Tests for Proportions

One-sample and two-sample z-tests

Hypothesis Tests for Proportions

One-Sample z-Test for Proportion

Test: Does sample provide evidence that population proportion differs from claimed value?

Hypotheses:

H₀: p = p₀
Hₐ: p ≠ p₀ (or p > p₀ or p < p₀)

Conditions:

Random sample
np₀ ≥ 10 and n(1-p₀) ≥ 10 (use p₀, not p̂!)
n < 10% of population

Test Statistic

$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Note: Use p₀ (null value) in SE, not p̂

Under H₀: z follows standard normal distribution

P-Value Calculation

Two-sided (Hₐ: p ≠ p₀): P-value = 2 × P(Z ≥ |z|)

Right-sided (Hₐ: p > p₀): P-value = P(Z ≥ z)

Left-sided (Hₐ: p < p₀): P-value = P(Z ≤ z)

Calculator: normalcdf

Example 1: Two-Sided Test

Claim: Coin is fair. Flip 100 times, get 58 heads. Test at α = 0.05.

STATE:

Parameter: p = true proportion of heads
H₀: p = 0.5
Hₐ: p ≠ 0.5
α = 0.05

PLAN:

One-sample z-test for proportion
Random: Assume ✓
Normal: 100(0.5) = 50 ≥ 10, 100(0.5) = 50 ≥ 10 ✓
Independent: 100 < all possible flips ✓

DO:

$\hat{p} = \frac{58}{100} = 0.58$

$z = \frac{0.58 - 0.5}{\sqrt{\frac{0.5(0.5)}{100}}} = \frac{0.08}{0.05} = 1.6$

P-value = 2 × P(Z ≥ 1.6) = 2(0.0548) ≈ 0.1096

CONCLUDE: P-value = 0.1096 > 0.05, fail to reject H₀. Insufficient evidence coin is unfair.

Example 2: One-Sided Test

Company claims > 80% customer satisfaction. Survey 200, find 168 satisfied.

STATE:

p = true proportion satisfied
H₀: p = 0.8
Hₐ: p > 0.8
α = 0.05

PLAN:

One-sample z-test for proportion
Conditions: ✓ (check all three)

DO:

$\hat{p} = \frac{168}{200} = 0.84$

$z = \frac{0.84 - 0.8}{\sqrt{\frac{0.8(0.2)}{200}}} = \frac{0.04}{0.0283} \approx 1.41$

P-value = P(Z ≥ 1.41) ≈ 0.079

CONCLUDE: P-value = 0.079 > 0.05, fail to reject H₀. Insufficient evidence satisfaction exceeds 80%.

Two-Sample z-Test for Proportions

Compare two proportions:

Hypotheses:

H₀: p₁ = p₂ (or p₁ - p₂ = 0)
Hₐ: p₁ ≠ p₂ (or p₁ > p₂ or p₁ < p₂)

Test Statistic:

$z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

Where pooled proportion:

$\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}$

Key: Pool data under assumption p₁ = p₂ (H₀)

Conditions for Two-Sample Test

Both samples:

Random/independent
n₁p̂c ≥ 10, n₁(1-p̂c) ≥ 10
n₂p̂c ≥ 10, n₂(1-p̂c) ≥ 10
n₁ < 10%N₁, n₂ < 10%N₂

Example 3: Two-Sample Test

Treatment vs Placebo:

Treatment: 45/100 improved
Placebo: 30/100 improved

STATE:

p₁ = proportion improved with treatment
p₂ = proportion improved with placebo
H₀: p₁ = p₂
Hₐ: p₁ > p₂
α = 0.05

PLAN:

Two-sample z-test
Conditions: ✓

DO:

$\hat{p}_1 = 0.45, \quad \hat{p}_2 = 0.30$

$\hat{p}_c = \frac{45 + 30}{100 + 100} = \frac{75}{200} = 0.375$

$z = \frac{0.45 - 0.30}{\sqrt{0.375(0.625)(\frac{1}{100} + \frac{1}{100})}} = \frac{0.15}{\sqrt{0.0047}} \approx 2.19$

P-value = P(Z ≥ 2.19) ≈ 0.014

CONCLUDE: P-value = 0.014 < 0.05, reject H₀. Sufficient evidence treatment proportion exceeds placebo.

