Statistics assignment 10

Business Statistics (BUS 505) Assignment 10
Page 1 of 25
Hypothesis testing
[Page: 342; 1-15]
Answer the question number 01
Given that,
0   16  x  .4  n 16 X = 15.84  =10%
Our hypothesis is Ho:   0  =16
And, alternative is 1 H : < 0  =16
The Decision rule is reject Ho in favor of 1H , if  Z
 
n
x
/
0

< - Z
So,  Z
 
n
x
/
0

< - Z
=
16 84. 15 
16 / 4.
< -1.285
= -1.6 < -1.285
So, our null Hypothesis is rejected.
Given that,
0  50  x  3  n 9 X = 48.2  =10%
Our hypothesis is Ho:  = 0  = 50
And, alternative is 1 H : < 0  = 50
Decision rule is reject Ho in favor of 1 H , if
 Z
 
n
x
/
0

< - Z
=
48.2  50
3/ 9
< - 10. Z
= -1.8<-1.285
According to the decision rule, our significant level  for which Z = -1.285 is bigger than -
1.8 .So our null hypothesis is rejected.
Given that,
0  =.03  x  .004  n 64 X = 3.07
Our hypothesis is Ho:  = 0  =.03
And, alternative is 1 H :  > 0  =.03
Decision rule is reject Ho in favor of 1 H , if  Z
 
n
x
/
0

> Z
(a) when 5% Significance level ,then
Z 
.0307 .03
.004 / 64
> .05 Z
=1.4>1.645

According to the decision rule, our significant level  for which Z = 1.645 is bigger so
Page 2 of 25
should maintain our null hypothesis.
(b) P-Value: Z =1.4
So, ) (z F .9192
1-  = .9192
 =1-.9192
=.0808=8.08%
(c)  Z
 
n
x
/
0

> 2 /  Z or Z
 
n
x
/
0

< - Z / 2
P value is higher because p value two sided so, .0808+.0808= .1616
(d) Here, alternative is more than (>) so we are consider just one side test.
Given that,
0   3  Sx 1.8  n 100 X =2.4
Our hypothesis is Ho:   0  =3
And, alternative is 1 H : < 0  =3
Decision rule is reject Null Hypothesis Ho in favor of 1H , if
x
0  
Sx
/
n <- Z Z=3.33
2.4  3
1.8 / 100
= -3.33 ) (z F .9996=1- 
 =1-.9996
=.0004=.04%
Given that,
0  =4  Sx 1.32  n 1562 X =4.27  =1%=.01  2 /  .005
Our hypothesis is Ho:  = 0  =4
And, alternative is 1 H :  ≠ 0  =4
Decision rule is reject Null Hypothesis Ho, in favor of 1H , if
x
0 
Sx /
n
>  / 2 Z Or,
x
0 
Sx /
n
<-  / 2 Z
4.27  4
=
1.32 / 1562
> .005 Z
= 8.084 >2.575
According to the decision rule, our significant level  /2 for which  / 2 Z = 2.575 is not 8.084.
So our null hypothesis is rejected.
Given that,
0  =0 Sx .021 n  76 X =.078

Page 3 of 25
Our hypothesis is Ho: x   0  =0
And, alternative is 1 H : ≠ 0  P-Value Test: Z=3.38
Z=
x
0 
Sx
/
n =
0 078 . 
76 / 201 .
=3.38 ) (z F .9996=1- 
 =1-.9996
=.0004 2
=.0008=.08%
Given that,
0  =3.0  Sx .70  n 172 X =3.31  =1%
Our hypothesis is Ho: x   0  = 3.0
and, alternative is 1H :   0  =3.0
=
x
0 
Sx
/
n >  Z
=
0. 3 31. 3 
172 / 70.
> 01. Z
=5.808>2.325
So, we reject the null hypothesis.
Given that,
0  =0  Sx 11.33  n 170 X = - 2.91
Z =
x
0  
Sx
/
n <- Z Z=3.35 ) (z F .9996=1- 
=
 2.91 0
11.33/ 170
 =1-.9996
=.0004
= - 3.35
Given that,
0  =125.32  Sx 25.41  n 16 X =131.78  =5%=.05  2 /  .025
Our hypothesis is Ho:  = 0  =125.32
And, alternative is 1 H :  ≠ 0  =125.32
Decision rule is reject Null Hypothesis Ho, in favor of 1 H , if

Page 4 of 25
tv =
x
0 
Sx
/
n > 2/ ,v t Or, tv =
x
0 
Sx
/
n <- 2/ ,v t
=
32. 125 78. 131 
16 / 41. 25
> 025,.15t
= 1.02 >2.131
According to the decision rule, Null hypothesis is maintained at level 5%.
Given that,
0  =20  n 10 X =
882
10
=8.82  =5%=.05  / 2  .025
Sx= 2.4
Our hypothesis is Ho:  = 0 =10
tv =
x
0 
Sx
/
n >tv, / 2 Or, tv =
x
0 
Sx
/
n <- tv, / 2
=
10 82. 8 
10 / 4. 2
<- 025,. 9t
= -1.55 < 2.62
Given that,
0  =20  n 9 X =
2. 183
9
=20.35556  =5%=.05  2 /  .025
Sx= .612
tv =
x
0 
Sx /
n
>tv, / 2 Or, tv =
x
0 
Sx /
n
<- tv, / 2
=
20.35556  20
.612 / 9
>t8,.025
= 1.74 >2.306
Given that,
0  =78.5 n  8 X =74.5  =10%=.10  / 2  .05
Sx= 6.2335

