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# Business Statistics - Nonparametric and Parametric Tests

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1) For the problem, you may not make any assumptions concerning normality, etc

The following data represent the number of wins as of March 7, 2005 for the 30 NBA (National Basketball Association) teams

Eastern Conference Western Conference
Atlantic Division Southwest Division
Boston 31 San Antonio 46
Philly 29 Dallas 38
New Jersey 26 Houston 34
New York 25 Memphis 33
Toronto 25 New Orleans 13
Southeast Division Northwest Division
Miami 45 Seattle 41
Washington 33 Denver 30
Orlando 31 Minnesota 30
Charlotte 12 Portland 22
Atlanta 10 Utah 20
Central Division Pacific Division
Detroit 36 Phoenix 45
Cleveland 31 Sacramento 37
Chicago 29 Lakers 29
Indiana 29 Clippers 26
Milwaukee 24 Golden State 18

a) Sportscasters often state that the Western Conference is stronger than the Eastern Conference. Use the appropriate nonparametric
rank test to see if this is true.

b) Similarly, test to see if the 8 teams with the most wins in the Western Division are stronger than the top 8 teams in the Eastern
Division.

2) The following data are given for 20 automobiles.

MPG Horsepower Weight Car's Age Teenage Driver
17 240 4485 2 Yes
21 225 3860 4 No
26 160 2780 1 No
22 225 3520 5 No
20 170 3745 4 No
18 220 3915 4 No
22 184 3390 1 No
19 184 3575 2 No
14 225 4715 3 Yes
20 205 3640 2 No
21 205 3880 1 No
15 185 4230 6 Yes
19 275 4070 4 No
20 180 3495 2 No
22 170 3050 3 No
12 285 5590 3 Yes
20 127 3055 1 Yes
14 270 4660 1 Yes
19 185 3990 2 No
21 250 3620 1 No

a) Develop the regression for this data that attempts to relate the mileage to the other variables
b) Which coefficients are significant?
c) Predict the mileage for a 210 horsepower car weighing 4000 pounds which is 3 years old and driven by a teenager.
d) How much of the variation in mileage is due to the independent variables?

3) I In the analysis for this problem, you may make necessary assumptions such as normality, etc. It has
been stated that the arrival of Southwest Airlines in Philadelphia has had an impact in lowering the prices
of airline tickets in Philadelphia. To test this assumption, the following data was obtained. These data are
the round-trip costs of a coach ticket for travel on March 30 and returning on April 6.

Five cities services by Southwest from Philadelphia were randomly selected. The cost data are:

SW USAIR DELTA AA CONT
HOU 218 204 212 215 203
LAX 178 178 231 231 191
KC 214 216 308 138 188
NASHVILLE 195 182 233 213 152
CHI 158 158 146 178 177

Next, five cities not serviced by Southwest were chosen to represent flights similar to the flights serviced by Southwest
For example, the inclusion of LAX (Los Angeles) resulted in the selection of SFO (San Francisco). The cities selected
here corresponded to a city in the first data set in terms of similar mileage (given below) and direction from Philadelphia.

USAIR DELTA AA CONT
DFW 218 339 218 261
SFO 284 468 295 300
MSP 178 356 189 424
MEMPHIS 191 258 185 265
MILWAULEE 187 188 188 148

LAS 2717 SFO 2878
HOU 1550 DFW 1469
CHI 759 MIL 857
NASH 804 MEMP 1016
K.C. 1129 MSP 1158

a) Test to see if Southwest has had an impact in lowering prices.
b) What other data might be important in testing the impact of Southwest? (The data 2 years ago before Southwest appeared would be helpful).

Â© BrainMass Inc. brainmass.com December 24, 2021, 5:13 pm ad1c9bdddf

## SOLUTION This solution is FREE courtesy of BrainMass!

See attached file for complete solution

1) For the problem, you may not make any assumptions concerning normality, etc

The following data represent the number of wins as of March 7, 2005 for the 30 NBA (National Basketball Association) teams

Eastern Conference Western Conference
Atlantic Division Southwest Division
Boston 31 San Antonio 46
Philly 29 Dallas 38
New Jersey 26 Houston 34
New York 25 Memphis 33
Toronto 25 New Orleans 13
Southeast Division Northwest Division
Miami 45 Seattle 41
Washington 33 Denver 30
Orlando 31 Minnesota 30
Charlotte 12 Portland 22
Atlanta 10 Utah 20
Central Division Pacific Division
Detroit 36 Phoenix 45
Cleveland 31 Sacramento 37
Chicago 29 Lakers 29
Indiana 29 Clippers 26
Milwaukee 24 Golden State 18

a) Sportscasters often state that the Western Conference is stronger than the Eastern Conference. Use the appropriate nonparametric rank test to see if this is true.

b) Similarly, test to see if the 8 teams with the most wins in the Western Division are stronger than the top 8 teams in the Eastern Division.

