Statistical Reasoning Lecture 10

Introduction to Hypothesis Testing

• Overview: The lecture discusses hypothesis testing in the context of comparing proportions and incidence rates between two populations, focusing on methods such as the two-sample z-test, chi-square test, and Fisher's exact test.

Learning Objectives

• Upon completion, students will be able to:

  • Estimate and interpret p-values when comparing proportions with the two-sample z-test.

  • Explain the relationship between risk difference (RD), relative risk (RR), and odds ratio (OR), emphasizing why only one p-value is necessary despite the multiple measures of association.

Two-Sample z-Test for Comparing Proportions

Concept

• The two-sample z-test is similar to the t-test for comparing means but is specifically for proportions. Its methodology includes:

  • Setting up null (
    ) and alternative hypotheses.

  • Assuming the null hypothesis is true to compute how far the observed sample estimate diverges from expectations under the null hypothesis.

  • Translating this distance into a p-value for decision-making.

Example: Treatment Response to ART in HIV+ Individuals

Data Summary

• A study conducted on 1,000 HIV-positive patients:

  • CD4 < 250: 127 out of 503 (25% response)

  • CD4 ≥ 250: 79 out of 497 (16% response)

  • Overall response: 206 out of 1,000.

Measures of Association

• Risk difference (RD): 0.09 (95% CI: 0.04, 0.14)

• Relative risk (RR): 1.56 (95% CI: 1.20, 2.01)

• Odds ratio (OR): 1.75 (95% CI: 1.27, 2.41)

Hypothesis Setup

• Competing hypotheses:

  • Null: H0: pC<250 = pC≥250

  • Alternative: HA: pC<250 ≠ p_C≥250

p-value Calculation

1. Compute the standardized distance:

   • z = rac{(p̂C<250 - p̂C≥250) - 0}{SE(p̂C<250 - p̂_C≥250)}

   • Here, SE(p̂C<250 - p̂C≥250) = ext{sqrt}igg( rac{p̂C<250(1-p̂C<250)}{nC<250} + rac{p̂C≥250(1-p̂C≥250)}{n_C≥250}igg)

2. The resulting z-value was approximately 3.6, suggesting a significant difference in response rates (above the null of 0).

3. p-value results:

   • Calculate: $p = 2<em>(1 - pnorm(3.6)) ext{ or } 2</em>pnorm(-3.6)$

   • Result: p-value < 0.001, indicating strong evidence against the null hypothesis.

Chi-Square Test for Comparing Proportions

• The chi-square test yields similar results to the two-sample z-test for comparing proportions.

• General procedure:

  • Setup null and alternative hypotheses, assume null is true, calculate the distance from observed to expected frequencies, convert this distance into a p-value.

Example: Treatment Response in HIV+ Individuals

Competing Hypotheses (Chi-Square)

• H0: pC<250 = pC≥250

• HA: pC<250 ≠ p_C≥250

Expected Frequencies Calculation

• Under the null hypothesis:

  • Calculate expected counts for each outcome based on observed proportions.

• Apply the chi-square statistic formula:

  • $χ^2 = rac{(O<em>i - E</em>i)^2}{E_i}$

Results and Interpretation

• Resulting p-value from the chi-square test was approximately identical to that from the z-test, underscoring the tests' equivalence.

Fisher’s Exact Test

• Fisher's exact test is utilized for small sample sizes and provides an exact p-value when comparing two groups.

• Comparisons suggest the test's application in scenarios where either group is less than 5 in frequency.

• Illustrative example:

  • Using treatment response from the same HIV study:

  • p-values derived from Fisher’s Exact Test were consistent with results from z-test and chi-square approach, reaffirming the findings across methods.

Conclusion on p-values and Hypothesis Testing

• Similar logic applies across various tests. The p-value signifies the probability of observing data as extreme as the sample under the null hypothesis.

• In practice, researchers generally use rejection levels at 0.05 and test norms with 95% confidence intervals. Different tests may yield slightly divergent p-values, yet they interpret identically in the context of studies comparing binary outcomes.