Page 5 of 25
Our hypothesis is Ho:  = 0 =78.5
And, alternative is 1 H :  ≠ 0  =78.5
tv =
x
0 
Sx
/
n > 2/ ,v t Or, tv =
x
0  
Sx
/
n <- 2/ ,v t
=
5. 78 5. 74 
8 / 2335. 6
<- 05,.7t
= -1.8149<-1.895
Given that,
0  =50  n 20 X =41.3 Sx= 12.2
Our hypothesis is Ho:  > 0 =50
And, alternative is 1 H :  < 0  =50
tv =
x
0 
Sx /
n
<- , t
41.350
=
12.2/ 20
<- 05,.19t
= -3.187<-1.729
According to the decision rule, Null hypothesis is rejected at level 5%.
Given that,
0  =400  n 15 X =381.35 Sx= 48.60
(a) Our hypothesis is Ho:  > 0  =400
tv =
x
0 
Sx /
n
<- tv,
=
381.35  400
48.60/ 15
<- t14,.05
= -1.486<-1.761
Given that,
0  =400 n  50 X =381.35 Sx= 48.60
(b) Our hypothesis is Ho:  > 0  =400

Page 6 of 25
Z =
x
0 
Sx
/
n <-Z Z=2.71 F(Z)= .9966
=
400 35. 381 
50 / 60. 48
<- Z 1-=.9966
= -2.71348<- Z = .0034
 =.34%
Hypothesis Testing of the Variance
[Page: 350; 16-27]
Given that, 8  n 500 2

4931
   0 375 . 616
8
n
xi
x
933.983
6537 .878
xi 
x
2  
7
( ) 2
1

 
n
sx
So, our null hypothesis is: : 500 2
0
2
0 H   
And the alternative is: 2
0
2
1 :    H
The decision rule is to reject H in favor of H if:  
0 1,
 
2
2
0
2
2 ( 1)
v
x
v
n s



13.076
2 (8 1)933.983 
500

 v 
2
v = 7,.1 12.02
At 10% significance level test: 1.   so,   ,
2  
So, 13.076 is bigger than 12.02. According to our decision rule, the null hypothesis is rejected
at 10% significant level.
Given that, n 10

2276
   a) 227.6
10
n
xi
x
5.138
46.24
xi 
x
2  
9
( )2
1

 
n
sx
b) Given that, 500 2
0     .05
So, our null hypothesis is: : 2.25 2
0
2
0 H   
0
2
1 H : 
The decision rule is to reject H in favor of H if:   0 1 ,


2
2
0
2
2 ( 1)
v
x
v
n s



20.552
2 (10 
1)5.138 
2.25
 v 
2
v = 9,.05 16.92
At 5% significance level test:   .05so,  ,
2  
[xi = 618, 660, 638, 625, 571,
598, 539, 582]
2 (xi  x) = 2.641 + 1903.141 + 467.641 + 74.391 +
2058.891 + 337.641 + 511.891 + 1181.641
= 6537.878
[xi =226, 226, 232, 227, 225, 228, 225, 228, 229, 230]
2 ) (  x xi = 2.56 + 2.56 + 19.36 + .36 + 6.76 + .16 + 6.76 +
.16 + 1.96 + 5.76 = 46.24

Therefore, 20.552 is bigger than 16.92. According to our decision rule, the null hypothesis is
rejected at 5% significant level.
v = 72 . 45 025 ,. 29
Page 7 of 25
Given that, 30  n 300 2
0   780 2  x s
0
2
0 H   
0
2
1 :    H
The decision rule is to reject 0 H in favor of 1H if:
, / 2
2
 v
2
0
2
2 ( 1)
 

x
v
n s


2

 or    v
,1 
/ 2
2
0
2
2 ( 1)
 

x
v
n s
46.4
2 (30 
1)480 
300
 v 
2
At 5% significance level test: 025 . 2 / 05 .      so, 2/ ,
2  
Therefore, according to our decision rule, the null hypothesis is rejected at 5%
Given that, n  20 2.0 2
 0  sx  2.36
0
2
H0   
And the alternative is:
2
0
2
H1 :  =2
 ,
 

2
2
0
2
2 ( 1)
v
x
v
n s



26.4556
(30 1)(2.36)
2
2
2

2 
 v 
2
v = 29,.05 30.14
At 5% significance level test: so,   ,
2  
Therefore, according to our decision rule, the null hypothesis is maintained at 5%
Given that, n  25 18.2 331.24 2
0 0     15.3 234.09 2    x x s s
So, our null hypothesis is: : 18.2 2
0
2
H0   
And the alternative is: : 18.2 2
0
2
H1   

