Wiki Page, Math 13 - Perez Sandbox : Module 21 - Hypothesis Test for a Population Proportion (2a of 3)

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2026-01-17T21:15:33-08:00

On the previous page, we looked at determining hypotheses for testing a claim about a population proportion. On this page, we look at how to determine P-values.

As we learned earlier, the P-value for a hypothesis test is based on a probability model, either a simulation or a normal curve. The probability model represents what occurs with the sample proportions in repeated random sampling from the population.

What are the normality conditions for this hypothesis test?

To conduct a hypothesis test we need to create a probability model to represent the distribution of sample proportions that comes from a population for which the null hypothesis is true. In other words, the null hypothesis says p is equal to some value. We use the symbol to represent this value, so : . We will create a probability distribution based on repeated random sampling from a population with proportion .

The normal curve is an appropriate model for the distribution of sample proportions if the expected number of success and failures are both at least 10. Here the expected number of successes and failures are based on the null hypothesis. Using the symbols for the population proportion and sample size, a normal curve is a reasonable model if the following conditions are met: and .

EXAMPLE Health insurance coverage

Let's return to the health insurance coverage example we discussed previously.

According to the Government Accountability Office, 80% of all college students (ages 18 to 23) had health insurance in 2006. The Patient Protection and Affordable Care Act of 2010 allowed young people under age 26 to stay on their parents' health insurance policy.

Has the proportion of college students (ages 18 to 23) who have health insurance increased since 2006?

A survey of 800 randomly selected college students (ages 18 to 23) indicated that 83% of them had health insurance. Use a 0.05 level of significance.

Step 1: Determine the hypotheses.

We did this on the previous page. The hypotheses are:

:

:

where p is the proportion of college students ages 18 to 23 who have health insurance now.

Step 2: Collect the data.

In this random sample of 800 college students, 83% have health insurance. If 80% of all college students have health insurance, is this 3% difference statistically significant or due to chance? We need to find a P-value to answer this question; therefore, if we are to use a normal curve to represent the probability model, then we must determine if the data meet normality conditions.

First note that the data are from a random sample. That is essential.

Now we need to determine if a normal model is a good fit for the sampling distribution. Since we assume that the null hypothesis is true, we build the sampling distribution with the assumption that 0.80 is the population proportion. We check the following conditions, using 0.80 for :

and

Because these are both more than 10, we can use the normal model to find the P-value.

Step 3: Assess the evidence.

Now that we know that the normal distribution is an appropriate model for the sampling distribution, our next goal is to determine the P-value. The first step is to determine the z-score for the observed sample proportion (the data).

The sample proportion is 0.83. We learned previously how to calculate the z-score. In this formula .

For this example, we calculate:

This z-score is also called the test statistic. It tells us the sample proportion of 0.83 is about 2.12 standard errors above the population proportion given in the null hypothesis. We use this statistic to find the P-value. The P-value describes the strength of the evidence against the null hypothesis.

We use the Normal Distribution applet that we use previously to determine the P-value.

Normal Distribution Calculator

Here are our results:

In this situation, the P-value is the area to the right of 0.83 because the alternative hypothesis is a “greater-than” statement. Since we are using the standard normal curve to find probabilities, the P-value is the area to the right of the corresponding Z = 2.12. So the area is approximately 0.0170.

In symbols, this is .

Interpreting the P-value: If the null hypothesis is true (p = 0.80), then sample proportions greater than 0.83 occur only about 1.7% of the time in random samples of this size (n = 800).

The P-value is a measure of whether random chance can explain the deviation of the data from the null hypothesis. In this case, the probability is too small to be due to chance.

Step 4: State a conclusion.

To determine our conclusion, we compare the P-value to the level of significance, . If our data are predicted to occur by chance less than 5% of the time, we have reason to reject the null hypothesis and accept the alternative.

Since our P-value of 0.017 is less than 0.05, we reject the null hypothesis. We state our conclusion in terms of the alternative hypothesis. We also state it in context.

Conclusion: The data from this study provides strong evidence that the proportion of all college students who have health insurance is now greater than 0.80 (P-value = 0.017). The 0.03 increase in the proportion who have health insurance since 2008 is statistically significant at the 0.05 level.

Alternatively, we can give the conclusion using the percentage rather than the decimal:

The data from this study provides strong evidence that the percentage of all college students who have health insurance is now greater than 80% (P-value = 0.017). The 3% increase in the percentage who have health insurance since 2008 is statistically significant at the 5% level.

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