Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Application Statistical Methods II - Solved Questions for Assignment 7 | STAT 776, Assignments of Data Analysis & Statistical Methods

Material Type: Assignment; Professor: Hu; Class: Appl Statistical Methods II; Subject: Statistics; University: Wichita State University; Term: Unknown 2000;

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

koofers-user-wo2
koofers-user-wo2 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Stat 776 HW07
1. p261 5.1 (c) [5th edition: p260 5.1 (c)] by p-value and by rejection region
I by p-value
H0:µ=µ7
11 Ha:µ6=µ7
11
Test Statistic: T2= (Xµ0)0(S
n)1(Xµ0), µ0=µ7
11
p-value: P(F(2, n 2) >n2
(n1)2 T2
ob
T2= (1,1)[1
4µ83.3333
3.3333 2 ]1µ1
1= 13.64
p-value: P(F(2, n 2) >n2
(n1)2 T2
ob =P(F(2,2) >4.55) >0.10
(F0.10(2,2) = 9.00)
Fail to reject H0.
II by rejection region
H0:µ=µ7
11 Ha:µ6=µ7
11
Test Statistic: T2= (Xµ0)0(S
n)1(Xµ0), µ0=µ7
11
Reject H0if T2>(n1)p
n2Fα(p, n p) = 3 ×19 = 57
T2= 13.64
Fail to reject H0.
2. p338 6.5 (a) [5th edition: p333 6.5 (a)] using rejection region
Let Cbe µ11 0
1 0 1
H0: = 0 Ha:C µ 6= 0
Test Statistic: T2= (CX )0(C SC0
n)1(CX )
Rejection region: T2>(n1)q
nqFα(q, n q) = (401)×2
402F0.05(2,38) = 6.63
T2
ob = 90.5
Reject H0
1
pf2

Partial preview of the text

Download Application Statistical Methods II - Solved Questions for Assignment 7 | STAT 776 and more Assignments Data Analysis & Statistical Methods in PDF only on Docsity!

Stat 776 HW

  1. p261 5.1 (c) [5th edition: p260 5.1 (c)] by p-value and by rejection region

I by p-value H 0 : μ =

Ha : μ 6 =

Test Statistic: T 2 = (X − μ 0 )′( Sn )−^1 (X − μ 0 ), μ 0 =

p-value: P (F (2, n − 2) > (^) (nn−−1)2^2 T (^) ob^2

T 2 = (− 1 , −1)[ (^14)

]−^1

p-value: P (F (2, n − 2) > (^) (nn−−1)2^2 T (^) ob^2 = P (F (2, 2) > 4 .55) > 0. 10 (F 0. 10 (2, 2) = 9.00) Fail to reject H 0.

II by rejection region H 0 : μ =

Ha : μ 6 =

Test Statistic: T 2 = (X − μ 0 )′( Sn )−^1 (X − μ 0 ), μ 0 =

Reject H 0 if T 2 > (nn−−1) 2 pFα(p, n − p) = 3 × 19 = 57 T 2 = 13. 64 Fail to reject H 0.

  1. p338 6.5 (a) [5th edition: p333 6.5 (a)] using rejection region

Let C be

H 0 : Cμ = 0 Ha : Cμ 6 = 0 Test Statistic: T 2 = (CX)′( CSC

′ n )

− 1 (CX)

Rejection region: T 2 > (nn−−1)q qFα(q, n − q) = (40 40 −−1) 2 × 2 F 0. 05 (2, 38) = 6. 63 T (^) ob^2 = 90. 5 Reject H 0

  1. p267 5.18 (a) [5th edition: p267 5.18 (a)]

H 0 : μ =

 (^) Ha : μ 6 =

Test Statistic: T 2 = (X − μ 0 )′( Sn )−^1 (X − μ 0 ), μ 0 =

p-value: P (F (p, n − p) > (^) p(nn−−p1) T (^) ob^2 ) T (^) ob^2 = (n − 1) (H − L trace)= 86 × 2 .5966 = 223. 31 p-value: P (F (3, 84) > 72 .71) < 0. 0001 Reject H 0.