Hypothesis Tests for Proportions
Learn how to test claims about population proportions using the normal approximation
Lesson Objectives
By the end of this lesson, you will be able to:
- Formulate hypotheses for population proportions
- Check conditions for valid proportion tests
- Calculate test statistics for proportions
- Conduct hypothesis tests using both critical value and p-value approaches
- Apply proportion tests to real-world scenarios
1. When to Test Proportions
We test hypotheses about population proportions when our data is:
- Categorical/Qualitative (yes/no, success/failure, pass/fail)
- Can be converted to a proportion or percentage
Real-World Examples
- Quality Control: "Less than 5% of products are defective"
- Politics: "More than 50% of voters support the candidate"
- Medicine: "The cure rate is 80%"
- Education: "At least 70% of students pass the exam"
- Marketing: "30% of customers will buy the new product"
Key Notation
| Symbol | Meaning |
|---|---|
| p | Population proportion (parameter, unknown) |
| p₀ | Hypothesized population proportion (claimed value in H₀) |
| p̂ (p-hat) | Sample proportion (statistic, calculated from data) |
| n | Sample size |
| x | Number of successes in sample |
Sample Proportion
Where x = number of successes, n = sample size
2. Conditions for Proportion Tests
IMPORTANT: Check These Conditions First!
Before conducting a hypothesis test for proportions, verify these conditions:
Required Conditions:
- Random Sample: The sample must be randomly selected from the population
-
Independence: Observations must be independent
- If sampling without replacement: n ≤ 0.05N (sample is at most 5% of population)
-
Normal Approximation (Success-Failure Condition):
np₀ ≥ 10 AND n(1 - p₀) ≥ 10
This ensures the sampling distribution of p̂ is approximately normal so we can use the z-distribution.
Example 1: Checking Conditions
Scenario: A company claims 25% of customers prefer their new product. We survey 200 customers.
Check conditions:
- Random sample: (assume survey was random)
- Independence: (200 is less than 5% of all customers)
- Normal approximation:
np₀ = 200(0.25) = 50 ≥ 10
n(1-p₀) = 200(0.75) = 150 ≥ 10
Conclusion: All conditions are met. We can proceed with the test!
3. Test Statistic for Proportions
Test Statistic for One-Sample Proportion Test
Where:
p̂ = sample proportion
p₀ = hypothesized population proportion (from H₀)
n = sample size
Hypothesis Forms
| Test Type | Null Hypothesis | Alternative Hypothesis |
|---|---|---|
| Two-tailed | H₀: p = p₀ | Hₐ: p ≠ p₀ |
| Right-tailed | H₀: p = p₀ | Hₐ: p > p₀ |
| Left-tailed | H₀: p = p₀ | Hₐ: p < p₀ |
4. Examples: Hypothesis Tests for Proportions
Example 2: Right-Tailed Test (Critical Value Approach)
Problem: A politician claims that more than 60% of voters support a new policy. A poll of 400 voters shows 256 support it. Test at α = 0.05.
Step 1: State hypotheses
- H₀: p = 0.60 (60% support, as claimed)
- Hₐ: p > 0.60 (more than 60% support) — right-tailed test
- α = 0.05
Step 2: Check conditions
n(1-p₀) = 400(0.40) = 160 ≥ 10
Conditions met!
Step 3: Calculate sample proportion and test statistic
z = (p̂ - p₀) / √(p₀(1-p₀)/n)
z = (0.64 - 0.60) / √(0.60(0.40)/400)
z = 0.04 / √(0.24/400)
z = 0.04 / √0.0006
z = 0.04 / 0.0245
z = 1.63
Step 4: Find critical value
Right-tailed test, α = 0.05 → critical value = 1.645
Step 5: Make decision
z = 1.63 < 1.645 (not in critical region)
Decision: Fail to reject H₀
Step 6: Conclusion
There is not sufficient evidence at the 0.05 significance level to support the politician's claim that more than 60% of voters support the policy. While the sample showed 64% support, this difference could be due to sampling variability.
