Homework Problems 3 - Applied Multivariate Analysis I | STAT 8108, Assignments of Descriptive statistics

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Homework Problems 3

1. Ex 6.17 a, b and c based on the Number of Parity Data in Table 6.8.
2. Ex. 6.19 a, b and c based on data in Table 6.10 (Milk Transportation Data)

6.17(a), p.

Test for treatment effects using a repeated measures design. Set  =.05.

Null hypothesis,

O

H : there are NO treatment effects.

Alternate hypothesis,

A

H : there ARE treatment effects.

3 4 1 2

(  +  ) - (  + )= Number Format contrast, representing the difference between the presence and

absence of Arabic format numbers

1 3 2 4

(  +  ) - ( + )

= Parity Type contrast, representing the difference between the presence and absence

of “different parity”

1 4 2 3

(  +  ) - (  +  )

= Contrast representing the influence of Arabic format on parity differences (i.e.

Format-Parity interaction

With



’=[

1

 ,

2

 ,

3

 ,

4

 ], the contrast matrix C is

C

 

 

 

  

 

 

 

 

The data (see Appendix 6.17) give:

x

 

 

 



 

 

 

and

36178.35 25650.31 17707.06 14637.

25650.31 22302.16 13759.3 13367.

17707.06 13759.3 17865.71 11027.

14637.81 13367.56 11027.05 12182.

S

 

 

 



 

 

 

Stat 8108: Multivariate Dr. Sarkar

C x

 

 

 

 

 

 

  

 

 

   

 

 

 

( 1967.56) ( 1876.89) (1828.63) (1716.63)

(1967.56) ( 1876.89) (1828.63) ( 1716.63)

(1967.56) ( 1876.89) ( 1828.63) (1716.63)

    

 

 

     

 

      

 

C x



 

 

 

 

 

 

36178.35 25650.31 17707.06 14637.

25650.31 22302.16 13759.3 13367.

' 1 1 1 1

17707.06 13759.3 17865.71 11027.

14637.81 13367.56 11027.05 12182.

CSC

 

 

 

 

 

 

  

 



   



 

 

  

 

  

 

  



  



 

1 1 1 1 *

 

 

 

  

 

   

 

36178.35( 1)+25650.31( 1)+17707.06(1)+14637.81(1)36178.

25650.31( 1) 22302.16( 1) 13759.3(1) 13367.56(1)

17707.06( 1) 13759.3( 1) 17865.71(1) 11027.05(1)

14637.81( 1) 13367.56( 1) 11027.05(1) 12182.03(1)

 

    

    

    

35(1)+25650.31( 1)+17707.06(1)+14637.81( 1 )36178.35(1)+

25650.31(1) 22302.16( 1) 13759.3(1) 13367.56( 1)

17707.06(1) 13759.3( 1) 17865.71(1) 11027.05( 1)

14637.81(1) 13367.56( 1) 11027.05(1) 12182.03( 1)

 

    

    

    

25650.31( 1)+17707.06( 1)+14637.81(1)

25650.31(1) 22302.16( 1) 13759.3( 1) 13367.56(1)

17707.06(1) 13759.3( 1) 17865.71( 1) 11027.05(1)

14637.81(1) 13367.56( 1) 11027.05( 1) 12182.03(1)

   

 

    

 

      

 

    

 

1 1 1 1 *

 

 

 

  

 

   

 

-36178.35-25650.31+17707.06+14637.81 36178.35-25650.31+17707.06-14637.

-25650.31-22302.16+13759.3 13367.56 25650.31-22302.16 13759

-17707.06-13759.3 17865.71 11027.

-14637.81-13367.56 11027.05 12182.

 

 

 

36178.35-25650.31 17707.06+14637.

.3 13367.56 25650.31-22302.16-13759.3 13367.

17707.06 13759.3 17865.71 11027.05 17707.06-13759.3-17865.71 11027.

14637.81 13367.56 11027.05-12182.0314637.81-13367.



 

   

  -11027.05 12182.

 

 

 

 

 



 

29483.79 13597.29 7458.

20825.61 3739.89 2956.

1 1 1 1 *

-2573.60 10786.42 -2890.

-4796.29 115.27 2425.



