BA 240 Yamasaki
Solutions to Practice 5
Temperature of Decrease in Pulse rate(Y)
Water (X)
in beats per minute
68
65
70
62
60
55
58
2
5
1
10
9
13
10
438
50
X2
4624
4225
4900
3844
3600
3025
3364
27582
XY
136
325
70
620
540
715
580
2986
Sxy
Sxx
= 2986 -(438*50)/7
= 27582 -(438*438)/7
=
=
-142.5714
175.7143
Syy
= 480 -(50*50)/7
=
122.8571
S2y/x
= 122.8571-((-142.571)2/175.7143)
7-2
Sy/x
√1.43557
=
1.1982
Sy/x/√Sxx
= 1.1982/√175.7143
=
0.090388
=
DCSI Solutions to Practice 5 ( 14 pages)
Y2
4
25
1
100
81
169
100
480
1.4356
2
b1 

xy 





x2 
 x
y
n
 x 2
n
438  50
2986 
7
4382
27582 
7
2986  3128.57
27582  27406.29
 142.57
175.71
.8114
b0 

n
y
 b1

x
n
50
438

 ( .8114)
7
7
 7.14  50.77
 57.913
i  57.913.811xi
y
DCSI Solutions to Practice 5 ( 14 pages)
3
Scatterdiagram of Water Temp vs
Decrease in Pulse Rate
Decrease in Pulse Rate
14
12
10
8
6
4
2
R² = 0.9416
0
30
40
50
60
70
80
Wate r Te mp (F)
c.
ŷ i  57.913  .811x i
at x  60
ŷ = 57.913 - .811(60) = 57.913 - 48.66 = 9.25
DCSI Solutions to Practice 5 ( 14 pages)
4
d.
H 0 :  1  1.00
H A :  1  1.00
df = 7 - 2 = 5
 t13.05  2.015
t =
- .8114 - (-1)
1.1982
175.7143

.1886
 2.086
.0904
Since 2.086  - 2.015 cannot reject H 0 . Data is insufficie nt
to indicate the population slope is  - 1
e.
Coefficient
of determination
S 
2
r
2

xy
S xx S yy
( 142.5714) 2

 .9416
175.7143  122.8571
DCSI Solutions to Practice 5 ( 14 pages)
5
f. Partition Sum of Squares
SST = Sy2 – (Sy)2/n = Syy
SSE = Sy2 – b0Sy – b1Sxy
Regression (ANOVA table)
Source
SS
Regression
SSR
Error
SSE
Total
SST
df
1
N-2
N-1
MS
F
MSR MSR/MSE
MSE
SSE = 480 - ( 57.913)*(50) –(-.8114)*(2986) = 7.1904
SST = Syy = 122.8571
Regression (ANOVA table)
Source
SS
Regression
115.6631
Error
7.1904
Total
122.8571
df
1
5
6
DCSI Solutions to Practice 5 ( 14 pages)
MS
115.6631
1.4381
F
80.43
6
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.97035
R Square
0.94158
Adjusted R Square
0.92990
Standard Error
1.19810
Observations
7.00000
ANOVA
df
Regression
Residual
Total
Intercept
Temp of Water (X)
1
5
6
SS
115.67991
7.17724
122.85714
MS
115.67991
1.43545
F
80.58806
Significance F
0.000286124
Coefficients
57.91220
-0.81138
Standard Error
5.67354
0.09038
t Stat
10.20742
-8.97709
P-value
0.00015
0.00029
Lower 95%
43.32790282
-1.043720786
RESIDUAL OUTPUT
Upper 95%
72.49648742
-0.579043442
PROBABILITY OUTPUT
Predicted
Observation
Decrease in Pulse rate(Y) in beats
Residuals
per minute
Standard Residuals
1
2.73821
-0.73821
-0.67496
2
5.17236
-0.17236
-0.15759
3
1.11545
-0.11545
-0.10556
4
7.60650
2.39350
2.18842
5
9.22927
-0.22927
-0.20962
6
13.28618
-0.28618
-0.26166
7
10.85203
-0.85203
-0.77903
DCSI Solutions to Practice 5 ( 14 pages)
Percentile
Decrease in Pulse rate(Y) in beats per minute
7.142857143
1
21.42857143
2
35.71428571
5
50
9
64.28571429
10
78.57142857
10
92.85714286
13
7
Residuals
Temperature of Water (X)
Residual Plot
4.00000
2.00000
0.00000
0
-2.00000
20
40
60
80
Temperature of Water (X)
Decrease in
Pulse
rate(Y) in
beats per
minute
Temperature of Water (X)
Line Fit Plot
Decrea s e i n Pul s e
ra te(Y) i n bea ts per
mi nute
20
10
0
0
50
100 Predi cted Decrea s e
i n Pul s e ra te(Y) i n
Temperature of Water (X) bea ts per mi nute
Normal Probability Plot
15
Decrease in Pulse
rate(Y) in beats
per minute
10
5
0
0
20
40
60
80
100
Sample Percentile
DCSI Solutions to Practice 5 ( 14 pages)
8
Simple Linear Regression
Model:
yi

