Why do we use unpooled SE for the CI but pooled SE for the test?

Answer: The test assumes p₁ = p₂ under H₀, so we pool; the CI does not assume equality. Under H₀ (p₁ = p₂), pooling gives a better estimate of the common proportion. CIs do not assume the proportions are equal.

What is the 95% CI for pA − pB?

Answer: (0.007, 0.193) Difference 0.10 ± 1.96·0.0476 = 0.10 ± 0.0933 ≈ (0.007, 0.193).

SE for the CI is approximately:

Answer: 0.0693 SE = sqrt(0.0024 + 0.0024) = sqrt(0.0048) ≈ 0.0693.

AP Statistics · Topic 6.8

Confidence Intervals for the Difference of Two Proportions Practice

Part of Inference for Categorical Data: Proportions.(UNC-4.C)

Practice questions

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Sample questions

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Sample 1difficulty 3/5
$Constructing a two-sample z-interval for p₁ - p₂.$
$Why do we use unpooled SE for the CI but pooled SE for the test?$
- A
  $The test assumes p₁ = p₂ under H₀, so we pool; the CI does not assume equality.$
  check_circle
- B
  $Pooled SE is always larger and more accurate.$
- C
  $It's an arbitrary convention without theoretical basis.$
- D
  $The CI requires more conservative estimates, so unpooled.$
Why
$Under H₀ (p₁ = p₂), pooling gives a better estimate of the common proportion. CIs do not assume the proportions are equal.$
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Sample 2difficulty 3/5
$Group A: 80 successes in 200. Group B: 60 successes in 200. Conditions met.$
$What is the 95% CI for p_A - p_B?$
- A
  $(0.020, 0.180)$
- B
  $(0.050, 0.150)$
- C
  $(0.007, 0.193)$
  check_circle
- D
  $(-0.093, 0.293)$
Why
$Difference 0.10 \pm 1.96\cdot0.0476 = 0.10 \pm 0.0933 \approx (0.007, 0.193).$
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Sample 3difficulty 3/5
$p̂₁ = 0.6 (n₁=100), p̂₂ = 0.4 (n₂=100).$
$SE for the CI is approximately:$
- A
  $0.0245$
- B
  $0.1000$
- C
  $0.0490$
- D
  $0.0693$
  check_circle
Why
$SE = sqrt(0.0024 + 0.0024) = sqrt(0.0048) \approx 0.0693.$
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Sample 4difficulty 3/5
$Sample A: 25 of 50. Sample B: 15 of 50.$
$What is the 95% CI for p_A - p_B?$
- A
  $(0.105, 0.295)$
- B
  $(0.000, 0.400)$
- C
  $(0.013, 0.387)$
  check_circle
- D
  $(-0.087, 0.487)$
Why
$Difference 0.20 \pm 1.96\cdot0.0954 = 0.20 \pm 0.187 \approx (0.013, 0.387).$
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Sample 5difficulty 3/5
$p̂₁ = 0.50 (n₁=200), p̂₂ = 0.40 (n₂=200). 90% CI for p₁ - p₂.$
$What is the 90% CI?$
- A
  $(0.019, 0.181)$
  check_circle
- B
  $(0.049, 0.151)$
- C
  $(0.000, 0.200)$
- D
  $(0.004, 0.196)$
Why
$0.10 \pm 1.645\cdot0.0492 = 0.10 \pm 0.0809 \approx (0.019, 0.181).$
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