Calculators & Converters

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

    One & Two Way ANOVA Calculator

    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 sets1234TotalSquare
    x18755 ∑x1 = 25 (∑x1)2 = 625
    x27448 ∑x2 = 23 (∑x2)2 = 529
    x33456 ∑x3 = 18 (∑x3)2 = 324
    Total18151419 G = 66 ∑Ti2 = 1478
    Square324225196361∑Tj2 = 1106
    Individual Squares :
    Data sets1234Total
    x1264492525163
    x2249161664145
    x32916253686
    Total1228166125 ∑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 variationdfSSMSSF-Ratio
    between Treatmentsk - 1SSRSSR/k-1

    = MST
    MST/MSE

    = FR
    between Varietiesh - 1SSCSSC/h-1

    = MSV
    MSV/MSE

    = FC
    Error(h - 1)(k - 1)SSESSE/(k-1)(h-1)

    = MSE
     
    TotalN - 1   
    ANOVA Table
    Sources of variationd.fSSMSSF-Ratio
    between Treatments3 - 1 = 26.56.5/2

    = 3.25
    3.25/3.1389

    = 1.0354
    between Varieties4 - 1 = 35.66675.6667/3

    = 1.8889
    1.8889/3.1389

    = 0.6018
    Error(3 - 1)(4 - 1) = 618.833318.8333/6

    = 3.1389
     
    Total12 - 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.

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