Calculators & Converters

    CHI-squared Test Example for df = 3 at α =

    Chi-squared (χ2) Test CalculatorSolved example work with steps, formula & summary to estimate χ²-statistic (χ²0), critical (table) value (χ²e) for given degrees of freedom & hypothesis test (H0) at a stated level of significance (α = ) for observed frequencies O = {95, 35, 75 & 25} & expected frequencies E = {110, 56, 70 & 23α=0.01} to check if the result of this probability & statistics experiment is statistically significant.
    Calculation Summary
    Observed Frequency (O){95, 35, 75 & 25}
    Expected Frequency (E){110, 56, 70 & 23α=0.01}
    Significance Level (α)
    χ²010.4515
    χ²e0

    Workout for O = {95, 35, 75 & 25} & E = {110, 56, 70 & 23α=0.01}

    The below is the solved example with step by step calculation shows how to estimate the chi-squared statistic (χ²0), critical (table) value (χ²e) for degrees of freedom & hypothesis test (H0) at a stated level of significance (α = ) for observed frequencies O = {95, 35, 75 & 25} & expected frequencies E = {110, 56, 70 & 23α=0.01} to check the quality of variances between two or more samples which are not uniformly distributed, may help learners or grade school students to solve the similar chi-squared test (χ²0, χ²e & H0) worksheet problems efficiently.

    Workout :
    step 1 Address the formula input parameters and values
    observed frequency = 95, 35, 75 & 25
    expected frequency = 110, 56, 70 & 23α=0.01
    Significance Level (α) =

    step 2 Refer the below χ² table to find χ²-statistic (χ²0)

    χ² = ∑(Oi - Ei)2Ei

    Observed Frequency (O)Expected Frequency (E)(O - E)2(O - E)2E
    951102252.0455
    35564417.875
    7570250.3571
    2523α=0.0140.1739
    (O - E)2E = 10.4515
    χ²0 = 10.4515

    step 3 Find the degrees of freedom
    df = n - 1
    df = 4 - 1
    df = 3

    step 4 Find the critical value of χ²e for df = 3 at a stated level of significance α = from χ²-distribution table.
    χ²e = 0

    Inference
    for χ²0 > χ²e
    There is significance difference between sample and population
    since χ²0 = 10.4515 is greater than χ²e = 0.
    Therefore the null hypothesis H0 is rejected.

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