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Stat 875 HW
- Let M =
(^1) n 1 · · · 0 .. .
0 · · · (^1) na
.^ Express the following matrices using
M , 1k.
(a) (M ′M )−^1 M ′ (^1) n = 1a
(b) M (M ′M )−^1 M ′ (^1) n = 1n
(c) M (M ′M )−^1 M ′ (^1) n(1′ n (^1) n)−^11 ′ n = 1n(1′ n (^1) n)−^11 ′ n
- Using the results in (a) to show
(a) [I − M (M ′M )−^1 M ′]Y and [M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n]Y are perpendicular.
〈[I − M (M ′M )−^1 M ′]Y, [M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n]Y 〉 = Y ′[I − M (M ′M )−^1 M ′][M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n]Y = Y ′ 0 Y = 0
(b) M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n is idempotent.
[M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n][M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n] = M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n
(c) If μ = 1a · c, then ‖[M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n](M μ)‖^2 = 0.
‖[M (M ′M )−^1 M ′^ − (^1) n(1′ n (^1) n)−^11 ′ n](M μ)‖^2 = ‖(1n − (^1) n) · c‖^2 = 0
- p118 3.23 (c) Report on your test of H 0 : μ 1 = μ 3 = μ 4.
H 0 : μ 1 = μ 3 = μ 4 ⇔ C′μ = 0 with C = (C(1), C(2)) =
Let D(1)^ = C(1); D(2)^ ∝ C(2)^ − 〈C
(2),D(1)〉 〈D(1),D(1)〉 D
(1) ∝ −C(1) + 2C(2)
Let D(2)^ = −C(1)^ + 2C(2). So D =
SSC = (Y^1 −Y^3 )
2 n^1 1 +^ n^1 3
+ (Y^1 +Y^3 −^2 Y^4 )
2 n^1 1 +^ n^1 3 +^ n^4 4
Source DF SS MS F p Model 3 1.0307 0.3436 3.33 0. Error 8 0.8249 0. C. Total 11 1. Contrasts 2 0.3853 0.1927 1.8688 0.
- H 0 : μ 1 = μ 3 and μ 1 = μ 4 H 1 : At least one equation in H 0 is false
- Test Statistic F = M SCM SE
- p-value: P (F (2, n − 4) > Fob)
- From samples Fob = 1.8688, p-value: P (F (2, 8) > 1 .8688) = 0. 216
- Fail to reject H 0
