Week 9
Today’s Plan
In December 2020, the Pfizer-BioNTech COVID-19 vaccine received Emergency Use Authorization.
Here is the key result from the Phase 3 clinical trial:
| Group | n | COVID-19 cases |
|---|---|---|
| Vaccine | 18,198 | 8 |
| Placebo | 18,325 | 162 |
Vaccine efficacy = 1 − (risk in vaccine group / risk in placebo group) = 1 − (8/18198)/(162/18325) ≈ 95%
But is 95% a parameter or a statistic? How confident should we be in that number?
So far, we’ve focused on inference for means (µ) using t-tests.
But many scientific questions are really about proportions (p):
The logic is exactly the same — we just need a different standard error.
| Symbol | Meaning |
|---|---|
| p | Population proportion (parameter) |
| p̂ | Sample proportion (statistic) |
| n | Sample size |
| p₀ | Hypothesized proportion (in H₀) |
Example: In the placebo group, \(\hat{p} = 162/18325 = 0.00884\) (about 0.88% infection rate).
We want to make inferences about the true infection rate p in an unvaccinated population.
Just like \(\bar{x}\), the sample proportion \(\hat{p}\) varies from sample to sample.
When conditions are met, the sampling distribution of \(\hat{p}\) is approximately normal:
\[\hat{p} \sim N\!\left(p,\ \sqrt{\frac{p(1-p)}{n}}\right)\]
The standard deviation of \(\hat{p}\) is called the standard error of a proportion:
\[SE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]
Before we compute anything, we must check three conditions:
1. Independence: Observations must be independent. → Usually satisfied if sampling randomly, or if treated/control groups are assigned independently.
2. Success/Failure (Large Sample) Condition: \[n\hat{p} \geq 10 \quad \text{AND} \quad n(1-\hat{p}) \geq 10\]
We need at least 10 “successes” and 10 “failures” in the sample.
3. Sample size / Population size: n < 10% of the population (if sampling without replacement).
For the placebo group in the Pfizer trial:
Check Success/Failure:
Independence: Participants were randomly assigned to placebo or vaccine ✅
10% condition: 18,325 is far less than 10% of all adults worldwide ✅
Conditions met — we can proceed.
\[\hat{p} \pm z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]
where \(z^*\) is the critical value from the standard normal:
| Confidence Level | \(z^*\) |
|---|---|
| 90% | 1.645 |
| 95% | 1.960 |
| 99% | 2.576 |
Notice: We use \(z^*\) (normal), not \(t^*\) (t-distribution), because proportions have a known theoretical SE formula.
Placebo group: \(\hat{p} = 162/18325 \approx 0.00884\), \(n = 18325\)
Standard Error: \[SE = \sqrt{\frac{0.00884 \times 0.99116}{18325}} = \sqrt{0.000000478} \approx 0.000692\]
95% CI: \[0.00884 \pm 1.96 \times 0.000692 = 0.00884 \pm 0.00136 = (0.00748,\ 0.01020)\]
Interpretation: We are 95% confident that the true infection rate in the unvaccinated population is between 0.75% and 1.02%.
[Poll Everywhere — respond now!]
In the Pfizer trial, the vaccine group had: \(n = 18{,}198\) participants, 8 COVID cases.
Discuss with your neighbor (2 min):
→ On Poll Everywhere: Type one word that describes what these two CIs tell us about vaccine effectiveness.
We want to test a specific claim about p.
Step 1: State hypotheses \[H_0: p = p_0 \qquad H_a: p \neq p_0 \text{ (or } < \text{ or } >)\]
Step 2: Check conditions (same as for CI)
Step 3: Compute the test statistic (z-score) \[z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}\]
Important: Under \(H_0\), we know p = p₀, so we use p₀ (not p̂) in the SE.
Step 4: Find the p-value using the standard normal distribution.
Step 5: Conclude in context.
When computing CIs, we don’t know p, so we estimate it with \(\hat{p}\).
When testing, we’re asking: “Assuming H₀ is true (p = p₀), how surprising is our result?”
So we use p₀ in the SE under H₀:
\[SE_{H_0} = \sqrt{\frac{p_0(1-p_0)}{n}}\]
This is a subtle but important distinction from the CI formula.
A hospital claims its surgical infection rate is only 2% (the national benchmark). You audit 350 surgeries and find 11 infections.
Hypotheses: \(H_0: p = 0.02\) vs. \(H_a: p > 0.02\) (one-sided — the concern is rates above benchmark)
Check conditions: \(n p_0 = 350 \times 0.02 = 7 < 10\) ⚠️
The success/failure condition is borderline — we should note this caveat before proceeding.
\(\hat{p} = 11/350 = 0.0314\)
\[z = \frac{0.0314 - 0.02}{\sqrt{(0.02)(0.98)/350}} = \frac{0.0114}{0.00748} = 1.52\]
p-value = P(Z > 1.52) = 0.064
Decision: p-value = 0.064 > 0.05, so we fail to reject H₀.
Conclusion: There is not sufficient evidence at the 5% significance level to conclude the true infection rate exceeds the 2% national benchmark.
But wait — practical significance matters too!
Statistical non-significance ≠ evidence that rates are equal
Coming up:
What if we want to compare two proportions — like the vaccine group vs. placebo? We need a test that handles two independent groups.
