This lesson explains how to conduct a hypothesis test for a difference between two population proportions. This type of test is known as a two-proportion z-test.

**Checking Conditions**

Before we can conduct a hypothesis test for a difference between two population proportions, we first need to make sure the following conditions are met to ensure that our hypothesis test will be valid:

**Random:**A random sample or random experiment should be used to collect data for both samples.**Independence:**The two samples are independent.**Successes & Failures:**Each sample includes at least 10 successes and 10 failures.**Size:**The size of each sample is at most 5% of its population.

If these conditions are met, we can then conduct the hypothesis test. The following two examples show how to conduct a hypothesis test for a difference between two population proportions.

**Hypothesis Test for a Difference in Proportions (Two-Tailed)**

A superintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school cafeterias is the same for school 1 and school 2. To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences. He finds that 70% of students prefer chocolate milk in school 1 and 68% of students prefer chocolate milk in school 2.

**Based on these results, can we reject the superintendent’s claim that the percentage of students who prefer chocolate milk is the same for school 1 and school 2? Use a .05 level of significance. **

**Step 1. State the hypotheses. **

The null hypothesis (H0): P_{1} = P_{2}

The alternative hypothesis: (Ha): P_{1} ≠ P_{2}

**Step 2. Determine a significance level to use.**

The problem tells us that we are to use a .05 level of significance.

**Step 3. Find the test statistic.**

First, find the pooled sample proportion p:

p = (p_{1} * n_{1} + p_{2} * n_{2}) / (n_{1} + n_{2})

p = (.70*100 + .68*100) / (100 + 100) = .69

Then use p in the following formula to find the test statistic z:

z = (p_{1}-p_{2}) / √p * (1-p) * [ (1/n_{1}) + (1/n_{2})]

z = (.70-.68) / √.69 * (1-.69) * [ (1/100) + (1/100)] = .02 / .0654 = **.306**

Use the Z Score to P Value Calculator with a z score of .306 and a two-tailed test to find that the p-value = **0.759**.

**4. Reject or fail to reject the null hypothesis.**

Since the p-value is not less than our significance level of .05, we fail to reject the null hypothesis.

**5. Interpret the results. **

Since we failed to reject the null hypothesis, we do not have sufficient evidence to say that the percentage of students who prefer chocolate milk is different for school 1 and school 2.

**Hypothesis Test for a Difference in Means (One-Tailed)**

A superintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school 1 is less than or equal to the percentage in school 2. To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences. He finds that 70% of students prefer chocolate milk in school 1 and 68% of students prefer chocolate milk in school 2.

**Based on these results, can we reject the superintendent’s claim that the percentage of students who prefer chocolate milk in school 1 is less than or equal to the percentage in school 2? Use a .05 level of significance. **

**Step 1. State the hypotheses. **

The null hypothesis (H0): P_{1} ≤ P_{2}

The alternative hypothesis: (Ha): P_{1} > P_{2}

**Step 2. Determine a significance level to use.**

The problem tells us that we are to use a .05 level of significance.

**Step 3. Find the test statistic.**

First, find the pooled sample proportion p:

p = (p_{1} * n_{1} + p_{2} * n_{2}) / (n_{1} + n_{2})

p = (.70*100 + .68*100) / (100 + 100) = .69

Then use p in the following formula to find the test statistic z:

z = (p_{1}-p_{2}) / √p * (1-p) * [ (1/n_{1}) + (1/n_{2})]

z = (.70-.68) / √.69 * (1-.69) * [ (1/100) + (1/100)] = .02 / .0654 = **.306**

Use the Z Score to P Value Calculator with a z score of .306 and a two-tailed test to find that the p-value = **0.379**.

**4. Reject or fail to reject the null hypothesis.**

Since the p-value is not less than our significance level of .05, we fail to reject the null hypothesis.

**5. Interpret the results. **

Since we failed to reject the null hypothesis, we do not have sufficient evidence to say that the percentage of students who prefer chocolate milk in school 1 is *not *less than or equal to the percentage in school 2.