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🎯⭐ INTERACTIVE LESSON

Tests for Proportions

Learn step-by-step with interactive practice!

Tests for Proportions - Complete Interactive Lesson

Part 1: Inference for Proportions Basics

📊 Inference for Proportions

Part 1 of 7 — Inference for Proportions Basics


The Setting

We have a sample proportion p^\hat{p} and want to make inferences about the population proportion pp.

Conditions for Inference

  1. Random: Data from a random sample or experiment
  2. Normal: np≥10np \geq 10 and n(1−p)≥10n(1-p) \geq 10 (use p^\hat{p} for CIs)
  3. Independent: Sample <10%< 10\% of population (10% condition)

Standard Error

SE(p^)=p^(1−p^)nSE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Key Distinction

PurposeFormula for SD
Confidence intervalSE=p^(1−p^)/nSE = \sqrt{\hat{p}(1-\hat{p})/n}
Hypothesis testSE=p0(1−p0)/nSE = \sqrt{p_0(1-p_0)/n} (use H0H_0 value)

Concept Check 🎯

Conditions Check 🧮

In a random sample of 200 voters, 120 support a candidate. p^=0.60\hat{p} = 0.60.

1) np^=?n\hat{p} = ?

2) n(1−p^)=?n(1-\hat{p}) = ?

3) Is the Normal condition met? (yes/no)

Part 2: Confidence Intervals for Proportions

📏 Confidence Intervals for Proportions

Part 2 of 7 — One-Sample Z Interval


Formula

p^±z∗p^(1−p^)n\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Common Critical Values

Confidence Levelz∗z^*
90%1.645
95%1.960
99%2.576

Interpretation

“We are [C]% confident that the true proportion of [context] is between [lower] and [upper].”

Example

n=400n = 400, p^=0.35\hat{p} = 0.35, 95% CI:

0.35±1.960.35×0.65400=0.35±0.04670.35 \pm 1.96\sqrt{\frac{0.35 \times 0.65}{400}} = 0.35 \pm 0.0467

CI: (0.303,0.397)(0.303, 0.397)

Concept Check 🎯

Confidence Interval 🧮

n=500n = 500, p^=0.40\hat{p} = 0.40, 95% CI.

1) SE=0.40×0.60/500SE = \sqrt{0.40 \times 0.60 / 500} = ? (round to 4 decimal places)

2) Margin of error = 1.96×SE1.96 \times SE = ? (round to 4 places)

3) Lower bound of CI? (round to 3 places)

Part 3: Hypothesis Tests for Proportions

⚖️ Hypothesis Tests for Proportions

Part 3 of 7 — One-Sample Z Test


Steps

  1. State hypotheses: H0:p=p0H_0: p = p_0 vs. Ha:p≠p0H_a: p \neq p_0 (or << or >>)
  2. Check conditions (Random, Normal, Independent)
  3. Calculate the test statistic:

z=p^−p0p0(1−p0)/nz = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}

  1. Find the p-value
  2. Conclude in context

P-Value Decision Rules

If p-valueDecision
≤α\leq \alphaReject H0H_0
>α> \alphaFail to reject H0H_0

Example

Claim: p=0.5p = 0.5. Sample: p^=0.56\hat{p} = 0.56, n=200n = 200.

z=0.56−0.500.50×0.50/200=0.060.0354=1.70z = \frac{0.56 - 0.50}{\sqrt{0.50 \times 0.50 / 200}} = \frac{0.06}{0.0354} = 1.70

Concept Check 🎯

Hypothesis Test 🧮

H0:p=0.30H_0: p = 0.30, Ha:p>0.30H_a: p > 0.30. n=150n = 150, p^=0.36\hat{p} = 0.36.

1) SE=0.30×0.70/150SE = \sqrt{0.30 \times 0.70 / 150} = ? (round to 4 places)

2) z=(0.36−0.30)/SEz = (0.36 - 0.30) / SE = ? (round to 2 places)

3) Is this a one-tailed or two-tailed test?

Part 4: Two-Proportion Inference

📊 Two-Proportion Inference

Part 4 of 7 — Comparing Two Proportions


Confidence Interval for p1−p2p_1 - p_2

(p^1−p^2)±z∗p^1(1−p^1)n1+p^2(1−p^2)n2(\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}

Hypothesis Test for p1−p2p_1 - p_2

H0:p1=p2H_0: p_1 = p_2 (or p1−p2=0p_1 - p_2 = 0)

Use the pooled proportion: p^c=x1+x2n1+n2\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}

z=(p^1−p^2)−0p^c(1−p^c)(1n1+1n2)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)}}


Key Difference

  • CI: Use individual p^1\hat{p}_1 and p^2\hat{p}_2 in the SE
  • Test: Use the pooled p^c\hat{p}_c (assuming H0:p1=p2H_0: p_1 = p_2 is true)

Concept Check 🎯

Two-Proportion Test 🧮

Group 1: x1=45x_1 = 45, n1=100n_1 = 100. Group 2: x2=30x_2 = 30, n2=100n_2 = 100.