Research question: Is the infection rate significantly lower in the vaccine group than the placebo group?
Setup:
| Group | n | Cases | \(\hat{p}\) |
|---|---|---|---|
| Vaccine | 18,198 | 8 | 0.000440 |
| Placebo | 18,325 | 162 | 0.008839 |
Hypotheses: \[H_0: p_V = p_P \quad \text{(vaccine has no effect)}\] \[H_a: p_V < p_P \quad \text{(vaccine reduces infection)}\]
For a confidence interval, we use both sample proportions:
\[SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\]
For a hypothesis test (where \(H_0: p_1 = p_2\)), we pool the proportions:
\[\hat{p}_{pool} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{8 + 162}{18198 + 18325} = \frac{170}{36523} \approx 0.00465\]
\[SE_{pool} = \sqrt{\hat{p}_{pool}(1-\hat{p}_{pool})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\]
\[\hat{p}_{pool} = \frac{170}{36523} = 0.004654\]
\[SE_{pool} = \sqrt{0.004654 \times 0.99535 \times \left(\frac{1}{18198} + \frac{1}{18325}\right)} = \sqrt{5.07 \times 10^{-7}} \approx 0.000712\]
\[z = \frac{(0.000440 - 0.008839) - 0}{0.000712} = \frac{-0.008399}{0.000712} \approx -11.8\]
p-value ≈ 0 (essentially zero for a one-sided test with z = −11.8)
Conclusion: There is overwhelming statistical evidence that the true infection rate is lower in the vaccinated group than in the placebo group (z = −11.8, p ≈ 0).
In R, we use prop.test(). Here is what the output looks like:
2-sample test for equality of proportions
data: c(8, 162) out of c(18198, 18325)
X-squared = 139.23, df = 1, p-value < 2.2e-16
alternative hypothesis: less
95 percent confidence interval:
-1.0000000 -0.0073064
sample estimates:
prop 1 prop 2
0.00043961 0.00883889Notes on the output:
prop.test() uses a chi-square statistic (\(X^2 = z^2 = 139.23 \approx 11.8^2\)) ✅[Poll Everywhere — respond now!]
Consider this R output from a different vaccine study:
2-sample test for equality of proportions
data: c(45, 68) out of c(900, 920)
X-squared = 4.21, df = 1, p-value = 0.040
alternative hypothesis: two.sided
95 percent confidence interval:
-0.0461 -0.0006
sample estimates:
prop 1 prop 2
0.050000 0.073913Discuss (2 min):
→ Poll Everywhere: True or False — If p-value < 0.05, the effect must be large.
\[(\hat{p}_1 - \hat{p}_2) \pm z^* \cdot \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\]
For the Pfizer trial:
\[\hat{p}_V - \hat{p}_P = 0.000440 - 0.008839 = -0.008399\]
\[SE = \sqrt{\frac{0.000440 \times 0.999560}{18198} + \frac{0.008839 \times 0.991161}{18325}} \approx 0.000712\]
95% CI:
\[-0.008399 \pm 1.96 \times 0.000712 = (-0.00979,\ -0.00701)\]
We are 95% confident that the true difference in infection rates (vaccine − placebo) is between −0.98 and −0.70 percentage points.
Check these for each group separately:
1. Independence: Groups are independent of each other ✅ (Random assignment ensures this)
2. Success/Failure in each group:
For the Pfizer vaccine group: \(n_1 \hat{p}_1 = 18198 \times 0.000440 = 8 < 10\) ⚠️
Note: With rare events (very small p), even large samples may not meet this condition perfectly. The z-test is approximate; exact methods exist for such cases.
[Poll Everywhere — respond now!]
A public health researcher studies HPV vaccination in two counties:
Discuss (2 min):
→ Poll Everywhere: What is \(\hat{p}_{pool}\) for this study? (Round to 2 decimal places)
| Single Proportion | Two Proportions | |
|---|---|---|
| Statistic | \(\hat{p}\) | \(\hat{p}_1 - \hat{p}_2\) |
| SE (CI) | \(\sqrt{\hat{p}(1-\hat{p})/n}\) | \(\sqrt{\hat{p}_1(1-\hat{p}_1)/n_1 + \hat{p}_2(1-\hat{p}_2)/n_2}\) |
| SE (test) | \(\sqrt{p_0(1-p_0)/n}\) | \(\sqrt{\hat{p}_{pool}(1-\hat{p}_{pool})(1/n_1 + 1/n_2)}\) |
| Distribution | Normal (z) | Normal (z) |
| Conditions | np̂ ≥ 10, n(1-p̂) ≥ 10 | Both groups separately |
CI vs. Test SE: They use different formulas!
Pooling: We pool for two-proportion tests because under H₀, both groups have the same true p — so we combine all data to estimate it.
Conditions matter: Always check success/failure for each group. When counts are small, interpret results cautiously.
Thursday: We extend contingency table analysis to the chi-square test — a more flexible approach when we have multiple categories.
HW8 / Practice Final: Covers all material from Week 5 onward with emphasis on interpretation. You’ll be given R output and asked to explain what it means.
Key reminder for the final: No distribution tables. No software. Focus on:
STAT 7 · Winter 2026