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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