AP Statistics · Topic 6.10

Tests for the Difference of Two Proportions Practice

Part of Inference for Categorical Data: Proportions.(VAR-6.E)

Practice questions

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Sample questions

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  1. Sample 1difficulty 2/5

    Hₐ: p ≠ p₀ at α = 0.05.

    −1.96 1.96

    The rejection region is:

    • A

      z < −1.96

    • B

      |z| > 1.645

    • C

      z > 1.96

    • D

      |z| > 1.96

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    Why

    For two-sided test at α = 0.05, reject if |z| > 1.96.

  2. Sample 2difficulty 3/5

    For two-sample inference about proportions.

    CI: unpooled SE Test: pooled SE (under H₀: p₁ = p₂)

    Why pool for the test but not the CI?

    • A

      The null assumes p₁ = p₂; pooling gives best estimate of the common proportion.

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    • B

      Pooling violates assumptions for CIs.

    • C

      Pooling reduces df.

    • D

      Unpooled is always more conservative.

    Why

    Under H₀, the proportions are equal, so we use pooled p̂_p. The CI does not assume equality.

  3. Sample 3difficulty 3/5

    Sample 1: 110 of 200. Sample 2: 90 of 200. Test H₀: p₁ = p₂ vs Hₐ: p₁ ≠ p₂.

    p̂_p = 200/400 = 0.50 SE = sqrt(0.5·0.5·(1/200+1/200)) = 0.05

    What is z?

    • A

      1.00

    • B

      4.00

    • C

      2.00

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    • D

      0.50

    Why

    z = (0.55 − 0.45)/0.05 = 2.00.

  4. Sample 4difficulty 3/5

    Two independent random samples are used to compare proportions p₁ and p₂.

    Two-sample conditions 1. Independent random samples 2. n₁p̂_p ≥ 10, n₁(1−p̂_p) ≥ 10 (same for n₂) 3. 10% condition for each

    For a two-sample z-test for proportions, the Large Counts condition uses:

    • A

      Each sample's own p̂ only.

    • B

      Only n₁ and the average proportion.

    • C

      The smaller of p̂₁ and p̂₂.

    • D

      The pooled proportion p̂_p applied to both n₁ and n₂.

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    Why

    For the test (under H₀: p₁ = p₂), conditions use the pooled proportion applied to each sample size.

  5. Sample 5difficulty 3/5

    Test H₀: p = 0.20 with sample n = 100, p̂ = 0.25.

    SE₀ = sqrt(p₀(1−p₀)/n) = sqrt(0.20·0.80/100) z = (p̂ − p₀)/SE₀

    What is the standard error used in the test?

    • A

      0.0433

    • B

      0.020

    • C

      0.050

    • D

      0.040

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    Why

    SE₀ uses p₀: sqrt(0.20·0.80/100) = sqrt(0.0016) = 0.040.