Example Workout: Two Way ANOVA Table for N = 12, h=4, k = 3 & α = 5%

ANOVA Table for Two-Way Classification
between Treatment	2	6.5	6.5/2 = 3.25	3.25/3.1389 = 1.0354
between Varieties	3	5.6667	5.6667/3 = 1.8889	1.8889/3.1389 = 0.6018
Error	6	18.8333	18.8333/6 = 3.1389
Total	11

The below is the example work with steps shows how to generate the two way ANOVA table for the different means of treatments (rows) & subjects (columns) to find the F test statistic for test hypothesis (H₀) at a 5% or 0.05 significance level for three sample means x₁, x₂ & x₃.

Workout :
step 1 Address the formula input parameters and values
x₁ = 8, 7, 5 & 5
x₂ = 7, 4, 4 & 8
x₃ = 3, 4, 5 & 6

step 2 Form the below table to carry out the analysis of variance

Data sets	1	2	3	4	Total	Square
x₁	8	7	5	5	∑x₁ = 25	(∑x₁)² = 625
x₂	7	4	4	8	∑x₂ = 23	(∑x₂)² = 529
x₃	3	4	5	6	∑x₃ = 18	(∑x₃)² = 324
Total	18	15	14	19	G = 66	∑T_i² = 1478
Square	324	225	196	361	∑T_j² = 1106

Individual Squares :

Data sets	1	2	3	4	Total
x₁²	64	49	25	25	163
x₂²	49	16	16	64	145
x₃²	9	16	25	36	86
Total	122	81	66	125	∑x_ij² = 394

step 3 Find Correction factor (CF)
N = sum of Values of all items = 4 + 4 + 4
N = 12
Correction factor (CF) = G²N
= 66²12
= 435612
CF = 363

step 4 Find the sum of squares of all individual items (∑x_ij) and then total sum of squares (TSS)

Total sum of squares (TSS) =k∑i = 1 k∑j = 1 x_ij² - CF
= 394 - 363
= 31

step 5 Find sum of squares between rows (SSR)
SSR = ∑T_i²h - CF
where h is the number of observation in each row
= 14784 - 363
= 369.5 - 363
= 6.5

step 6 Find sum of squares between columns (SSC)
SSC = ∑T_j²k - CF
where k is the number of observation in each column
= 11063 - 363
= 368.6667 - 363
= 5.6667

step 7 Find Sum of square due to error (SSE)
SSE = TSS - SSR - SSC
= 31 - 6.5 - 5.6667
= 18.8333

step 8 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.

Sources of variation	df	SS	MSS	F-Ratio
between Treatments	k - 1	SSR	SSR/k-1 = MST	MST/MSE = F_R
between Varieties	h - 1	SSC	SSC/h-1 = MSV	MSV/MSE = F_C
Error	(h - 1)(k - 1)	SSE	SSE/(k-1)(h-1) = MSE
Total	N - 1

ANOVA Table

Sources of variation	d.f	SS	MSS	F-Ratio
between Treatments	3 - 1 = 2	6.5	6.5/2 = 3.25	3.25/3.1389 = 1.0354
between Varieties	4 - 1 = 3	5.6667	5.6667/3 = 1.8889	1.8889/3.1389 = 0.6018
Error	(3 - 1)(4 - 1) = 6	18.8333	18.8333/6 = 3.1389
Total	12 - 1 = 11

step 9 Find Critical values F_e or Table values of F

The critical value variance between treatments is obtained from F table for
[(k –1), (k – 1) (h – 1)] degrees of freedom at 5% level of significance.
The critical value variance between varieties is obtained from F table for [(h – 1), (k –1) (h – 1)] degrees of freedom at 5% level of significance.

Critical values of F from F-distribution table

For treatments, the critical value F_e(2,6) from F-distribution table at 5% of significance level is 5.1433
For varieties, the critical value F_e(3,6) from F-distribution table at 5% of significance level is 4.7571

Inference

There is no significance difference between treatments, since the calculated value of F₀ = 1.0354 is smaller than the table value of F_e = 5.1433. Therefore the null hypothesis H₀ is accepted.
There is no significance difference between varieties or subjects, since the calculated value of F₀ = 0.6018 is smaller than the table value of F_e = 4.7571. Therefore the null hypothesis H₀ is accepted.

Two Way ANOVA Table for N = 12, k = 3, h = 4 & α = 5%

Work with Steps for N = 12, k = 3, h = 4 & α = 5%