We will use Mann-Whitney U test for this problem

a) Sportscasters often state that the Western Conference is stronger than the Eastern Conference. Use the appropriate nonparametric rank test to see if this is true.

Writing the data together

Eastern Conference Western Conference

Miami 45 E San Antonio 46 W
Detroit 36 E Phoenix 45 W
Washington 33 E Seattle 41 W
Boston 31 E Dallas 38 W
Orlando 31 E Sacramento 37 W
Cleveland 31 E Houston 34 W
Philly 29 E Memphis 33 W
Chicago 29 E Denver 30 W
Indiana 29 E Minnesota 30 W
New Jersey 26 E Lakers 29 W
New York 25 E Clippers 26 W
Toronto 25 E Portland 22 W
Milwaukee 24 E Utah 20 W
Charlotte 12 E Golden State 18 W
Atlanta 10 E New Orleans 13 W

And On arranging all the data in ascending order and ranking the number of wins in order from lowest to highest

S No Team No of wins E/W Rank Modified rank after breaking the tie
1 Atlanta 10 E 1 1
2 Charlotte 12 E 2 2
3 New Orleans 13 W 3 3
4 Golden State 18 W 4 4
5 Utah 20 W 5 5
6 Portland 22 W 6 6
7 Milwaukee 24 E 7 7
8 New York 25 E tie 8 8.5
9 Toronto 25 E tie 8 8.5
10 New Jersey 26 E tie 10 10.5
11 Clippers 26 W tie 10 10.5
12 Philly 29 E tie 12 13.5
13 Chicago 29 E tie 12 13.5
14 Indiana 29 E tie 12 13.5
15 Lakers 29 W tie 12 13.5
16 Denver 30 W tie 16 16.5
17 Minnesota 30 W tie 16 16.5
18 Boston 31 E tie 18 19
19 Orlando 31 E tie 18 19
20 Cleveland 31 E tie 18 19
21 Washington 33 E tie 18 21.5
22 Memphis 33 W tie 18 21.5
23 Houston 34 W 23 23
24 Detroit 36 E 24 24
25 Sacramento 37 W 25 25
26 Dallas 38 W 26 26
27 Seattle 41 W 27 27
28 Miami 45 E tie 28 28.5
29 Phoenix 45 W tie 28 28.5
30 San Antonio 46 W 30 30

Tie is handled by calculating the average of the serial no

eg
Serial No Rank
8 8
9 8
Average =(8+9)/2= 8.5

Serial No Rank
12 12
13 12
14 12
15 12

Average =(12+13+14+15)/2= 13.5

Separating the ranks of Eastern Conference and Western Conference

Team E/W Modified rank after breaking the tie
Atlanta E 1 209
Charlotte E 2 256
Milwaukee E 7
New York E 8.5
Toronto E 8.5
New Jersey E 10.5
Philly E 13.5
Chicago E 13.5
Indiana E 13.5
Boston E 19
Orlando E 19
Cleveland E 19
Washington E 21.5
Detroit E 24
Miami E 28.5

New Orleans W 3
Golden State W 4
Utah W 5
Portland W 6
Clippers W 10.5
Lakers W 13.5
Denver W 16.5
Minnesota W 16.5
Memphis W 21.5
Houston W 23
Sacramento W 25
Dallas W 26
Seattle W 27
Phoenix W 28.5
San Antonio W 30

Eastern Conference
Team No of wins Modified rank breaking the tie
Atlanta 10 1
Charlotte 12 2
Milwaukee 24 7
New York 25 8.5
Toronto 25 8.5
New Jersey 26 10.5
Philly 29 13.5
Chicago 29 13.5
Indiana 29 13.5
Boston 31 19
Orlando 31 19
Cleveland 31 19
Washington 33 21.5
Detroit 36 24
Miami 45 28.5
Total ranks= 209
R1

Western Conference
Team No of wins Modified rank breaking the tie
New Orleans 13 3
Golden State 18 4
Utah 20 5
Portland 22 6
Clippers 26 10.5
Lakers 29 13.5
Denver 30 16.5
Minnesota 30 16.5
Memphis 33 21.5
Houston 34 23
Sacramento 37 25
Dallas 38 26
Seattle 41 27
Phoenix 45 28.5
San Antonio 46 30
Total ranks= 256
R2

n1=no of teams in Eastern Conference= 15
n2=no of teams in Western Conference= 15
R1=sum of ranks of Eastern Conference= 209
R2=sum of ranks of Western Conference= 256

Calculating the U statistic

U= n1*n2+1/2*(n2*(n2+1))-R2= 89 =15*15+ 0.5*15*(15+1)-256

Testing the hypothesis:

Null Hypothesis: Ho:Î¼1=Î¼2
There is no difference between the mean no of wins of Western Conference and Eastern Conference
Alternative Hypothesis: H1: Î¼1<Î¼2
Mean no of wins of Western Conference is greater than the mean no of wins of Eastern Conference
Since we are finding out whether one of the means is greater than the other
this is a two tailed test.
Let us test the hypothesis at a significance level of Î±= 0.05
No of tails= 1
From the tables:
n1= 15
n2= 15
U= 89
level of significance= 0.172842 P (one-tailed)
The two samples are not significantly different (P >= 0.05, one-tailed test).

Thus we would accept the null hypothesis
There is no difference between the mean no of wins of Western Conference and Eastern Conference

b) Similarly, test to see if the 8 teams with the most wins in the Western Division are stronger than the top 8 teams in the Eastern Division.

Writing the data together for the top 8 teams

Eastern Conference Western Conference

Miami 45 E San Antonio 46 W
Detroit 36 E Phoenix 45 W
Washington 33 E Seattle 41 W
Boston 31 E Dallas 38 W
Orlando 31 E Sacramento 37 W
Cleveland 31 E Houston 34 W
Philly 29 E Memphis 33 W
Chicago 29 E Denver 30 W

And On arranging all the data in ascending order and ranking the number of wins in order from lowest to highest

S No Team No of wins E/W Rank Modified rank after breaking the tie
1 Philly 29 E tie 1 1.5
2 Chicago 29 E tie 1 1.5
3 Denver 30 W 3 3
4 Boston 31 E tie 4 5
5 Orlando 31 E tie 4 5
6 Cleveland 31 E tie 4 5
7 Washington 33 E tie 7 7.5
8 Memphis 33 W tie 7 7.5
9 Houston 34 W 9 9
10 Detroit 36 E 10 10
11 Sacramento 37 W 11 11
12 Dallas 38 W 12 12
13 Seattle 41 W 13 13
14 Miami 45 E tie 14 14.5
15 Phoenix 45 W tie 14 14.5
16 San Antonio 46 W 16 16
136

Separating the ranks of Eastern Conference and Western Conference

Team E/W Modified rank after breaking the tie
Miami E 14.5 50
Detroit E 10 86
Washington E 7.5
Boston E 5
Orlando E 5
Cleveland E 5
Philly E 1.5
Chicago E 1.5

San Antonio W 16
Phoenix W 14.5
Seattle W 13
Dallas W 12
Sacramento W 11
Houston W 9
Memphis W 7.5
Denver W 3

Eastern Conference
Team No of wins Modified rank breaking the tie
Miami 45 14.5
Detroit 36 10
Washington 33 7.5
Boston 31 5
Orlando 31 5
Cleveland 31 5
Philly 29 1.5
Chicago 29 1.5
Total ranks= 50
R1

Western Conference
Team No of wins Modified rank breaking the tie
San Antonio 46 16
Phoenix 45 14.5
Seattle 41 13
Dallas 38 12
Sacramento 37 11
Houston 34 9
Memphis 33 7.5
Denver 30 3
Total ranks= 86
R2

n1=no of teams in Eastern Conference= 8
n2=no of teams in Western Conference= 8
R1=sum of ranks of Eastern Conference= 50
R2=sum of ranks of Western Conference= 86

Calculating the U statistic

U= n1*n2+1/2*(n2*(n2+1))-R2= 14 =8*8+ 0.5*8*(8+1)-86

Testing the hypothesis:

Null Hypothesis: Ho: Î¼1= Î¼2
There is no difference between the mean no of wins of 8 top teams of Western Conference and Eastern Conference
Alternative Hypothesis: H1: Î¼2> Î¼1
Mean no of wins of 8 top teams of Western Conference is greater than the mean no of wins of 8 top teams of Eastern Conference

Since we are finding out whether one of the means is greater than the other
this is a two tailed test.
Let us test the hypothesis at a significance level of &#945;= 0.05
No of tails= 1

From the tables
n1= 8
n2= 8
U= 14
level of significance= 0.0324786 P (one-tailed)
The two samples are significantly different (P <= 0.05, one-tailed test).

