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    Z-Test (Z0, Ze & H0) for p1 = 0.75 & p2 = 0.85 with unknown P Values

    Z-Test Calculator

    Work with steps, formula & calculation summary to estimate the Z-statistic (Z0), critical (table) value (Ze) for degrees of freedom & hypothesis test (H0) at a stated level of significance for difference between two sample proportions p1 = 0.75 & p2 = 0.85 with unknown population probability for sample size n1 = 90 & n2= 100. The below is the calculation summary for Z-statistic for the test of significance or hypothesis for difference between two sample proportions p1 = 0.75 & p2 = 0.85.

    Calculation Summary
    Sample proportion p10.75
    Sample proportion p20.85
    Sample size n190
    Sample size n2100
    Z01.7291

    Z-Test Work with Steps for p1 = 0.75, p2 = 0.85

    The below is the work with step by step calculation shows how to estimate the Z-statistic (Z0), critical (table) value (Ze) for degrees of freedom & hypothesis test (H0) at a stated level of significance for difference between two sample proportions p1 = 0.75 & p2 = 0.85 with unknown P values for sample size n1 = 90 & n2 = 100 may help grade school students to solve the similar Z-test statistic (Z0) worksheet problems efficiently.

    Workout :
    step 1 Address the formula input parameters and values
    Sample Proportion p1 = 0.75
    Sample Proportion p2 = 0.85
    Sample Size n1 = 90
    Sample Size n2 = 100
    z score value (z) = 0.78
    Formula
    Z0=p1 - p2
    P̄Q̄ (1 / n1 + 1 / n2)



    step 2 Find estimated proportion
    P̄ = n1p1 + n2p2n1 + n2
    P̄ = (90 x 0.75 + 100 x 0.85)/(90 + 100)
    = (67.5 + 85)/190
    = 152.5/190
    Probability P̄ = 0.8026
    Q̄ = 1 - P̄ = 1 - 0.8026
    Q̄ = 0.1974

    step 3 Substitute p1, p2, P̄, Q̄, n1 & n2 value in the below test significance for difference between two proportion formula
    =0.75 - 0.85
    0.8026x0.1974 {(1 / 90) + (1 / 100)}


    step 4 Simplify the above expression
    =0.1
    0.8026x0.1974 {(1 / 90) + (1 / 100)}


    =0.1
    0.8026x0.1974 (0.0111 + 0.01)


    =0.1
    0.8026 x 0.1974 x 0.0211


    =0.1
    0.0033


    =0.10.0578

    Z0 = 1.7291

    Inference
    There is significance difference
    since the calculated value of Z0 = 1.7291 is greater than the table value of Ze = 0.78.
    Therefore the null hypothesis H0 is rejected.

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