dependent
var iable


0

Intercept


1

xi


Slope independent
var iable
Assume :1) i  N ( 0, 
2

i
random
error
)
2) i ' s are independent
To estimate regression line, use least squares method.
DCSI Solutions to Practice 5 ( 14 pages)
9
b1 
S xy
S xx
where S xy 

S xx 

  xi

 x   yi  y 
xi yi 
  xi

  xi  
 x
xi2 
yi 
n
2
  xi  2
n
b0  y  b1 x
Estimate of  2  S y2/ x
where S yy 

  yi



  S yy 


 y
yi2 

S xy

S xx
2





n  2
2

yi 
2
n
DCSI Solutions to Practice 5 ( 14 pages)
10
Example:
A real estate agent would like to predict selling price of a single family home based
on size of house. She takes a random sample of 15 recently sold homes.
House size(100sq ft)
20
14.8
20.5
12.5
18.0
14.3
27.5
16.5
24.3
20.2
22.0
19.0
12.3
14.0
16.7
Selling Price($1000)
89.5
79.9
83.1
56.9
66.6
82.5
126.3
79.3
119.9
87.9
112.6
120.8
78.5
74.3
74.8
DCSI Solutions to Practice 5 ( 14 pages)
11
Preliminary calculations:
Sx = 272.6
Sx2 = 5222.24
Sy = 1332.6
Sy2 = 124618.42
b1 

S xy
S xx
25257.97 
Sxy = 25257.97
272.6 1332.6 
15
2

272.6 
5222.24 
15

1040.18
 3.88
268.19
 1332.6 
 272.6 
b0  y  b1 x  
  3.88

15


 15 
 18.34
Our least squares estimate of the regression line is
ŷ = 18.34 + 3.88x
So if x = 13.5 (i.e. house is 1350 sq ft) our predicted
price would be 18.34 + 3.88(13.5) = 70.72 (i.e. $70,720).
S yy 
y
 y 

2
2
i
i
n
2

1332.6 
 124618.42 
15
 6230.24

S
2
y/x

2 

S xy

 S yy 
S xx





n2


1040.18 2

 6230.24
268.19







13
 2195.88 13  168.91
DCSI Solutions to Practice 5 ( 14 pages)
12
Tests of hypotheses about 1
H0 : 1  
H A : 1  (  ,  ) 
t 
b1  
Sy/ x
S xx
Suppose that historically houses in the area were selling at $3,000
per 100sq feet.
Due to the recent demand the real estate agent believed
that houses were now selling for more.
Is there sufficicient evidence
to substantiate her belief? ( =.05)
H0 : 1  3
H A : 1  3
df
=
n - 2
=
13
t13.05  1.771
t =
3.88 - 3
13.00
268.19

.88
 111
.
.794
Since 1.11 < 1.771 cannot reject H 0 .
Data is insufficient
to substantiate her belief
DCSI Solutions to Practice 5 ( 14 pages)
13
Coefficient of determination
r
2


S xy

2
S xx S yy
1040.182

268.19  6230.24
 .6475
So 64.75% of the variation in house prices due to differences
in size.
0  r2  1
Correlation Coefficient (  )
S xy
Estimate r =
S xx
Syy
1040.18

268.19
6230.24
 .8047
H0 :   0
H A:   (, )0
 = .05
t 
r
n  2
1  r2
So reject H 0 .

.8047
15  2
1  .8047 
2
 4.89  t13.025  2.16
Conclude that there is a significant linear
association between size and p rice.
Note that this test formula works to H 0 :   0.
DCSI Solutions to Practice 5 ( 14 pages)
14