## Two Proportion Z Test

**Overview:**

A Two-Proportion Hypothesis Test allows you to test a statistical hypothesis about the difference between two proportions in a population. You use a two prop z-test when you are analyzing two samples, and you are making hypothesis based on two proportions.

**Steps to Conduct a Two Prop z-Test**

- Test the conditions
- State the hypothesis
- Perform the mechanics
- State the conclusion

**Conditions**

- Sampling method for each population is selected using randomization.
- Each sample has at least 10 successes and 10 failures
- Each sample has to be less than 10% of the population

Hypotheses Each hypothesis test should have two hypotheses. The first is the null hypothesis (H0), which is usually p1-p2=0 or p1=p2. The other hypothesis is the alternative hypothesis (HA), which is the expressions that are reasonable for p1 and p considering H0 is false. The alternative hypothesis will be one of the following:

**Mechanics**

To perform a one prop z-test, use your calculator and follow these steps:

1. Push the STAT button, tab over to TESTS and select 6: 2-PropZTest

2. Enter the observed value for sample 1 for x1, the sample size for sample 1 for n1, the observed value for sample 2 for x2, the sample size for sample 2 for n2, and the alternative hypothesis (HA) for p1.

3. Tab down to calculate and hit enter. The p value will be displayed next to p=

4. For an FRQ record the test, z-score, and p-value.

**Conclusion**

Your conclusion should follow this basic format: With a p-value of (insert p value here) I (reject/fail to reject) the null hypothesis. There (is/is not enough) evidence to conclude that (Insert HA here).

Whether the null hypothesis is accepted or rejected is based on the p-value. Remember this phrase:

**If p is high, let it fly… if p is low, reject that null.**Generally a low p-value is one that is below .05.

**Example:**

Suppose the Theta Drug Company develops a new drug, designed to prevent the flu. The company states that the drug is more effective for women than for men. To test this claim, they choose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.

At the end of the study, 38% of the women caught a cold; and 51% of the men caught the flu. Based on these findings, can we conclude that the drug is more effective for women than for men? Use a 0.01 level of significance.

*Solution:*

The null hypothesis is that the drug is just as effective for women and men, as in Ho:p1=p2 where p1 is women and p2 is men.

The alternative hypothesis is p1<p2.

Push the STAT button, tab over to TESTS and select 6: 2-PropZTest. Insert the data.

X1:38

n1:100

X2:102

N2:200

p1<p2

You should get a p-value of .0167.

This is above the indicated level of significance of .01. The conclusion should look something like this:

With a p-value of

**.0167**I

**fail**

**to reject**the null hypothesis. There

**is not enough**evidence to conclude that

**the drug is more effective for women**.