Calculator Commands (TI-83/84)

One-sample: STAT → TESTS → 5:1-PropZTest

p₀, x, n, direction
Calculate

Two-sample: STAT → TESTS → 6:2-PropZTest

x₁, n₁, x₂, n₂, direction
Calculate

Common Mistakes

❌ Using p̂ instead of p₀ in SE for one-sample
❌ Not pooling for two-sample test
❌ Checking conditions with p̂ instead of p₀
❌ Wrong P-value for one-sided vs two-sided
❌ Forgetting to check conditions

When to Use

One-sample: Comparing proportion to claimed value

Two-sample: Comparing two independent groups

Paired: If data paired, analyze differences (not proportions)

Quick Reference

One-sample:

Test statistic: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$
Use p₀ in SE

Two-sample:

Test statistic uses pooled p̂c
Pool assuming H₀: p₁ = p₂ is true

Conditions: Random, normal (np ≥ 10, n(1-p) ≥ 10), independent

Remember: For proportions, use z-test (not t). Check conditions with null hypothesis values!

📚 Practice Problems

1Problem 1easy

❓ Question:

A company claims 30% of customers prefer their new product. In a random sample of 200 customers, 48 prefer it. Test at α = 0.05 if the true proportion is less than 30%.

💡 Show Solution

Step 1: Set up hypotheses Claim: p = 0.30 Suspect: p < 0.30

H₀: p = 0.30 Hₐ: p < 0.30 (one-tailed, left)

Step 2: Check conditions n = 200, p₀ = 0.30

RANDOM: Random sample ✓ NORMAL: np₀ = 200(0.30) = 60 ≥ 10 ✓ n(1-p₀) = 200(0.70) = 140 ≥ 10 ✓ INDEPENDENT: 200 ≤ 0.10N (assume) ✓

Can use z-test!

Step 3: Calculate sample proportion p̂ = 48/200 = 0.24

Step 4: Calculate test statistic z = (p̂ - p₀)/√(p₀(1-p₀)/n) = (0.24 - 0.30)/√(0.30(0.70)/200) = -0.06/√(0.21/200) = -0.06/√0.00105 = -0.06/0.0324 ≈ -1.85

Step 5: Find p-value (one-tailed) Left-tailed test (Hₐ: p < 0.30)

From z-table: P(Z < -1.85) = 0.0322

p-value = 0.0322

Step 6: Make decision p-value = 0.0322 α = 0.05

Is 0.0322 < 0.05? YES

REJECT H₀

Step 7: State conclusion At the α = 0.05 significance level, there is sufficient evidence to conclude that the true proportion who prefer the new product is less than 30%.

The sample data (24%) provides convincing evidence that preference is lower than the company's claim.

Step 8: Interpret in context Evidence suggests:

Fewer than 30% prefer new product
Company's claim may be overstated
Sample result (24%) is significantly lower
Not just due to random chance

Answer: H₀: p = 0.30, Hₐ: p < 0.30 Test statistic: z = -1.85 P-value: 0.032 Decision: Reject H₀ at α = 0.05 Conclusion: Sufficient evidence that true proportion is less than 30%

2Problem 2easy

❓ Question:

In 2020, 45% of voters supported a candidate. In 2024, a poll of 400 voters shows 200 support the candidate. Test if support has changed from 45%.

💡 Show Solution

Step 1: Set up hypotheses Previous: p = 0.45 Question: Has it changed?

H₀: p = 0.45 (support unchanged) Hₐ: p ≠ 0.45 (support has changed)

TWO-TAILED test!

Step 2: Check conditions n = 400, p₀ = 0.45

RANDOM: Assume random sample ✓ NORMAL: np₀ = 400(0.45) = 180 ≥ 10 ✓ n(1-p₀) = 400(0.55) = 220 ≥ 10 ✓ INDEPENDENT: 400 ≤ 0.10N ✓

Step 3: Calculate sample proportion p̂ = 200/400 = 0.50

Step 4: Calculate standard error SE = √(p₀(1-p₀)/n) = √(0.45(0.55)/400) = √(0.2475/400) = √0.00061875 ≈ 0.0249

Step 5: Calculate test statistic z = (p̂ - p₀)/SE = (0.50 - 0.45)/0.0249 = 0.05/0.0249 ≈ 2.01

Step 6: Find p-value (two-tailed!) Right tail: P(Z > 2.01) ≈ 0.0222 Left tail: P(Z < -2.01) ≈ 0.0222

Two-tailed p-value: p-value = 2 × 0.0222 = 0.0444

Step 7: Make decision p-value = 0.0444 α = 0.05

Is 0.0444 < 0.05? YES (barely!)

REJECT H₀

Step 8: State conclusion At the α = 0.05 significance level, there is sufficient evidence that support has changed from 45%. The current support appears to be different from 2020 levels.

Step 9: Additional interpretation Current support: 50% Previous support: 45% Change: +5 percentage points

This increase is statistically significant:

Not just random variation
Real change in support
Though p-value (0.044) is close to α (0.05)
Borderline significance

Answer: H₀: p = 0.45, Hₐ: p ≠ 0.45 Test statistic: z = 2.01 P-value: 0.044 (two-tailed) Decision: Reject H₀ at α = 0.05 Conclusion: Support has changed significantly from 45% (appears higher at 50%)

3Problem 3medium

❓ Question:

A quality control manager wants to test if the defect rate is greater than 2%. She samples 500 items and finds 15 defects. Conduct the test at α = 0.01.

