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Tests for Proportions

Perform one-sample and two-sample z-tests for proportions.

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🔍 Tests of Significance for Proportions

Hypothesis Tests for Proportions

A hypothesis test for a proportion uses sample data to evaluate claims about a population proportion.

Key question: Is the observed proportion significantly different from the hypothesized value?

One-Sample z-Test for a Proportion

When to use: One sample, testing whether p=p0p = p_0 (null hypothesis)

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

Where:

  • p^\hat{p} = sample proportion
  • p0p_0 = hypothesized population proportion
  • nn = sample size

Conditions (must check all):

  1. Random sample: Data collected randomly
  2. Independence: Observations are independent (or n≤0.1Nn \leq 0.1N)
  3. Large counts: Both np0≥10np_0 \geq 10 and n(1−p0)≥10n(1-p_0) \geq 10

Worked Example: A company claims 80% of customers are satisfied. In a random sample of 150 customers, 115 were satisfied. Test at α=0.05\alpha = 0.05.

  • p^=115/150=0.767\hat{p} = 115/150 = 0.767
  • z=0.767−0.800.80(0.20)150=−0.0330.0327=−1.01z = \frac{0.767 - 0.80}{\sqrt{\frac{0.80(0.20)}{150}}} = \frac{-0.033}{0.0327} = -1.01
  • From z-table: p-value = 0.312 (two-tailed)
  • Conclusion: Fail to reject H0H_0 (insufficient evidence)

Two-Sample z-Test for Difference of Proportions

When to use: Comparing two populations; testing p1−p2=0p_1 - p_2 = 0

Test statistic: 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)}}

Where:

  • p^c=x1+x2n1+n2\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2} (pooled proportion)

Conditions:

  1. Random samples from both populations
  2. Independence: Both samples independent; each ≤10%\leq 10\% of population
  3. Large counts: n1p^c≥10n_1\hat{p}_c \geq 10, n1(1−p^c)≥10n_1(1-\hat{p}_c) \geq 10, etc.

Common Mistakes

❌ Using p^\hat{p} instead of p0p_0 in SE for one-sample test ❌ Forgetting to pool proportions in two-sample test ❌ Using t-distribution for proportions (always use z) ❌ Not checking conditions before testing

Decision Rule

  • If ∣z∣>zα/2|z| > z_{\alpha/2}, reject H0H_0
  • If p-value <α< \alpha, reject H0H_0

AP Exam Tip

State conditions first. Graders award partial credit for checking them. Name the test: "z-test for a proportion" or "two-sample z-test for difference of proportions."

📚 Practice Problems

1Problem 1easy

❓ Question:

A survey of 400 students finds 120 prefer online learning. State the null and alternative hypotheses for testing whether the proportion differs from 0.30.

💡 Show Solution

Null hypothesis: H0:p=0.30H_0: p = 0.30 (the proportion is 0.30). Alternative hypothesis: Ha:p≠0.30H_a: p \ne 0.30 (the proportion differs from 0.30). This is a two-tailed test. The sample proportion is p^=120/400=0.30\hat{p} = 120/400 = 0.30, which equals the hypothesized value, but we'll test for statistical significance.

2Problem 2medium

❓ Question:

In a one-sample z-test for proportions with p^=0.65\hat{p} = 0.65, p0=0.60p_0 = 0.60, n=200n = 200, calculate the test statistic and verify conditions.

💡 Show Solution

Conditions check: • Random: assumed ✓ • 10% condition: n=200<10%(N)n = 200 < 10\%(N) ✓ • Large Counts: np0=200(0.60)=120≥10np_0 = 200(0.60) = 120 \ge 10 ✓ and n(1−p0)=200(0.40)=80≥10n(1-p_0) = 200(0.40) = 80 \ge 10 ✓ All conditions met. Test statistic: z=p^−p0p0(1−p0)/n=0.65−0.600.60(0.40)/200=0.050.0012=0.050.0346≈1.44z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.65 - 0.60}{\sqrt{0.60(0.40)/200}} = \frac{0.05}{\sqrt{0.0012}} = \frac{0.05}{0.0346} ≈ 1.44.

3Problem 3hard

❓ Question:

A city claims 75% of residents support a new park. A sample of 150 residents shows 105 support it (p^=0.70\hat{p} = 0.70). Test at α=0.05\alpha = 0.05. Conclude in context.

💡 Show Solution

State: H0:p=0.75H_0: p = 0.75 vs Ha:p≠0.75H_a: p \ne 0.75, α=0.05\alpha = 0.05. Plan/Check: Random sample, 150<10%(N)150 < 10\%(N), np0=150(0.75)=112.5≥10np_0 = 150(0.75) = 112.5 \ge 10, n(1−p0)=37.5≥10n(1-p_0) = 37.5 \ge 10. Conditions met. Do: z=0.70−0.750.75(0.25)/150=−0.050.0354≈−1.41z = \frac{0.70 - 0.75}{\sqrt{0.75(0.25)/150}} = \frac{-0.05}{0.0354} ≈ -1.41. Two-tailed p-value ≈0.158≈ 0.158. Conclude: Since p-value=0.158>0.05p \text{-value} = 0.158 > 0.05, we fail to reject H0H_0. Sufficient evidence that 75% support the park.

Explain using:

⚠️ Common Mistakes: Tests for Proportions

Avoid these 3 frequent errors

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Inference for Two Sample Proportions (CI and Test)
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📌 Related Topics in Unit 6: Inference for Categorical Data — Proportions

Confidence Intervals for ProportionsInference for Two Sample Proportions (CI and Test)

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Perform one-sample and two-sample z-tests for proportions.
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