AP Statistics · Topic 5.6
Sampling Distributions for Differences in Sample Proportions Practice
Part of Sampling Distributions.(UNC-3.G)
Practice questions
4
Sample questions
4 of 4 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 3/5
Two independent samples: n₁ = 100 from population with p₁ = 0.5; n₂ = 200 from population with p₂ = 0.4.
What is the standard deviation of p̂₁ − p̂₂?
- A
≈ 0.005
- Bcheck_circle
≈ 0.060
- C
≈ 0.10
- D
≈ 0.030
Why
SD = √(0.5·0.5/100 + 0.4·0.6/200) = √(0.0025 + 0.0012) = √0.0037 ≈ 0.060.
- A
Sample 2difficulty 3/5
Two independent samples: n₁ = 50, p̂₁ from p₁ = 0.5; n₂ = 50, p̂₂ from p₂ = 0.5.
What is μ_(p̂₁ − p̂₂)?
- A
0.25
- B
1
- Ccheck_circle
0
- D
0.5
Why
Mean of difference = p₁ − p₂ = 0.
- A
Sample 3difficulty 4/5
Conditions for normal approximation of p̂₁ − p̂₂
- Acheck_circle
Each sample has at least 10 successes AND 10 failures, plus independent samples
- B
Equal sample sizes
- C
p₁ = p₂
- D
Just n large
Why
"10 successes, 10 failures" applied to each sample, plus independence.
- A
Sample 4difficulty 4/5
SD of p̂₁ − p̂₂ for independent samples is
- A
Cannot be computed
- B
p₁(1−p₁) + p₂(1−p₂)
- Ccheck_circle
√(p₁(1−p₁)/n₁ + p₂(1−p₂)/n₂)
- D
(p₁ − p₂)/√n
Why
Variances of p̂s add; SD is the square root.
- A