 

 

 

 



 

 

  

 

 

 

 

 

 

 

Stat 8108: Multivariate Dr. Sarkar

1( 29483.79) 1( 20825.61) 1( 2573.60) 1( 4796.29) 1(13597.29) 1(3739.89) 1(10786.42) 1(115.27) 1(7458.79) 1(2956.41) 1( 2890.90) 1(2425.23)

1( 29483.79) 1( 20825.61) 1( 2573.60) 1( 4796.29) 1(13597.

                

        ) 1(3739.89) 1(10786.42) 1(115.27) 1(7458.79) 1(2956.41) 1( 2890.90) 1(2425.23)

1( 29483.79) 1( 20825.61) 1( 2573.60) 1( 4796.29) 1(13597.29) 1(3739.89) 1(10786.42) 1(115.27) 1(7458.79) 1(2956.41) 1( 2890

      

             .90) 1(2425.23)

 

 

 

  

 

29483.79 20825.61 2573.60 4796.29 13597.29 37 39.89 10786.42 115.27) 7458.79 2956.41 2890.9 0 2425.

29483.79 20825.61 2573.60 4796.29 13597.29 37 39.89 10786.42 115.27) 7458.79 2956.41 2890.9 0 2425.

2

          

          

 9483.79 20825.61 2573.60 4796.29) 13597.29 37 39.89 10786.42 115.27) 7458.79 2956.41 2890.9 0 2425.

 

 

 

          

 

42939.51 6435.49 10880.

' 6435.49 20528.55 813.

10880.87 813.75 9818.

CSC

 

 

 

   

 

   

 

2 1

T n C x ( ) '( CSC ') ( C x )





 

2 1

32 -299.19 202.67 -21.33 ( ') 202.

T CSC



 

 

 

 

 

 

Even though I spent a lot of time calculating

CSC '

, calculating

1

( CSC ')



would also be time-consuming,

so let’s just assume that SAS calculated

1

( CSC ')



correctly and also calculated

2

T

correctly.

2

 T 135.

(See Appendix 6.17.)

SAS also shows us that:

2

1, 1

( 1)( 1)

( )

( 1)

q n q

n q

c F

n q



  

 



 

4 1,32 4 1

(32 1)(4 1)

(0.05)

(32 4 1)

F

  

 



 

3,

(31)(3)

(0.05)

(29)

 F

From SAS (Appendix 6.17), we see that F=2.

2

 c 3.2069* 2.93 9.

=the Critical Value of F.

Since SAS determines that

2

T

=135.85, which is > the Critical Value for F of 9.4, we REJECT the

O

H

that there are no treatment effects.



We conclude that there ARE treatment effects.

Stat 8108: Multivariate Dr. Sarkar

6.17(b), p.

Construct 95% (simultaneous) confidence intervals for the contrasts representing the number

format effect, the parity type effect and the interaction effect. Interpret the resulting intervals.

To see which of the contrasts are responsible for the rejection of the null hypothesis, we construct 95%

simultaneous confidence intervals.

Here 1

c ' is the 1

st

row of the contrast matrix C.

The contrast 1 3 4 1 2

c '  (  +  ) - (  + ) = Number Format influence; is estimated by the confidence

interval:

( x 3 + x 4 ) - ( x 1 + x 2 )

2 1 1

c 'Sc

c *

n



1 1

3 4 1 2

1, 1

( 1)( 1) c 'Sc

=( + ) - ( + ) ( ) *

( 1)

q n q

n q

x x x x F

n q n



  

 



 

which = -411.4998 to -186.8802, as calculated by SAS (see Appendix 6.17). This can be

expressed as -299.



112.3098. (-299.19 is the average of -411.4998 and -186.8802, and

112.3098 is half the difference between -411.4998 and -186.8802) Since this entire range is

negative, we conclude that there is a Number Format effect in that the presence of an Arabic

number format reduces the judgment time.

The contrast 1 3 2 4

(  +  ) - (  +  )

=Parity Type influence; is estimated by the confidence interval:

2 2 2

1 3 2 4

c 'Sc

( x + x ) - ( x + x ) c *

n



2 2

1 3 2 4

1, 1

( 1)( 1) c 'Sc

=( + ) - ( + ) ( ) *

( 1)

q n q

n q

x x x x F

n q n



  

 



 

which = 125.01524 to 280.32476 (see Appendix 6.17). This can be expressed as 202.67 

77.6548, which is an entirely positive interval, so we conclude that there is a Parity Type effect

in that the presence of different parity prolongs the judgment time.

