Hypothesis Testing: Two Proportions (aka Difference of Proportions)

Introduction

Two Proportions hypothesis tests are used when...

• You are comparing two different populations
• You have TWO proportions from TWO INDEPENDENT random samples

For example, as a researcher, you might want to know if there is a difference in the proportion of males who use Facebook and the proportion of females who use Facebook. A quality control specialist might also want to know if there is a difference in the percentage of defective items produced by two different machines.

A few symbols need to be defined before we dive in:

• $$\hat{p_1}$$ and $$\hat{p_2}$$ refer to the sample proportions that you will use to disprove the null.
• $$\hat{p_c}$$ refers to the combined proportion (formula down below ↓ )
• $$\hat{q_c}$$ refers to 1 minus the combined proportion, i.e. $$1 - \hat{p_c}$$
• $$n_1$$ and $$n_2$$ refer to the sample sizes.

Example

A columnist claims that women are more safety-conscious than men when it comes to driving. A recent survey on use of seatbelts was done among a random sample of 150 men and 250 women. Based on the results, 105 men said they always wear seatbelts when driving and 186 women said the same. Using a 0.05 level of significance, do the results of the survey support the columnist’s claim?

Step 1: Name Test: 2-Proportions / Difference of Proportions

Step 2Define Test:

The null hypothesis assumes that the proportions are equal  ($$H_0 : p_1 = p_2$$)

With this null hypothesis, the options for the alternative hypothesis are as follows:

 Left-Sided Test Two-Sided Test Right-Sided Test $$H_0: p_1 = p_2$$ $$H_A: p_1 < p_2$$ $$H_0: p_1 = p_2$$ $$H_A: p_1 \neq p_2$$ $$H_0: p_1 = p_2$$ $$H_A: p_1 > p_2$$

In this case, let's call the proportion of men who wear seatbelts $$p_M$$ and the proportion of women who wear seatbelts $$p_W$$. If the alternative hypothesis is that women are more safety-conscious than men, then women should have a higher seatbelt usage and $$p_W > p_M$$

$$H_0 : p_W = p_M$$

$$H_A : p_W > p_M$$

Step 3: Assume $$H_0$$ is true and define its normal distribution. Then check the conditions.

1. The data is drawn from TWO independent random samples.

2a.  From Sample 1: $$N_1 ≥ 10n_1$$

2b.  From Sample 2: $$N_2 ≥ 10n_2$$

3a.  From Sample 1: $$n_1 \hat{p_1} ≥ 10$$ and $$n_1 \hat{q_1} ≥ 10$$

3b.  From Sample 2: $$n_2 \hat{p_2} ≥ 10$$ and  $$n_2 \hat{q_2} ≥ 10$$

Step 4: Using the normal distribution, calculate the test statistics and p-value.

Although the full formula is  $$z = {( \hat{p_1} - \hat{p_2} ) - ( p_1 - p_2) \over \sqrt { {\hat{p_c}\hat{q_c} \over n_1} + {\hat{p_c}\hat{q_c} \over n_2} } }$$ , it can be simplified. Recall that the null is $$H_0 : p_1 = p_2$$. Thus, $$p_1 - p_2 = 0$$ . This leaves us with the formula below:

Test Statistic (Difference of 2 Proportions):  $$z = {( \hat{p_1} - \hat{p_2} ) \over \sqrt { {\hat{p_c}\hat{q_c} \over n_1} + {\hat{p_c}\hat{q_c} \over n_2} } }$$

Now, let's consider how to calculate the combined proportion $$\hat{p_c}$$. Recall that the proportion $$\widehat{p}$$ of a sample having a certain attribute is given by $$\widehat{p} = {x \over n}$$ , where $$x$$ is the number of elements in the sample possessing that certain attribute and $$n$$ is the sample size. Thus, the combined proportion $$\hat{p_c}$$ is calculated as follows:

Combined Proportion:  $$\hat{p_c} = {x_1 + x_2 \over n_1 + n_2}$$

Test Statistic:

$$\hat{p_W} = {186 \over 250} = 0.744$$  and  $$\hat{p_M} = {105 \over 150} = 0.70$$

$$\hat{p_c} = {186 + 105 \over 250 + 150} = 0.7275$$

→    $$z = {( 0.744 - 0.70 ) \over \sqrt { {(0.7275)(0.2725) \over 250} + {(0.7275)(0.2725) \over 150} } }$$  →  $$z = 1.045$$

P-Value:

The p-value will be found by using the normal cdf function on your calculator:

• lower limit: $$z$$
• upper limit: 999
• distribution center: 0
• standard deviation: 1
• All together, it looks like this: normalcdf ($$z$$, 999, 0, 1)

*Note: If it was a left-sided test and the test statistic was negative (z < 0), then your lower limit would be -999 and your upper limit would be the test statistic ($$z$$).

In this case, we do normalcdf (1.045, 999, 0, 1) to get a p-value of 0.15.

Step 5: Analyze your results and determine if they are statistically significant.

We calculated a p-value of 0.15. This p-value is greater than the significance level of 0.05. Therefore, we FAIL to reject the null hypothesis. The data does NOT support the columnist’s claim that there is a difference between the proportion of men and women who always use seatbelt when driving.