The decision rule is to reject H in favor of H if:     0 1 ,
1 


2
2
0
2
2 ( 1)
v
x
v
n s
16.96
2 (24 
1)234.09 
331.24
 v 

Page 8 of 25
2
v = 85. 1395 ,. 24
At 5% significance level test: 05.   so,   1,
2  
Therefore, 16.96 is not bigger than 13.85. According to our decision rule, the null hypothesis
is maintained at 5% significant level.
105
  
n
Given that, 361n 29 .
361
x
Px
So, our null hypothesis is: H0 : Px  P0  .25
And the alternative is: x P H : 1 >P0=.25
p 
p
Z x 0

The decision rule is to reject 0 H in favor of 1H if:  Z
p 
p
n

0(1 0)
.29 
.25
Z  Z
.25(1 
.25)

361
.04
=  Z
.02
=2
Z=2
F(2)=.9772 1-=.9772 =.0228 =2.28%
173
  
n
Given that, n=998 x= (998X17.3%)=173 .17335
998
x
Px
So, our null hypothesis is: H0 : Px  P0  .25
And the alternative is: x P H : 1 <P0=.25
p 
p
Z x 0
 
The decision rule is to reject 0 H in favor of 1H if:  Z
p 
p
n

0(1 0)
.05
.17335 
.25
Z  Z
.25(1 
.25)
998

.07665
= 1.645
.01371
 

=5.59  1.645
So, we reject that null hypothesis.
72
Given that, n=160 x= 72 .45
  
n
160
x
Px


Z x 0
 
p p
 Or / 2

Z x 0
 
p p
 Or / 2
Page 9 of 25
So, our null hypothesis is: H0 : Px  P0  .5
And the alternative is: x P H : 1  P0=.5
/ 2

Z x 0

p p
p p
(1 
)
0 0
Z
n
p p
(1 
)
0 0
Z
n


Z  Z
. / 2
.45 .5
.5(1 
.5)
160

.05
 Z

= / 2
.039528
=1.2654  Z / 2
Z= 2654 . 1 
F(1.27)=.8980 1- 2 / =.8980  2 / =.102  =.204  =20.4%
104
  
n
Given that, n=199 x= 104 .522613
199
x
Px
And the alternative is: x P H : 1  P0=.5
/ 2

Z x 0

p p
p p
(1 
)
0 0
Z
n
p p
(1 
)
0 0
Z
n

.05
.522613 .5
Z  Z
.5(1 
.5)
199


.022613
Z= 1.645
.035444

Z=.63799>1.645
So, we are maintained null hypothesis at level 10% level.
Z=.64
F(.64)=.73891- / 2=.7389 / 2=.2611 =.5222 =52.22%
28
Given that, n=50 x= 28 .56
  
n
50
x
Px
And the alternative is: H1 : Px >P0=.5

Page 10 of 25
Z
p 
p
Z x 0

p 
p
n

0(1 0)
.56 
.5
Z  Z
.5(1 
.5)

50
Z=.8485>  Z
Z=.8485
F(.85)=.8023 1-=.8023 =.1977 =19.77%
118
Given that, n=172 x= 118 .6860
  
n
172
x
Px
And the alternative is: x P H : 1 <P0=.75
Z
p 
p
Z x 0
 
p 
p
n

0(1 0)
.6860 
.75
Z  Z
.75(1 
.75)

172
Z=1.94  Z
Z=1.94
F(1.94)=.9738 1-=.9738=.0262 =2.62%
140
Given that, n=202 x= 140 .6931
  
n
202
x
Px
So, our null hypothesis is: H0 : Px  P0  .75
And the alternative is: H1 : Px <P0=.75
The decision rule is to reject 0 H in favor of 1 H if:
Z
p 
p
Z x 0
 
p 
p
n

0(1 0)

Page 11 of 25
.0569 
.75
Z  Z
.75(1 
.75)

202
Z= Z   86. 1
Z=1.86
F(1.86)=.9686 1-=.9686 =.0314  =3.14%
F-Distribution
[Page: 375; 43-47]
Given that, 30  x n 770. 451 2  x s
30  y n 208. 1614 2  y s
Our null hypothesis is, x y H    : 0
And, alternative hypothesis is, 2 2
1 : y x H  
2
y
s
F  
v v F
The decision rule is to reject H in favor of H if: 0 11, 2 2 v 1, v
2,
x
s
1614.208
2
s
 y
3.57
1, 2   
451.770
2
x
v v
s
F
At 1% significance level test: 01.  so, 477 . 2 1, 2, 29,29,.01 F  F  v v 
Given that,  4 x n 114.09 2  x s 05.  
 7 y n 16.08 2  y s
Our null hypothesis is, x y H :  0
1 : x y H  
2
x
s
The decision rule is to reject H in favor of H if: F  
F
0 1 v 1, v 2 2 v 1, v
2,
y
s
114.09
2
s
 x
7.095
1, 2   
16.08
2
y
v v
s
F
At 5% significance level test:   .05so, 4.76 1, 2, 3,6,.05 F  F  v v 
[2.49-1/6(2.49-2.41)
= 2.49-.013 = 2.477]

