Z test for difference in proportions

• Assumptions
  

1. Method of simple random sampling is used
   
2. Two samples are independent
3. There are only two outcomes, p (probability of success) or q (probability of failure). Also, p + q = 1
4. If n1 p1≥10 and n1 q1≥10 for sample 1 and n2 p2≥10 and n2 q2≥10 for sample 2, then normal approximation to binomial is valid

• 5-Step Hypothesis
  

Null and Alternative Hypothesis

• Ho: there is no significant difference in the population proportion of two groups. p1-p2=0

• Ha1: there is significant difference in the population proportion of two groups. p1-p2≠0 (Two tailed test)

Ha2: population mean of sample 1 is less than that of sample 2. μ1-μ2<0  (Left tailed test)

Ha3: population mean of sample 1 is greater than that of sample 2. μ1-μ2>0  (Right tailed test)

Test Statistic

Critical value

• -z(a/2), z(a/2) (Two tailed test)
• -z(a) (Left tailed test)
• z(a) (Right tailed test)

P-value

• 2*(1-P(Z≤|z|)  (Two tailed test)
• P(Z≤z)  (Left tailed test)
• P(Z≥z)  (Right tailed test)

Decision rule

• Reject Ho if |z| > z(a/2) or p-value < alpha (two tailed test)
• Reject Ho if –z < -z(a) or p-value < alpha (left tailed test)
• Reject Ho if z > z(a) or p-value < alpha (right tailed test)

• Confidence Interval

Standard error (SE) and margin of error (ME) is given by:

100(1-alpha)% Confidence interval for the population proportions difference is given by:

This implies I am 100(1-alpha)% confident that estimated population mean difference between two samples lies in the obtained interval. If confidence interval contains 0, I fail to reject null hypothesis ho and conclude that there is no significant difference in the means of two samples.

z test for difference in proportions CALCULATOR