(a) Compose the articulation for the assessed model.

(b) Contend that in a model of this sort, for a particular estimation of x. 1 III b2] x x2 [

Gasp = 9 = [b0

(c) Utilize the assessed model to anticipate the mean estimation of y when x = 2.5.

Question 18

Let C = [3 1. Leave Yi and 14 alone autonomous arbitrary factors with E[Y1] = 5, E[Y2] = 10, 1 Var NCI = Var 1/2 = 6.

(a) Discover Erg where y = [ Y2

(b) Discover Var Y.

(c) Discover Var CY.

Question 19

Consider the model aurixx, = 01X1/32x2 By definition

SSE = [Yi (Bo + Bixii + B2x2r)]2 r=

(a) Square the term on the privilege and whole over I to get SSE = [n_yma 1-] - FB,x-F B2X21)2/700 Bix If + B2x2)]

(b) Show that E [(B0 B X B2X2r) 2 - Yao Bi X I B2X2M 1=1

IZ

= (Bo + B1 + B2x2i Y;)/= I

+ B1 (Boxii + Bixii + B2x11x21 xliYi) 1=1 + B2 (BoX2i B Xi IX 2i + B2 - X2,1) r=1

(c) Utilize the ordinary conditions for the numerous direct relapse model given in Sec. 12.1 to contend that every one of the segments on the privilege of the condition to some degree (b) is equivalent to 0.

('0 Show that n SSE = I 11 - B0 x it; - BlExu11 B2Exullf I-1 1=1 r=1 e=1

subsequently incompletely checking the calculations used to discover SSE.

Question 20

The three essential primary components of an information handling framework are documents, streams, and cycles. Documents are assortments of lasting records in the framework, streams are information interfaces between the framework and the climate, and cycles are practically characterized sensible controls of information. An examination of the expense of creating programming as identified with records, streams, and cycles was explored. The accompanying information depend on that review.

Cost (in units of 1000) Documents Streams Cycles (Y) (x1) (x2) (x3) 22.6 4 44 18 15.0 2 33 15 78.1 20 80 28.0 6 24 21 80.5 6 227 50 24.5 3 20 18 20.5 4 41 13 147.6 16 187 137 4.2 4 19 15 48.2 6 50 21 20.5 5 as 17

Think about the model

gtixi, x2, x, = PO + P 'XI P2x2 03x3 (a) Track down the model particular grid.

(b) For these information

.3197263 - .0408268 - .00202208 .005305965 - .0408268 .0140738 .0003717104 - .00224159 (r)_t = - .00202208 .0003717104 .00005188447 - .000113861 .005305965 - .00224159 - .000113861 .0004938527

489.7 = 5259 59088.4 33845.2

s2 = 98.37516961 Utilize this data to assess the model.