Page 12 of 25
[From exercise 34] Given that,  33 x n 22.93 525.78 2    x x s s
36  y n 55 . 759 56 . 27 2    y y s s
Our null hypothesis is, x y H    : 0
1 : x y H   
/ 2
1
1 2
2
F x
v v  
2
1, 2
s
sy Fv v 
525.78
2
sx
 6922 .
1, 2   
759.55
2
sy
Fv v
At 2% significance level test:   .02 / 2  .01.
Now,
1
1
Fv 1 v 2  / 2
= 2
. 277=.439
Therefore, .6922 is not smaller than 439. According to our decision rule, we will maintain
the null hypothesis at 2% significant level.
[From exercise 36] Given that, 10  x n 4439449 2107 2    x x s s
 10 y n 1681 2825761 2    y y s s
1 : x y H  
2
x
s
F
0 1 v 1, v 2 2 v 1, v 2, 
/ 2
y
s
4439449
2
s
 x
1.571
1, 2   
2825761
2
y
v v
s
F
At 2% significance level test:   .02 / 2  .01so, 5.35 1, 2, / 2 9,9,.01 F  F  v v 
Therefore, 1.571 is not bigger than 5.35. According to our decision rule, we will maintain the
null hypothesis at 2% significant level.
[From sxample: 9.8] Given that,  4 x n 24.4 595.36 2    x x s s
 4 y n 20.2 408.04 2    y y s s
1 : x y H  

Page 13 of 25
2
x
s
F
01v 1, v 2 2
v 1 , v
2 , y
s
595.36
2
s
 x
1
. 459 1, 2   
408.04
2
y
v v
s
F
At 1% significance level test: 01.  so, 46 . 29 1, 2, 3,3,.01 F  F  v v 
Therefore, 1.459 is not bigger than 29.46. According to our decision rule, we will maintain
the null hypothesis at 1% significant level.

X Y XY X2 Y2 Ẏ Y Y  ( Y Y  )2 Ẏ- Y (Ẏ- Y )2
81 76 6156 6561 5776 81.60 -0.7 0.49 4.9 24.01
62 71 4402 3844 5041 68.33 -5.7 32.49 -8.37 70.06
74 69 5106 5476 4761 76.71 -7.7 59.29 .01 .0001
78 76 5928 6084 5776 79.50 -0.7 0.49 2.8 7.84
93 87 8091 8649 7569 89.98 10.3 106.09 13.28 176.36
69 62 4218 4761 3844 73.22 -14.7 216.09 -3.48 12.11
72 80 5760 5184 6400 75.31 3.3 10.89 -1.39 1.93
83 75 6225 6889 5625 82.99 -1.7 2.89 6.29 39.56
90 92 8280 8100 8464 87.88 15.3 234.09 11.28 124.99
84 79 6636 7056 6241 83.69 2.3 5.29 6.99 48.86
786 767 60862 62604 59497 799.21 0 668.1 0 505.7245
Page 14 of 25
Answer the question number 01 [page: 438]
78.6 76.7
The Correlation is: r =

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
60862 
10*78.6*76.7
2 2  
(62604 10*78.6 )(59497 10*76.7 )
=
575.8
824.4*668.1
=
8. 575
15. 742
=.776
Here r = .776 so we can say that there are some positive relationship.
Hypothesis testing: Here, n=10 r = .776 v=10-2=8
So, our null hypothesis is: H0 : Pxy  P0  0
And the alternative is: xy P H : 1  P0=0
We reject H0 if, tv =
r
1 2
2

n
r

> tv, 2 /  or tv =
r
1 2
2

n
r

< - tv, 2 / 
=
.776
1 .776
10 2
2


> t8,.025
=3.48>2.306
So we can reject null hypothesis at 5% level, and we can accept r = .776.

Page 15 of 25
Linear Regression:
Y= a + b x
Y=25.03+.6984 X
a = Y - b X = 78.6 - .6984(76.7)=25.03
b =


XiYi

n X Y Xi
2 2 
n X =
60862 
10 *78.6*76.7
62604 
10 *6177.96
=
8. 575
4. 824
=.6984
R2=
SSR
SST
=
7245 . 505
1. 668
=.7569=75.69%
1.5 14.9 22.35 2.25 222.01 9.89 3.08 9.4864 -1.93 3.72
.2 -9.2 -1.84 .04 84.64 12.52 -21.02 441.8404 .7 .49
-.1 19.6 -1.96 .01 384.16 13.13 7.78 60.5284 1.31 1.7161
2.8 20.3 56.84 7.84 412.09 7.246 8.48 71.9104 -4.574 20.92
2.2 -3.7 -8.14 4.84 13.69 8.464 -15.52 240.8704 -3.356 11.26
-1.6 27.7 -44.32 2.56 767.29 16.178 15.88 252.1744 4.358 18.99
-1.3 22.6 -29.38 1.69 510.76 15.57 10.78 116.2084 3.75 14.06
5.6 2.3 12.88 31.36 5.29 1.562 -9.52 90.6304 -10.26 105.23
-1.4 11.9 -16.66 1.96 141.61 15.77 .08 .0064 3.95 15.60
1.4 27.0 37.8 1.96 729 10.09 15.18 230.43 -1.73 2.99
1.5 -4.3 -6.45 2.25 18.49 9.88 -16.12 259.85 -1.94 3.76
-4.7 20.3 -95.41 22.09 412.09 22.47 8.48 71.91 10.65 113.42
1.1 4.2 4.62 1.21 17.64 10.69 -7.62 58.0644 -1.13 1.277
7.2 153.6 -69.67 80.06 3718.76 153.46 0 1903.91 0 313.433
.55 11.82

