This lesson describes how to calculate probabilities associated with the difference between two proportions.

**Difference Between Proportions**

Often in statistics we’re interested in comparing the difference between two population proportions. To do so, we obtain sample data from both populations and compare the two sample proportions. But before we can compare the difference between proportions, we first need to make sure the following conditions are met:

**Size:**The size of the population is large relative to the sample being drawn from the population.**Expectation:**The sample sizes need to be big enough so that we can use a normal distribution to model the difference between the sample proportions. Specifically, we need to make sure that n_{1}P_{1}__>__10, n_{1}(1 -P_{1})__>__10, n_{2}P_{2}__>__10, and n_{2}(1 – P_{2})__>__10.**Independence:**The two samples need to be independent.

If these conditions are met, we can use the following formulas to quantify the difference between the two proportions:

The expected value of the difference between all possible sample proportions is:

**E(p _{1} – p_{2})** = P

_{1}– P

_{2}

where p_{1} is the sample 1 proportion, p_{2 }is the sample 2 proportion, P_{1} is the population 1 proportion, and P_{2} is the population 2 proportion.

The standard deviation of the difference between sample proportions is:

**σ _{d}** = √ [P

_{1}(1 – P

_{1}) / n

_{1}] + [P

_{2}(1 – P

_{2}) / n

_{2}]

Let’s walk through an example of how to use these formulas to answer a question about probability.

**Example: Finding the Difference Between Proportions**

In school 1, 70% of students prefer pizza over ice cream. In school 2, 68% of students prefer pizza over ice cream. Researchers obtain a simple random sample of 100 students from each school and survey them about their preferences. **What is the probability that the survey will show that a greater percentage of students in school 2 prefer pizza compared to school 1?**

**Solution:**

**Step 1: Find the expected difference in proportions between school 1 and school 2.**

**E(p _{1} – p_{2})** = P

_{1}– P

_{2}=.70 – .68 =

**.02**

**Step 2: Find the standard deviation of the difference.**

**σ _{d}** = √ [.7(1 – .7) / 100] + [.68(1 – .68) / 100] =

**.065**

Step 3:

**Step 3: Find the z-score for when the proportion of school 1 is less than the proportion of school 2 (i.e. when. P _{1} – P_{2} < 0)**

z = (x – μ)/σ = (0 – .02)/.065 =** -0.31**

**Step 4: Find the cumulative probability associate with the z-score.**

In this case, we want to know the probability that proportion 1 is *less *than the proportion 2, so we want to find the area to the left of z = -0.31 in the z table, which turns out to be **.3783**:

Thus, the probability that the survey will show that a greater percentage of students in school 2 prefer pizza compared to school 1 is **0.3783**.