Setting Up and Carrying Out a Test For a Population Proportion

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Last Updated: September 23, 2024

In AP Statistics, testing for a population proportion involves determining if a sample proportion significantly differs from a hypothesized population proportion. This process includes stating hypotheses, checking conditions, calculating the test statistic, determining the p-value, and making a decision. Understanding this method is essential for analyzing real-world data and drawing valid conclusions about population parameters. Mastering these steps ensures a solid foundation for success in the AP Statistics exam and beyond.

Learning Objectives

You will be able to identify and state the null and alternative hypotheses for a population proportion test. You will learn to check necessary conditions, including randomness, the 10% condition, and large counts. You will be guided to calculate the test statistic and determine the p-value using the standard normal distribution. You will be expected to compare the p-value to the significance level and make appropriate decisions about the hypotheses. Lastly, you will be prepared to state clear conclusions in the context of the problem.

Steps for Setting Up and Carrying Out a Test for a Population Proportion

Step 1: State the Hypotheses

  • Null Hypothesis (H_0​): The null hypothesis states that the population proportion is equal to a specific value. It is usually denoted as \(H_0: p = p_0\)​, where \(p_0\) is the hypothesized population proportion.
  • Alternative Hypothesis (H_a): The alternative hypothesis states that the population proportion is different from the hypothesized value. It can be one of the following:
    • \(H_a: p \neq p_0) (two-tailed test)\)
    • \(H_a: p > p_0) (right-tailed test)\)
    • \(H_a: p < p_0) (left-tailed test)\)

Step 2: Check the Conditions

Before proceeding with the test, ensure the following conditions are met:

  • Random Sample: The sample should be randomly selected.
  • 10% Condition: The sample size should be less than 10% of the population if sampling without replacement.
  • Large Counts Condition: Both \(np_0) and (n(1 – p_0)\) should be at least 10, where n is the sample size.

Step 3: Calculate the Test Statistic

The test statistic for a population proportion is calculated using the following formula: \(z = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0(1 – p_0)}{n}}} \) where:

  • \(\hat{p}\)​ is the sample proportion
  • p_0​ is the hypothesized population proportion
  • n is the sample size

Step 4: Determine the P-Value

  • The p-value is the probability of observing a sample proportion as extreme as (or more extreme than) the observed \( \hat{p} \)​, assuming the null hypothesis is true.
  • Use the standard normal distribution (z-distribution) to find the p-value corresponding to the calculated test statistic.

Step 5: Make a Decision

  • Compare the p-value to the significance level (α\alphaα):
    • If p-value \( \leq \alpha \), reject the null hypothesis (\( H_0 )\).
    • If p-value \( > \alpha \), fail to reject the null hypothesis.

Step 6: State the Conclusion

  • Clearly state whether you reject or fail to reject the null hypothesis in the context of the problem.

Examples

Example 1

Scenario: A company claims that 60% of its customers are satisfied with their service. A random sample of 150 customers shows that 85 are satisfied. Test the company’s claim at the 5% significance level.

Solution:

  1. Hypotheses:
    • \(&H_0: p = 0.60\
    • &H_a: p \neq 0.60\)
  2. Check Conditions:
    • Random Sample: Assumed
    • 10\% Condition: \( 150 < 10\% \text{ of the population} \)
    • Large Counts: \( np_0 = 150 \times 0.60 = 90 \), \( n(1 – p_0) = 150 \times 0.40 = 60 \) \(Both ( \geq ) 10\)
  3. Calculate Test Statistic:
  4. \(\hat{p} = \frac{85}{150} = 0.567\)\(z = \frac{0.567 – 0.60}{\sqrt{\frac{0.60(1 – 0.60)}{150}}} = \frac{-0.033}{0.040} = -0.825\)
  5. Determine P-Value:
    • \(\text{P-value for } z = -0.825 \text{ (two-tailed)}: 2 \times P(Z < -0.825) = 2 \times 0.2043 = 0.4086\)
  6. Make Decision:
    • \(0.4086 > 0.05\)
  7. Conclusion:
    • There is not enough evidence to reject the company’s claim that 60% of its customers are satisfied.

Example 2

Scenario: A researcher claims that less than 30% of people exercise regularly. A survey of 200 people finds that 52 exercise regularly. Test this claim at the 1% significance level.

Solution:

  1. Hypotheses:
    • \(H_0: p = 0.30\)
    • \(H_a: p < 0.30\)
  2. Check Conditions:
    • Random Sample: Assumed
    • 10% Condition: \( 200 < 10\% \text{ of the population} \)
    • Large Counts:
    • \( np_0 = 200 \times 0.30 = 60 ), ( n(1 – p_0) = 200 \times 0.70 = 140 ) (Both ( \geq ) 10\)
  3. Calculate Test Statistic: \(\hat{p} = \frac{52}{200} = 0.26\)\(z = \frac{0.26 – 0.30}{\sqrt{\frac{0.30(1 – 0.30)}{200}}} = \frac{-0.04}{0.0324} = -1.235\)
  4. Determine P-Value:
    • \(\text{P-value for } z = -1.235 \text{ (left-tailed)}: P(Z < -1.235) = 0.1085\)
  5. Make Decision:
    • \(0.1085 > 0.01\)
  6. Conclusion:
    • There is not enough evidence to support the claim that less than 30% of people exercise regularly.

