Chapter 13 Anova Experimental Design

7/27/2019 Chapter 13 Anova Experimental Design

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ANALYSIS OF VARIANCE (ANOVA)AND EXPERIMENTAL DESIGN


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ANOVA can be used to test forequality of three or morepopulation means.


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Exp:

National Computers Product,Inc.manufactures printers and fax machinesat plants located in Atlanta, Dallas, andSeatle.

To measure how much employees atthese plants know about total qualitymanagement, a random sample of six

employees was selected from each plantand given a quality awarenessexamination (table 13.1)STT CH 13 (ANOVA & EXP.DESIGN).xlsx
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1. For each population, the responsevariables is normally distributed.Implication: In the NCP example, the

examination scores (response variable)must be normally distributed at eachplant.

2. The variance of the response variable,denoted , is the same for all of thepopulations.

3. The observation must be independent.


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If the means for the threepopulations are equal, we wouldexpect the three sample means tobe close together.

If the variability among the samplemeans is Small, it supports Ho

If the variability among the samplemeans is Large, it supports Ha


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HYPOTHESES:


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FORMULA FOR THE SAMPLE MEAN ANDSAMPLE VARIANCE FOR TREATMENT j:


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The overall sample mean:

If the size of each sample is n, nT= kn;


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If n1 n2 n3... nk


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The estimate of : Mean Square Due toTreatments (MSTR)

If Ho is true, MSTR provides an unbiased

estimate of . However, if the means of the kpopulations are not equal, MSTR is not anunbiased estimate of


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The estimate of : Mean Square Due to Error(MSE)

MSE is based on the variation within each of the

treatments; its not influenced by whether thenull hypothesis is true. Thus, MSE alwaysprovides an unbiased estimate of


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Test statistic for The Equality ofkPopulation

Means.

F = 258/ 28,67 = 9


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F table:

F = 9

Because F = 9 is greater than 6,36, the area inthe upper tail at F = 9 is less than 0,01. Thus,the p-value is less than 0,01=> Ho is rejected.

Area in Upper Tail 0,10 0,05 0,025 ,01F Value (fd1=2,df2=15)

2,70 3.68 4,77 6,36


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The test provides sufficientevidence to conclude that the

means of the three populationsare not equal. In other words,

analysis of variance supportsthe conclusion that thepopulation mean examination

scores at the three NCP plantsare not equal.


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The general form of an interval estimate of apopulation mean is

t0,025 = 2,131

In the analysis of variance the best estimate ofis provided by the square root of MSE or thePooled StDev.


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Thus, the individual 95% confidenceinterval for the Atlanta plant goes from

79-4,66 = 73,34 to 79+4,66 = 83,66

Because the sample size are equal for

the NCP example, the individualconfidence intervals for the Dallas arealso constructed by adding andsubtracting 4,66 from each samplemean.


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FISHERS LSD (Least Significant Difference):

determine where the difference occur.

Do the mean of population 1 and 2 differ?




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Summary of Fishers LSD Prosedure

Test statistic


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Rejection Rule at a Level of Significance

Sample Differences Significant?Method A Method B = 62-66 = - 4 No

Method A Method C = 62-52 = 10 Yes

Method B Method C = 66-52 = 4 Yes


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Rejection Rule:

p-value approach:

Reject Ho if p-value

Critical value approach:Reject Ho if t - t/2

or t t/2

df: nT k

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Chapter 13 Anova Experimental Design