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(MATH244)final01F_sol.pdf
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1. (a) X=5.5 , Y =50
777
X2 =218.75 , Y2 =18154 , XY =1991.5
iiii
i=1 i=1 i=1

.7 5.5 2 .()( ) 2
S=218.75 ()( ) =7 , S =18154 7 50 =654
xx yy
Sxy = .()( )( ) =66.5

1991.5 7 5.5 50
Sxy 66.5

==9.5 , aY .bX 50 9.5 5.5 2.25
b= = = .()( ) =.
Sxx 7
%
Fitted regression line : Y =.2.25 +9.5X
(b) SST =S yy =654
R b Sxx =()() 7 631.75
SS = 2 9.5 2 =
SSE =654 .631.75 =22.25

ANOVA table :
Source SS d.f. MS F-ratio
Regression 631.75 1 631.75 141.97
Error 22.25 5 4.45
Total 654

2 SSR 631.75
(c) R == =96.60%
SST 654
%
(d) Point prediction of Y0 at X0 =5 : Y0 =.2.25 +9.55 =45.25
A 95% prediction interval for Y0 at X0 =5 is given by
. 2 .
2
%
1 (X 0 .X ). 1 (5 .5.5).
Y t MS .1++ .=45.25 (2.571)( 4.45).1++ .
0 5,0.025 E
. nS ..77 .
. xx . .. =45.25 5.8879 =[39.3621, 51.1379]
(e)
Yes. The prediction based on a single observation is less accurate than the prediction based on all the data, especially when the regression effect is so significant.

(f)
We use the linear regression model Y =+X + to model the relationship between the variables X and Y only in the range of X from 4.0g to 7.0g. We wouldnt extrapolate this


relationship to X = 0g which is out of the observed range. Therefore even when is not equal to zero, the model is still reasonable and not ridiculous as stated by the experimenter.
p. 1
2. (a) For circuit (i),
2
Pr(circuit functions properly)=Pr(both amplifiers function)=0.9 =0.81
For circuit (ii),
Pr(circuit functions properly)=1.Pr(circuit failed)
=1.Pr(both parallelcomponentsfailed)
=1.(1.0.81)2 =0.9639
For circuit (iii),
Pr(circuit functions properly)=Pr(both components in series function)

22
=(1.(1.0.9))=0.9801
Circuit (iii) is the most reliable circuit.

(b) X~ b( , EX 4 0.9 3.6 , Var X 4 0.9 0.1 0.36
4,0.9) ()= = ()= =
.4. 22
(c) ( )=( )Pr(X =2) ()()0.1 =Pr two broken Pr two functions == 0.9 0.0486
.
.

2
..
The circuit can function with only 2 functioning amplifiers in one of the following cases : 1,2 functions and 3,4 failed; 1,4 functions and 2,3 failed; 2,3 functions and 1,4 failed; 3,4 functions and 1,2 failed. Hence
22
Pr(two amplifiers broken circuit functions)=4 0.9 0.1 =0.0324
0.0324 2
Pr(circuit functions | two amplifiers broken)= ==0.6667
0.0486 3
3. (a) Team: Nominal scale Points: Ratio scale
(b) Sorted points: 33, 40, 45, 45, 45, 46, 49, 49, 51, 56, 57, 59, 60, 61, 66, 68
QL =X ( )=X 4 +0.25(X () .X 4 )=45 +0.25(45 .45)=45

4.25 () 5 ()
80th percentile = X ()13.6 X +0.6(() X )=60 +0.6(61.60)60.6
= ()13 X 14 . ()13 =
(c) Test H0: = vs H1: .
BD BD
For team B, m =8, X =48.75 , Sx 2 =67.0714
For team D, n =8, Y =55, Sy 2 =109.7143
2 ()8 .1 (67.0714)( )+8 .1 (109.7143)

S ==88.3929
pool
8 +8 .2
X .Y 48.75 .55

T =
= =.1.3295
S pool m + n (88.3929)(18 +)

T
=1.3295 <2.145 =t
14,0.025
Do not reject H0 a