Example of CRD Factorial

Applied Example

Experiment: There are 3 types of materials for the manufacture of batteries (A, B, C) tried at 3 temperatures (15^oF, 70^oF, 125^oF). From the experiment, you want to know whether the type of material and temperature affect battery life? Is a certain type of material suitable for a certain temperature?  From the experiment, battery life data were obtained as follows:

Table 22.  Battery Life Data From 3 Types of Materials at Three Different Temperatures

Solution:

Treatment Table:

 Step 1: Calculate the Correction Factor

 $$\begin{matrix}CF=\frac{Y...^2}{rab}=\frac{3799^2}{4\times3\times3}=400900.028\\\\\end{matrix}$$

Step 2: Calculate the Sum of The Total Squares

 $$\begin{matrix}SSTOT=\sum_{i,j,k}{Y_{ijk}}^2-CF\\=(130^2+74^2+....+104^2+60^2)-400900.028\ \\=478547.000\\\end{matrix}$$

Step 3: Calculate the Sum of Squares of Treatment

 $$\begin{matrix}SS(A)=\sum_{i}\frac{{Y_{i..}}^2}{rb}-CF\\=\frac{(998^2+1300^2+1501^2)}{4\times3}-400900.028\ \ \\=10683.722\\\end{matrix}$$

 $$\begin{matrix}SS(B)=\sum_{j}\frac{{Y_{.j.}}^2}{ra}-CF\\=\frac{(1738^2+1291^2+770^2)}{4\times3}-400900.028\ \ \\=39118.722\\\end{matrix}$$

 $$\begin{matrix}SS(AB)=\sum_{i,j}\frac{{Y_{ij.}}^2}{r}-CF-SS(A)-SS(B)\\=\frac{(539^2+229^2+...+583^2+342^2)}{4}-400900.028-10683.722-39118.722\\=9613.778\\\end{matrix}$$

Step 4: Calculate the Sum of Squares of Errors

 $$\begin{matrix}SSE=SSTOT-SS(A)-SS(B)-SS(AB)\\=18230.750\\\end{matrix}$$

Step 5: Create a Variance Analysis Table with its F-Values table

Table 23.  Battery Life Analysis of variance

F_{(0.05,2,2 7)} = 3,354

 F_{(0.0 1,2,2 7})  = 5,488

 F_{(0.0 5,4,2 7})  = 2,728

 F_{(0.0 1,4,2 7})  = 4,106

Step 6: Make a Conclusion

Material (A)

Since F-stat (7.91) > 3,354 then we reject H₀: μ₁ = μ₂ = μ₃ at a confidence level of 95% (usually given one asterisk mark (*), which indicates a significant difference)

Since F-stat (7.91) > 5,488 then we reject H₀: μ₁ = μ₂ = μ₃ at a confidence level of 99% (usually given two asterisk marks (**), which indicate a very noticeable difference)

Temperature (B)

Since F-stat (28.97) > 3,354 then we reject H₀: μ₁ = μ₂ = μ₃ at a confidence level of 95% (usually given one asterisk mark (*), which indicates a significant difference)

Since F-stat (28.97) > 5,488 then we reject H₀: μ₁ = μ₂ = μ₃ at a confidence level of 99% (usually given two asterisk marks (**), which indicates a very noticeable difference)

Material Interaction x Temperature (AxB)

Since F-stat (3.56) > 2,728 then we reject H₀: μ₁ = μ₂ = μ₃ at a 95% confidence level (usually given one asterisk sign (*), which indicates a significant difference)

Since the calculation (3.56) ≤ 4,106 then we fail to reject H₀: μ₁ = μ₂ = μ₃ at a confidence level of 99%

First, we check whether the Interaction Effect is significant or not? If it is significant, then examine the simple effect of the interaction, and ignore the main effect (the self-reliance), even if the main effect is significant! Why? Take another look at the discussion about the effect of interactions and main effect!  Primary ( if significant) effect testing is only performed when the effect of the interaction is not significant.

The value of F_{0.05(df1=4, df2=27)} = 2.728.  The value (F_interaction = 3.56) > F_{0.05(df1=4, df2=27)}, therefore at the significant level of α = 5 % we can conclude that the effect of the interaction between the material and the significant temperature. The effect of non-free material and temperature on the average battery life.  This means that the effect of certain specific materials on different temperature levels. Because of the effect of significant interactions, we do not need to test the main effect.