Example 3

Scenario: In a certain city, it is believed that 40% of the population supports a new law. A random sample of 120 residents shows that 58 support the law. Test the belief at the 10% significance level.

Solution:

  1. Hypotheses:
    • \(&H_0: p = 0.40\
    • &H_a: p \neq 0.40\)
  2. Check Conditions:
    • Random Sample: Assumed
    • 10% Condition: \( 120 < 10\% \text{ of the population} \)
    • Large Counts: \( np_0 = 120 \times 0.40 = 48 ), ( n(1 – p_0) = 120 \times 0.60 = 72 ) (Both ( \geq ) 10\)
  3. Calculate Test Statistic: \(\hat{p} = \frac{58}{120} = 0.483\) \(z = \frac{0.483 – 0.40}{\sqrt{\frac{0.40(1 – 0.40)}{120}}} = \frac{0.083}{0.0458} = 1.81\)
  4. Determine P-Value:
    • \(
    • \text{P-value for } z = 1.812 \text{ (two-tailed)}: 2 \times P(Z > 1.812) = 2 \times 0.0350 = 0.070
    • \)
  5. Make Decision:
    • \(
    • 0.070 < 0.10
    • \)
  6. Conclusion:
    • There is sufficient evidence to reject the belief that 40% of the population supports the new law.

Example 4

Scenario: A health organization claims that 25% of adults are smokers. A survey of 250 adults finds that 70 are smokers. Test this claim at the 5% significance level.

Solution:

  1. Hypotheses:
    • \(&H_0: p = 0.25\
    • &H_a: p \neq 0.25\)
  2. Check Conditions:
    • Random Sample: Assumed
    • 10% Condition: \( 250 < 10\% \text{ of the population} \)
    • Large Counts: \( np_0 = 250 \times 0.25 = 62.5 ), ( n(1 – p_0) = 250 \times 0.75 = 187.5 ) (Both ( \geq ) 10\)
  3. Calculate Test Statistic:
  4. (\hat{p} = \frac{70}{250} = 0.28\)
  5. \(z = \frac{0.28 – 0.25}{\sqrt{\frac{0.25(1 – 0.25)}{250}}} = \frac{0.03}{0.0274} = 1.095\)
  6. Determine P-Value:
    • \(\text{P-value for } z = 1.095 \text{ (two-tailed)}: 2 \times P(Z > 1.095) = 2 \times 0.1366 = 0.2732\)
  7. Make Decision:
    • [0.2732 > 0.05\)
  8. Conclusion:
    • There is not enough evidence to reject the health organization’s claim that 25% of adults are smokers.

Example 5

Scenario: A tech company claims that 70% of its employees are satisfied with their job. A random sample of 180 employees finds that 125 are satisfied. Test this claim at the 5% significance level.

Solution:

  1. Hypotheses:
    • \( \begin{aligned} &H_0: p = 0.25\
    • &H_a: p \neq 0.25 \end{aligned} \)
  2. Check Conditions:
    • Random Sample: Assumed
    • 10% Condition: \(180 < 10\% \text{ of the population}\)
    • Large Counts: \(np_0 = 180 \times 0.70 = 126), (n(1 – p_0) = 180 \times 0.30 = 54) (Both (\geq) 10\)
  3. Calculate Test Statistic: \(\hat{p} = \frac{125}{180} = 0.694\)\(z = \frac{0.694 – 0.70}{\sqrt{\frac{0.70(1 – 0.70)}{180}}} = \frac{-0.006}{0.0343} = -0.175\)
  4. Determine P-Value:
    • \(\text{P-value for } z = -0.175 \text{ (two-tailed)}: 2 \times P(Z < -0.175) = 2 \times 0.4302 = 0.8604\)
  5. Make Decision: \(0.8604 > 0.05\)
  6. Conclusion:
    • There is not enough evidence to reject the tech company’s claim that 70% of its employees are satisfied with their job.

Multiple-Choice Questions (MCQs)

Question 1

A study claims that 45% of college students prefer online classes. A random sample of 200 college students reveals that 92 prefer online classes. What is the p-value for testing the claim at the 5% significance level?

a) 0.036

b) 0.087

c) 0.126

d) 0.173

Answer: b) 0.087

Explanation: \(\hat{p} = \frac{92}{200} = 0.46\)
\(z = \frac{0.46 – 0.45}{\sqrt{\frac{0.45(1 – 0.45)}{200}}} = \frac{0.01}{0.035} = 0.286\)
\(P-value for (z = 0.286) (two-tailed): \(2 \times P(Z > 0.286) = 2 \times 0.0435 = 0.087\)

Question 2

Which of the following is a condition that must be met before performing a hypothesis test for a population proportion?

a) The population standard deviation must be known.

b) The sample size must be at least 30.

c) The sample proportion must be normally distributed.

d) The sample should be randomly selected.

Answer: d) The sample should be randomly selected.

Explanation: One of the key conditions for a hypothesis test for a population proportion is that the sample should be randomly selected to ensure that the sample is representative of the population.

Question 3

If the p-value is less than the significance level (α\alphaα), what decision should be made regarding the null hypothesis?

a) Accept the null hypothesis.

b) Reject the null hypothesis.

c) Fail to reject the null hypothesis.

d) Increase the sample size.

Answer: b) Reject the null hypothesis.

Explanation: If the p-value is less than the significance level, we reject the null hypothesis because there is sufficient evidence to support the alternative hypothesis.