Thus we would reject the null hypothesis
There is difference between the mean no of wins of top 8 teams of Western Conference and Eastern Conference

2) The following data are given for 20 automobiles.

MPG Horsepower Weight Car's Age Teenage Driver Teenage Driver
17 240 4485 2 Yes 1
21 225 3860 4 No 0
26 160 2780 1 No 0
22 225 3520 5 No 0
20 170 3745 4 No 0
18 220 3915 4 No 0
22 184 3390 1 No 0
19 184 3575 2 No 0
14 225 4715 3 Yes 1
20 205 3640 2 No 0
21 205 3880 1 No 0
15 185 4230 6 Yes 1
19 275 4070 4 No 0
20 180 3495 2 No 0
22 170 3050 3 No 0
12 285 5590 3 Yes 1
20 127 3055 1 Yes 1
14 270 4660 1 Yes 1
19 185 3990 2 No 0
21 250 3620 1 No 0

a) Develop the regression for this data that attempts to relate the mileage to the other variables
b) Which coefficients are significant?
c) Predict the mileage for a 210 horsepower car weighing 4000 pounds which is 3 years old and driven by a teenager.
d) How much of the variation in mileage is due to the independent variables?

Since teenage driver is a dummy variable we will take Yes=1 and No=0

a) Develop the regression for this data that attempts to relate the mileage to the other variables

The regresion equation is Y= 34.028 + 0.0126 X1 -0.0043 X2 -0.1474 X3 -1.9762 X4
(see calculations below for details)
where
Y= MPG
X1= Horsepower
X2= Weight
X3= Car's Age
X4= Teenage Driver

b) Which coefficients are significant?

We have used a significance level &#945;= 0.05
See step 5 below for Testing whether the coefficients are different from zero

The results are reprodued below

Abs(bi/Sbi) t P value
b1= 0.0126 1.947685531 2.1314 0.070419466 b1=is zero
b2= -0.0043 10.43964786 2.1314 2.82872E-08 b2=is non zero
b3= -0.1474 0.822795213 2.1314 0.423512456 b3=is zero
b4= -1.9762 3.455635513 2.1314 0.003530772 b4=is non zero

Thus the coefficients that are significant are b2 and b4

This means that variables
Weight and Teenage Driver are significant

c) Predict the mileage for a 210 horsepower car weighing 4000 pounds which is 3 years old and driven by a teenager.

Y= 34.028 + 0.0126 X1 -0.0043 X2 -0.1474 X3 -1.9762 X4

a= 34.028
b1= 0.0126 X1= 210
b2= -0.0043 X2= 4000
b3= -0.1474 X3= 3
b4= -1.9762 X4= 1

Therefore Y= 17.06 MPG =34.028+0.0126*210-0.0043*4000-0.1474*3-1.9762*1

d) How much of the variation in mileage is due to the independent variables?

Step 2 gives the value of
Coefficient of determination r 2 = 0.9028

Thus 90.28% of the variation in mileage is due to the independent variables

(Calculations for the answers above)

Four independent variables k= 4 significance level &#945;= 0.05

Determine regression equation

Y=a+b1 X1+b2X2+b3X3+b4X4= 34.028 + 0.0126 X1 -0.0043 X2 -0.1474 X3 -1.9762 X4

First the regression equation is calculated ( see below ) then Y predicted is calculated and the R2, standard error values are calculated.

"SS error=
SSE" "SS regression=
SSR" "SS total=
SST"
Y X1 X2 X3 X4 Y pred (Y-pred Y)2 (Y pred - Y average)2 (Y-av Y)2 X1Y X2Y X3Y X4Y X1X2 X1X3 X1X4 X2X3 X2X4 X3X4 X12 X22 X32 X42
MPG Horsepower Weight Car's Age Teenage Driver
17 240 4485 2 1 15.5 2.25 12.96 4.41 4080 76245 34 17 1076400 480 240 8970 4485 2 57600 20115225 4 1
21 225 3860 4 0 19.68 1.7424 0.3364 3.61 4725 81060 84 0 868500 900 0 15440 0 0 50625 14899600 16 0
26 160 2780 1 0 23.94 4.2436 23.4256 47.61 4160 72280 26 0 444800 160 0 2780 0 0 25600 7728400 1 0
22 225 3520 5 0 20.99 1.0201 3.5721 8.41 4950 77440 110 0 792000 1125 0 17600 0 0 50625 12390400 25 0
20 170 3745 4 0 19.48 0.2704 0.1444 0.81 3400 74900 80 0 636650 680 0 14980 0 0 28900 14025025 16 0
18 220 3915 4 0 19.38 1.9044 0.0784 1.21 3960 70470 72 0 861300 880 0 15660 0 0 48400 15327225 16 0
22 184 3390 1 0 21.62 0.1444 6.3504 8.41 4048 74580 22 0 623760 184 0 3390 0 0 33856 11492100 1 0
19 184 3575 2 0 20.68 2.8224 2.4964 0.01 3496 67925 38 0 657800 368 0 7150 0 0 33856 12780625 4 0
14 225 4715 3 1 14.17 0.0289 24.3049 26.01 3150 66010 42 14 1060875 675 225 14145 4715 3 50625 22231225 9 1
20 205 3640 2 0 20.66 0.4356 2.4336 0.81 4100 72800 40 0 746200 410 0 7280 0 0 42025 13249600 4 0
21 205 3880 1 0 19.78 1.4884 0.4624 3.61 4305 81480 21 0 795400 205 0 3880 0 0 42025 15054400 1 0
15 185 4230 6 1 15.31 0.0961 14.3641 16.81 2775 63450 90 15 782550 1110 185 25380 4230 6 34225 17892900 36 1
19 275 4070 4 0 19.4 0.16 0.09 0.01 5225 77330 76 0 1119250 1100 0 16280 0 0 75625 16564900 16 0
20 180 3495 2 0 20.97 0.9409 3.4969 0.81 3600 69900 40 0 629100 360 0 6990 0 0 32400 12215025 4 0
22 170 3050 3 0 22.61 0.3721 12.3201 8.41 3740 67100 66 0 518500 510 0 9150 0 0 28900 9302500 9 0
12 285 5590 3 1 11.16 0.7056 63.0436 50.41 3420 67080 36 12 1593150 855 285 16770 5590 3 81225 31248100 9 1
20 127 3055 1 1 20.37 0.1369 1.6129 0.81 2540 61100 20 20 387985 127 127 3055 3055 1 16129 9333025 1 1
14 270 4660 1 1 15.27 1.6129 14.6689 26.01 3780 65240 14 14 1258200 270 270 4660 4660 1 72900 21715600 1 1
19 185 3990 2 0 18.91 0.0081 0.0361 0.01 3515 75810 38 0 738150 370 0 7980 0 0 34225 15920100 4 0
21 250 3620 1 0 21.46 0.2116 5.5696 3.61 5250 76020 21 0 905000 250 0 3620 0 0 62500 13104400 1 0

&#931;= 382 4170 77265 52 6 20.5948 191.7668 211.8 78219 1438220 970 92 16495570 11019 1332 205160 26735 16 902266 306590375 178 6
Mean 19.1 208.5 3863.25 2.6 0.3
Mean Y Mean X1 Mean X2 Mean X3 Mean X4 "SS error=
SSE" "SS regression=
SSR" "SS total=
SST" X12 X22 X32 X42

n= 20
k= 4
n-k-1= 15 degrees of freedom

Step 1 Calculation of regression coefficients (b1 , b2, ----- , bp are the regression coefficients (or B coefficients) a is the intercept)
using the equations

ÏƒY=na+b1ÏƒX1+b2ÏƒX2+b3ÏƒX3+b4ÏƒX4
382 = 20 a + 4170 b1 + 77265 b2 + 52 b3 + 6 b4

ÏƒX1Y=aÏƒX1+b1ÏƒX12+b2ÏƒX1X2+b3ÏƒX1X3+b4ÏƒX1X4
78219 = 4170 a + 902266 b1 + 16495570 b2 + 11019 b3 + 1332 b4

ÏƒX2Y=aÏƒX2+b1ÏƒX1X2+b2ÏƒX22+b3ÏƒX2X3+b4ÏƒX2X4
1438220 = 77265 a + 16495570 b1 + 306590375 b2 + 205160 b3 + 26735 b4

