statistics for managers using microsoft excel/spss · 10/02/2016 · anova and other c-sample tests...

© 1999 Prentice-Hall, Inc. Chap. 10 - 1

Statistics for Managers

Using Microsoft Excel/SPSS

Chapter 10

ANOVA and Other C-Sample

Tests With Numerical Data


• The Completely Randomized Model: One-Factor Analysis of Variance

F-Test for Difference in c Means

The Tukey-Kramer Procedure

ANOVA Assumptions

• The Factorial Design Model: Two-Way Analysis of Variance

Examine Effect of Factors and Interaction

• Kruksal-Wallis Rank Test for Differences in c

Medians

Chapter Topics


One-Factor

Analysis of Variance

• Evaluate the Difference Among the

Means of 2 or More (c) Populations

e.g., Several Types of Tires, Oven Temperature Settings

• Assumptions:

Samples are Randomly and Independently Drawn

(This condition must be met.)

Populations are Normally Distributed

(F test is Robust to moderate departures from

normality.)

Populations have Equal Variances


One-Factor ANOVA Test Hypothesis

H0: m1 = m2 = m3 = ... = mc

•All population means are equal

•No treatment effect (NO variation in means

among groups)

H1: not all the mk are equal

•At least ONE population mean is different

(Others may be the same!)

•There is treatment effect

Does NOT mean that all the means are different:

m1 m2 ... mc


One-Factor ANOVA: No Treatment Effect

m1 = m2 = m3

H0: m1 = m2 = m3 = ... = mc

H1: not all the mk are equal

The Null

Hypothesis is

True


One Factor ANOVA: Treatment Effect Present

m1 = m2 m3

H0: m1 = m2 = m3 = ... = mc

H1: not all the mk are equal The Null

Hypothesis is

NOT True


One-Factor ANOVA

Partitions of Total Variation

Variation Due to

Treatment SSA

Variation Due to Random

Sampling SSW

Total Variation SST

Commonly referred to as:

Sum of Squares Within, or

Sum of Squares Error, or

Within Groups Variation

Commonly referred to as:

Sum of Squares Among, or

Sum of Squares Between, or

Sum of Squares Model, or

Among Groups Variation

= +


Total Variation

2

1 1

)XX(SSTc

j

n

iij

j

== =

n

X

X

c

j

n

iij

j

== =1 1 the overall or grand mean

Xij = the ith observation in group j

nj = the number of observations in group j

n = the total number of observations in all groups

c = the number of groups


Among-Group Variation

2

1

)XX(nSSA j

c

jj =

=

nj = the number of observations in group j

c = the number of groups

the sample mean of group j

the overall or grand mean

mi mj

Variation Due to Differences Among Groups.

1=

c

SSAMSA

Xj

X

_

_ _


Within-Group Variation

2

1 1

)XX(SSW j

c

j

n

iij

j

== =

=ijX the ith observation in group j

=jX the sample mean of group j

m j

Summing the variation within

each group and then adding

over all groups.

cn

SSWMSW

=


Within-Group Variation

m j

)n()n()n(

S)n(S)n(S)n(

cn

SSWMSW

c

cc

111

111

21

2222

211

=

=

For c = 2, this is the

pooled-variance in the

t-Test.

•If more than 2 groups,

use F Test.

•For 2 groups, use t-Test.

F Test more limited.


One-Way ANOVA

Summary Table

Source of Variation

Degrees of

Freedom

Sum of Squares

Mean Square

(Variance)

Among (Factor)

c - 1 SSA MSA = SSA/(c - 1)

MSA

MSW

Within (Error)

n - c SSW MSW = SSW/(n - c)

Total n - 1 SST = SSA+SSW

F Test

Statistic

=


One-Factor ANOVA

F Test Example

As production manager, you

want to see if 3 filling

machines have different mean

filling times. You assign 15

similarly trained &

experienced workers, 5 per

machine, to the machines. At

the .05 level, is there a

difference in mean filling

times?

Machine1 Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40


One-Factor ANOVA

Example: Scatter Diagram

27

26

25

24

23

22

21

20

19

• •

• •

•

• •

•

• •

• • • •

•

X

X

x

x

X = 24.93 X = 22.61 X = 20.59

X = 22.71

Machine1 Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

_ _

_ _

_ _ _

_

_

_


One-Factor ANOVA

Example Computations

X1 = 24.93

X2 = 22.61

X3 = 20.59

X = 22.71

SSA = 5 [(24.93 - 22.71) 2+ (22.61 - 22.71)2 + (20.59 - 22.71) 2]

= 47.164

SSW = 4.2592+3.112 +3.682 = 11.0532

MSA = SSA/(c-1) = 47.16/2 = 23.5820

MSW = SSW/(n-c) = 11.0532/12 = .9211

nj =5

c = 3

n = 15

Machine1 Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

_

_

_

_

_


Summary Table

Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

(Variance)

F =

= 25.60

Among (Machines)

3 - 1 = 2 47.1640 23.5820

Within (Error)

15 - 3 = 12 11.0532 .9211

Total 15 - 1 = 14 58.2172

MSA

MSW


F 0 3.89

One-Factor ANOVA

Example Solution H0: m1 = m2 = m3

H1: Not All Equal

a = .05

df1= 2 df2 = 12

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Reject at a = 0.05

There is evidence that at least

one m i differs from the rest.

a = 0.05

F MSA

MSW = = =

23 5820

9211 25 6

.

