# Two Way ANOVA Table for N = 12, k = 3, h = 4 & α = 5% One way ANOVA table or summary to estimate the test of hypothesis (H0) at 5% or 0.05 significance level for sample means by using F-distribution table

ANOVA Table for Two-Way Classification
Sources of VariationdfSSMSSF-Ratio
between Treatment26.56.5/2 = 3.253.25/3.1389 = 1.0354
between Varieties 35.66675.6667/3 = 1.88891.8889/3.1389 = 0.6018
Error618.833318.8333/6 = 3.1389
Total11

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

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 ∑Ti2 = 1478 Square 324 225 196 361 ∑Tj2 = 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 ∑xij2 = 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 (∑xij) and then total sum of squares (TSS)

Total sum of squares (TSS) =ki = 1 kj = 1 xij2 - CF
= 394 - 363
= 31

step 5 Find sum of squares between rows (SSR)
SSR = ∑Ti2h - 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 = ∑Tj2k - 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 = FR between Varieties h - 1 SSC SSC/h-1 = MSV MSV/MSE = FC 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 Fe or Table values of F
1. 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.
2. 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
1. For treatments, the critical value Fe(2,6) from F-distribution table at 5% of significance level is 5.1433
2. For varieties, the critical value Fe(3,6) from F-distribution table at 5% of significance level is 4.7571

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