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    F-Test Solved Example for x̄ = 14.7143, ȳ = 15.5714

    F-Test CalculatorSolved example with steps, formula & summary to estimate the F-statistic (F0), F-critical (table) value (Fe) for given degrees of freedom & hypothesis test (H0) at a stated level of significance for sample x = {11, 13, 14, 16, 18, 11 & 20} & y = {20, 13, 12, 14, 15, 17 & 18} to check if the results of analysis between two variances is statistically significant.
    Calculation Summary
    Data set value x{11, 13, 14, 16, 18, 11 & 20}
    Data set value y{20, 13, 12, 14, 15, 17 & 18}
    F01.4368
    Fe20.0891

    F0, Fe & H0 Workout for x̄ = 14.7143, ȳ = 15.5714 & α = 0.001

    The below is the work with step by step calculation shows how to estimate the F-statistic (F0), critical (table) value (Fe) for degrees of freedom & hypothesis test (H0) at a stated level of significance (α = 0.001) for x = 11, 13, 14, 16, 18, 11 & 20 and y = 20, 13, 12, 14, 15, 17 & 18 to check the quality of variances among two or more sample dataset may help learners or grade school students to solve the similar F-test (F0, Fe & H0) worksheet problems efficiently.

    Workout :

    step 1 Address the formula, input parameters and values
    x = {11, 13, 14, 16, 18, 11 & 20}
    y = {20, 13, 12, 14, 15, 17 & 18}
    α = 0.001

    step 2Refer the below F-table to conduct the test of significance

    x̄ = ∑xn1 & ȳ = ∑yn2

    x̄ = 1037& ȳ = 1097

    x̄ =14.7143 & ȳ =15.5714

    xx - x̄(x - x̄)2yy - ȳ(y - ȳ)2
    11-3.714313.79602449204.428619.61249796
    13-1.71432.9388244913-2.57146.61209796
    14-0.71430.5102244912-3.571412.75489796
    161.28571.6530244914-1.57142.46929796
    183.285710.7958244915-0.57140.32649796
    11-3.714313.79602449171.42862.04089796
    205.285727.93862449182.42865.89809796
    ∑x = 103∑(x-x̄) = -9.9999999997991E-5∑(x-x̄)2 = 71.42857143∑y = 109∑(y-ȳ) = 0.00019999999999598∑(y-ȳ)2 = 49.71428572

    step 3 Substitute the above table to below formulas

    S12 = ∑(x - x̄)2n1-1

    S12 = 71.428571437-1 = 71.428571436

    S12 = 11.904761905


    S22 = ∑(y - ȳ)2n2-1

    S12 = 49.714285727-1 = 49.714285726

    S22 = 8.2857142866667

    step 4 Find F-statistic (F0) using the below formula

    F0= S12S22

    It should be noted that the numerator is always greater than the denominator in F0
    F0 = 11.904761905/8.2857142866667

    F0 = 1.4368

    step 5 Find critical value of F (Fe) using stated level of significance (α = 0.001) & degrees of freedom (ν1, ν2) from F-distribution table
    ν1 = n1 - 1
    ν2 = n2 - 1
    ν1 = 7 - 1 = 6
    ν2 = 7 - 1 = 6

    The critical value F(6, 6) from F-distribution table at 0.001 of significance level is 20.0891
    F e= 20.0891

    Inference
    There is no significance difference
    since the calculated value of F0 = 1.4368 is smaller than the table value Fe = 20.0891
    Therefore the null hypothesis H0 is accepted.

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