The contrast 1 4 2 3

(  +  ) - (  +  )

=Format-Parity interaction; is estimated by the confidence

interval:

2 3 3

1 4 2 3

c 'Sc

( x + x ) - ( x + x ) c *

n



Stat 8108: Multivariate Dr. Sarkar

3 3

1 4 2 3

1, 1

( 1)( 1) c 'Sc

=( + ) - ( + ) ( ) *

( 1)

q n q

n q

x x x x F

n q n



  

 



 

which = 75.03463 to 32.37463 (see Appendix 6.17) i.e. -21.



53.7046, which includes the

value within the interval i.e. this interval is not significantly different from zero, so we conclude

that there is an absence of interaction.

6.17(c), p.

The absence of interaction supports the M model of numerical cognition, while the presence of

interaction supports the C and C model of numerical cognition. Which model is supported in this

experiment?

The M model of numerical cognition is supported since there is an absence of interaction.

Appendix 6.

N

32

P

4

XBAR VAR F C S

967.56 36178.35 2.93 9.4 36178.35 25650.31 17707.06 14637.

876.89 22302.16 25650.31 22302.16 13759.3 13367.

828.63 17865.71 17707.06 13759.3 17865.71 11027.

716.63 12182.03 14637.81 13367.56 11027.05 12182.

T

HOTELLING

Reject Null Hypothesis

Stat 8108: Multivariate Dr. Sarkar

NOTE

Number Format effect

INTERVAL

-411.4998 -186.

NOTE

Parity Type effect

INTERVAL

125.01524 280.

NOTE

Format-Parity interaction

INTERVAL

-75.03463 32.

6.19(a), p.

Test for differences in the mean cost vectors. Set  =.01.

Null hypothesis, O

H

: cost vectors are EQUAL.

Alternate hypothesis, A

H

: cost vectors are NOT equal.

The summary statistics are:

1

x

 

 



 

 

 

,

1

23.01 12.37 2.

12.37 17.54 4.

2.91 4.77 13.

S

 

 



 

 

 

,

1

n  36

2

x

 

 



 

 

 

,

2

4.36 0.76 2.

0.76 25.85 7.

2.36 7.69 46.

S

 

 



 

 

 

,

2

n  23

1 2

1 2

1 2 1 2

pooled

n n

S S S

n n n n

 

 

   

Stat 8108: Multivariate Dr. Sarkar

23.01 12.37 2.91 4.36 0.76 2.

12.37 17.54 4.77 0.76 25.85 7.

2.91 4.77 13.96 2.36 7.69 46.

   

 

   

 

   

   

   

   

23.01 12.37 2.91 4.36 0.76 2.

12.37 17.54 4.77 0.76 25.85 7.

2.91 4.77 13.96 2.36 7.69 46.

   

   

 

   

   

   

15.81 7.89 2.

7.89 20.75 5.

2.7 5.9 26.

pooled

S

 

 

 

 

 

 

as calculated by SAS (Appendix 6.19).

1 2

2 1 2

, 1

1 2

( 2)

( )

p n n p

n n p

c F

n n p



  

 



  

3,36 23 3 1

(36 23 2)

(0.01)

F

  

 



  

3,

(57)

(0.01)

 F

(57)

(4.13)





2

c

=12.93, as calculated by SAS (Appendix 6.19).

The likelihood ratios test (Johnson, p.285) of 0 1 2 0

H     

is based on the square of the statistical

distance,

2

T

. We Reject

0

H

if

1

2 2

1 2 1 2

0 0

1 2

( ) ' ( ) ( )

pooled

T x x S x x c

n n

 



 

      

 

 

For our purposes, let 0



=0.



We Reject

0

H

if

1

2 2

1 2 1 2

1 2

( ) ' ( ) ( )

pooled

T x x S x x c

n n



 

    

 

 

Is

1

2

12.22 10.11 15.81 7.89 2.7 12.22 10.

( 8.11 10.76 ) ' ( ) 7.89 20.75 5.9 ( 8.11 10.76 ) 12. 93

9.59 18.17 2.7 5.9 26.58 9.59 18.

T



 

         

 

         

    

 

         

 

         

         

 

Stat 8108: Multivariate Dr. Sarkar

Is

 

1

2

1.1266 0.5622 0.1924 2.

