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STAT5002: Final Exam
Q1: a – d Q2: a – b Q3: a – d Q4: a – c Q5: a – d Q6: a – f
Topics: ▪ Inferences about population proportion such as estimation (point and interval) and hypothesis testing
np ≥ 5 and nq ≥ 5 => the sample size is large enough to invoke the CLT. By the CLT, p̂ is approximately normal with E(̂) = p, se(̂) = √ pq n Estimate p using interval estimation method: p̂ ± Zα/2√ p̂q̂ n One-sample Z test about p: Η0: P = P0 against H1: P (>, <, ≠) P0
Ζ = p̂−p √ pq n ≈ N(0, 1) under H0
▪ Chi-square test of independence (how to conduct the Chi-square and its assumptions)
▪ Paired difference t test
▪ Sign test (H0: p+ = 0.5 versus H1: p+ (>, < , ≠) 0.5
X = number of + signs ~ Bin(n, p+) x = observed number of + signs Need to calculate p-value H1: p+ < 0.5, p-value = P(X ≤ x) H1: P+ > 0.5, p-value = P(X ≥ x) H1: P+ ≠ 0.5, p-value = 2P(X ≥ x) for x > n 2 , p-value = 2P(X ≤ x) for x < n 2 Test for proportion: Η0: P = P0 against H1: P < P0 X = number of success ~ Bin(n, P0) p-value = P(X ≤ x)
Η0: P = P0 against H1: P > P0 X = number of success ~ Bin(n, P0) p-value = P(X ≥ x)
Η0: P = P0 against H1: P ≠ P0 X = number of success ~ Bin(n, P0) p-value = P(|X – nP0| ≥ |x – nP0|)
If P0 = 0.5, p-value = 2P(X ≥ x) for x > n 2
p-value = 2P(X ≤ x) for x < n 2 ▪ Four graphs: identify 3 issues and give an remedy ▪ Logistic regression (issues with linear probability model) ln(odds) => exp{ln(odds)} => odds = P 1−P => P = odds 1+odds ▪ Log-linear regression: ln(Y) = β0 + β1X1 + β2X2 + … + βkXk + ε Y = exp(β0 + β1X1 + β2X2 + … + βkXk + ε) Y = exp(β0 + β1X1 + β2X2 + … + βkXk)exp(ε) E(Y|X) = exp(β0 + β1X1 + β2X2 + … + βkXk)E[exp(ε)|X] If ε ~ N(0, σ2), then exp(ε) ~ log-normal distribution Hence, E[exp(ε)] = exp( σ2 2 ) E(Y|X) = exp(β0 + β1X1 + β2X2 + … + βkXk)exp( σ2 2 ) σ̂2 = MSE = RSS n−p Example: ln(y)̂ = 1.592 + 0.044X, RSS = 0.079374, n = 50 If X = 10, Ŷ = ? σ̂2 = MSE = RSS n−p = 0.079374 50−2 = 0.001653625 ln(y)̂ = 1.592 + 0.044(10) = 2.032 ̂ = exp(2.032)exp( σ̂2 2 ) = exp(2.032)exp( 0.001653625 2 ) = 7.6356