10. Two Proportions (Normal)

Math 115 - Examples for Topic 10

Author

Yurk

These worked examples demonstrate inference for comparing two proportions using the normal distribution (model-based approach).

Example 1: Hypothesis Test Using Normal Model

Scenario: A university wants to know if providing free tutoring services affects whether students pass introductory statistics. They randomly assigned 180 students to two groups:

  • Tutoring group: Offered free tutoring sessions (90 students)
  • Control group: No additional support offered (90 students)

At the end of the semester, they recorded whether each student passed the course.

Group Passed Failed Total
Tutoring 76 14 90
Control 63 27 90
Total 139 41 180

Research question: Is there evidence that the pass rate differs between the tutoring and control groups?

Task: Conduct a hypothesis test at the \(\alpha = 0.05\) level using the normal model.

Step 1: Identify the parameter and calculate sample statistic

Let \(p_T\) = the true pass rate for students offered tutoring

Let \(p_C\) = the true pass rate for students in the control group

The parameter of interest is \(p_T - p_C\), the difference in pass rates between groups.

Sample proportions:

  • Tutoring group: \(\hat{p}_T = \frac{76}{90} = 0.844\)
  • Control group: \(\hat{p}_C = \frac{63}{90} = 0.7\)
  • Observed difference: \(\hat{p}_T - \hat{p}_C = 0.844 - 0.7 = 0.144\)

Step 2: State the hypotheses

Since we want to test whether the pass rate differs (in either direction), this is a two-sided test:

  • \(H_0: p_T - p_C = 0\) (no difference in pass rates)
  • \(H_A: p_T - p_C \neq 0\) (pass rates differ)

Step 3: Check conditions for using normal model

Independence:

  • Randomized experiment with independent groups ✓

Success-Failure Condition: For a hypothesis test, we use the pooled proportion to calculate expected counts.

Pooled proportion:

\[\hat{p}_{pool} = \frac{\text{total passed}}{\text{total students}} = \frac{139}{180} = 0.772\]

Expected counts using pooled proportion:

  • Tutoring group:
    • Expected successes: \(90 \times 0.772 = 69.5\) ≥ 10 ✓
    • Expected failures: \(90 \times 0.228 = 20.5\) ≥ 10 ✓
  • Control group:
    • Expected successes: \(90 \times 0.772 = 69.5\) ≥ 10 ✓
    • Expected failures: \(90 \times 0.228 = 20.5\) ≥ 10 ✓

All conditions are met, so we can use the normal model.

Step 4: Calculate the test statistic

Standard error (using pooled proportion):

\[SE = \sqrt{\hat{p}_{pool}(1-\hat{p}_{pool})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.772 \times 0.228 \times \left(\frac{1}{90} + \frac{1}{90}\right)} = 0.0625\]

Z-score:

\[Z = \frac{(\hat{p}_T - \hat{p}_C) - 0}{SE} = \frac{0.144 - 0}{0.0625} = 2.31\]

Step 5: Calculate the p-value using Jamovi

Randomize module → Model-Based Inference

To calculate the p-value in Jamovi:

  1. Under Distribution, choose Standard normal (Z)
  2. Check “Calculate area (p-value)”
  3. Enter the Z-score: 2.31
  4. Select tail direction: Both (for two-sided test)

Jamovi calculates the two-sided p-value using the “double the smaller tail” method:

  • Area to the right of Z = 2.31: 0.0104
  • Area to the left of Z = 2.31: 0.9896
  • Smaller tail: 0.0104
  • Two-sided p-value = 2 × 0.0104 = 0.0209

Step 6: Interpret the p-value

Interpretation: If there were truly no difference in pass rates between the tutoring and control groups, there is approximately a 2.09% chance of observing a difference as extreme as 0.144 (or more extreme) in a random sample.

Step 7: Make a decision

Using the significance level α = 0.05:

  • P-value = 0.0209
  • Compare: Is 0.0209 < 0.05? YES

Decision: We reject \(H_0\) at the α = 0.05 level.

Step 8: State the conclusion in context

The data provide convincing evidence that there is a difference in pass rates between the tutoring and control groups. Since this was a randomized experiment, we can conclude that offering tutoring causes an increase in pass rates. The tutoring group had a pass rate 14.4 percentage points higher than the control group (Z = 2.31, p-value = 0.0209).

Example 2: Confidence Interval Using Normal Model

Scenario: Using the same tutoring study from Example 1, construct a 95% confidence interval for the true difference in pass rates between the tutoring and control groups.

Recall from Example 1:

  • Tutoring group: 76 out of 90 passed
  • Control group: 63 out of 90 passed

Task: Use the normal model to construct a 95% confidence interval for \(p_T - p_C\).

