Hypothesis Test of Two Population Proportions

Comparing Two Population Proportions with Dependent Samples

One of the assumptions you will need is that your samples are independent.

In the example involving employees, picking a person who is a full-time employee doesn't influence who you pick as a part-time employee, so they are independent. That is very different from dependent samples.

If you were doing a study on twins then picking a twin for one sample would automatically put the other twin in the second sample. Twins are a common example of dependent samples. This is called matched-pair data, and it requires a different form of hypothesis testing than you will see here.

Forming Your Hypothesis

There are many ways that \(p_1\) can be different from \(p_2\). It might be that \(p_1 < p_2\), or that \(p_1>p_2\). Rather than try and list all of the ways they are different and do a hypothesis test for each, you can look at the difference between the two population proportions. In fact, a hypothesis test for two population proportions is often called a hypothesis test for the difference between two population proportions for this very reason!

In this kind of hypothesis test, your null hypothesis will almost always be that the two population proportions are the same. If you state that in terms of their difference you get:

\[ H_0:\; p_1 - p_2 = 0.\]

Then there are three varieties of alternative hypotheses outlined in the next table.

Let's go back to the example from the start of this article.

Next, let's look at the test statistic for this type of hypothesis test.

Significance Test Statistic for Two Population Proportions

It is important that your samples are independent, or the test statistic will be different from the one shown here. Since you are using independent samples, remember that

\[ \mu_{\hat{p}_1 - \hat{p}_2} = p_1 - p_2.\]

For the standard deviation,

\[ \sigma_{\hat{p}_1 - \hat{p}_2} = \sqrt{ \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} }.\]

So far you have only assumed that the samples are independent. For the next part, you will need to assume that the sample sizes are large enough. If they are, you can use the Central Limit Theorem to get that your sampling distribution \(\hat{p}_1 - \hat{p}_2 \) is approximately normal.

How do you know if your samples are large enough? If all four of the following conditions are satisfied, then your samples are large enough for the sampling distribution \(\hat{p}_1 - \hat{p}_2 \) to be approximately normal:

• \[n_1\hat{p_1} \ge 10\].

• \[n_2\hat{p_2} \ge 10\].

• \[n_1(1-p_1) \ge 10\]. and

• \[n_2(1-p_2) \ge 10\].

It isn't too hard to check that the sample sizes in the savings example are large enough for the sampling distribution to be approximately normal.

The last condition to use this type of hypothesis test is that your sample is less than \(10\%\) of the overall population. In this case, the sample size is certainly less than \(10\%\) of all of the people in your country, so this condition is satisfied as well.

Z-test for Difference in Population Proportions

When doing a hypothesis test for the difference in population proportions, a \(z\)-test is used. To do this, you will need to calculate the test statistic, which uses the difference in the two proportions. To make calculations a little easier, it is helpful to find:

\[ \begin{align}\hat{p}_c &= \frac{\text{number of successes in the two samples} }{\text{total of the two sample sizes}} \\ &= \frac{n_1\hat{p_1} + n_2\hat{p_2} }{n_1 + n_2} \end{align}\]

Combining counts to get an overall proportion is called pooling, and \(p_c\) is called the pooled (or combined) proportion.

As long as your null hypothesis is \(H_0:\; p_1 -p_2 = 0 \), the test statistic can be calculated using the formula:

\[ z = \frac{\hat{p_1} - \hat{p_2} }{\sqrt{ \dfrac{\hat{p}_c (1-\hat{p}_c) }{n_1} +\dfrac{\hat{p}_c (1-\hat{p}_c) }{n_2} } }\]

Let's finish up the hypothesis test for the savings example. No significance level was given, so you will need to consider the Type I and Type II error consequences. See Errors in Hypothesis Testing for more information and examples. In this example, a Type I error would be deciding that the savings proportions are not the same for the two groups when in fact they are the same.

