Z-Test (Z0, Ze & H0) for p1 = 0.65 & p2 = 0.72 with P Values 0.34 & 0.39

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 = 0.65 & p2 = 0.72, P values of population probability P1 = 0.34 & P2 = 0.39 for sample size n1 = 89 & n2= 95 The below is the calculation summary for Z-statistic for the test of significance or hypothesis for difference between two sample proportions p1 = 0.65 & p2 = 0.72
Calculation Summary | |
---|---|
Sample proportion p1 | 0.65 |
Sample proportion p2 | 0.72 |
Population probability P1 | 0.34 |
Population probability P2 | 0.39 |
Sample size n1 | 89 |
Sample size n2 | 95 |
Z0 | 0.9874 |
Z-Test Work with Steps for p1 = 0.65, p2 = 0.72
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 = 0.65 & p2 = 0.72 with known P values of population probability P1 = 0.34 & P2 = 0.39 for sample size n1 = 89 & n2= 95 may help grade school students to solve the similar Z-test statistic (Z0) worksheet problems efficiently.
step 1 Address the formula input parameters and values
Sample Proportion p1 = 0.65
Sample Proportion p2 = 0.72
Population Probability P1 = 0.34
Population Probability P2 = 0.39
Sample Size n1 = 89
Sample Size n2 = 95
z score value (z) = 0.7
Formula
Z0=p1 - p2
√
step 2 Find Q1 and Q2 from P1 and P2 respectively
Q1 = 1 - P1
Q1 = 0.66
Q2 = 1 - P2
Q2 = 0.61
step 3 Substitute p1, p2, P1, P2, Q1, Q2, n1 & n2 value in the below formula for test of significance for difference between two proportion having known P1 & P2 values.
=0.65 - 0.72
√
step 4 Simplify the above expression
=0.07
√
=0.07
√
=0.07
√
=0.07
√
=0.070.0709
Z0 = 0.9874
Inference
There is significance difference
since the calculated value of Z0 = 0.9874 is greater than the table value of Ze = 0.7.
Therefore the null hypothesis H0 is rejected.
