1G118 STAT 202 Activity # 3 Analysis of Variance: 1- Given the data below, test
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Question
1G118 STAT 202 Activity # 3 Analysis of Variance: 1- Given the data below, test the claim that the four samples come from the same pop with the same mean at the significance level of ulation 1%. Injuries to Car Crash Test Dummies- LEFT FEMUR LOAD (lb) Subcompact Cars: Ford Escort Honda Clvic Hyundai Accent Nissan Sentra Saturn SL4 595 1063 885 519 422 Compact Cars: Chevrolet Cavalier Dodge Neon Mazda 626 DX Pontiac Sunfire Subaru Legacy 1051 1193 946 984 584 Midsize Cars: Chevrolet Camaro Dodge Intrepid Ford Mustang Honda Accord Volvo S70 629 1686 880 181 645 Full-size Cars: Audi A8 Cadillac Deville Ford Crown Victoria Oldsmobile Aurora Pontiac Bonneville 1085 971 996 804 1376Explanation / Answer
Solution
This is a direct application of One-way ANOVA (4 groups) with equal (5) number of observations per group.
Let xij represent the observation on the jth car in the ith row (group – type of car),
j = 1,2,…,n; i = 1,2,……,r
Then the ANOVA model is: xij = µ + ?i + ?ij, where µ = common effect, ?i = effect of ith row, and ?ij is the error component which is assumed to be Normally Distributed with mean 0 and variance ?2.
Null hypothesis: H0: ?1 = ?2 = ?3 = ?4 = 0 [Claim that the four groups come from the same population with the same mean]
Vs
Alternative: HA: H0 is false [at least one ?i is different from others]
NOTE
Final ANOVA TABLE and Conclusions are given below. Details of Excel Calculations and Back-up Theory follow at the end.
ANOVA
? =
0.01
Source
DF
SS
MS
F
Fcrit
p-value
Row
3
343275.8
114425.3
0.96193
5.29221
0.43468
Error
16
1903262
118953.9
Total
19
2246538
Since F < Fcrit, [also p-value > ?] null hypothesis is accepted. =>
Conclusion: There is sufficient evidence to suggest that the claim that the four samples come from the same population is valid. ANSWER.
Details of Excel Calculations
Data
Car Type
Car1
Car2
Car3
Car4
Car5
1.Sub-compact
595
1063
885
519
422
2.Compact
1051
1193
946
984
584
3.Mid-size
629
1686
880
181
645
4.Full Size
1085
971
946
804
1376
Calculations
i
xi.
sumxij^2
1
3484
2714664
G
17445
2
4758
4732078
C
2E+07
3
4021
4461423
SST
2E+06
4
5182
5554774
SSR
343276
SSE
2E+06
Back-up Theory
Suppose we have data of a 1-way classification ANOVA, with r rows/treatments/groups and n observations per row/treatment/group.
Let xij represent the jth observation in the ith row, j = 1,2,…,n; i = 1,2,……,r
Then the ANOVA model is: xij = µ + ?i + ?ij, where µ = common effect, ?i = effect of ith row, and ?ij is the error component which is assumed to be Normally Distributed with mean 0 and variance ?2.
Null hypothesis: H0: ?1 = ?2 = ?3 = 0 Vs Alternative: HA: H0 is false [at least one ?i is different from others]
Now, to work out the solution,
Terminology:
Row total = xi.= sum over j of xij
Grand total = G = sum over i of xi.
Correction Factor = C = G2/N, where N = total number of observations = r x n =
Total Sum of Squares: SST = (sum over i,j of xij2) – C
Row Sum of Squares: SSR = {(sum over i of xi.2)/(n)} – C
Error Sum of Squares: SSE = SST – SSR
Mean Sum of Squares = Sum of squares/Degrees of Freedom
Fcrit: upper ?% point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for Row and n2 is the DF for Error
Significance: Fobs is significant if Fobs > Fcrit
ANOVA TABLE
Source of
Variation
Degrees of Freedom (DF)
Sum of squares (SS)
Mean Sum
of squares
(MS = SS/DF)
Fobs
Fcrit*
Significance**
Row
r - 1
SSR
MSR/MSE
Error
rn - r
SSE
Total
rn - 1
SST
NOTE:
* Fcrit: upper ?% point of F-Distribution with degrees of freedom n1 and n2, where n1
is the DF for Row and n2 is the DF for Error
** Significance: Fobs is significant if Fobs > Fcrit
DONE
ANOVA
? =
0.01
Source
DF
SS
MS
F
Fcrit
p-value
Row
3
343275.8
114425.3
0.96193
5.29221
0.43468
Error
16
1903262
118953.9
Total
19
2246538
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