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Chapter 7
Analysis of ariance
Variation
Data without dispersion
information is false data.
— Kaoru Ishikawa
• Inherent or Natural Variation
Due to
the cumulative effect of many small
unavoidable causes. Also referred to as
noise
• Special or Assignable Variation Due to
a) improperly adjusted machine
b) operator error
c) defective raw material
Analysis Of
Variance
ANOVA is often used for studying the relationship
between a response variable (Y) and one or more
explanatory or predictor variables (X’s). The
predictor variables are also called factors or
treatments.
While the response is quantitative, the predictors
may be either quantitative or qualitative.
However, quantitative predictors are analyzed as
if they are qualitative (or categorical).
ANOVA —
Application
ANOVA can be used to
Determine the statistical significance of effects
 To identify sources of variability in Y
。
 To determine the signification of a regression
equation.
 To determine which factors affect the output in a
DOE.
•
ANOVA —
Assumptions
The observations are mutually
independent.
– Stat  Nonparametrics  Runs Test
• The k groups exhibit homogeneity of
variance.
i.e. 1² = 2² =  = k²
– Stat  ANOVA  Test for Equal Variances
• The data from each of the k groups is
normally distributed.
i.e.Factor Level i ~ N (i,i²)
– Stat  Basic Statistics  Normality Test
ANOVA —
Principle
组内

2
Within

1
k

k
i 1

2
 Total
  2Within   2Between
2
i
N(4,4²)
20
Response
N(1,1²)
N(3,3²)
15
N(2,2²)
10
1
2
3
Factor Level
4

2
Between

1
k 1
 
k
i 1
 
2
i
ANOVA — Hypothesis
Testing
1 = 2 =  = k
H0 :
all group means are equal
Ha :
i  j
for some i  j
at least one pair of group means is not equal
ANOVA verifies the null hypothesis by comparing
the variance between the groups against the
variation within a group mean:
F*

Variation between groups
Variation within group

2between
 2within
*
F
 F;  between ,  within
The null hypothesis is rejected if

One-way Analysis of Variance
• 1. Satisfy level of measurement
requirements
– Dependent variable is interval (ordinal)
– Independent variable designates groups
• 2. Satisfy assumption of normality
– Skewness and kurtosis
– Central Limit Theorem
One-way Analysis of Variance
• 3. Test assumption of equal variances
among groups
– Levene test of equality of variances
• 4. Make decision about null
hypothesis based on
– Probability of F-statistic <= alpha  reject null
hypothesis
– Probability of F-statistic > alpha  fail to
reject null hypothesis
One-way Analysis of Variance
• 5. Draw conclusion about research
hypothesis based on decision about null
hypothesis
– Reject null hypothesis  support research
hypothesis
– Fail to reject null hypothesis  do not support
research hypothesis
One-Way ANOVA
Treatment 1
Treatment 2
Treatment 3
89
84
79
98
77
81
97
92
80
94
79
88
1. （Minitab: H0： Data is Normal; Ha： Data is NOT Normal） 。
2. （Minitab: H0： 1= 2 = 3 Ha： at lease one is different） 。
.
–Stat  Basic Statistics  Normality Test
H0： Data is Normal; Ha： Data is NOT Normal
Hardness Normality Test
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
78
88
98
Hardness
Average: 86.5
StDev: 7.50151
N: 12
Anderson-Darling Normality Test
A-Squared: 0.409
P-Value: 0.290
P -Value
Stat  ANOVA  Test for Equal Variances
H0： 1= 2 = 3 Ha： at lease one is different
P -Value
Hardness of Equal Variable
95% Confidence Intervals for Sigmas
Factor Levels
Treatment 1
Bartlett's Test
Test Statistic: 0.929
P-Value
: 0.628
Treatment 2
Levene's Test
Test Statistic: 0.670
P-Value
Treatment 3
0
10
20
30
40
: 0.535
Welcome to Minitab, press F1 for help.
There is only a 1.2%
chance ….
One-way ANOVA: Hardness versus Treatment
Analysis of Variance for Hardness
Source
DF
SS
MS
Treatmen
2
386.0
193.0
Error
9
233.0
25.9
Total
11
619.0
Within
Level
Treatmen
Treatmen
Treatmen
N
4
4
4
Pooled StDev =
Mean
94.50
83.00
82.00
5.09
StDev
4.04
6.68
4.08
F
7.45
P
0.012
Individual 95% CIs For Mean
Based on Pooled StDev
--+---------+---------+---------+---(-------*-------)
(--------*-------)
(-------*-------)
--+---------+---------+---------+---77.0
84.0
91.0
98.0
Individual Value Plot of Hardness vs teratment
100
Hardness
95
90
85
80
Treatment1
Treatment2
teratment
Treatment3
Boxplot of Hardness by teratment
100
Hardness
95
90
85
80
Treatment1
Treatment2
teratment
Treatment3
Residual Plots for Hardness
Normal Probability Plot of the Residuals
Residuals Versus the Fitted Values
99
10
Residual
Percent
90
50
10
1
5
0
-5
-10
-5
0
Residual
5
10
85
Histogram of the Residuals
10
Residual
Frequency
95
Residuals Versus the Order of the Data
3
2
1
0
90
Fitted Value
5
0
-5
-5.0
-2.5
0.0
2.5
5.0
Residual
7.5
10.0
1
2
3
4
5
6 7 8
9
Observation Order
10 11 12
Normal Probability Plot of the Residuals
(response is Hardness)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-10
-5
0
Residual
5
10
Normal probability plot - indicates whether the data are normally
distributed, other variables are influencing the response, or outliers exist in
the data。
Histogram of the Residuals
(response is Hardness)
3.0
Frequency
2.5
2.0
1.5
1.0
0.5
0.0
-5.0
-2.5
0.0
2.5
Residual
5.0
7.5
10.0
Histogram - indicates whether the data are skewed or outliers exist in the
data
Residuals Versus the Fitted Values
(response is Hardness)
10.0
7.5
Residual
5.0
2.5
0.0
-2.5
-5.0
82
84
86
88
90
Fitted Value
92
94
96
Residuals versus fitted values - indicates whether the variance is
constant, a nonlinear relationship exists, or outliers exist in the data
Residuals Versus the Order of the Data
(response is Hardness)
10.0
7.5
Residual
5.0
2.5
0.0
-2.5
-5.0
1
2
3
4
5
6
7
8
Observation Order
9
10
11
12
Residuals versus order of the data - indicates whether there are
systematic effects in the data due to time or data collection order
Thanks for Your Attention