Step 1: Identify the parameter and point estimate

The parameter of interest is \(p_T - p_C\), the true difference in pass rates between students offered tutoring and students in the control group.

From Example 1:

  • \(\hat{p}_T = 0.844\)
  • \(\hat{p}_C = 0.7\)
  • Observed difference: \(\hat{p}_T - \hat{p}_C = 0.144\)

Step 2: Check conditions for using normal model

Independence: Randomized experiment with independent groups ✓

Success-Failure Condition: For a confidence interval, we use observed counts (not pooled proportion).

  • Tutoring: 76 passed, 14 failed (both ≥ 10) ✓
  • Control: 63 passed, 27 failed (both ≥ 10) ✓

Conditions are met for using the normal model.

Step 3: Calculate the standard error

For confidence intervals, use separate proportions (not pooled):

\[SE = \sqrt{\frac{\hat{p}_T(1-\hat{p}_T)}{n_T} + \frac{\hat{p}_C(1-\hat{p}_C)}{n_C}}\]

\[SE = \sqrt{\frac{0.844 \times 0.156}{90} + \frac{0.7 \times 0.3}{90}} = 0.0616\]

Note: This is a different SE formula than the one used for the hypothesis test!

Step 4: Find the critical value using Jamovi

Randomize module → Model-Based Inference

To find \(z^*\) in Jamovi:

  1. Under Distribution, choose Standard normal (Z)
  2. Check “Calculate CI multiplier”
  3. Enter confidence level: 95

Jamovi returns: \(z^* = 1.960\)

Step 5: Calculate the confidence interval

Margin of error:

\[ME = z^* \times SE = 1.960 \times 0.0616 = 0.1207\]

Confidence interval:

\[\text{CI} = (\hat{p}_T - \hat{p}_C) \pm ME = 0.144 \pm 0.121\]

\[95\% \text{ CI} = (0.024, 0.265)\]

Step 6: Interpret the confidence interval

Interpretation: We are 95% confident that the true difference in pass rates (tutoring − control) is between 0.024 and 0.265.

In practical terms:

  • Both bounds are positive, indicating the tutoring group has a higher pass rate
  • We’re confident tutoring increases pass rates by somewhere between 2.4 and 26.5 percentage points

Step 7: Connect to hypothesis test

Does the 95% CI include 0?

  • CI: (0.024, 0.265)
  • 0 is NOT in the interval

This is consistent with our hypothesis test result: since 0 is not a plausible value, we rejected \(H_0\).

Example 3: Full Inference Workflow — Sulphinpyrazone Trial

Scenario: The Anturane Reinfarction Trial (1980) examined whether sulphinpyrazone, an anti-clotting drug, prevents sudden cardiac death in patients who had recently survived a myocardial infarction (heart attack). Researchers randomly assigned 1,475 MI survivors to receive sulphinpyrazone or a placebo, then tracked deaths through follow-up. Patients were volunteers recruited from clinical settings — not a random sample.

Note

The original paper’s primary endpoint was sudden cardiac death, and the published headline finding was a significant reduction in that specific endpoint on an “analyzable” patient subgroup. The dataset used here (openintro::sulphinpyrazone) records all-cause deaths for the full 1,475-patient cohort, and the analysis below is therefore an analysis of that outcome — not a replication of the paper’s primary finding. The FDA’s contemporaneous reanalysis of the trial’s full cohort reached a conclusion closer to what we compute here.

Group Died Lived Total
Control (placebo) 60 682 742
Treatment (sulphinpyrazone) 41 692 733
Total 101 1374 1475

Research question: Does sulphinpyrazone affect death rates after a heart attack?

Task: Conduct a two-sided hypothesis test at \(\alpha = 0.05\) AND construct a 95% confidence interval for \(p_T - p_C\). State the scope of inference and give a final conclusion in context.

Step 1: Set up the parameter and state hypotheses

Let \(p_T\) = the true death rate among MI survivors taking sulphinpyrazone.

Let \(p_C\) = the true death rate among MI survivors taking placebo.

The parameter of interest is \(p_T - p_C\), the difference in death rates between treatment and placebo.

Since we test whether the rates differ (in either direction), this is a two-sided test:

  • \(H_0: p_T - p_C = 0\) (no difference in death rates)
  • \(H_A: p_T - p_C \neq 0\) (death rates differ)

Step 2: Calculate the sample statistics

  • Treatment: \(\hat{p}_T = \dfrac{41}{733} = 0.0559\)
  • Control: \(\hat{p}_C = \dfrac{60}{742} = 0.0809\)
  • Observed difference: \(\hat{p}_T - \hat{p}_C = 0.0559 - 0.0809 = -0.0249\)

The treatment group’s death rate is about 2.5 percentage points lower than placebo. We now ask whether this gap is statistically significant.

