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STATISTICAL INFERENCE I
Math 5080, Fall 2020 Homework 5 Version: September 28, 2020 7:02pm Z Due Monday, October 5th at 11:59 PM Instructions (Read before you begin) • Turn in solutions to the questions marked in bold red • Only one of these questions will be graded and is out of 10 points • Upload solutions directly to Canvas • All uploads must be in PDF format so that I can mark them directly • I have also included the TeX file if you want to type up your solutions, but this is not required • Question numbers are from the Bain and Englehardt book, unless otherwise indicated 8.8 Suppose that X ∼ χ2(m), Y ∼ χ2(n) and X and Y are independent. Is Y −X ∼ χ2 if n > m? If so what are the degrees of freedom? 8.9 Supposes that X ∼ χ2(m), S = X + Y ∼ χ2(m + n), and X and Y are independent. Is S −X ∼ χ2(n)? 8.14 If T ∼ t(ν), give the distribution of T 2. 8.15 Suppose that Xi ∼ N(µ, σ2) for i = 1, 2, . . . , n, and Zi ∼ N(0, 1) for i = 1, 2, . . . , k, and that all variables are independent. State the distribution of each of the following variables if it is a “named” distribution, or otherwise state “unknown”. (a) X1 −X2 (b) X2 + 2X3 (c) X1−X2 σSZ √ 2 , where SZ indicates the sample standard deviation of the Z terms (d) Z21 (e) √ n(X̄ − µ)/(σSZ) (f) Z21 + Z 2 2 (g) Z21 − Z22 (h) Z1/ √ Z22 (i) Z21/Z2 1 (j) Z1/Z2 (k) X̄/Z̄ (l) √ nk(X̄ − µ) σ √∑k i=1 Z 2 i (m) ∑n i=1 (Xi−µ)2 σ2 + ∑k i=1(Zi − Z̄)2 (n) X̄/σ2 + 1k ∑k i=1 Zi (o) kZ̄2 (p) (k − 1) ∑n i=1(Xi − X̄)2 (n− 1)σ2 ∑k i=1(Zi − Z̄)2 8.18 Assume that Z, V1, and V2 are independent random variables with Z ∼ N(0, 1), V1 ∼ χ2(5), and V2 ∼ χ2(9). Find the following (a) P(V1 + V2 < 8.6)="" (b)="" p(z/="" √="" v1/5="">< 2.015)="" (c)="" p(z=""> 0.611 √ V2) (d) P(V1/V2 < 1.450)="" (e)="" the="" value="" b="" such="" that="" p="" (="" v1="" v1+v2="">< b ) = 0.90 8.19 if t ∼ t(1) then show the following: (a) the cdf of t is f (t) = 1/2 + arctan(t)/π (b) the 100× γth percentile is tγ(1) = tan[π(γ − 1/2)] 2 b="" )="0.90" 8.19="" if="" t="" ∼="" t(1)="" then="" show="" the="" following:="" (a)="" the="" cdf="" of="" t="" is="" f="" (t)="1/2" +="" arctan(t)/π="" (b)="" the="" 100×="" γth="" percentile="" is="" tγ(1)="tan[π(γ" −="" 1/2)]=""> b ) = 0.90 8.19 if t ∼ t(1) then show the following: (a) the cdf of t is f (t) = 1/2 + arctan(t)/π (b) the 100× γth percentile is tγ(1) = tan[π(γ − 1/2)] 2>