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 Variation | df | SS | MSS | F-Ratio |
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 |
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
Individual Squares :
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) =k∑i = 1 k∑j = 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.
ANOVA Table
step 9 Find Critical values Fe or Table values of F
Critical values of F from F-distribution table
Inference
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 |
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 |
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) =k∑i = 1 k∑j = 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 |
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
- 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 Fe(2,6) from F-distribution table at 5% of significance level is 5.1433
- For varieties, the critical value Fe(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 Fe = 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 Fe = 4.7571. Therefore the null hypothesis H0 is accepted.
