Chapter 7Analysis of ariance

Variation

• Inherent or Natural Variation Due to the cumulative effect of many small unavoidable causes. Also referred to as noise

• Special or Assignable Variation Due toa) improperly adjusted machine b) operator error

c) defective raw material

Data without dispersion information is false data.

— Kaoru Ishikawa

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

4321

20

15

10

Factor Level

Res

pons

e

N(1,1²)

N(2,2²)N(3,3²)

N(4,4²)

2

Between

2

Within

2

Total k

1i

2

ik12

Within

组内

k

1i

2

i1k12

Between组间

ANOVA — Hypothesis

TestingH0 : 1 = 2 = = k

all group means are equal

Ha : i j for some i jat 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:

2within

2between*

groupwithinVariationgroupsbetweenVariation

F

withinbetween* ,;FF 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 ） 。.

P-Value: 0.290A-Squared: 0.409

Anderson-Darling Normality Test

N: 12StDev: 7.50151Average: 86.5

988878

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

babi

lity

Hardness

Hardness Normality Test

–Stat Basic Statistics Normality Test H0 ： Data is Normal; Ha ： Data is NOT Normal

P -Value

0 10 20 30 40

95% Confidence Intervals for Sigmas

Bartlett's Test

Test Statistic: 0.929

P-Value : 0.628

Levene's Test

Test Statistic: 0.670

P-Value : 0.535

Factor Levels

Treatment 1

Treatment 2

Treatment 3

Hardness of Equal Variable P -Value

Stat ANOVA Test for Equal VariancesH0 ： 1= 2 = 3 Ha ： at lease one is different

Welcome to Minitab, press F1 for help.

One-way ANOVA: Hardness versus Treatment

Analysis of Variance for HardnessSource DF SS MS F PTreatmen 2 386.0 193.0 7.45 0.012Error 9 233.0 25.9Total 11 619.0 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev --+---------+---------+---------+----Treatmen 4 94.50 4.04 (-------*-------) Treatmen 4 83.00 6.68 (--------*-------) Treatmen 4 82.00 4.08 (-------*-------) --+---------+---------+---------+----Pooled StDev = 5.09 77.0 84.0 91.0 98.0

Within

There is only a 1.2% chance ….

teratment

Hard

ness

Treatment3Treatment2Treatment1

100

95

90

85

80

Individual Value Plot of Hardness vs teratment

teratment

Hard

ness

Treatment3Treatment2Treatment1

100

95

90

85

80

Boxplot of Hardness by teratment

Residual

Perc

ent

1050-5-10

99

90

50

10

1

Fitted Value

Resi

dual

959085

10

5

0

-5

Residual

Fre

quency

10.07.55.02.50.0-2.5-5.0

3

2

1

0

Observation Order

Resi

dual

121110987654321

10

5

0

-5

Normal Probability Plot of the Residuals Residuals Versus the Fitted Values

Histogram of the Residuals Residuals Versus the Order of the Data

Residual Plots for Hardness

Residual

Perc

ent

1050-5-10

99

95

90

80

70

605040

30

20

10

5

1

Normal Probability Plot of the Residuals(response is Hardness)

Normal probability plot - indicates whether the data are normally distributed, other variables are influencing the response, or outliers exist in the data。

Residual

Frequency

10.07.55.02.50.0-2.5-5.0

3.0

2.5

2.0

1.5

1.0

0.5

0.0

Histogram of the Residuals(response is Hardness)

Histogram - indicates whether the data are skewed or outliers exist in the data

Residuals versus fitted values - indicates whether the variance is constant, a nonlinear relationship exists, or outliers exist in the data

Fitted Value

Resi

dual

9694929088868482

10.0

7.5

5.0

2.5

0.0

-2.5

-5.0

Residuals Versus the Fitted Values(response is Hardness)

Residuals versus order of the data - indicates whether there are systematic effects in the data due to time or data collection order

Observation Order

Resi

dual

121110987654321

10.0

7.5

5.0

2.5

0.0

-2.5

-5.0

Residuals Versus the Order of the Data(response is Hardness)

Thanks for Your Attention