ÏƒX3Y=aÏƒX3+b1ÏƒX1X3+b2ÏƒX2X3+b3ÏƒX32+b4ÏƒX3X4
970 = 52 a + 11019 b1 + 205160 b2 + 178 b3 + 16 b4

ÏƒX4Y=aÏƒX4+b1ÏƒX1X4+b2ÏƒX2X4+b3ÏƒX3X4+b4ÏƒX42
92 = 6 a + 1332 b1 + 26735 b2 + 16 b3 + 6 b4

Use matrix and inverse of matrix to calculate the coefficients

20 4170 77265 52 6 a = 382
4170 902266 16495570 11019 1332 b1 = 78219
77265 16495570 306590375 205160 26735 b2 = 1438220
52 11019 205160 178 16 b3 = 970
6 1332 26735 16 6 b4 = 92

Inverse of the matrix

2.679706864 0.002014628 -0.000842481 0.009365339 0.602026202 382 a
0.002014628 8.82882E-05 -5.63717E-06 0.000168879 0.003053351 X 78219 = b1
-0.000842481 -5.63717E-06 5.71579E-07 -3.02459E-05 -0.000372271 1438220 b2
0.009365339 0.000168879 -3.02459E-05 0.025498156 0.019919269 970 b3
0.602026202 0.003053351 -0.000372271 0.019919269 0.49245751 92 b4

Solving

a= 34.0280
b1= 0.0126
b2= -0.0043
b3= -0.1474
b4= -1.9762

Hence regression equation= 34.028 + 0.0126 X1 -0.0043 X2 -0.1474 X3 -1.9762 X4

Step 2 Coefficient of determination r 2

The formula for R squared is

SS error= Ïƒ(Y-Y predicted)2
SS total= Ïƒ(Y-Y average)2

r2=1-(Ïƒ(Y-Y predicted)2)/(Y- Y average)2= 1-(20.5948 / 211.8 )= 0.9028

Step 3 Adjusted r squared

adjusted R squared. This statistic adjusts for the degrees of freedom in the model, and penalizes an unnecessarily complex linear model. The formula is

=1-((20.5948 / 15)/(211.8/19))= 0.876833113

ADJUSTED R-SQUARED is also computed as

adjusted R squared= =1-(1-0.9028)*(19/ 15) 0.87688

Step 3 standard error of estimate Se

standard error= Se=âˆš{Ïƒ(Y-predicted Y)2/(n-k-1)}= =sqrt( 20.5948 / 15 )= 1.172

Step 4 standard error of regression coefficient= Sb1, Sb2,Sb3 ,Sb4

standard error of regression coefficient= Sb1=Se /âˆš{Ïƒ(X12)-n*(Mean X1)2}
standard error of regression coefficient= Sb2=Se /âˆš{Ïƒ(X22)-n*(Mean X2)2}
standard error of regression coefficient= Sb3=Se /âˆš{Ïƒ(X32)-n*(Mean X3)2}
standard error of regression coefficient= Sb4=Se /âˆš{Ïƒ(X42)-n*(Mean X4)2}

Sb1 = =1.172/ square root of (902266-20*208.5^2)= 0.006469217
Sb2 = =1.172/ square root of (306590375-20*3863.25^2)= 0.000411891
Sb3 = =1.172/ square root of (178-20*2.6^2)= 0.179145427
Sb4 = =1.172/ square root of (6-20*0.3^2)= 0.571877443

Step 5 Testing whether the coefficients are different from zero

Significance level 0.05
degrees of freedom 15
t value= 2.1314 corresponding to degrees of freedom and significance level

Null Hypothesis: b i s are zero
if abs(bi/sbi)<t null hypothesis is true
Abs(bi/Sbi) t P value
Sb1 = 0.006469217 b1= 0.0126 1.947685531 2.1314 7.04% b1=is zero
Sb2 = 0.000411891 b2= -0.0043 10.43964786 2.1314 0.00% b2=is non zero
Sb3 = 0.179145427 b3= -0.1474 0.822795213 2.1314 42.35% b3=is zero
Sb4 = 0.571877443 b4= -1.9762 3.455635513 2.1314 0.35% b4=is non zero

Step 6 F test on the regression as a whole

F=((SSR/k)/(SSE/(n-k-1)) Significance F=P value
F= =((191.7668/4)/(20.5948/15)= 34.91781906 0.00%

F value for F for significance level=&#945;= 0.05 is 3.0556

Since Computed F value= 34.91781906 is greater than the critical value = 3.0556
and P value is close to zero
The regression as a whole is significant

3) I In the analysis for this problem, you may make necessary assumptions such as normality, etc. It has been stated that the arrival of Southwest Airlines in Philadelphia has had an impact in lowering the prices of airline tickets in Philadelphia. To test this assumption, the following data was obtained. These data are the round-trip costs of a coach ticket for travel on March 30 and returning on April 6.