💡 Show Solution

Step 1: Set up hypotheses Claim: p = 0.02 (2% defect rate) Concern: p > 0.02 (higher than acceptable)

H₀: p = 0.02 Hₐ: p > 0.02 (one-tailed, right)

Step 2: Check conditions n = 500, p₀ = 0.02

RANDOM: Assume random sample ✓ NORMAL: np₀ = 500(0.02) = 10 ≥ 10 ✓ (exactly!) n(1-p₀) = 500(0.98) = 490 ≥ 10 ✓ INDEPENDENT: 500 ≤ 0.10N ✓

Conditions met (barely for np₀)

Step 3: Calculate sample proportion p̂ = 15/500 = 0.03 (3%)

Step 4: Calculate SE SE = √(p₀(1-p₀)/n) = √(0.02(0.98)/500) = √(0.0196/500) = √0.0000392 ≈ 0.00626

Step 5: Calculate test statistic z = (p̂ - p₀)/SE = (0.03 - 0.02)/0.00626 = 0.01/0.00626 ≈ 1.60

Step 6: Find p-value Right-tailed test (Hₐ: p > 0.02)

P(Z > 1.60) ≈ 0.0548

p-value = 0.0548

Step 7: Make decision p-value = 0.0548 α = 0.01

Is 0.0548 < 0.01? NO

FAIL TO REJECT H₀

Step 8: State conclusion At the α = 0.01 significance level, there is insufficient evidence to conclude that the defect rate exceeds 2%.

The observed 3% defect rate could reasonably occur by chance if the true rate is 2%.

Step 9: Context and interpretation Observed: 3% defects Standard: 2% defects Difference: 1 percentage point

This difference is:

Not statistically significant at α = 0.01
p = 0.055 > 0.01
Could be random variation
Not enough evidence for concern

Step 10: Additional notes If α = 0.05 instead:

0.0548 > 0.05 (still fail to reject, barely!)
Borderline case
Might warrant further monitoring

Conservative conclusion:

Insufficient evidence of problem
But keep monitoring
Sample evidence suggestive but not conclusive

Answer: H₀: p = 0.02, Hₐ: p > 0.02 Test statistic: z = 1.60 P-value: 0.055 Decision: Fail to reject H₀ at α = 0.01 Conclusion: Insufficient evidence that defect rate exceeds 2%

4Problem 4medium

❓ Question:

A researcher finds p̂ = 0.58 from n = 100. She wants to test H₀: p = 0.50 vs Hₐ: p > 0.50. Check the conditions and determine if a z-test is appropriate.

💡 Show Solution

Step 1: List the conditions for z-test For one-sample z-test for proportion:

RANDOM: Random sample
NORMAL: np₀ ≥ 10 AND n(1-p₀) ≥ 10
INDEPENDENT: n ≤ 0.10N

Step 2: Check RANDOM condition Statement: Need random sample

Given: "researcher finds" - not stated Assume: Random sample ✓

Note: Should be explicitly stated!

Step 3: Check NORMAL condition Use p₀ = 0.50 (from H₀, not p̂!)

np₀ = 100(0.50) = 50 ≥ 10 ✓ n(1-p₀) = 100(0.50) = 50 ≥ 10 ✓

Both satisfied! Sampling distribution approximately normal

Step 4: Check INDEPENDENT condition Need: n ≤ 0.10N

n = 100 Need: 100 ≤ 0.10N N ≥ 1000

If population has at least 1000: Independent ✓

Assumption: Reasonable for most populations

Step 5: Common mistake to avoid DON'T use p̂ to check normal condition! USE p₀ (from null hypothesis)

WRONG: np̂ = 100(0.58) = 58 RIGHT: np₀ = 100(0.50) = 50

Why? Because we're testing under assumption H₀ is true!

Step 6: Is z-test appropriate? YES, if: ✓ Sample is random ✓ np₀ = 50 ≥ 10 ✓ n(1-p₀) = 50 ≥ 10 ✓ Population reasonably large (N ≥ 1000)

All conditions met! Can proceed with z-test

Step 7: Additional consideration Sample size n = 100:

Moderate sample size
Sufficient for normal approximation
SE will be reasonably small

SE = √(p₀(1-p₀)/n) = √(0.50(0.50)/100) = √0.0025 = 0.05

Step 8: Conduct the test (since appropriate) z = (p̂ - p₀)/SE = (0.58 - 0.50)/0.05 = 0.08/0.05 = 1.60

p-value = P(Z > 1.60) ≈ 0.0548

Step 9: When z-test might NOT be appropriate Fails if: ✗ np₀ < 10 (too few successes expected) ✗ n(1-p₀) < 10 (too few failures expected) ✗ Not random sample ✗ Sample from small population without replacement

Example of failure: n = 15, p₀ = 0.10 np₀ = 1.5 < 10 ✗ Cannot use z-test!

