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CI for difference of two proportions

Short answer

Compute p̂₁=0.77, p̂₂=0.6675. Difference 0.1025. SE = √(p̂₁q̂₁/n₁ + p̂₂q̂₂/n₂) = 0.03015. Using z=1.96, margin=0.0591. 95% CI is (0.043, 0.162). Since interval excludes zero, the proportion favoring higher limit is significantly greater in Maricopa County.

A random sample of 500 adult residents of Maricopa County found that 385 were in favor of increasing the highway speed limit to 75 mph, while another sample of 400 adult residents of Pima County found that 267 were in favor of the increased speed limit. Construct 95% confidence interval on the difference in the two proportions.

For two large independent samples, the confidence interval for p₁ − p₂ is (p̂₁ − p̂₂) ± z·√(p̂₁q̂₁/n₁ + p̂₂q̂₂/n₂).

Maricopa: n₁ = 500, p̂₁ = 385/500 = 0.7700, q̂₁ = 0.2300
Pima:     n₂ = 400, p̂₂ = 267/400 = 0.6675, q̂₂ = 0.3325

p̂₁ − p̂₂ = 0.7700 − 0.6675 = 0.1025
SE = √(0.77×0.23/500 + 0.6675×0.3325/400)
   = √(0.0003542 + 0.0005549) = 0.03015
z (95%) = 1.960
margin = 1.960 × 0.03015 = 0.0591
limits = 0.1025 ± 0.0591 = (0.0434, 0.1616)

Answer: the 95% confidence interval for p₁ − p₂ is (0.043, 0.162).

Since the interval does not contain zero, the proportion in favour of the higher speed limit is significantly greater in Maricopa County than in Pima County.

✓ Answer checked by Claude Sonnet 5.5

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