1) p^1=?\hat{p}_1 = ?

2) Pooled p^c=(45+30)/(100+100)=?\hat{p}_c = (45 + 30)/(100 + 100) = ?

3) p^1−p^2=?\hat{p}_1 - \hat{p}_2 = ?

Part 5: Sample Size Determination

📐 Sample Size Determination

Part 5 of 7 — Planning a Study


Finding the Required Sample Size

For a desired margin of error MEME at confidence level z∗z^*:

n=(z∗ME)2p^(1−p^)n = \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p})

If no prior estimate of pp exists, use p^=0.5\hat{p} = 0.5 (maximizes nn, conservative).

n=(z∗ME)2(0.25)n = \left(\frac{z^*}{ME}\right)^2 (0.25)


Example

Want a 95% CI with margin of error ≤0.03\leq 0.03:

n=(1.960.03)2(0.25)=(65.33)2(0.25)=4268.4(0.25)=1067.1n = \left(\frac{1.96}{0.03}\right)^2 (0.25) = (65.33)^2(0.25) = 4268.4(0.25) = 1067.1

Round up: n=1068n = 1068

🔑 Always round UP to the next whole number when computing sample size.

Concept Check 🎯

Sample Size Calculation 🧮

Desired: 95% CI, margin of error ≤0.04\leq 0.04, no prior estimate of pp.

1) What value of p^\hat{p} should you use?

2) n=(1.96/0.04)2×0.25=?n = (1.96/0.04)^2 \times 0.25 = ? (round to nearest integer)

3) What nn do you report? (remember rounding rule)

Part 6: Problem-Solving Workshop

🏆 Problem-Solving Workshop

Part 6 of 7 — AP-Style Practice


AP FRQ Template for Inference

  1. State: Name the procedure and define parameters
  2. Plan: Check conditions (Random, Normal, Independent)
  3. Do: Show calculations
  4. Conclude: Interpret in context

Common Mistakes to Avoid

  • Using p^\hat{p} in the test statistic SE (should use p0p_0)
  • Using p0p_0 in the CI SE (should use p^\hat{p})
  • Saying “accept H0H_0” instead of “fail to reject H0H_0”
  • Forgetting to check conditions
  • Not interpreting in context

Concept Check 🎯

AP Practice 🧮

A poll finds 52% of 1000 voters favor a candidate. Test H0:p=0.50H_0: p = 0.50 vs. Ha:p>0.50H_a: p > 0.50 at α=0.05\alpha = 0.05.

1) z=(0.52−0.50)/0.25/1000z = (0.52 - 0.50)/\sqrt{0.25/1000} = ? (round to 2 places)

2) p-value ≈P(Z>z)≈?\approx P(Z > z) \approx ? (round to 3 places)

3) Decision at α=0.05\alpha = 0.05? (reject/fail to reject)

Part 7: Mixed Review

📝 Mixed Review

Part 7 of 7 — Comprehensive Review


Quick Reference

ProcedureSE FormulaWhen to Use
1-prop CIp^(1−p^)/n\sqrt{\hat{p}(1-\hat{p})/n}Estimating pp
1-prop testp0(1−p0)/n\sqrt{p_0(1-p_0)/n}Testing H0:p=p0H_0: p = p_0
2-prop CIp^1(1−p^1)n1+p^2(1−p^2)n2\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}Estimating p1−p2p_1 - p_2
2-prop testp^c(1−p^c)(1/n1+1/n2)\sqrt{\hat{p}_c(1-\hat{p}_c)(1/n_1 + 1/n_2)}Testing H0:p1=p2H_0: p_1 = p_2

AP Exam Tips

  • Always state your hypotheses using the parameter pp, not p^\hat{p}
  • Check all three conditions: Random, Normal, Independent
  • Give a conclusion IN CONTEXT

Concept Check 🎯

Final Challenge 🧮

n=800n = 800, p^=0.65\hat{p} = 0.65, 95% CI.

1) Margin of error =1.96×0.65×0.35/800≈?= 1.96 \times \sqrt{0.65 \times 0.35/800} \approx ? (round to 3 places)

2) Lower bound of CI?

3) Upper bound of CI?