Step 10: Summary of condition checking Given information: n = 100 ✓ p̂ = 0.58 p₀ = 0.50

Conditions: Random: Assumed ✓ Normal: 50 ≥ 10 and 50 ≥ 10 ✓ Independent: Assumed N ≥ 1000 ✓

Conclusion: Z-test IS appropriate

Answer: YES, z-test is appropriate.

CONDITIONS CHECK:

Random: Assume random sample ✓
Normal: np₀ = 100(0.50) = 50 ≥ 10 ✓ n(1-p₀) = 100(0.50) = 50 ≥ 10 ✓
Independent: Assume N ≥ 1000 so n ≤ 0.10N ✓

All conditions satisfied. Can use z-test for proportion.

Key: Use p₀ (not p̂) when checking normal condition!

5Problem 5hard

❓ Question:

A coin is flipped 100 times, resulting in 60 heads. Test at α = 0.05 if the coin is biased. Then, determine how many heads would be needed for the result to be significant.

💡 Show Solution

PART 1: Test with 60 heads

Step 1: Set up hypotheses H₀: p = 0.50 (coin is fair) Hₐ: p ≠ 0.50 (coin is biased)

Two-tailed test!

Step 2: Check conditions n = 100, p₀ = 0.50

Random: Assume random flips ✓ Normal: np₀ = 50 ≥ 10 ✓ n(1-p₀) = 50 ≥ 10 ✓ Independent: Flips independent ✓

Step 3: Calculate p̂ = 60/100 = 0.60

SE = √(0.50(0.50)/100) = √0.0025 = 0.05

z = (0.60 - 0.50)/0.05 = 0.10/0.05 = 2.00

Step 4: Find p-value P(Z > 2.00) = 0.0228 Two-tailed: p = 2(0.0228) = 0.0456

Step 5: Decision p = 0.0456 < 0.05 REJECT H₀

Conclusion: Coin appears biased!

PART 2: How many heads needed?

Step 6: Find critical values For α = 0.05, two-tailed: α/2 = 0.025 in each tail

z* = ±1.96

Step 7: Set up inequality Reject H₀ if: |z| ≥ 1.96

|(p̂ - 0.50)/0.05| ≥ 1.96

Step 8: Solve for p̂ (upper tail) (p̂ - 0.50)/0.05 ≥ 1.96 p̂ - 0.50 ≥ 0.098 p̂ ≥ 0.598

Step 9: Convert to number of heads p̂ = x/100 ≥ 0.598 x ≥ 59.8 x ≥ 60 (since x must be integer)

Step 10: Check lower tail (p̂ - 0.50)/0.05 ≤ -1.96 p̂ - 0.50 ≤ -0.098 p̂ ≤ 0.402

x/100 ≤ 0.402 x ≤ 40.2 x ≤ 40

Step 11: Rejection regions Reject H₀ if: x ≥ 60 heads (too many) OR x ≤ 40 heads (too few)

Fail to reject if: 41 ≤ x ≤ 59 (consistent with fair coin)

Step 12: Verify with 60 x = 60 is exactly at boundary p̂ = 0.60 z = 2.00 p-value = 0.0456 < 0.05 ✓

Just barely significant!

Step 13: Verify with 59 x = 59 p̂ = 0.59 z = (0.59 - 0.50)/0.05 = 1.80 p-value = 2(0.0359) = 0.0718 > 0.05 Not significant ✓

Step 14: Interpretation For 100 flips, α = 0.05:

Significantly too many heads: ≥ 60 Significantly too few heads: ≤ 40 Consistent with fair coin: 41-59

Pretty wide range! Only extreme results reject fairness

Step 15: Connection to confidence interval 95% CI for fair coin: p̂ ± 1.96(SE) 0.50 ± 1.96(0.05) 0.50 ± 0.098 (0.402, 0.598) (40.2, 59.8) heads

Values outside this = significant!

Answer: PART 1: With 60 heads: z = 2.00, p-value = 0.046 Reject H₀ at α = 0.05 Coin appears biased (barely significant)

PART 2: Need for significance: ≥ 60 heads (too many) OR ≤ 40 heads (too few) Range 41-59 is consistent with fair coin

60 heads is the minimum to show bias on the high side at α = 0.05 level.

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics