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Confidence Intervals for Proportions

Construct and interpret confidence intervals for a population proportion.

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📐 Confidence Intervals for Proportions

Confidence Interval Formula

A confidence interval for a population proportion p is:

p^±z∗⋅SE(p^)\hat{p} \pm z^* \cdot SE(\hat{p})

where:

  • p^\hat{p} = sample proportion (point estimate)
  • z∗z^* = critical z-value (depends on confidence level)
  • SE(p^)=p^(1−p^)nSE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Common critical values:

  • 90% CI: z∗=1.645z^* = 1.645
  • 95% CI: z∗=1.96z^* = 1.96
  • 99% CI: z∗=2.576z^* = 2.576

Conditions for Validity

Before constructing CI for proportions, verify:

  1. Random sample: Data collected randomly
  2. Independence: Sampling without replacement; use 10% rule (n ≤ 0.10N)
  3. Success/failure rule: Both np^≥10n\hat{p} \geq 10 and n(1−p^)≥10n(1-\hat{p}) \geq 10

If these fail, don't use the standard formula.

Worked Example

A poll of 400 likely voters finds 220 support Candidate A. Construct a 95% CI for the population proportion.

Step 1: p^=220400=0.55\hat{p} = \frac{220}{400} = 0.55

Step 2: Check conditions:

  • Random sample ✓
  • n = 400 ≤ 0.10(population) ✓ (assume population is large)
  • np^=400(0.55)=220≥10n\hat{p} = 400(0.55) = 220 \geq 10 ✓
  • n(1−p^)=400(0.45)=180≥10n(1-\hat{p}) = 400(0.45) = 180 \geq 10 ✓

Step 3: Calculate SE: SE=0.55⋅0.45400=0.2475400=0.00061875≈0.0249SE = \sqrt{\frac{0.55 \cdot 0.45}{400}} = \sqrt{\frac{0.2475}{400}} = \sqrt{0.00061875} \approx 0.0249

Step 4: Calculate margin of error (ME): ME=1.96×0.0249≈0.0488ME = 1.96 \times 0.0249 \approx 0.0488

Step 5: Confidence interval: 0.55±0.0488=(0.5012,0.5988)0.55 \pm 0.0488 = (0.5012, 0.5988)

Interpretation: We are 95% confident that between 50.1% and 59.9% of likely voters support Candidate A.

Margin of Error

The margin of error (ME) represents the uncertainty in the point estimate:

ME=z∗⋅SE(p^)=z∗p^(1−p^)nME = z^* \cdot SE(\hat{p}) = z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Larger ME means wider CI (less precise). Smaller ME means narrower CI (more precise).

Sample Size for Desired Margin of Error

To find the sample size needed for margin ME:

n=(z∗)2⋅p(1−p)ME2n = \frac{(z^*)^2 \cdot p(1-p)}{ME^2}

If p is unknown, use p = 0.5 (gives maximum sample size; most conservative).

Example: You want 95% CI with ME = 0.03. How large must the sample be?

n=(1.96)2⋅0.5(0.5)(0.03)2=3.8416⋅0.250.0009=0.96040.0009≈1068n = \frac{(1.96)^2 \cdot 0.5(0.5)}{(0.03)^2} = \frac{3.8416 \cdot 0.25}{0.0009} = \frac{0.9604}{0.0009} \approx 1068

Common Mistakes

  1. Wrong formula: Using z∗z^* instead of t∗t^* (t-values are for means, not proportions)
  2. Forgetting to check conditions: Missing the success/failure rule or 10% rule
  3. Using p instead of p^\hat{p}: The SE formula uses the sample proportion, not the population proportion

AP Exam Tip

Know the margin of error formula cold. Many free-response questions ask you to find the sample size needed to achieve a specific margin. Always verify conditions before calculating; if conditions fail, state which ones and explain why the interval is not valid.

📚 Practice Problems

1Problem 1easy

❓ Question:

In a random sample of 400 voters, 220 support a proposition. Construct a 95% confidence interval for the true proportion of voters who support the proposition.

💡 Show Solution

Step 1: Identify the information n = 400 (sample size) x = 220 (number of successes) p̂ = 220/400 = 0.55 (sample proportion)

Confidence level: 95%

Step 2: Check conditions for proportion CI RANDOM: Sample is random ✓ NORMAL: np̂ ≥ 10 and n(1-p̂) ≥ 10 400(0.55) = 220 ≥ 10 ✓ 400(0.45) = 180 ≥ 10 ✓ INDEPENDENT: n ≤ 0.10N 400 ≤ 0.10(all voters) - assume yes ✓

All conditions met!

Step 3: Find critical value 95% confidence → α = 0.05 z* = 1.96 (from table for 95% CI)

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

Step 5: Calculate margin of error ME = z* × SE = 1.96 × 0.0249 ≈ 0.0488

Step 6: Construct confidence interval CI = p̂ ± ME = 0.55 ± 0.049 = (0.501, 0.599)

Or: (0.50, 0.60) rounded

Step 7: Interpret the interval We are 95% confident that the true proportion of voters who support the proposition is between 0.50 and 0.60 (or 50% and 60%).

This means:

  • If we repeated this sampling process many times
  • About 95% of intervals would contain true p
  • This specific interval either contains p or doesn't
  • But the process is reliable 95% of the time

Answer: 95% CI for p: (0.50, 0.60)

We are 95% confident that between 50% and 60% of all voters support the proposition.

2Problem 2easy

❓ Question:

A quality control inspector finds 8 defects in a sample of 200 items. Construct a 90% confidence interval for the defect rate.

💡 Show Solution

Step 1: Calculate sample proportion n = 200 x = 8 p̂ = 8/200 = 0.04

Step 2: Check conditions RANDOM: Assume random sample ✓ NORMAL: np̂ = 200(0.04) = 8 < 10 ✗ n(1-p̂) = 200(0.96) = 192 ≥ 10 ✓

Condition fails! But let's proceed with caution. (In practice, might use exact binomial method)

Step 3: Find z* for 90% confidence 90% confidence → z* = 1.645

Step 4: Calculate SE SE = √[p̂(1-p̂)/n] = √[0.04(0.96)/200] = √[0.0384/200] = √0.000192 ≈ 0.0139

Step 5: Calculate ME ME = 1.645 × 0.0139 ≈ 0.023

Step 6: Construct CI CI = 0.04 ± 0.023 = (0.017, 0.063) = (1.7%, 6.3%)

Step 7: Interpret with caution We are 90% confident the true defect rate is between 1.7% and 6.3%.

Note: This interval may not be as reliable since np̂ < 10.

Answer: 90% CI: (0.017, 0.063) or (1.7%, 6.3%)

Caution: The success-failure condition is marginally violated (only 8 successes), so this normal-based interval may not be fully reliable.

3Problem 3medium

❓ Question:

A researcher wants to estimate the proportion of defective items with a margin of error no more than 0.03 at 90% confidence. How large a sample is needed if no prior estimate exists?

💡 Show Solution

Step 1: Identify what we need ME = 0.03 Confidence level = 90% → z* = 1.645 No prior estimate → use p̂ = 0.5

Step 2: Use sample size formula n = (z*)²p̂(1-p̂)/ME²

Step 3: Calculate n = (1.645)²(0.5)(0.5)/(0.03)² = 2.706(0.25)/0.0009 = 0.6765/0.0009 ≈ 751.67

Step 4: Round UP Always round UP to ensure ME is no larger than desired n = 752

Step 5: Why use p̂ = 0.5? The product p̂(1-p̂) is maximized at p̂ = 0.5 This gives the most conservative (largest) sample size Guarantees ME ≤ 0.03 regardless of true p

Answer: n = 752

Need a sample of at least 752 items to achieve a margin of error no more than 0.03 at 90% confidence.

4Problem 4medium

❓ Question:

Two polls: Poll A (n=500, p̂=0.52) and Poll B (n=1000, p̂=0.51). Both use 95% confidence. Which poll has a smaller margin of error? Calculate both.

💡 Show Solution

Step 1: Recall margin of error formula ME = z*√[p̂(1-p̂)/n]

For 95% CI: z* = 1.96

Step 2: Calculate ME for Poll A p̂ = 0.52, n = 500

ME_A = 1.96√[0.52(0.48)/500] = 1.96√[0.2496/500] = 1.96√0.0004992 = 1.96(0.0223) ≈ 0.044

Step 3: Calculate ME for Poll B p̂ = 0.51, n = 1000

ME_B = 1.96√[0.51(0.49)/1000] = 1.96√[0.2499/1000] = 1.96√0.0002499 = 1.96(0.0158) ≈ 0.031

Step 4: Compare Poll A: ME ≈ 0.044 or 4.4% Poll B: ME ≈ 0.031 or 3.1%

Poll B has smaller margin of error!

Step 5: Why is Poll B better? Larger sample size (1000 vs 500) ME ∝ 1/√n Doubling n reduces ME by factor of √2 ≈ 1.41

500 × 2 = 1000 ME_A/ME_B = √(1000/500) = √2 ≈ 1.41 0.044/0.031 ≈ 1.42 ✓

Step 6: Effect of p̂ Poll B also has p̂ closer to 0.5 But this increases ME slightly Effect of larger n dominates

Answer: Poll B has smaller ME (0.031 vs 0.044)

Poll B's larger sample size (1000 vs 500) gives more precision despite having p̂ closer to 0.5.

5Problem 5hard

❓ Question:

Explain why we can't construct a valid confidence interval for a proportion when the sample proportion is 0 or 1.

💡 Show Solution

Step 1: Recall CI formula CI = p̂ ± z*√[p̂(1-p̂)/n]

SE = √[p̂(1-p̂)/n]

Step 2: What happens when p̂ = 0? SE = √[0(1)/n] = 0 CI = 0 ± 0 = (0, 0)

This says we're 100% certain p = 0 Unreasonable from a sample!

Step 3: What happens when p̂ = 1? SE = √[1(0)/n] = 0 CI = 1 ± 0 = (1, 1)

This says we're 100% certain p = 1 Also unreasonable!

Step 4: Normal approximation fails Need: np̂ ≥ 10 AND n(1-p̂) ≥ 10

When p̂ = 0: np̂ = 0 < 10 ✗ When p̂ = 1: n(1-p̂) = 0 < 10 ✗

Can't use normal-based method!

Step 5: What to do instead Use: Wilson score interval, Agresti-Coull, or exact binomial methods These give reasonable intervals even with extreme values

Answer: When p̂ = 0 or 1, SE = 0, giving a degenerate interval. Normal approximation conditions fail. Alternative methods should be used.

Explain using:

⚠️ Common Mistakes: Confidence Intervals for Proportions

Avoid these 3 frequent errors

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📌 Related Topics in Unit 6: Inference for Categorical Data — Proportions

Tests for ProportionsInference for Two Sample Proportions (CI and Test)

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Construct and interpret confidence intervals for a population proportion.
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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