pooled variance t test tests means of 2 independent populations having equal variances parametric...
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Pooled Variance t Test
Pooled Variance t Test
• Tests means of 2 independent populations having equal variances
• Parametric test procedure• Assumptions
– Both populations are normally distributed– If not normal, can be approximated by normal distribution
(n1 30 & n2 30 )
– Population variances are unknown but assumed equal
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Two Independent Populations Examples
Two Independent Populations Examples
• An economist wishes to determine whether there is a difference in mean family income for households in 2 socioeconomic groups.
• An admissions officer of a small liberal arts college wants to compare the mean SAT scores of applicants educated in rural high schools & in urban high schools.
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Pooled Variance t Test ExamplePooled Variance t Test Example
You’re a financial analyst for Charles Schwab. You want to see if there a difference in dividend yield between stocks listed on the NYSE & NASDAQ. NYSE NASDAQNumber 21 25
Mean 3.27 2.53
Std Dev 1.30 1.16
Assuming equal variances, isthere a difference in average yield ( = .05)?
© 1984-1994 T/Maker Co.
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Pooled Variance t Test Solution
Pooled Variance t Test Solution
H0: 1 - 2 = 0 (1 = 2)
H1: 1 - 2 0 (1 2)
.05
df 21 + 25 - 2 = 44
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
t0 2.0154-2.0154
.025
Reject H 0 Reject H 0
.025
t0 2.0154-2.0154
.025
Reject H 0 Reject H 0
.025
2.03
251
211
1.510
2.533.27t
Reject at Reject at = .05 = .05
There is evidence of a There is evidence of a difference in meansdifference in means
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Test StatisticSolution
Test StatisticSolution
tX X
Sn n
Sn S n S
n n
P
P
FHG
IKJ
FH IK
1 2 1 2
2
1 2
2 1 12
2 22
1 2
2 2
1 1
3 27 2 53 0
1510121
125
2 03
1 1
1 1
21 1 130 25 1 116
21 1 25 11510
c ha f a f af
a f a fa f a f
a fa f a fa fa f a f
. .
..
. ..
tX X
Sn n
Sn S n S
n n
P
P
FHG
IKJ
FH IK
1 2 1 2
2
1 2
2 1 12
2 22
1 2
2 2
1 1
3 27 2 53 0
1510121
125
2 03
1 1
1 1
21 1 130 25 1 116
21 1 25 11510
c ha f a f af
a f a fa f a f
a fa f a fa fa f a f
. .
..
. ..
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You’re a research analyst for General Motors. Assuming equal variances, is there a difference in the average miles per gallon (mpg) of two car models ( = .05)?
You collect the following:
Sedan Van
Number 15 11Mean 22.00 20.27Std Dev 4.77 3.64
Pooled Variance t Test Thinking Challenge
Pooled Variance t Test Thinking Challenge
AloneAlone GroupGroup Class Class
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Test StatisticSolution*
Test StatisticSolution*
tX X
Sn n
Sn S n S
n n
P
P
FHG
IKJ
FH IK
1 2 1 2
2
1 2
2 1 12
2 22
1 2
2 2
1 1
22 00 20 27 0
18 7931
151
11
100
1 1
1 1
15 1 4 77 11 1 3 64
15 1 11 118 793
c ha f a f af
a f a fa f a f
a fa f a fa fa f a f
. .
..
. ..
tX X
Sn n
Sn S n S
n n
P
P
FHG
IKJ
FH IK
1 2 1 2
2
1 2
2 1 12
2 22
1 2
2 2
1 1
22 00 20 27 0
18 7931
151
11
100
1 1
1 1
15 1 4 77 11 1 3 64
15 1 11 118 793
c ha f a f af
a f a fa f a f
a fa f a fa fa f a f
. .
..
. ..
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One-Way ANOVA F-TestOne-Way ANOVA F-Test
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2 & c-Sample Tests with Numerical Data
2 & c-Sample Tests with Numerical Data
2 & C -SampleTests
#Samples
Median VarianceMean
C CF Test(2 Samples)
Kruskal-Wallis Rank
Test
WilcoxonRank Sum
Test
#Samples
PooledVariance
t Test
One-WayANOVA
22
2 & C -SampleTests
#Samples
Median VarianceMean
C CF Test(2 Samples)
Kruskal-Wallis Rank
Test
WilcoxonRank Sum
Test
#Samples
PooledVariance
t Test
One-WayANOVA
22
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ExperimentExperiment
• Investigator controls one or more independent variables– Called treatment variables or factors– Contain two or more levels (subcategories)
• Observes effect on dependent variable – Response to levels of independent variable
• Experimental design: Plan used to test hypotheses
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Completely Randomized DesignCompletely Randomized Design
• Experimental units (subjects) are assigned randomly to treatments– Subjects are assumed homogeneous
• One factor or independent variable– 2 or more treatment levels or classifications
• Analyzed by: – One-Way ANOVA– Kruskal-Wallis rank test
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Factor (Training Method)Factor levels(Treatments)
Level 1 Level 2 Level 3
Experimentalunits
Dependent 21 hrs. 17 hrs. 31 hrs.
variable 27 hrs. 25 hrs. 28 hrs.