Example 3: Left-Tailed Test (P-Value Approach)
Problem: A manufacturer claims their defect rate is 3%. Quality control inspects 500 items and finds 10 defects. Test if the defect rate is less than claimed (α = 0.01).
Step 1: State hypotheses
- H₀: p = 0.03 (defect rate is 3%)
- Hₐ: p < 0.03 (defect rate is less than 3%) — left-tailed test
- α = 0.01
Step 2: Check conditions
n(1-p₀) = 500(0.97) = 485 ≥ 10
Conditions met!
Step 3: Calculate test statistic
z = (0.02 - 0.03) / √(0.03(0.97)/500)
z = -0.01 / √(0.0291/500)
z = -0.01 / 0.00763
z = -1.31
Step 4: Find p-value
For left-tailed test: p-value = P(Z < -1.31)
From z-table: p-value = 0.0951
Step 5: Make decision
p-value (0.0951) > α (0.01)
Decision: Fail to reject H₀
Step 6: Conclusion
There is not sufficient evidence to conclude the defect rate is less than 3%. The observed 2% defect rate could reasonably occur by chance if the true rate is 3%.
Example 4: Two-Tailed Test
Problem: A university claims 70% of graduates find jobs within 6 months. A survey of 300 recent graduates finds 195 employed. Test if the actual rate differs from the claim (α = 0.05).
Step 1: State hypotheses
- H₀: p = 0.70 (employment rate is 70%)
- Hₐ: p ≠ 0.70 (employment rate differs from 70%) — two-tailed test
- α = 0.05
Step 2: Check conditions
n(1-p₀) = 300(0.30) = 90 ≥ 10
Step 3: Calculate test statistic
z = (0.65 - 0.70) / √(0.70(0.30)/300)
z = -0.05 / √(0.21/300)
z = -0.05 / 0.0265
z = -1.89
Step 4: Decision (both approaches)
Critical value: Two-tailed, α = 0.05 → ±1.96
|z| = 1.89 < 1.96 → Fail to reject H₀
P-value: For z = -1.89, two-tailed p-value = 2(0.0294) = 0.0588
0.0588 > 0.05 → Fail to reject H₀
Step 5: Conclusion
There is not sufficient evidence to conclude the employment rate differs from the claimed 70%. Although only 65% of the sample were employed, this could be due to sampling variability (p = 0.0588 is close to, but above, α = 0.05).
Check Your Understanding
Question 1: A candy company claims 20% of their candies are red. In a bag of 100 candies, you find 15 red ones. Check if the normal approximation conditions are met.
Check conditions:
n(1-p₀) = 100(0.80) = 80 ≥ 10
Conclusion: Yes, conditions are met. We can use the normal approximation (z-test) for proportions.
Question 2: Calculate the test statistic: n = 200, x = 50, p₀ = 0.20
Step 1: Calculate p̂
Step 2: Calculate z
z = (0.25 - 0.20) / √(0.20(0.80)/200)
z = 0.05 / √(0.16/200)
z = 0.05 / √0.0008
z = 0.05 / 0.0283
z = 1.77
Question 3: For a two-tailed test at α = 0.05, if z = -2.15, what is your decision?
Critical values: ±1.96
Comparison: |z| = |-2.15| = 2.15 > 1.96
Decision: Reject H₀
Reasoning: The test statistic falls in the critical region (more extreme than ±1.96), so we have sufficient evidence to reject the null hypothesis at the 0.05 significance level.
Summary
- Use proportion tests for categorical data (yes/no, success/failure)
- Check conditions: Random sample, independence, and np₀ ≥ 10 AND n(1-p₀) ≥ 10
- Test statistic: z = (p̂ - p₀) / √(p₀(1-p₀)/n)
- Use p₀ (hypothesized value) in standard error, not p̂
- Critical values come from standard normal (z) distribution
- Always interpret results in the context of the real-world problem