Dunnet - Post Hoc Test

Dunnett Test

In some specific experimental cases, perhaps we are only interested in comparisons between controls and other treatments.  For example, comparing a local variance or a standard chemical with a new one.  For those cases, we can use the Dunnet test.  Dunnet developed this test and popularized it in 1955.  The Dunnet test maintains MEER at a level that is no more than the specified significant level, e.g. α= 0.05.  In this method, it only requires one comparative value to be used to compare the control with other treatments.  The formula is like LSD, but in this test, the t-value used is not the t-student used in the LSD test.  Dunnet uses its own t table, which is usually attached to experimental design books.

The formula for calculating the DLSD value is as follows:

$$\begin{matrix}DLSD={t^\ast}_{\alpha/2(p,dfe)}.s_{\bar{Y}};with\ s_{\bar{Y}}=\sqrt{\frac{2KTG}{r}}\ \\if\ number\ of\ replication\ different:\\DLSD={t^\ast}_{\alpha/2(p,dfe)}\sqrt{KTG(\frac{1}{r_i}+\frac{1}{r_j})}\ \\\end{matrix}$$

where r is the sum of the multiplications, MSE = Mean Square of Error obtained from the analysis of variance, α = significant level, p = number of treatments, excluding control (p = t-1), dfe = error-degree of freedom.  The value of t^* is the value obtained from the table t-Dunnet at a significant level α with the degree of freedom = dfe (In t-Dunnet tables it is usually predetermined for two-way testing, so α in the table the actual value of α/2). 

Calculation Procedure:

As an illustration of the use of the Dunnet 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.

Dunnet Test Example

1. Calculate the DLSD value:
   • Determine the value of the MSE and its degree of freedom obtained from the Analysis of variance Table.
     • MSE = 11.7887
     • df = 24
   • Specify the value of t-Dunnet.
     • There are three parameters needed to determine the value of t-Dunnet, namely the significant level (α), the number of treatments to be compared, excluding the control (p), and the error-degree of freedom (df).  In this example, the value df= 24 (see df error in the Analysis of variance table), p=t-1=6-1=5, and α=0.05. Next, specify the value of t*_{(0.05/2, 24)}.
     • To find the value of t*_{(0.05/2, 24)} we can look at it in the table The spread of t-Dunnet at a significant level of 0.05 with a degree of freedom of 24, and p = 5.  Consider the following image to specify t-Dunnet.
       
     • From the table we get the value of the value of t_{(0.05/2, 5, 24)} = 2.70
     • Calculate the DLSD value by using the following formula:
       $\begin{matrix}DLSD=t\ast_{0.05/2;5;24}\sqrt{\frac{2MSE}{r}}\\=2.70\times\sqrt{\frac{2(11.79)}{5}}\\=5.86\ \ mg\\\end{matrix}$
2. Test criteria:
   • Compare the absolute value of the difference between the two means that we will see the difference with the LSD value with the following test criteria:
     $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>DLSD_{0.025}\ \mathbf{reject}\ H_0,\ two\ means\ are\ significantly\ different\\\le DLSD_{0.025}\ \mathbf{accept}\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$
   • In this case, we only compared the difference between the controls (Combined) and the mean of the other treatments.

The results are as follows:

Only the 3Dok1 treatment is different from the Combined.

Calculation using SmartstatXL Excel Add-In