. .



• Tells Which Population Means

Are Significantly Different

e.g., m1 = m2 m3

• Post Hoc (a posteriori)

Procedure

Done after rejection

of equal means in

ANOVA

• Ability for Pairwise Comparisons:

Compare absolute mean differences with

‘critical range’

X

f(X)

m 1 = m

2 m

3

2 groups whose means may be

significantly different.


The Tukey-Kramer Procedure:

Example 1. Compute absolute mean

differences:

Machine1 Machine2 Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40 02259206122

34459209324

32261229324

32

31

21

...XX

...XX

...XX

==

==

==

2. Compute Critical Range:

3. Each of the absolute mean difference is greater. There is a

significance difference between each pair of means.

=

'jj

)cn,c(Unn

MSWQRangeCritical

11

2= 1.618


Two-Way ANOVA

• Examines the Effect of:

Two Factors on the Dependent Variable

e.g., Percent Carbonation and Line Speed on

Soft Drink Bottling Process

Interaction Between the Different Levels of

these Two Factors

e.g., Does the effect of one particular

percentage of Carbonation depend on

which level the line speed is set?


Two-Way ANOVA

Assumptions

• Normality

Populations are normally distributed

• Homogeneity of Variance

Populations have equal variances

• Independence of Errors

Independent random samples are drawn


Two-Way ANOVA

Total Variation Partitioning

Variation Due to

Treatment A

Variation Due to

Random Sampling

Variation Due to

Interaction

SSE

SSFA +

SSAB + SST =

Variation Due to

Treatment B SSFB +

Total Variation


Two Way ANOVA: The F Test Statistic

F Test for Factor A Effect

MSFA MSE

F =

F Test for Factor B Effect

F = MSFB MSE

F Test for Interaction Effect

F = MSFAB MSE

H0: m1 .= m2 . = ••• = mr .

H1: Not all mi . are equal

H0: ABij = 0 (for all i and j)

H1: ABij 0

H0: m. 1 = m. 2 = ••• = m. c

H1: Not all mi are equal

Reject if

F > FU

Reject if

F > FU

Reject if

F > FU


Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square F Statistic:

A (Row)

r - 1 SSFA MSFA MSFA

MSE

B (Column)

c - 1 SSFB MSFB MSFB

MSE

AB (Interaction)

(r-1)(c-1) SSAB MSAB MSAB

MSE

Error r·c·(n’-1) SSE MSE

Total r·c·n’ - 1 SST

Two-Way ANOVA

Summary Table

=

=

=


Kruskal-Wallis Rank

Test for c Medians

• Extension of Wilcoxon Rank Sum Test

Tests the equality of more than 2 (c)

population medians

• Distribution-free test procedure

• Used to analyze completely randomized

experimental designs

• Use c2 distribution to approximate if each

sample group size nj > 5, df = c - 1


Kruskal-Wallis Rank

Test • Assumptions:

Independent random samples are drawn

Continuous dependent variable

Data may be ranked both within and among samples

Populations have same variability

Populations have same shape

• Robust with regard to last 2 conditions

Use F Test in completely randomized designs and when the more stringent assumptions hold.


Kruskal-Wallis Rank

Test Procedure •Obtain Rank combined data values. In event of tie, each of the tied values gets their average rank.

•Add the ranks for data from each of the c groups.

Square to obtain Tj2

n = n1 + n2 + … + nc nj = # of observations

in jth sample

•Compute Test

Statistic: )n(n

T

)n(nH

c

j j

j13

1

12

1

2

=

=


Kruskal-Wallis Rank

Test Procedure

•Test Statistic, H may be approximated by Chi-

square distribution (df = c -1)

•Critical Value for a given a: Upper tail

•Decision Rule: Reject H0: M1 = M2 = ••• = Mc

if Test Statistic H >

otherwise do not reject H0

2cU

2cU


As production manager, you

want to see if 3 filling

machines have different

median filling times. You

assign 15 similarly trained &

experienced workers,

5 per machine, to the machines.

At the .05 level, is there a

difference in median filling

times?

Kruksal-Wallis Rank Test:

Example

Machine1 Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40


Example Solution: Step 1

Obtaining a Ranking

Raw Data Ranks

65 38 17

Machine1 Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

Machine1 Machine2

Machine3

14 9

2 15 6

7

12 10 1

11 8 4

13 5 3


Example Solution: Step 2

Test Statistic Computation

)n(c

j jn

jT

)n(nH 13

1

2

112

=

=

)()(

11535

17

5

38

5

65

11515

12222

=

= 11.58


H0: M1 = M2 = M3

H1: Not all equal

a = .05

df = c - 1 = 3 - 1 = 2

Critical Value(s):

Reject at

c 20 5.991 c 20 5.991

H = 1158.H = 1158.

Kruskal-Wallis Test

Example Solution

Test Statistic:

Decision:

Conclusion:

There is evidence that

population medians are

not all equal.

a = .05

a = .05


• Described The Completely Randomized Model: One-Factor Analysis of Variance

F-Test for Difference in c Means


ANOVA Assumptions

• Discussed The Factorial Design Model: Two-Way Analysis of Variance

Examined effect of Factors and Interaction

• Addressed the Kruskal-Wallis Rank Test for Differences in c Medians

Chapter Summary

statistics for managers using microsoft excel/spss · 10/02/2016 · anova and other c-sample tests...

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