Interpreting a P-value And Justifying a Claim About a Population Proportion

AP Statistics emphasizes the critical understanding of interpreting a P-value, a key concept in hypothesis testing. A P-value measures the probability of obtaining results as extreme as those observed, assuming the null hypothesis is true. This concept is particularly important when justifying claims about a population proportion, where students must evaluate evidence to either support or refute a hypothesis. Mastery of P-value interpretation is essential for success in AP Statistics, ensuring accurate conclusions based on statistical data.

Learning Objectives

You will be able to interpret P-values effectively in the context of hypothesis testing by the end of this topic. A deeper understanding of how to justify or refute claims about population proportions will be developed. You will also be equipped to compare P-values with significance levels to make informed decisions in statistical analysis. The ability to apply these skills will be strengthened, enhancing your overall competence in AP Statistics.

What is a P-value?

What is a P-value
  • Definition: A P-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true.
  • Range: P-values range from 0 to 1.
    • A small P-value \(typically \leq 0.05\) indicates strong evidence against the null hypothesis, leading to its rejection.
    • A large P-value \(> 0.05\) suggests weak evidence against the null hypothesis, so it is not rejected.

Hypothesis Testing for a Population Proportion

Hypothesis Testing for a Population Proportion
  • Null Hypothesis (H₀): It generally states that there is no effect or no difference. For a population proportion, H₀ typically asserts that the population proportion equals a specific value \(e.g., H_0: p = p_0\).
  • Alternative Hypothesis (H₁): It represents what you aim to support.
    For example, \(H_1: p \neq p_0\) (two-tailed test), \(H_1: p > p_0\) (right-tailed test), or \(H_1: p < p_0 \) (left-tailed test).

Steps for Hypothesis Testing

Steps for Hypothesis Testing
  1. State the Hypotheses: Clearly define the null and alternative hypotheses.
  2. Choose a Significance Level (α): Commonly used significance levels are 0.05, 0.01, and 0.10.
  3. Calculate the Test Statistic: Use the sample data to compute the test statistic, typically a z-score in the case of population proportions.
  4. Find the P-value: Determine the P-value associated with the test statistic.
  5. Make a Decision:
    • If the P-value ≤ α, reject the null hypothesis.
    • If the P-value > α, fail to reject the null hypothesis.
  6. Interpret the Results: Draw a conclusion in the context of the problem.

Interpreting the P-value

Interpreting the P-value
  • P-value Interpretation: The P-value tells you the probability of obtaining results at least as extreme as those observed, assuming the null hypothesis is true.
    • Low P-value (< α): The observed data is unlikely under H₀. This suggests that the alternative hypothesis may be true.
    • High P-value (≥ α): The observed data is consistent with H₀, so you don’t have enough evidence to support H₁.

Justifying a Claim About a Population Proportion

To justify a claim about a population proportion using a P-value:

  1. Compare the P-value to the significance level (α).
  2. Reject or fail to reject the null hypothesis based on the comparison.
  3. State the conclusion in the context of the problem. If you reject H₀, you are justifying the claim that the population proportion is different from the hypothesized value.

Examples

  1. Example 1:
    • Problem: A company claims that 60% of its customers are satisfied with their service. A survey of 100 customers finds that 65% are satisfied. Is there enough evidence at the 0.05 significance level to support the claim that more than 60% of customers are satisfied?
    • Solution:
      • \(H_0: p = 0.60\)
      • \( $H_1: p > 0.60\)
      • Test statistic: \(z = 1.14\)
      • \(P-value = 0.1271\)
      • \(Since P-value $> 0.05$, fail to reject $H_0$\). There is not enough evidence to support the claim.
  2. Example 2:
    • Problem: A school administrator claims that 75% of students are in favor of the new school policy. A random sample of 150 students shows that 110 favor the policy. Test the claim at the 0.01 significance level.
    • Solution:
      • \(H_0: p = 0.75\)
      • \(H_1: p \neq 0.75\)
      • Test statistic: \(z = -2.02\)
      • P-value = 0.0433
      • \(Since P-value > 0.01$, fail to reject $H_0$​\). The evidence is insufficient to refute the claim.
  3. Example 3:
    • Problem: In a survey, 55 out of 200 people said they prefer tea over coffee. A researcher believes that the true proportion of tea lovers is less than 30%. Test this claim at the 0.05 level.
    • Solution:
      • \( $H_0: p = 0.30\)
      • \($H_1: p < 0.30\)
      • Test statistic: \(z = -1.52\)
      • \(P-value = 0.0643 \)
      • Since P-value \(> 0.05\), fail to reject \(H_0\)​. The evidence is not strong enough to support the researcher’s claim.
  4. Example 4:
    • Problem: A pharmaceutical company claims that the success rate of a new drug is 80%. A study involving 400 patients shows that 310 were successfully treated. Is the company’s claim valid at a 0.01 significance level?
    • Solution:
      • \(H_0: p = 0.80\)
      • \(H_1: p \neq 0.80\)
      • Test statistic: \(z = -2.19\)
      • \( P-value = 0.0284\)
      • Since P-value \(> 0.01\), fail to reject \(H_0\). There is not enough evidence to reject the company’s claim.
  5. Example 5:
    • Problem: A dietitian believes that only 15% of adults eat the recommended five servings of fruits and vegetables per day. A survey of 500 adults shows that 85 eat the recommended servings. Test the dietitian’s belief at a 0.10 significance level.
    • Solution:
      • \(H_0: p = 0.15\)
      • \(H_1: p \neq 0.15\)
      • Test statistic: \(z = -0.85\)
      • \(P-value = 0.3962\)
      • Since P-value \(> 0.10\), fail to reject \(H_0\). The survey does not provide enough evidence to refute the dietitian’s belief.

Multiple-Choice Questions (MCQs)

MCQ 1: A company claims that 40% of its products are made from recycled materials. A random sample of 200 products shows that 90 are made from recycled materials. What is the P-value if you test the company’s claim at the 0.05 significance level?

A) 0.0455

B) 0.0789

C) 0.0232

D) 0.0567
Answer: B) 0.0789
Explanation: The test statistic z is calculated, and the P-value corresponding to z is approximately 0.0789. Since the P-value > 0.05, there is insufficient evidence to reject the company’s claim.

MCQ 2: If a P-value is 0.02 and the significance level is 0.05, what conclusion can be drawn?

A) Reject the null hypothesis; there is strong evidence against it.

B) Fail to reject the null hypothesis; there is strong evidence for it.

C) Fail to reject the null hypothesis; there is weak evidence against it.

D) Reject the null hypothesis; there is weak evidence against it.
Answer: A) Reject the null hypothesis; there is strong evidence against it.
Explanation: Since the P-value is less than the significance level, you reject the null hypothesis, indicating strong evidence against it.

MCQ 3: Which of the following correctly interprets a P-value of 0.35 in a hypothesis test?

A) There is a 35% chance that the null hypothesis is true.

B) There is a 35% chance that the alternative hypothesis is true.

C) There is a 35% probability of obtaining the observed results, assuming the null hypothesis is true.

D) There is a 35% probability that the null hypothesis is false.
Answer: C) There is a 35% probability of obtaining the observed results, assuming the null hypothesis is true.
Explanation: The P-value represents the probability of observing the test results under the assumption that the null hypothesis is true.