Five cities services by Southwest from Philadelphia were randomly selected. The cost data are:

SW USAIR DELTA AA CONT
HOU 218 204 212 215 203
LAX 178 178 231 231 191
KC 214 216 308 138 188
NASHVILLE 195 182 233 213 152
CHI 158 158 146 178 177

Next, five cities not serviced by Southwest were chosen to represent flights similar to the flights serviced by Southwest.
For example, the inclusion of LAX (Los Angeles) resulted in the selection of SFO (San Francisco).The cities selected here corresponded to a city in the first data set in terms of similar mileage (given below) and direction from Philadelphia.

USAIR DELTA AA CONT
DFW 218 339 218 261
SFO 284 468 295 300
MSP 178 356 189 424
MEMPHIS 191 258 185 265
MILWAULEE 187 188 188 148

LAS 2717 SFO 2878
HOU 1550 DFW 1469
CHI 759 MIL 857
NASH 804 MEMP 1016
K.C. 1129 MSP 1158

a) Test to see if Southwest has had an impact in lowering prices.
b) What other data might be important in testing the impact of Southwest? (The data 2 years ago before Southwest appeared would be helpful)

a) Test to see if Southwest has had an impact in lowering prices.

CHI 759 MIL 857
NASH 804 MEMP 1016
K.C. 1129 MSP 1158
HOU 1550 DFW 1469
LAS 2717 SFO 2878

Thus distance wise

DFW is equivalent to HOU
SFO is equivalent to LAS
MSP is equivalent to KC
MEMPHIS No equivalence
MILWAULEE is equivalent to NASH

Therefore we will compare the fares of

DFW and HOU
SFO and LAS
MSP and KC
MILWAULEE and NASH

to see whether there is any difference between the fares where Southwest Airlines operates and where it does not

Average fare where Southwest does not operate
USAIR DELTA AA CONT Average fare Distance
DFW 218 339 218 261 259.00 1469
SFO 284 468 295 300 336.75 2878
MSP 178 356 189 424 286.75 1158
MILWAULEE 187 188 188 148 177.75 857

Average fare where Southwest operates
SW USAIR DELTA AA CONT Average fare Distance
HOU 218 204 212 215 203 210.40 1550
LAX 178 178 231 231 191 201.80 2717
KC 214 216 308 138 188 212.80 1129
NASHVILLE 195 182 233 213 152 195.00 804

This is a matched pair design
Testing differences between means with dependent samples
At significance level= 0.05

Average fare where South west airlines does not operate
Average fare where South west airlines operates "Difference
"

259.00 210.40 48.6
336.75 201.80 134.95
286.75 212.80 73.95
177.75 195.00 -17.25

Î¼= 60.0625 =Mean of difference
Ïƒ= 63.0069 =Standard deviation of difference
n= 4 =sample size

Ïƒ;x=standard error of mean=Ïƒ/&#8730;n=
=63.0069/ sqrt ( 4 ) 31.5035

Null Hypothesis: Ho:Î¼1=Î¼2 There is no difference between the means
Alternative Hypothesis: H1: Î¼1>Î¼2 There is difference between the means

degrees of freedom=n-1= 3
Significance level= 0.05

t-value= 2.35336 corresponding to level of significance= 0.05 and degrees of freedom= 3

Î¼H0 = Hypothesized difference of mean= 0
Upper limit of acceptance region=&#956;H0+t*Ïƒx= 74.1391 = 0 + 2.35336 * 31.5035

Since the mean of difference = 60.06 is within the acceptance region = 74.1391
Accept null hypothesis; The difference of means is equal to 0

Thus at a significance level of 0.05 or 5% we cannot conclude that Southwest Airlines has lowered the price

b) What other data might be important in testing the impact of Southwest? (The data 2 years ago before Southwest appeared would be helpful)

Other data could be the number of airlines on different routes (ie the level of competition).
A more competitive route would see more competition and hence lowering of prices.

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