Step 7: Calculate the Coefficient of Variance (CV)

 $$\begin{matrix}CV=\frac{\sqrt{MSE}}{\bar{Y}..}\times100\%=\frac{\sqrt{675.213}}{105.528}\times100\%\\=24.62\% \end{matrix}$$

Post-Hoc

Based on the analysis of variance, the effect of the interaction between Matter and Temperature is significant, so we need to test its simple effect which are logical consequences of factorial experimental models in research.  This is done to get a more comprehensive conclusion and not just to state that the effect of the interaction is significant and busy with the testing of the main effect of the factors being tried.

In this follow-up test, the differences between the average pairs of treatments were carried out using the Duncan test.

1. Step 1: Calculate the shortest significant region value (R_p):

• Determine the value of the MSE and its degree of freedom obtained from the Variance Analysis Table.
• MSE = 675.213
• ν = df = 27
• 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 r α(p,df)}, 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, the value df = 27 (see the error df in the Variance Analysis table) and α = 0.05. Next, determine the value of r_{0.05(p, 27)}.
• To find the value of r_{0.05(p, 27)} we can see it in the table Significant Ranges for Duncan's Multiple Range Test at a significant level α = 0.05 with p = 2.3 and degree of freedom (v)= 27.  Consider the following image to specify the r-table.
• 
• From the table we get the values of r: 2.905 and 3.050
• Calculate the shortest significant region (R_p):

•  Test criteria:
• Compare the absolute value of the difference between the two averages with the value of the shortest significant area (Rp), with the following test criteria:
•  $ if \left|\mu_i-\mu_j\right| \left\langle\begin{matrix}\mathrm{>Rp\ \gg\ test\ results\ are\ significant.}\\\mathrm{\le Rp\ \gg\ test\ results\ are\ not\ significant.}\\\end{matrix}\right.$

2. Step 2: Sort the average table of treatment from small to large or vice versa. In this example, the average treatment is sorted from small to large

Differences of two material averages at the same temperature level:

Simple effect testing difference of two material averages at a temperature of 15 ^oC:

Notes:

(1)    superscript number [(2); (3)] indicates the rating (p) to compare with the difference between the two averages corresponding to the rating (Remember! the average treatment is pre-sorted). For example, the difference between A vs C compares with Rp(2) = 37,743, because A and C are not separated by other treatments (neighboring), while the difference between A vs B compares with Rp(3) = 39,627.

(2)    ns = not significant; * = significant at a significant level of 5%

(3)    the same line shows no difference between the average pairs of treatments.  When annotated, black lines are notated with letters, while red lines are ignored, because they are already represented by black lines.

Simple effect testing differences of two material averages at a temperature of 70 ^oC:

Simple effect testing of differences of two material averages at a temperature of 115 ^oC:

Difference of two average Temperatures at the same Material level:

Simple effect testing of differences in two average Temperatures on Material A:

Simple effect testing of differences in two average Temperatures on Material B:

Simple effect testing of differences in two average Temperatures on Material C:

The presentation of a simple effect test on such experiments can be summarized in the form of a two-way table as shown in the following table:

Remarks:

The numbers followed by the same letter did not differ markedly according to Duncan's test at a significant level of 5%.  Lowercase letters are read in the vertical direction (column) and capital letters are read in the horizontal direction (row)

Alternative: Interaction Effect (combination of factor level)

If the interaction is significant, a simple effect should be examined. However, if you want to compare the combination (AxB), then the number of treatments to be compared (p) is different from the previous calculation. The number of p = t-1 = 9-1 = 8.

Table of combination levels of the two factors

Comparison (Duncan)

Matrix Table of differences in the differences of the average pairs of AxB (after being sorted in ascending order)

Remarks:

Compare the difference between the pairs of two averages with the corresponding comparison values based on the distance rating between the two averages (in the example above, to facilitate the understanding of comparing the difference in averages with the corresponding ratings marked in the same color code between the difference and the comparator)

The presentation of the AxB interaction effect test on the experiment can be summarized in the form of a two-way table as shown in the following table:

Remarks:
The numbers followed by the same letter did not differ markedly according to Duncan's test at a significant level of 5%.

Which material is the best? Materials C? In the table above, it is difficult to interpret, because there are interdependencies between factors.

What temperature is best? 15 C?

Try to compare the results of the interaction effect test with the simple effect test!

In a simple effect test, interdependencies can be interpreted.

Remarks:
The numbers followed by the same letter did not differ markedly according to Duncan's test at a significant level of 5%.  Lowercase letters are read in the vertical direction (column) and capital letters are read in the horizontal direction (row)

Which material is the best? Materials C? No! All materials do not differ when the temperature is low (15) or high (125). Materials B and C are good when the temperature is 70 C.

What temperature is best? 15 C? Yes, if material A is used, but if material B or C is used, the suitable temperatures are 15 C and 75 C.

Calculation Using SmartstatXL Excel Add-In