A Type II error would be not thinking there is a difference in the population proportion between the two groups when in fact they are not the same. Neither error is very bad (unlike in a medical trial where the type of error is of much more importance) so choosing a significance level of \(\alpha = 0.05\) would be fine.

Remember that this is a two-tailed test! So the \(P\)-value is twice the area under the \(z\)-curve and to the right of the \(z\)-value. In other words:

\[ \begin{align} P\text{-value} &= 2(\text{area under curve to the right of }0.63) \\ &= 2\cdot P(z>0.63) \\ &= 2(0.2643) \\ &\approx 0.529 \end{align} \]

The \(P\)-value is greater than the significance level of \(\alpha = 0.05\), so you will fail to reject the null hypothesis.

Why was there a difference in the population proportions then? It might have been from sampling variability. All you can say from the sample proportions is that you are not convinced there is a difference between the two sampling proportions.

Hypothesis Testing of Two Population Proportions Example

Let's look at another example of hypothesis testing for the difference in two population proportions.

Many bulldog owners report that their pet snores, and in fact, their bulldog snores more frequently as it gets older.

Sleeping bulldog puppy.

You have decided to do a test to see if this is actually true or maybe just a matter of perception. So you break down bulldogs into two groups, those under three years of age and those over three years of age, and choose a random sample of \(700\) bulldog owners to ask them about their dog's snoring. From the survey responses (not everyone responds to surveys), you create the following table:

Before going any further, let's check to make sure that the conditions for doing a hypothesis test for two population proportions are satisfied. First, the samples are independent since a bulldog can't be both under \(3\) years old and over \(3\) years old at the same time. In addition, there are certainly far more than \(591\) people worldwide that own bulldogs, so the number of bulldog owners sampled is less than \(10\%\) of the overall population of people who own bulldogs. Also,

• \(n_1\hat{p_1} = 300(0.26)=78 \ge 10\),

• \(n_2\hat{p_2} = 291(0.392) = 114 \ge 10\).

• \(n_1(1-p_1) = 300(1-0.26) = 222 \ge 10\)

• \(n_2(1-p_2) = 291(1-0.392) = 176.9 \ge 10\).

so all of the conditions for applying the test are met.

The next step is deciding on the null and alternative hypotheses. The null hypothesis would be:

\[ H_0: \; p_2-p_1 = 0\]

or in other words that there is no difference in snoring between the two groups. The alternative hypothesis would be that there is a difference in the snoring rates of the two groups, so:

\[H_a:\; p_2-p_1 \ne 0\]

Calculating the pooled success rate (sometimes called the combined success rate):

\[ \begin{align}\hat{p}_c &= \frac{n_1\hat{p_1} + n_2\hat{p_2} }{n_1 + n_2} \\ &= \frac{300(0.26)+291(0.392)}{300+291} \\ &\approx 0.325 . \end{align}\]

Then the test statistic is:

\[\begin{align} z &= \frac{\hat{p_2} - \hat{p_1} }{\sqrt{ \dfrac{\hat{p}_c (1-\hat{p}_c) }{n_1} +\dfrac{\hat{p}_c (1-\hat{p}_c) }{n_2} } } \\ &= \frac{ 0.392 - 0.26 }{\sqrt{ \dfrac{0.325 (1-0.325) }{300} +\dfrac{0.325 (1-0.325) }{291} } } \\ &\approx 3.425 \end{align}\]

Remember that this is a two-tailed test! So the \(P\)-value is twice the area under the \(z\)-curve and to the right of the \(z\)-value. In other words:

\[ \begin{align} P\text{-value} &= 2(\text{area under curve to the right of }3.425) \\ &= 2\cdot P(z>3.425) \\ &\approx 2(0.0003) \\ &= 0.0006, \end{align} \]

where the value of \(P(z>3.425)\) can be found using a standard normal table or calculator.

So at a \(\alpha = 0.05\) significance level, you can reject the null hypothesis, and conclude that there is a difference in bulldog snoring based on age.