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
 
69.67 13*.55*11.82
2 2  
(80.06 13*.55 )(3719.39 13*11.8 )
=
154.18
380.63
=.4051
Here r =- .4051 so we can say that there are some nagative relationship.
Hypothesis testing: Here, n=13 r = -.40 v=13-2=11
And the alternative is: H1 : Pxy  P0=0

Page 16 of 25
r
1 2
2

n
r

> tv, 2 / or tv =
r
1 2
2

n
r

< - tv, 2 /
=
40.
1

(  . 40) 2 13
2


< - 05,.11t
=-1.44749 <- 1.796
So we are maintain null hypothesis at 10% level, and we can reject r = -.4051
Linear Regression:
Y= a + b x
Y=12.93+(-2.02533) X
a = Y - b X = 11.82 - (-2.02533)(.55)=12.93
b =


XiYi

n X Y Xi
2 2 
n X =
 
69.67 13*.55*11.82
80.06 
13*(.55)
2 =
183. 154 
1275. 76
=-2.02532
R2=
SSR
SST
313.433
=
1903.91
=.1646=16.46%
X Y XY X2 Y2 Ẏ Y Y (Y Y )2 Ẏ- Y (Ẏ- Y )2
2.8 2.6 7.28 7.84 6.76 2.51552 -.35 .1225 -.43448 .18877
3.7 2.9 10.73 13.69 8.41 2.88793 -.05 .0025 -.06207 .0039
4.4 3.3 14.52 19.36 10.89 3.17758 .35 .1225 .22758 .05179
3.6 3.2 11.52 12.96 10.24 2.84655 .25 .0625 -.10345 .01070
4.7 3.1 14.57 22.09 9.61 3.30172 .15 .0225 .35172 .12370
3.5 2.8 9.8 12.25 7.84 2.80517 -.15 .0225 -.14483 .02097
4.1 2.7 11.07 16.81 7.29 3.05345 -.25 .0625 .10345 .01070
3.2 2.4 7.68 10.24 5.76 2.68104 -.55 .3025 -.26896 .07233
4.9 3.5 17.15 24.01 12.25 3.38448 .55 .3025 .43448 .18877
4.2 3.0 12.60 17.64 9 3.09483 -.05 .0025 .14483 .02097
3.8 3.4 12.92 14.44 11.56 2.92931 .45 .2025 -.03069 .00094
3.3 2.5 8.25 10.89 6.25 2.72242 -.45 .2025 -.22758 .05179
46.2 35.4 138.09 182.22 105.86 35.4 0 1.43 0 .74533
3.85 2.95

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
138.09 
12*3.85*2.95
2 2  
(182.22 12*.85 )(105.86 12*2.95 )

Page 17 of 25
=
8 .1
49. 2
=.722
Here r =.722 so we can say that there are some positive relationship.
So, our null hypothesis is: 0: 0 0 PPHxy
And the alternative is: xyP H : 1 > P0=0
r
1 2
2

n
r

> tv,
=
722 .
) 722 (. 1
2 12
2


> 10,.10t = 3.30 > 1.372
So we are reject null hypothesis at 10% level, and we can accept r = .722
Linear Regression:
Y= a + b x
Y=1.35691+(.41379) X
a = Y - b X = 2.95 - (.41379)(3.85)=1.35691
b =


XiYi nXY

2 Xi 2 
nX
=
138.09 12*3.85*2.95
182.22 
12*(3.85)
2 
1.8
=
4.35
=0.41379
R2=
SSR
SST
=
74533.
43. 1
=.5212=52.12%
7.70 141.77 1091.63 59.29 20098.73 164.775 -18 324 5.005 25.05
4.17 96.97 404.36 17.39 9403.18 160.00 -62.8 3943.84 .23 .0529
1.52 163.92 249.16 2.31 26869.77 156.43 4.15 17.22 -3.34 11.156
10.04 154.70 1553.18 100.80 23932.09 167.93 -5.07 25.70 8.16 66.58
6.02 151.61 912.69 36.24 22985.59 162.51 -8.16 66.59 2.74 7.51
4.81 147.82 711.01 23.14 21850.75 160.87 -11.95 142.80 1.1 1.21
1.57 98.61 154.82 2.46 9723.93 156.50 -61.16 3740.55 -3.27 10.69
3.63 179.18 650.42 13.17 32105.47 159.28 19.41 376.75 -.49 .2401
1.57 125.19 196.55 2.46 15672.54 156.50 -34.58 1195.78 -3.27 10.6929
4.65 171.81 798.92 21.62 29518.68 160.66 12.04 144.96 .89 .7921
2.97 200.23 594.68 8.82 40092.05 158.39 40.46 1637.01 -1.38 1.9044
.98 120.49 118.08 .96 14517.84 155.70 -39.28 1542.92 -4.07 16.5649
4.18 95.83 400.57 17.47 9183.39 160.02 -63.94 4088.32 .25 .0625
6.09 196.67 1197.72 37.08 38679.09 162.60 36.9 1361.61 2.83 8.0089
3.09 275.97 852.75 9.55 76159.44 158.55 116.2 13502.44 -1.22 1.4884
3.08 289.59 891.94 9.49 83862.37 158.54 129.82 16853.23 -1.23 1.5129

