University:
York University
Course:
MATH 1131 | Introduction to Statistics I
Academic year:

2022
Views:

323

Pages:

4
Author:

Niklas

Two Populations We have considered inference for a single population. Now we will consider comparisons between different groups. Examples: Comparing average incomes of two regions. Comparing the effectiveness of a new drug to an existing one. Inference Tools To compare populations, we need inferential tools to construct confidence intervals and perform hypothesis testing. Example: Comparing Latency Time Assume we have two populations: Subjects with placebo: average latency time μ₁ Subjects with sleeping pill: average latency time μ₂ We observe two sample means: x̄₁ and x̄₂ These sample means are normally distributed under the assumption that they come from two normally distributed populations. Difference of Two Means We are interested in the mean difference μ₁ - μ₂. We want to test the hypothesis that the drug is inefficient: μ₁ = μ₂ or μ₁ - μ₂ = 0. The difference can be estimated with x̄₁ - x̄₂. Pivotal Quantity Hypothesis testing and confidence intervals on the mean of one population were based on: Z = (x̄ - μ) / (σ / √n) ~ N(0, 1) Two-Sided Significance Test for the Difference of Two Normal Pop Means Null hypothesis: H₀: μ₁ = μ₂ Alternative hypothesis: Hₐ: μ₁ ≠ μ₂ Test statistic: D = (x̄₁ - x̄₂) / (√(σ₁²/n₁ + σ₂²/n₂)) P-value: P(D > dₜ) = 1 - Φ(dₜ) One-Sided Significance Test for the Difference of Two Normal Pop Means Null hypothesis: H₀: μ₁ ≤ μ₂ Alternative hypothesis: Hₐ: μ₁ > μ₂ Test statistic: D = (x̄₁ - x̄₂) / (√(σ₁²/n₁ + σ₂²/n₂)) P-value: P(D > dₜ) = 1 - Φ(dₜ) Example Sample sizes: n₁ = 26, n₂ = 36 Sample means: x̄₁ = 1.95, x̄₂ = 2.35 Sample variances: s₁² = 2.997, s₂² = 5.9995 P-value: P(D > 1.26) = 0.000000006 Conclusion: There is strong evidence to reject the null hypothesis. The average marketing major salary is significantly higher than the average finance major salary. Two Populations with Unknown Variances Use a pooled variance estimate: S² = ((n₁ - 1)s₁² + (n₂ - 1)s₂²) / (n₁ + n₂ - 2) Test statistic: D = (x̄₁ - x̄₂) / (Sp * √(1/n₁ + 1/n₂)) P-value: P(D > dₜ) = 1 - Φ(dₜ) Assumptions: H₀: μ₁ - μ₂ = 0 Test Statistic: D = (x̄₁ - x̄₂) / (Sp * √(1/n₁ + 1/n₂)) ~ t_(n₁ + n₂ - 2) P-Value: P(D > dₜ) = 1 - Fₜ(dₜ, n₁ - 1, n₂ - 1) Rejection Region: Reject H₀ if |D| > t_(n₁ + n₂ - 2, α/2) Example Sample sizes: n₁ = 26, n₂ = 36 Sample means: x̄₁ = 1.95, x̄₂ = 2.35 Sample variances: s₁² = 2.997, s₂² = 5.9995 P-value: P(D > 1.26) = 0.000000006 Inference on the Mean for Large Samples If sample sizes are large, we can use the following simplifications: Normality assumption is not necessary due to the Central Limit Theorem. Unknown variances can be estimated without assuming equal variances. Difference of Two Means, Large n For large samples, we have: x̄₁ ~ N(μ₁, σ₁²/n₁) x̄₂ ~ N(μ₂, σ₂²/n₂) Test statistic: (x̄₁ - x̄₂) / (√(s₁²/n₁ + s₂²/n₂)) ~ N(0, 1) Significance Test for the Difference of Two Means, Large n Assuming H₀: μ₁ - μ₂ = 0, the p-value is: P(|x̄₁ - x̄₂| / √(s₁²/n₁ + s₂²/n₂)) > |x̄₁ - x̄₂| / √(s₁²/n₁ + s₂²/n₂)