Step 3: Hypothesis test — check conditions

Independence: Randomized experiment with independent groups ✓

Success-Failure (for the H-test, use pooled expected counts):

Pooled proportion:

\[\hat{p}_{pool} = \frac{\text{total deaths}}{\text{total patients}} = \frac{101}{1475} = 0.0685\]

Expected counts under \(H_0\):

  • Treatment: \(733 \times 0.0685 = 50.2\) deaths; \(733 \times 0.9315 = 682.8\) no-deaths. Both ≥ 10 ✓
  • Control: \(742 \times 0.0685 = 50.8\) deaths; \(742 \times 0.9315 = 691.2\) no-deaths. Both ≥ 10 ✓

All conditions met — we can use the normal model.

Step 4: Hypothesis test — calculate \(Z\)

Pooled standard error:

\[SE = \sqrt{\hat{p}_{pool}(1-\hat{p}_{pool})\left(\frac{1}{n_T} + \frac{1}{n_C}\right)} = \sqrt{0.0685 \times 0.9315 \times \left(\frac{1}{733} + \frac{1}{742}\right)} = 0.0132\]

Z-score:

\[Z = \frac{(\hat{p}_T - \hat{p}_C) - 0}{SE} = \frac{-0.0249}{0.0132} = -1.9\]

Step 5: Hypothesis test — p-value and decision

Two-sided p-value (course convention: double the smaller tail):

  • Area to the left of \(Z = -1.9\): \(0.029\)
  • Area to the right of \(Z = -1.9\): \(0.971\)
  • Smaller tail: \(0.029\)
  • p-value = \(2 \times 0.029 = 0.058\)

Compare to \(\alpha = 0.05\): is \(0.058 < 0.05\)? NO

Decision: Do not reject \(H_0\) at the \(\alpha = 0.05\) level.

Step 6: Confidence interval — check conditions

For a confidence interval, the success-failure check uses observed counts (not pooled):

  • Treatment: 41 died, 692 lived (both ≥ 10) ✓
  • Control: 60 died, 682 lived (both ≥ 10) ✓

Step 7: Confidence interval — compute SE, ME, and CI

SE for the CI (use separate proportions):

\[SE = \sqrt{\frac{\hat{p}_T(1-\hat{p}_T)}{n_T} + \frac{\hat{p}_C(1-\hat{p}_C)}{n_C}} = \sqrt{\frac{0.0559 \times 0.9441}{733} + \frac{0.0809 \times 0.9191}{742}} = 0.0131\]

Note: In this case, the pooled SE and the separate-proportion SE round to about the same value, because \(\hat{p}_T\) and \(\hat{p}_C\) are both close to \(\hat{p}_{pool}\). They are not identical formulas — they happen to produce similar numbers here.

Margin of error: \(ME = z^* \times SE = 1.96 \times 0.0131 \approx 0.0257\)

95% CI:

\[(\hat{p}_T - \hat{p}_C) \pm ME = -0.0249 \pm 0.0257 = (-0.0506,\ 8\times 10^{-4})\]

Step 8: Interpret the CI and connect to the H-test

Interpretation: We are 95% confident that the true difference in death rates (\(p_T - p_C\)) is between \(-0.0506\) and \(8\times 10^{-4}\). In plain terms: the treatment death rate is somewhere between about 5 percentage points lower than placebo and roughly equal to placebo.

Connection to the H-test: The CI just barely includes 0, which is consistent with our decision to fail to reject \(H_0\).

Step 9: Scope of inference (recall 8a / 8b)

  • Causation: Patients were randomly assigned to treatment or placebo, so the design supports causal claims. However, we failed to reject \(H_0\), so we make no causal claim about sulphinpyrazone here. (Had we rejected, the random assignment would have supported the claim that sulphinpyrazone caused the difference.)
  • Generalization: Patients were not a random sample — they were MI-survivor volunteers from clinical settings, screened for eligibility. Conclusions apply only to similar patient populations in similar clinical settings, not to the general public.

Step 10: Final conclusion in context

This full-cohort analysis of the trial data does not provide convincing evidence that sulphinpyrazone affects all-cause death rates among heart-attack survivors at \(\alpha = 0.05\) (\(Z = -1.9\), p-value \(= 0.058\)). The 95% confidence interval for the difference in death rates is \((-0.0506,\ 8\times 10^{-4})\), which just barely includes 0. The direction is consistent with a possible benefit of treatment — 2.5 percentage points fewer deaths in the sulphinpyrazone group than in placebo — but the evidence is not strong enough to rule out chance. Because patients were randomly assigned but were not a random sample, any future significant finding would support a causal claim limited to similar MI-survivor populations in similar clinical settings.

