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    F-Test Solved Example for x̄ = 78.8, ȳ = 76

    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 = {50, 72, 85, 96 & 91} & y = {55, 65, 77, 89 & 94} to check if the results of analysis between two variances is statistically significant.
    Calculation Summary
    Data set value x{50, 72, 85, 96 & 91}
    Data set value y{55, 65, 77, 89 & 94}
    F01.2867
    Fe4.1072

    F0, Fe & H0 Workout for x̄ = 78.8, ȳ = 76 & α = 0.1

    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 = 0.1) for x = 50, 72, 85, 96 & 91 and y = 55, 65, 77, 89 & 94 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 = {50, 72, 85, 96 & 91}
    y = {55, 65, 77, 89 & 94}
    α = 0.1

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

    x̄ = ∑xn1 & ȳ = ∑yn2

    x̄ = 3945& ȳ = 3805

    x̄ =78.8 & ȳ =76

    xx - x̄(x - x̄)2yy - ȳ(y - ȳ)2
    50-28.8829.4455-21441
    72-6.846.2465-11121
    856.238.447711
    9617.2295.848913169
    9112.2148.849418324
    ∑x = 394∑(x-x̄) = 1.4210854715202E-14∑(x-x̄)2 = 1358.8∑y = 380∑(y-ȳ) = 0∑(y-ȳ)2 = 1056

    step 3 Substitute the above table to below formulas

    S12 = ∑(x - x̄)2n1-1

    S12 = 1358.85-1 = 1358.84

    S12 = 339.7


    S22 = ∑(y - ȳ)2n2-1

    S12 = 10565-1 = 10564

    S22 = 264

    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 = 339.7/264

    F0 = 1.2867

    step 5 Find critical value of F (Fe) using stated level of significance (α = 0.1) & degrees of freedom (ν1, ν2) from F-distribution table
    ν1 = n1 - 1
    ν2 = n2 - 1
    ν1 = 5 - 1 = 4
    ν2 = 5 - 1 = 4

    The critical value F(4, 4) from F-distribution table at 0.1 of significance level is 4.1072
    F e= 4.1072

    Inference
    There is no significance difference
    since the calculated value of F0 = 1.2867 is smaller than the table value Fe = 4.1072
    Therefore the null hypothesis H0 is accepted.

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