1.76 105.71 186.05 3.09 11174.60 156.76 -54.06 2922.48 -3.01 9.0601
67.83 2716.07 10964.53 365.05 485829.5 2716.02 0 51886.2 0 172.58
3.99 159.77
Page 18 of 25

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
10964.53 
17*3.99*159.77
2 2  
(365.05 17*3.99 )(485829.51 17*159.77 )
=
3309 . 127
136 . 2213
=.05753
And the alternative is: xyP H : 1  P0=0
r
1 2
2

n
r

> tv, 2 / or tv =
r
1 2
2

n
r

< - tv, 2 /
=
.05753
1 (.05753)
17 2
2


> 025,.15t
= .22 >2.131
So we are maintain null hypothesis at 5% level, and we can reject r = .05753
Linear Regression:
Y= a + b x
Y=154.38+(1.35) X
a = Y - b X = 159.77 - (1.35)(3.99)=154.38
b =


XiYi nXY

2 Xi 2 
nX
=
10964.53 17 *3.99 *159.77
365.05 
17 *(3.99)
2 
127.33
=
94.41
=1.35
R2=
SSR
SST
172.58
=
51886.2
=.0033=.3326%

Page 19 of 25
2 1 2 4 1 3.68 -11.5 132.25 -8.82 77.79
4 2 8 16 4 5.36 -10.5 110.25 -7.14 50.98
1 3 3 1 9 2.84 -9.5 90.25 -9.66 93.32
13 4 52 169 16 12.92 -8.5 72.25 .67 .4489
5 5 25 25 25 6.2 -7.5 56.25 -6.3 39.69
7 6 42 49 36 7.88 -6.5 42.25 -4.62 21.34
6 7 42 36 49 7.04 -5.5 30.25 -5.46 29.81
8 8 64 64 64 8.72 -4.5 20.25 -3.78 14.29
3 9 27 9 81 4.52 -3.5 12.25 -7.98 63.68
9 10 90 81 100 9.56 -2.5 6.25 -2.94 8.64
17 11 187 289 121 16.28 -1.5 2.25 3.78 14.29
19 12 228 361 144 17.96 -.5 .25 5.46 29.81
12 13 156 144 169 12.08 .5 .25 -.42 .18
10 14 140 100 196 10.4 1.5 2.25 -2.1 4.41
15 15 225 225 225 14.6 2.5 6.25 2.1 4.41
18 16 288 324 256 17.12 3.5 12.25 4.62 21.34
11 17 187 121 289 11.24 4.5 20.25 -1.26 1.59
23 18 414 529 324 21.32 5.5 30.25 8.82 77.79
20 19 380 400 361 18.80 6.5 42.25 6.30 39.69
16 20 320 256 400 15.44 7.5 56.25 2.94 8.64
21 21 441 441 441 19.64 8.5 72.25 7.14 50.98
14 22 308 196 484 13.76 9.5 90.25 1.26 1.59
22 23 506 484 529 20.48 10.5 110.25 7.98 63.68
24 24 576 576 576 22.16 11.5 132.25 9.66 93.32
300 300 4711 4900 4900 300 0 1150 0 811.70
12.5 12.5

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
4711 
24*12.5*12.5
2 2  
(4900 24*12.5 )(4900 24*12.5 )
961
=
1150
=.8357
And the alternative is: H1 : Pxy > P0=0

Page 20 of 25
r
1 2
2

n
r

> tv,
=
8357 .
) 8357 (. 1
2 24
2


> 05,.22 t
= 7.137 > 1.717
Linear Regression:
Y= a + b x
Y=2+(.84) X
a = Y - b X = 12.5- (.84) (12.5) =2
b =