Example 4: Factors Affecting Strength of Evidence and CI Width

These questions ask you to think about what affects the strength of evidence against the null hypothesis and confidence interval width.

Part A: Sample Size and Strength of Evidence

Scenario: Dr. Martinez conducts a study comparing a new medication to a placebo. With 50 patients per group, she observes that 60% of the medication group improved compared to 48% of the placebo group. Her p-value is 0.21.

Dr. Martinez’s colleague, Dr. Chen, conducts an identical study but with 200 patients per group. He observes the exact same proportions: 60% improved in the medication group and 48% in the placebo group.

Question: Will Dr. Chen’s p-value be larger, smaller, or about the same as Dr. Martinez’s p-value of 0.21?

Dr. Chen’s p-value will be smaller (stronger evidence against \(H_0\)).

With a larger sample size, the standard error decreases, which makes the Z-score larger. A larger Z-score corresponds to a smaller p-value. Even though both studies observed the same 12 percentage point difference, the larger study provides stronger evidence that this difference is real and not due to chance.

Part B: Sample Size and CI Width

Scenario: Using the same two studies from Part A:

  • Dr. Martinez: 50 patients per group
  • Dr. Chen: 200 patients per group

Both construct 95% confidence intervals for the difference in improvement rates.

Question: Whose confidence interval will be wider?

Dr. Martinez’s CI will be wider.

Larger samples lead to smaller standard errors, which produce narrower confidence intervals. Dr. Chen’s larger study will give a more precise estimate of the true difference in improvement rates.

Part C: Effect Size and Strength of Evidence

Scenario: Two researchers study whether a new teaching method improves test scores compared to traditional instruction. Both have 100 students per group.

  • Professor Adams: New method: 72% passed, Traditional: 68% passed (4 percentage point difference)
  • Professor Baker: New method: 85% passed, Traditional: 65% passed (20 percentage point difference)

Question: Which professor’s study will provide stronger evidence against the null hypothesis?

Professor Baker’s study will provide stronger evidence (smaller p-value).

A larger observed difference (effect size) leads to a larger Z-score and thus a smaller p-value. Even with the same sample sizes, observing a 20 percentage point difference provides much stronger evidence against the null hypothesis than observing a 4 percentage point difference.

Part D: Confidence Level and CI Width

Scenario: A political poll of 500 likely voters finds that 52% support Candidate X in District A, compared to 47% in District B. The pollster wants to report a confidence interval for the difference in support between the two districts.

Question: If the pollster reports a 99% confidence interval instead of a 95% confidence interval, will the interval be wider or narrower?

The 99% CI will be wider.

Higher confidence requires a larger multiplier (\(z^*\)), which increases the margin of error. There’s a trade-off: if you want to be more confident that your interval captures the true value, you need a wider interval.

Part E: Proportions Near 50% vs. Extreme Proportions

Scenario: Two clinical trials compare treatment success rates between treatment and control groups, each with 150 patients per group.

  • Trial 1: Treatment: 52% success, Control: 42% success (10 percentage point difference)
  • Trial 2: Treatment: 92% success, Control: 82% success (10 percentage point difference)

Both trials observe the same 10 percentage point difference in success rates.

Question: Which trial will have stronger evidence (smaller p-value) against the null hypothesis?

Trial 2 will have a smaller p-value (stronger evidence).

Proportions near 50% create more variability (larger standard error) than proportions near 0% or 100%. Since Trial 1 has proportions closer to 50%, it has a larger SE, which leads to a smaller Z-score and larger p-value. Trial 2’s more extreme proportions result in less variability and stronger evidence for the same observed difference.

Part F: Confidence Level Trade-off

Scenario: A market researcher compares customer satisfaction rates between two store locations, surveying 200 customers at each location. She finds that 78% are satisfied at Location A compared to 71% at Location B.

She constructs both a 90% and a 99% confidence interval for the difference in satisfaction rates.

Questions:

  1. Which interval is more likely to contain the true difference?
  2. Which interval provides a more precise estimate?

  1. The 99% CI is more likely to contain the true difference (99% vs 90% probability).

  2. The 90% CI provides a more precise estimate because it is narrower.

This illustrates the fundamental trade-off: you can have higher confidence OR more precision, but not both (unless you increase the sample size).

References

  • Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin (Section 17.3).
  • Anturane Reinfarction Trial Research Group. (1980). Sulfinpyrazone in the Prevention of Sudden Death after Myocardial Infarction. New England Journal of Medicine, 302(5), 250–256. Dataset available as openintro::sulphinpyrazone.
Back to top