: |23.98 – 19.92| = 4.06. Compare this value with Wp(2) = 4.50.  Because 4.06 ≤ 4.50, which means it is no different, 3Dok7, in addition to being given the letter "b", is also given the letter "c" because it is not different from 3Dok5. 
• p = 3: 3Dok5 vs Combined: |23.98 – 18.70| = 5.28. Compare this value with Wp(3) = 5.43.  Since 5.28 ≤ 5.43, which means no different, then the Combined besides earlier has been given the letter "b", is also given the letter "c" because it is not different from 3Dok5. 

5. Step Forward #3:

• The starting point now starts from the smallest treatment that gets the letter "c", in this case on the Combined treatment (18.70).
• Compare the difference between Combined (18.70) and the next ranked treatment that has not been given letter notation which is 3Dok1 using the appropriate Wp value.
• Give the same letter (in this case "c") if the difference is ≤ the corresponding Wp
• p = 4: Combined vs 3Dok1: |18.70 – 28.82| = 10.12. Compare this value with Wp(3) = 6.00.  Since 10.12 > 6.00, which means a significant difference, then 3Dok1 is given a different letter, which is "d". 

6. Step Back:

• Now we compare the 3Dok1 treatment with the previous treatment located between 3Dok1 and combined.
• p = 2: 3Dok1 vs 3Dok5: |28.82 – 23.98| = 4.84. Compare this value with Wp(2) = 4.50.  Since 4.84 > 4.50, which means a significant difference, then 3Dok5 is not given the letter "d" and the comparison is complete here.

The result is as follows:

Summary:

Information:

↓ Comparison with next rank (step forward)

→ the granting of a new notation (move on to the next column)

↑ Comparison with previous ratings (step back)

Calculation using SmartstatXL Excel Add-In