{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MAT4375 HW#2, 1999

# MAT4375 HW#2, 1999 - MAT4375 HW#2 1999...

This preview shows pages 1–2. Sign up to view the full content.

MAT4375 HW#2, 1999 http://aix1.uottawa.ca/~mat4375/Solutions/hw2.html 1 of 3 10/2/2007 8:25 PM -3 0 3 0 1 -1 -3 1 2 X-X 1 = 9 -3/2 15/2 -3/2 1 -1/2 15/2 -1/2 7 S = MAT4375 HW#2, 1999 © A.R. Dabrowski, 1999 Due to html limitations, ``X- bar'' will be noted X . From the third edition of Johnson and Wichern Problem 3.4 Problem 3.6 Problem 4.4 Problem 4.14 (#4.19 in the fourth edition) Problem 4.20 (#4.27 in the fourth edition) Problems 4.21 and 4.22 (#4.28 and 4.29 in the fourth edition) Problem 3.4 For visual simplicity we work in millions of dollars rather than dollars. (a) Here p 1 = (x 1 ·1/ || 1 || ) 1/ || 1 || , which equals x 1 ·1 2.10121. (b) e 1 = y 1 -x i 1 (1.3967, .3843, -.3183, -.3757, -.4556, -.6314) . Note that (s 11 ) 1/2 = || y 1 -x 1 || /6 1/2 . (c) By construction p 1 and e 1 are orthogonal. We have that || e 1 || 1.717 and || p 1 || 5.147. We can sketch the relationship among y 1 , e 1 and p 1 as a right-angle triangle with hypotenuse y 1 , short side e 1 and long side p 1 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}