2.11 2.65 8.58 0.5622 1.4786 0.4204 2.65 12.

0.1924 0.4204 1.894 8.

T



 

   

 

   

    

 

   

 

    

   

 

…………………………

6.19(b), p.

If the hypothesis of equal cost vectors is rejected in Part (a), find the linear combination of mean

components most responsible for the rejection.

Since 1 2

( x  x )=

12.22 10.

8.11 10.

9.59 18.

   

   



   

   

   

=

 

 



 

 



 

, we can visually see that the Capital cost component on

the surface appears to have the biggest difference (between gasoline and diesel trucks). However, we do

not know if this Capital cost component difference is a statistically significant difference, and/or if Fuel is

also showing a statistically significant difference, and/or if Repair is also showing a statistically

significant difference, and/or if the sum of all three costs would show a statistically significant difference.

For testing 0 1 2

H :     0

, the linear combination â '( x 1  x 2 )

, with coefficient vector

1

1 2

â ( )

pooled

S x x



 

, quantifies the largest population difference. I.e., if

2

T

rejects

0

H

, then

1 2

â '( x  x )

will have a nonzero mean. If the absolute value

1 2

1 2

a '( x  x )  a '(   )

is

â( )

pooled

 c  S a

then mathematically it must follow that

1 2 1 2

1 2

a '( ) â( ) a '( ) a '( ) â( )

pooled pooled

x  x  c  S a      x  x  c  S a

Referring to the answer to question 6.19(c) below, we see that the Capital cost component is the only

statistically significant difference in cost vectors between gasoline and diesel trucks. Knowing this, we

can conclude that we should choose a vector â which will isolate the Capital cost component when â' is

multiplied by ( x 1  x 2 )

. This is obviously  

â'= 0 0 1

. Thus the linear combination is â '( x 1  x 2 )

=

 

 

 



 

  

 

.

6.19(c), p.

Stat 8108: Multivariate Dr. Sarkar

Construct 99% simultaneous confidence intervals for the pairs of mean components. Which costs,

if any, appear to be quite different?

: (12.22 10.11) 12.93 + *15.

Fuel

 

 

 

 

 2.11 3.5958 (0.02778 0.04348) *15.

 2.11 3.5958 (0.07126)*15.

 2.11 3.5958 1.

 2.11 3.5958*1.

 2.11 8.

 1.7066  Fuel 5.

Repair : (8.11 10.76) 12.93 + *20.

 

 

 

 

 2.65 3.5958 (0.07126)* 20.

 2.65 3.5958 1.

 2.65 3.5958*1.

 2.65 4.

 7.0224 Repair 1.

: (9.59 18.17) 12.93 + *26.

Capital

 

 

 

 

 8.58 3.5958 (0.07126)* 26.

 8.58 3.5958 1.

 8.58 3.5958*1.

 8.58 4.

Stat 8108: Multivariate Dr. Sarkar

 13.5287  Capital 3.

The confidence interval for the difference in the costs of the Capital component does not include zero.



Capital is quite a different cost between gasoline and diesel trucks.

Appendix 6.

N1 P1 XBAR1 S

36 3 12.22 23.01 12.37 2.

8.11 12.37 17.54 4.

9.59 2.91 4.77 13.

N2 P2 XBAR2 S

23 3 10.11 4.36 0.76 2.

10.76 0.76 25.85 7.

18.17 2.36 7.69 46.

SPOOL

15.81 7.89 2.

7.89 20.75 5.

2.7 5.9 26.

C

MEANDIFFERENCE INTERVAL

Stat 8108: Multivariate Dr. Sarkar

u11-u12 -1.71 5.

MEANDIFFERENCE INTERVAL

u11-u12 -7.02 1.

MEANDIFFERENCE INTERVAL

u11-u12 -13.53 -3.

EVAL EVEC

32.955088 0.3837284 0.4924704 0.

20.294232 0.5910466 0.5189794 -0.

9.8906808 0.7095184 -0.698665 0.

99% confidence ellipse:

INTERV

Stat 8108: Multivariate Dr. Sarkar

i d 1 2

y

0

1 0

2 0

3 0

x

0 1 0 2 0 3 0

References

Johnson RA and Wichern DW, “Applied Multivariate Statistical Analysis,” 2007.

Stat 8108: Multivariate Dr. Sarkar