XiYi

n X Y Xi
2 2 
n X =
4711 24*12.5*12.5
4900 
24*(12.5)2

961
=
1150
=.84
R2=
SSR
SST
=
70. 811
1150
=.7058=70.58%
X Y XY X2 Y2 Ẏ Y Y (Y Y )2 Ẏ- Y (Ẏ- Y )2
1 1 1 1 1 4.03 -9.5 90.25 -6.47 41.86
2 4 8 4 16 4.71 -6.5 42.25 -5.79 33.52
3 2 6 9 4 5.39 -8.5 72.25 -5.11 26.11
4 14 56 16 196 6.07 3.5 12.25 -4.43 19.62
5 5 25 25 25 6.75 -5.5 30.25 -3.75 14.06
6 9.5 57 36 90.25 7.43 -1 1 -3.07 9.42
7 12 84 49 144 8.11 1.5 2.25 -2.39 5.71
8 7 59.5 72.25 49 9.13 -3.5 12.25 -1.37 1.88
9 9.5 80.75 72.25 90.25 9.13 -1 1 -1.37 1.88
10 11 110 100 121 10.15 .5 .25 -.35 .1225
11 15 165 121 225 10.83 4.5 20.25 .33 .1089
12 3 36 144 9 11.51 -7.5 56.25 1.01 1.0201
13 8 104 169 64 12.19 -2.5 6.25 1.69 2.86
14 18 252 196 324 12.87 7.5 56.25 2.37 5.62
15 16 240 225 256 13.55 5.5 30.25 3.05 9.30
16 13 208 256 169 14.23 2.5 6.25 3.73 13.91
17 6 102 289 36 14.91 -4.5 20.25 4.41 19.45
18 20 360 324 400 15.59 9.5 90.25 5.09 25.91
19 17 323 361 289 16.27 6.5 42.25 5.77 33.29
20 19 380 400 361 16.95 8.5 72.25 6.45 41.60
210 210 2657.25 2869.5 2869.5 209.8 0 664.5 0 305.37
10.5 10.5

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 

Page 21 of 25
=
2657.25 
20*10.5*10.5
2 2  
(2869.5 20*10.5 )(2869.5 20*10.5 )
=
25. 452
5. 664
=.6806
And the alternative is: H1 : Pxy > P0=0
r
1 2
2

n
r

> tv,
=
6806 .
) 6806 (. 1
2 20
2


> 05,.18t
= 3.94 > 1.734
Linear Regression:
Y= a + b x
Y=3.35+(.68) X
a = Y - b X = 10.5- (.6806) (10.5) =3.35
b =


XiYi nXY

2 Xi 2 
nX
=
2657.25 
20*10.5*10.5
2869.5 
20*(10.5)
2 =
25. 452
5. 664
=.6806
R2=
SSR
SST
=
37. 305
5. 664
=.4595=45.95%
Answer the question number 13 (Page-455)
X Y XY X2 Y2 Ẏ Y Y  ( Y Y  )2 Ẏ-Y (Ẏ-Y )2
5.5 420 2310 30.25 176400 410.32 15 225 5.32 28.3027
6.0 380 2280 36 144400 389.03 -25 625 -15.97 255.0409
6.5 350 2275 42.25 122500 367.74 -55 3025 -37.26 1388.308
6.0 400 2400 36 160000 389.03 -5 25 -15.97 255.0409
5.0 440 2200 25 193600 431.61 35 1225 26.61 708.0921
6.5 380 2470 42.25 144400 367.74 -25 625 -37.26 1388.308
4.5 450 2025 20.25 202500 452.9 45 2025 47.9 2294.41
5.0 420 2100 25 176400 431.61 15 225 26.61 708.0921
45 3240 18060 257 1320200 3239.98 0 8000 0 7025.594
5.625 405

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 

Page 22 of 25
=
18060 
8*5.625*405
2 2  
(257 8*5.625 )(1320200 8*405 )
=
165 
07. 176
=-.9371
Here r = - .9371 so we can say that there are some negative relationship.
Hypothesis testing: Here, n=8 r = - .9371 v=8-2=6
And the alternative is: H1 : Pxy  P0=0
r
1 2
2

n
r

> tv, / 2 or tv =
r
1 2
2

n
r

< - tv, / 2
=
9371 .

) 9371. ( 1
2 8
2

 
<- 025 ,.6t
= -6.57 < -2.447
So we are reject null hypothesis at 5% level, and we can accept r = -.9371
Linear Regression:
Y= a + b x
Y=644.5125+ (-42.58) X
a = Y - b X = 405 - (-42.58) (5.625) =644.5125
b =


XiYi nXY

2 Xi 2 
nX
=
18060 
8*5.625 *405
257 
8*(5.625)
2 =
165 
875 . 3
=-42.58
R2=
SSR
SST
=
7025.594
8000
=.8782=87.82%
X Y XY X2 Y2 Ẏ Y Y  ( Y Y  )2 Ẏ-Y (Ẏ-Y )2
38 4.7 178.6 1444 22.09 5.286 0.276 0.076176 0.862 0.743044
24.5 4.7 115.16 600.25 22.09 4.07775 0.276 0.076176 -0.34625 0.119889
21.5 4 86 462.25 16 3.80925 -0.424 0.179776 -0.61475 0.377918
30.8 4.7 144.76 948.64 22.09 4.6416 0.276 0.076176 0.2176 0.04735
20.3 3 60.9 412.09 9 3.70185 -1.424 2.027776 -0.72215 0.521501
24 4.4 105.6 576 19.36 4.033 -0.024 0.000576 -0.391 0.152881
29.6 5 148 876.16 25 4.5342 0.576 0.331776 0.1102 0.012144
19.4 3.3 64.02 376.36 10.89 3.6213 -1.124 1.263376 -0.8027 0.644327
25.6 3.8 97.28 655.36 14.44 4.1762 -0.624 0.389376 -0.2478 0.061405
39.5 6.4 252.8 1560.25 40.96 5.42025 1.976 3.904576 0.99625 0.992514
23.3 3.3 76.89 542.89 10.89 3.97.035 -1.124 1.263376 -0.45365 0.205798
28 3.6 100.8 784 12.96 4.391 -0.824 0.678976 -0.033 0.001089
30.8 4.7 144.76 948.64 22.09 4.6416 0.276 0.076176 0.2176 0.04735
32.9 4.4 144.76 1082.41 19.36 4.82955 -0.024 0.000576 0.40555 0.164471
30.3 5.4 163.62 918.09 29.36 4.59685 0.976 0.952576 0.17285 0.029877

