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8/8/2019 Anova - Copy - Copy

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Quantitative Methods

Varsha Varde


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Example. Fourgroups of sales people for a magazine sales

agency were subjected to different sales trainingprograms. Because there were some dropouts during

the training program, the number of trainees varied

from program to program. At the end of the training

programs each salesperson was assigned a salesarea from a group of sales areas that were judged to

have equivalent sales potentials. The table below lists

the number of sales made by each person in each of

the four groups of sales people during the first weekafter completing the training program. Do the data

present sufficient evidence to indicate a difference in

the mean achievement for the four training programs?

Varsha Varde3


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Data

Training Group

1 2 3 465 75 59 94

87 69 78 89

73 83 67 80

79 81 62 88

81 72 83

69 79 76

90

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Definitions

1 be the mean achievement for the first group

2

be the mean achievement for the second group

3 be the mean achievement for the third group

4 be the mean achievement for the fourth group

X1 be the sample mean for the first group

X2 be the sample mean for the second group X3 be the sample mean for the third group

X4 be the sample mean for the fourth group

S21 be the sample variance for the first group S22 be the sample variance for the second group

S23 be the sample variance for the third group

S2

4 be the sample variance for the fourth groupVarsha Varde 5


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Hypothesis To be Tested

Test whether the means are equal or

not. That is

H0 : 1 = 2 = 3 = 4 Ha : Not all means are equal

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Data Training Group

1 2 3 4

65 75 59 94

87 69 78 89

73 83 67 80

79 81 62 88

81 72 83

69 79 76

90

n1 = 6 n2 = 7 n3 = 6 n4 = 4 n = 23

Ti 454 549 425 351 GrandTotal= 1779

Xi 75.67 78.43 70.83 87.75 S2i S21 S22 S23 S24

x= = 1779/23 =77.3478

Varsha Varde

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One Way ANOVA: Completely Randomized Experimental Design

ANOVA Table

Source of error df SS MS F p-value

Treatments 3 712.6 237.5 3.77

Error 19 1,196. 6 63.0

Totals 22 1 ,909.2

Inferences about population means

Test.

H0: 1 = 2 = 3 = 4

Ha : Not all means are equal

T.S. : F = MST/MSE = 3.77

where F is based on (k-1) and (n-k) df. RR: Reject H0 if F > F,k-1,n-k i.e. Reject H0 if F > F0.05,3,19 = 3.13

Conclusion: At 5% significance level there is sufficient statistical

evidence to indicate a difference in the mean achievement for the

four training programs. 8


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One Way ANOVA: Completely Randomized Experimental Design

Assumptions.

1. Sampled populations are normal 2. Independent random samples

3. All k populations have equal variances

Computations.

ANOVA Table Sources of error df SS MS F

Trments k-1 SST MST=SST/(k-1) MST/MSE

Error n-k SSE MSE=SSE/(n-k)

Totals n-1 TSS

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(i) Response Variable: variable of interest

or dependent variable (sales)

(ii) Factor or Treatment: categoricalvariable or independent variable (training

technique)

(iii) Treatment levels (factor levels):

methods of training( Number of groups or

populations) k =4

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Confidence Intervals.

Estimate of the common variance:

s = s2 = MSE = SSE/n-t CI fori:

Xi t/2,n-k(s/ni)

CI fori - j: (Xi- Xj) t/2,n-k { s2(1/ni+1/nj)}

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Training Group

1 2 3 4

x11 x21 x31 x41

x12 x22 x32 x42

x13 x23 x33 x43

x14 x24 x34 x44

x15 x25 x35

x16 x26 x36

x27

n1 n2 n3 n4 n

Ti T1 T2 T3 T4 GT

Xi X1 X2 X3 4

parameter 1 2 3 4 12


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FORMULAE Notation:

TSS: sum of squares of total deviation.

SST: sum of squares of total deviation between treatments. SSE: sum of squares of total deviation within treatments (error).

CM: correction for the mean

GT: Grand Total.

Computational Formulas for TSS, SST and SSE:

k ni CM = { xij } 2/n

i=1 j=1

k ni

TSS = x

ij2

- CM i=1 j=1

k

SST = Ti2/ni - CM

i=1

SSE = TSS - SST 13


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Calculations for the training example produce

k ni CM = { xij } 2/ n = 1, 7792/23 = 137, 601.8

i=1 j=1

k ni TSS = xij2 - CM = 1, 909.2

i=1 j=1

k

SST = Ti2/ni - CM = 712.6 i=1

SSE = TSS SST =1, 196.6

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ANOVA Table

Source of error df SS MS F p-value

Treatments 3 712.6 237.5 3.77

Error 19 1,196.6 63.0

Totals 22 1,909.2

Confidence Intervals.

Estimate of the common variance: s = s2 = MSE = SSE/n-t

CI fori:

Xi t/2,n-k (s/ni)

CI fori - j:

(Xi- Xj) t /2,n-k { s2(1/ni+1/nj)}

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