## Hypothesis Test – Single Proportion (p-value)

In this blog I will go over the steps for performing a hypothesis test for a single population proportion. I will use the p-value approach, and give directions for using StatCrunch.

Example: A physician claims that more than 10% of pregnant women smoke while pregnant. A survey of 400 randomly selected pregnant women revealed that 60 of them smoked while pregnant. Test the physician’s claim at the 0.05 level of significance.

This is a one-proportion test because the claim compares a single population proportion (the proportion of all pregnant women who smoke) to a certain number (10%). In addition we have one set of sample information of the form x out of n (60 out of 400).

Step 1

Since the claim is that the proportion of all pregnant women who smoke is above 10%, this tells us that the hypotheses for this test are:
$H_0: p=0.10\\ H_1: p>0.10$

Step 2

The level of significance stated in the problem is $\alpha=0.05$. If the p-value is less that 0.05, this is sufficient sample evidence to reject the null hypothesis.

Step 3

The appropriate test statistic for the one-proportion hypothesis test is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$. This formula is used to calculate a z-value associated with the sample data, which is in turn used to determine the p-value for the test.

Step 4

The decision rule for this test is to reject $H_0$ if the p-value is less than 0.05. When using the p-value approach, the decision rule is always to reject $H_0$ is &latex p-value<\alpha}\$.

Step 5

In StatCrunch, begin by following Stat > Proportions > One sample > with summary.
Enter the sample information for the number of successes (60) and the number of observations (400). Click the “Next” button.
The hypothesis testing option should be active. In the box for the null hypothesis, change the proportion to 0.10, and change the option for the alternate hypothesis to “>”. Click the “Calculate” button.

Here is the output from StatCrunch:

The value of the test statistic is $z=3.33$. (You could calculate this with a calculator using the formula in step 3.)

The p-value for this test is 0.0004. (You could determine this by finding the area under the normal curve to the right of $z=3.33$.)

Since the p-value is less than 0.05, we reject $H_0$. This supports the physician’s claim that more than 10% of all pregnant women smoke.

Summary

I hope that you find this useful. In future blogs I will explain how to use StatCrunch to perform other hypothesis tests. If you have any StatCrunch or statistical questions, you can reach me through the contact page on my website .

-George

I am a mathematics instructor at College of the Sequoias in Visalia, CA. Each Friday I often blog about technology (including StatCrunch), inside and outside of the classroom. Let me know if there are other topics you’d like me to cover by leaving a comment or by reaching me through the contact page on my website: georgewoodbury.com.