199 3 59.7 396.01 9 3.66605 -1.424 2.027776 -0.75795 0.574488
24.6 4.9 120.54 605.16 24.01 4.0867 0.476 0.226576 -0.3373 0.113771
32.3 5.2 167.96 1043.29 27.04 4.77585 0.776 0.602176 0.35185 0.123798
24.7 4.2 103.74 610.09 17.64 4.09565 -0.224 0.050176 -0.32835 0.107814
18.7 3.3 61.71 349.69 10.89 3.55865 -1.124 1.263376 -0.86535 0.748831
36.8 4.1 150.88 1354.24 16.81 5.1786 -0.324 0.104976 0.7546 0.569421
31.2 6 187.2 973.44 36 4.6774 1.576 2.483776 0.2534 0.064212
50.9 5.8 295.22 2590.81 33.64 6.44055 1.376 1.893376 2.01655 4.066474
30.7 4.9 150.43 942.49 24.01 4.63265 0.476 0.226576 0.20865 0.043535
20.3 3.8 77.14 412.09 14.44 3.70185 -0.624 0.389376 -0.72215 0.521501
708.6 110.6 3258.46 21464.7 509.86 110.5447 0 20.5656 0 11.0554
28.344 4.424
Page 23 of 25

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
3226.16 
25*28.344*4.424
2 2  
(21464.69 25*28.344 )(509.86 25*4.424 )
=
3136. 91
47. 168
=.5420
Hypothesis testing: Here, n=25 r = 0.5420 v=25-2=23
And the alternative is: xy PH: 1  P0=0
r
1 2
2

n
r

> tv, 2 /  or tv =
r
1 2
2

n
r

< - tv, 2 / 
=
0.5420
1 (0.5420)
25 2
2


> 025 ,.23 t
= 3.09 > 2.069
So we are reject null hypothesis at 5% level, and we can accept r = 0.5420
Linear Regression:
Y= a + b x
Y=1.88757+(0.0895) X
a = Y - b X = 4.424 - (0.0895) (28.34) =1.88757
b =


XiYi nXY

2 Xi 2 
nX
=
3258.46 25*28.34*4.424
21464.7 
25*(28.34)
2 
124.056
=
1385.81
=0.0895
R2=
SSR
SST
=
11.0554
20.5656
=.5376=53.76%

Page 24 of 25
55 10 550 3025 100 10.5 -6.875 47.26562 -6.375 40.641
60 12 720 3600 144 12.5 -4.875 23.76562 -4.375 19.141
85 28 2380 7225 784 22.5 11.125 123.76562 5.625 31.641
75 24 1800 5625 576 18.5 7.125 50.76562 1.625 2.641
80 18 1440 6400 324 20.5 1.125 1.26562 3.625 13.141
85 16 1360 7225 256 22.5 -0.875 0.76562 5.625 31.641
65 15 975 4225 225 14.5 -1.875 3.51562 -2.375 5.641
60 12 720 3600 144 12.5 -4.875 23.76562 -4.375 19.141
565 135 9945 40925 2553 0 274.875 0 163.628

XiYi 
nXY
2 2 2 2 Xi nX Yi nY
(  )( 
)
 
=
9945 
8*70.625*16.875
2 2  
(40925 8*70.625 )(2553 8*16.875 )
=
410.625
529.99
=.7748
And the alternative is: xy PH: 1  P0=0
r
1 2
2

n
r

> tv, 2 /  or tv =
r
1 2
2

n
r

< - tv, 2 / 
=
0.7748
1 (0.7748)
8 2
2


> t6,.025
= 3.00 > 2.447
So we are reject null hypothesis at 5% level, and we can accept r = 0.7748
Linear Regression:
Y= a + b x
Y=1.88757+(.4018) X

Page 25 of 25
a = Y - b X = 16.875 - (0.0895) (70.625) =1.88757
b =


XiYi

n X Y Xi
2 2 
n X =
9945 8*70.625 *16.875
40925 
8*(70.625)2

=
625. 410
875 . 1021
=0.4018
R2=
SSR
SST
=
628. 163
875. 274
=.5952=60%

Statistics assignment 10

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Statistics assignment 10