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