# Hypothesis Test for a Proportion

This lesson explains how to conduct a hypothesis test of a proportion, 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 sample includes at least 10 successes and 10 failures.
- The population size is at least 20 times as big as the sample size.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

## 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; and vice versa.

## Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

- Significance level. Often, researchers choose
significance levels
equal to
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.
- Test method. Use the one-sample z-test to determine whether the hypothesized population proportion differs significantly from the observed sample proportion.

## Analyze Sample Data

Using sample data, find the test statistic and its associated P-Value.

- Standard deviation. Compute the
standard deviation (σ)
of the sampling distribution.
σ = sqrt[ P * ( 1 - P ) / n ]

where P is the hypothesized value of population proportion in the null hypothesis, and n is the sample size. - Test statistic. The test statistic is a z-score (z) defined by
the following equation.
z = (p - P) / σ

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and σ is the standard deviation of the sampling distribution. - P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a z-score, use the Normal Distribution Calculator to assess the probability associated with the z-score. (See sample problems at the end of this lesson for examples of how this is done.)

## Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, 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, a one-tailed test.

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**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 four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) 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 ≠ 0.80Note 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.

**Formulate an analysis plan**. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.**Analyze sample data**. Using sample data, we calculate the standard deviation (σ) and compute the z-score test statistic (z).σ = sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.8 * 0.2) / 100] = sqrt(0.0016) = 0.04

z = (p - P) / σ = (.73 - .80)/0.04 = -1.75where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.75 or greater than 1.75.

We use the Normal Distribution Calculator to find P(z < -1.75) = 0.04, and 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 four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) 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 < 0.80Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

**Formulate an analysis plan**. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.**Analyze sample data**. Using sample data, we calculate the standard deviation (σ) and compute the z-score test statistic (z).σ = sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.8 * 0.2) / 100] = sqrt(0.0016) = 0.04

z = (p - P) / σ = (.73 - .80)/0.04 = -1.75where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -1.75. We use 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.

**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.