Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
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Chapter 7.2, Problem 39E

a.

To determine

Find the distribution of θ^=c1X¯1+c2X¯2+.......+ckX¯k.

Give the reasons.

a.

Expert Solution
Check Mark

Answer to Problem 39E

The distribution of θ^=c1X¯1+c2X¯2+.......+ckX¯k follows normal distribution with mean μ1+μ2+......+μk and variance c12σ2n1+c22σ2n2+....+ck2σ2nk.

The reasons are if Y1,Y2,......,Yn has a normal distribution with mean μ and variance σ2, then Y¯ follows normal distribution with mean μ, variance σ2n and sum of normal variables are also normal.

Explanation of Solution

From the given information, X1,X2,............,Xk follows normal distribution with mean μi and variance σ2.

Let θ^=c1X¯1+c2X¯2+.......+ckX¯k.

From theorem 7.1, if Y1,Y2,......,Yn has a normal distribution with mean μ and variance σ2, then Y¯ follows normal distribution with mean μ and variance σ2n.

By using theorem 7.1, as X¯1N(μ1,σ2n1), X¯2N(μ2,σ2n2),…….., X¯kN(μk,σ2n2), thus for constants c1,c2,...,ck it can be said that c1X¯1N(c1μ1,c12σ2n1), c2X¯2N(c2μ2,c22σ2n2),…………….., ckX¯kN(ckμk,ck2σ2nk).

Then,

θ^=c1X¯1+c2X¯2+.......+ckX¯kN(μ1+μ2+......+μk,c12σ2n1+c22σ2n2+....+ck2σ2nk)

Thus, the distribution of θ^=c1X¯1+c2X¯2+.......+ckX¯k follows normal distribution with mean μ1+μ2+......+μk and variance c12σ2n1+c22σ2n2+....+ck2σ2nk.

The reasons are if Y1,Y2,......,Yn has a normal distribution with mean μ and variance σ2, then Y¯ follows normal distribution with mean μ, variance σ2n and sum of normal variables are also normal.

b.

To determine

Find the distribution of SSEσ2.

Give the reasons.

b.

Expert Solution
Check Mark

Answer to Problem 39E

The distribution of SSEσ2 is chi-square with parameter (n1+n2+.......+nk)k df.

The reasons are if Y1,Y2,......,Yn has a normal distribution with mean μ and variance σ2, then (n1)S2σ2=1σ2i=1n(YiY¯)2 follows chi-square distribution with (n1) df and sum of chi-square distributions is also chi-square.

Explanation of Solution

From the given information, SSE=i=1k(ni1)Si2.

From the theorem 7.3, if Y1,Y2,......,Yn has a normal distribution with mean μ and variance σ2, then (n1)S2σ2=1σ2i=1n(YiY¯)2 follows chi-square distribution with (n1) df.

Then, by using theorem 7.3, (n11)S12σ2 follows chi-square distribution with (n11) df, (n21)S22σ2 follows chi-square distribution with (n21) df,…….and (nk1)Sk2σ2 follows chi-square distribution with (nk1) df.

Then, SSEσ2=i=1k(ni1)Si2σ2 follows chi-square distribution with (n11)+(n21)+.......+(nk1)=(n1+n2+.......+nk)k df.

Thus, the distribution of SSEσ2 is chi-square with parameter (n1+n2+.......+nk)k df.

The reasons are if Y1,Y2,......,Yn has a normal distribution with mean μ and variance σ2, then (n1)S2σ2=1σ2i=1n(YiY¯)2 follows chi-square distribution with (n1) df and sum of chi-square distributions is also chi-square.

c.

To determine

Find the distribution of θ^θ(c12n1+c22n2+......+ck2nk)MSE.

Give the reasons.

c.

Expert Solution
Check Mark

Answer to Problem 39E

The distribution of θ^θ(c12n1+c22n2+......+ck2nk)MSE has a t distribution with n1+n2+.......+nkk.

The reason is that if Y1,Y2,........,Yn follow normal distribution with mean μ and variance σ2, then T=Y¯μS/n has a t distribution with (n1) df.

Explanation of Solution

From the given information, MSE=SSEn1+n2+.......+nkk.

From the definition 7.2, if Y1,Y2,........,Yn follow normal distribution with mean μ and variance σ2, then T=Y¯μS/n has a t distribution with (n1) df.

Then, by using the definition 7.2, θ^θ(c12n1+c22n2+......+ck2nk)MSE has a t distribution with n1+n2+.......+nkk.

Thus, the distribution of θ^θ(c12n1+c22n2+......+ck2nk)MSE has a t distribution with n1+n2+.......+nkk.

The reason is that if Y1,Y2,........,Yn follow normal distribution with mean μ and variance σ2, then T=Y¯μS/n has a t distribution with (n1) df.

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Chapter 7 Solutions

Mathematical Statistics with Applications

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