Hypothesis Test for a Proportion

A hypothesis test for a proportion involves testing a claim about a population proportion (P) using sample data. Here's a step-by-step guide to performing a one-sample z-test for a proportion, which is the most common method for this type of test.

When to Use This Analysis

The approach described in this lesson is appropriate when the following conditions are met:

• The sampling method is simple random sampling.
• Each sample point can result in just two possible outcomes. (We call one of these outcomes a success and the other, a failure. The proportion of successes in the sample is denoted by p.)
• The sample includes at least 10 successes and 10 failures. (This condition is required to support an assumption that the sampling distribution of the proportion will be approximately normal in shape, which is necessary to justify the use of a one-sample z-test.)
• The population size is at least 20 times as big as the sample size. (This condition is required to justify using an approximate formula to compute the standard error of the sampling distribution.)

Before proceeding with a hypothesis test, ensure that these conditions are met.

General Procedure for Hypothesis Testing

To test any hypothesis, the same five-step procedure is used: (1) state the hypotheses, (2) choose the significance level, (3) compute the test statistic, (4) find the P-value, and (5) interpret results. Here, we apply the general procedure to hypothesis tests of proportions.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false, as shown below.

• Null hypothesis (H₀): The population proportion equals P₀.

  H₀: P = P₀

• Alternative hypothesis (H_a): The population proportion differs from P₀ in one of three possible ways.

  H_a: P ≠ P₀ (Two-tailed test checking for any difference from P₁)

  H_a: P > P₀ (One-tailed test checking if P is greater than P₁)

  H_a: P < P₀ (One-tailed test checking if P is less than P₁)

where P is the true population proportion and P₀ is the hypothesized population proportion.

Choose the Significance Level

The significance level is the probability of rejecting the null hypothesis when it is actually true. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10.

Compute the Test Statistic

Use the one-sample z-test to determine whether the hypothesized population proportion differs significantly from the observed sample proportion. The test statistic is a z-score (z) defined by the following equation.

z = (p - P₀) / SD;

where P₀ is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and SD is the standard deviation of the sampling distribution.

When the population size is at least 20 times bigger than the sample size, the standard deviation of the sampling distribution (SD) can be computed from this formula:

SD = sqrt[ P₀ * ( 1 - P₀ ) / n ]

where P₀ is the hypothesized value of population proportion in the null hypothesis, and n is the sample size.

Find the P-Value

The P-value is the probability of observing a sample statistic as extreme as the z-score test statistic. To assess the probability associated with the z-score, use an online calculator, a graphing calculator, or a normal distribution statistical table. (See sample problems at the end of this lesson for examples of how this is done with Stat Trek's Normal Distribution Calculator.)

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. This involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two hypothesis testing examples illustrate how to conduct a hypothesis test of a proportion. The first problem involves a a two-tailed test; the second problem, one-tailed test.

Problem 1: Two-Tailed Test

The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisified. Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.

Solution: The solution to this problem takes five steps: (1) state the hypotheses, (2) choose the significance level, (3) compute the test statistic, (4) find the P-value, and (5) interpret results. We work through those steps below:

• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

  Null hypothesis: P = 0.80

  Alternative hypothesis: P ≠ 0.80

  Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.
• Choose the significance level. For this analysis, the significance level is 0.05.
• Compute the test statistic. The one-sample z-test is an appropriate method to determine whether a hypothesized population proportion differs significantly from an observed sample proportion. The test statistic for a one-sample z-test is a z-score. We calculate the standard deviation (SD) and compute the z-score test statistic (z) from the formulas below:

  SD = sqrt[ P₀ * ( 1 - P₀ ) / n ]

  SD = sqrt [(0.8 * 0.2) / 100]

  SD = sqrt(0.0016) = 0.04

  z = (p - P₀) / SD = (.73 - .80)/0.04 = -1.75

  where P₀ is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

• Find the P-value. Since we have a two-tailed test, the P-value is the probability of observing a z-score more extreme than the absolute value of the test statistic (i.e., less than -1.75 or greater than 1.75). We use the Normal Distribution Calculator to find P(z < -1.75) = 0.04. Since the standard normal distribution is symmetric with a mean of zero, we know that P(z > 1.75) = 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.

• Interpret results. Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the sample included at least 10 successes and 10 failures, and the population size was at least 10 times the sample size.

Problem 2: One-Tailed Test

Suppose the previous example is stated a little bit differently. Suppose the CEO claims that at least 80 percent of the company's 1,000,000 customers are very satisfied. Again, 100 customers are surveyed using simple random sampling. The result: 73 percent are very satisfied. Based on these results, should we accept or reject the CEO's hypothesis? Assume a significance level of 0.05.

Solution: The solution to this problem takes five steps: (1) state the hypotheses, (2) choose the significance level, (3) compute the test statistic, (4) find the P-value, and (5) interpret results. We work through those steps below:

• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

  Null hypothesis: P = 0.80

  Alternative hypothesis: P < 0.80

  Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
• Choose the significance level. For this analysis, the significance level is 0.05.
• Compute the test statistic. The one-sample z-test is an appropriate method to determine whether a hypothesized population proportion differs significantly from an observed sample proportion. The test statistic for a one-sample z-test is a z-score. We calculate the standard deviation (SD) and compute the z-score test statistic (z) from the formulas below:

  SD = sqrt[ P * ( 1 - P₀ ) / n ] = sqrt [(0.8 * 0.2) / 100]

  SD = sqrt(0.0016) = 0.04

  z = (p - P₀) / SD = (.73 - .80)/0.04 = -1.75

  where P₀ is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

• Find the P-value. Since we have a one-tailed test , the P-value is the probability that the z-score is less than -1.75. In the previous problem, we used the Normal Distribution Calculator to find P(z ≤ -1.75) = 0.04. Thus, the P-value = 0.04.
• Interpret results. Since the P-value (0.04) is less than the significance level (0.05), we cannot accept the null hypothesis.

Problem 3

In Problem 1 and Problem 2 above, we used identical sampling plans (same sampling method, same sample size, etc.), the null hypothesis (H₀ = 0.80) was the same in both problems, and the P-value (-1.75) was the same in both problems. We rejected the null hypothesis in Problem 2 but not in Problem 1. Explain how this happened.

Solution: In both problems, the significance level was 0.05. But we conducted a two-tailed test in Problem 1 and a one-tailed test in problem 2. That makes a difference.

• Two-tailed test. The area of rejection is split between the two tails of the sampling distribution. Since the significance level was 0.05, 2.5% of the rejection area would be in each tail, and if the test statistic falls in either of these small tails, we reject the null hypothesis.
• One-tailed test. The rejection area is only on one side of the sampling distribution. So, all 5% of the rejection area is placed in the lower tail. If the test statistic falls in this larger area, we reject the null hypothesis.

Basically, the region of rejection in the lower tail was larger for the one-tailed test than for the two-tailed test. That allowed us to reject the null hypothesis in Problem 2, but not in Problem 1.