Tukey HSD - Post Hoc Test

Tukey HSD Test

The Tukey test is often also referred to as the honest significant difference test, introduced by Tukey (1953). The testing procedure is identical to LSD, which has one comparison and is used as an alternative to LSD if we want to test all the mean pairs of treatment without a plan. The Tukey test was used to compare all the mean pairs of treatments after the Analysis of variance test was performed.

Test procedure with Tukey HSD Test:

1. Testing steps:
   • Sort treatment mean (ascending/descending order)
   • Specify the Value of Tukey HSD () with the formula: $ \omega$
     $ \omega=q_\alpha(p,\nu)\sqrt{\frac{MSE}{r}}$
   • p = number of treatments = t
   • ν   = error-degree of freedoms
   • r = number of replications
   • α  = significant level
   • q_a(p, ν) = critical value obtained from the table of significant regions of the student
2. Test criteria:
   • Compare the absolute value of the difference between the two means that we will see the difference with the HSD value with the following test criteria:
     $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>HSD_{0.05}\ reject\ H_0,\ two\ means\ are\ significantly\ different\\\le HSD_{0.05}\ accept\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$

Examples of Using the Tukey Test

As an illustration, we reuse Red Clover's experimental data.

1. Step 1: Calculate the HSD value:
   • 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 the critical value from the table of the student's significant region. 
     • There are three parameters needed to determine the value of q  namely the significant level (α), p = the number of treatments to be compared, and the error-degree of freedom (df).  In this example, p=6, the value df=24 (see df error in its Analysis of variance table) and α=0.05. Next, specify the value of q_{0.05(6, 24)}.
     • To find the value of q_{0.05(6, 24)} we can see it in the table Studentized range distribution at a significant level α = 0.05 with p = 6 and degree of freedom (v)= 24.  Consider the following image to specify the q-table.
       
     • From the table we get the value of the value of q_{0.05(6, 24)} = 4.37
     • Calculate the HSD value by using the following formula:
       $\begin{matrix}\omega=q_\alpha(p,\nu)\sqrt{\frac{MSE}{r}}\\=4.37\times\sqrt{\frac{11.79}{5}}\\=6.71\\\end{matrix}$
   • Test criteria:
     • Compare the absolute value of the difference between the two means that we will see the difference with the HSD value with the following test criteria:
       $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>6.71\ reject\ H_0,\ two\ means\ are\ significantly\ different\\\le 6.71\ 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

3. The next step is to calculate the difference between the mean treatments to determine which treatment means are the same and which are different.  One way is to give letter notation to the mean value. 

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.

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 HSD value (6.71).  Give the code "ns" when the mean difference value is less or equal to the HSD value, and give the symbol "*"when the mean difference value is greater than the HSD value
3. 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 with the HSD value, if it is smaller than the HSD value (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 HSD 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 value of HSD = 6.71. Since 1.38 ≤ 6.71 which indicates no difference, then we give/draw the same line on both means.
     • 3Dok13 vs Combined: Compare the difference (5.44) with HSD = 6.71. Since 5.44 ≤ 6.71 (ns), which indicates no difference, then we continue to draw the same line from 3Dok13 to Combined.
     • 3Dok13 vs 3Dok7: Compare the difference (6.66) with HSD = 6.71. Since 6.66 ≤ 6.71 (ns), which indicates no difference, then we draw/give again the same line until the 3Dok7 treatment.
     • 3Dok13 vs 3Dok5: Compare the difference (10.72) with HSD = 6.71. Since 10.72 > 6.71 (*), stop! The same line is not given again. Continue comparison the treatment in the next column, in this case the 5th column is 3Dok4 vs other comparison.
   • 3Dok13 vs other treatment (comparison in the 5th column):
     • 3Dok4 vs Combined. Compare the difference (4.06) with HSD = 6.71. Since 4.06 ≤ 6.71 (ns) so give the same line to the column 3Dok4;
     • 3Dok4 vs 3Dok7: 5.28 ≤ 6.71 (ns); thus, giving the same line to the 3Dok4 column
     • 3Dok4 vs 3Dok5: 9.34 > 6.71 (*); stop! the same line is not given again. Continue with the comparison in the next column, the 6th column which is the Combined vs other treatment. etc. comes to the comparison in the 9th column.
4. Ignore (discard) Lines that are red, because they are already represented by the lines in the previous comparison!
5. 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 letter-code of red lines (lines already represented by other lines).

Second Method:

The full stages are as follows: 

1. Step Forward #1:

• Give the letter quality "a" to the mean value of the first smallest treatment, which is 13.26.
• add the value 13.26 with the HSD value: 13.26 + 6.71 = 19.97. 
• give the same letter for the mean value of the treatment that is less than or equal to the value of 19.97.

• It appears that 3Dok13 (13.26) is no different from 3Dok4 (14.64), Combined (18.70), and 3Dok7 (19.92), because the value ≤ 19.97, while 3Dok13 (13.26) is already different from 3Dok5 (23.98) because 23.98 > 19.97!
• Since the 3Dok5 treatment is different from 3Dok13, then give the letter "b" to the mean value of the treatment.

2. Step Back:

• We have already given the letter b to the 3Dok5 treatment, but we don't yet know if 3Dok5 is different from other treatments whose value lies between 3Dok13 and 3Dok5!  In this example, the treatment between 3Dok13 and 3Dok5 is the 3Dok4, Combined, and 3Dok7 treatment.  A more practical way is to check the step back, which is to compare 3Dok5 with these treatments whose value is smaller than 3Dok5 but greater than 3Dok13.
• Subtract mean value of 3Dok5 with HSD value: 23.98 – 6.71 = 17.27
• Give the same letter code as the 3Dok5 treatment, which is "b" at the mean value of the treatment whose value ≥ 17.27.
• In this case, it turns out that the Combined treatment and 3Dok7 have a value of ≥ 17.27, so that in addition to being coded with the letter "a", it is also given the letter "b" because it is no different from 3Dok5.

3. Step Forward #2:

• The starting point now starts from the smallest treatment that gets the letter "b", which is the Combined treatment (18.70).
• add the mean value of the Combined treatment (18.70) with the HSD value: 18.70 + 6.71 = 25.41. 
• give the same letter as the Combined treatment (in this case "b") for the mean value of the treatment that is less than or equal to the value of 25.41.  Incidentally, in this case, the letter "b" is only up to the treatment of 3Dok5 which previously had already received the letter "b" so the result is still the same as before.

• It appears that Combined (18.70) is different from 3Dok1 because 28.82 > 25.41.
• Because it is different, then give the letter "c" to the mean value of the treatment 3Dok1 (28.82).

4. Step Back:

• Subtract the mean value of 3Dok1 (28.82) with an HSD value: 28.82 – 6.71 = 22.11
• Give the same letter as 3Dok1, i.e. "c" to the mean value of the treatment whose value ≥ 22.11.
• In this case, it turns out that the 3Dok5 (23.98) treatment, whose value ≥ 22.11, so that in addition to being coded with the letter "b", it also gets the letter "c" because it is no different from 3Dok1.

5. Since there is no longer a treatment whose value is above 3Dok1 and the back check with a Step Back has been carried out, it means that the provision of letter notation has been completed and the final notation is as follows:

Summary:

Information:

↓ Comparison with the next mean value (step forward)

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

↑ Comparison with the previous mean value (step back)

Calculation using SmartstatXL Excel Add-In