(Response) 29 hrs. 20 hrs. 22 hrs.
Factor (Training Method)Factor levels(Treatments)
Level 1 Level 2 Level 3
Experimentalunits
Dependent 21 hrs. 17 hrs. 31 hrs.
variable 27 hrs. 25 hrs. 28 hrs.
(Response) 29 hrs. 20 hrs. 22 hrs.
Randomized Design ExampleRandomized Design Example
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One-Way ANOVA F-Test
One-Way ANOVA F-Test
• Tests the equality of 2 or more (c) population means
• Variables– One nominal scaled independent variable
• 2 or more (c) treatment levels or classifications– One interval or ratio scaled dependent variable
• Used to analyze completely randomized experimental designs
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One-Way ANOVA F-Test AssumptionsOne-Way ANOVA F-Test Assumptions
• Randomness & independence of errors– Independent random samples are drawn
• Normality– Populations are normally distributed
• Homogeneity of variance– Populations have equal variances
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One-Way ANOVA F-Test Hypotheses
One-Way ANOVA F-Test Hypotheses
• H0: 1 = 2 = 3 = ... = c
– All population means are equal
– No treatment effect
• H1: Not all j are equal– At least 1 population mean is
different– Treatment effect 1 2 ... c is wrong
X
f(X)
1 = 2 = 3
X
f(X)
1 = 2 3
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• Compares 2 types of variation to test equality of means
• Ratio of variances is comparison basis• If treatment variation is significantly greater
than random variation then means are not equal
• Variation measures are obtained by ‘partitioning’ total variation
One-Way ANOVA Basic Idea
One-Way ANOVA Basic Idea
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ANOVA Partitions Total Variation
ANOVA Partitions Total Variation
Variation due to treatment
Variation due to treatment
Variation due to random samplingVariation due to
random sampling
Total variationTotal variation
Sum of squares within Sum of squares error Within groups variation
Sum of squares among Sum of squares between Sum of squares model Among groups variation
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Total VariationTotal Variation
Group 1 Group 2 Group 3
Response, X
Group 1 Group 2 Group 3
Response, X
SST X X X X X Xn cc 11
2
21
2 2e j e j e jSST X X X X X Xn cc 11
2
21
2 2e j e j e j
XX
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Among-Groups VariationAmong-Groups Variation
Group 1 Group 2 Group 3
Response, X
Group 1 Group 2 Group 3
Response, X
SSA n X X n X X n X Xc c 1 1
2
2 2
2 2e j e j e jSSA n X X n X X n X Xc c 1 1
2
2 2
2 2e j e j e j
XXXX33
XX22XX11
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Within-Groups VariationWithin-Groups Variation
Group 1 Group 2 Group 3
Response, X
Group 1 Group 2 Group 3
Response, X
SSW X X X X X Xn c cc 11 1
221 1
2 2c h c h c hSSW X X X X X Xn c cc 11 1
221 1
2 2c h c h c h
XX22XX11
XX33
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One-Way ANOVA Test Statistic
One-Way ANOVA Test Statistic
• Test statistic– F = MSA / MSW
• MSA is Mean Square Among• MSW is Mean Square Within
• Degrees of freedom– df1 = c -1
– df2 = n - c• c = # Columns (populations, groups, or levels)• n = Total sample size
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One-Way ANOVA Summary Table
One-Way ANOVA Summary Table
Sourceof
Variation
Degreesof
Freedom
Sum ofSquares
MeanSquare
(Variance)
F
Among(Factor)
c - 1 SSA MSA =SSA/(c - 1)
MSAMSW
Within(Error)
n - c SSW MSW =SSW/(n - c)
Total n - 1 SST =SSA+SSW
Sourceof
Variation
Degreesof
Freedom
Sum ofSquares
MeanSquare
(Variance)
F
Among(Factor)
c - 1 SSA MSA =SSA/(c - 1)
MSAMSW
Within(Error)
n - c SSW MSW =SSW/(n - c)
Total n - 1 SST =SSA+SSW
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One-Way ANOVA Critical Value
One-Way ANOVA Critical Value
0
Reject H 0
Do NotReject H 0
F0
Reject H 0
Do NotReject H 0
FFU c n c( ; , ) 1FU c n c( ; , ) 1
If means are equal, If means are equal, FF = = MSAMSA / / MSWMSW 1. Only reject large 1. Only reject large FF!!
Always One-Tail!Always One-Tail!© 1984-1994 T/Maker Co.
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One-Way ANOVA F-Test Example
One-Way 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?
Mach1Mach1Mach2Mach2Mach3Mach325.4025.40 23.4023.40 20.0020.0026.3126.31 21.8021.80 22.2022.2024.1024.10 23.5023.50 19.7519.7523.7423.74 22.7522.75 20.6020.6025.1025.10 21.6021.60 20.4020.40
Mach1Mach1Mach2Mach2Mach3Mach325.4025.40 23.4023.40 20.0020.0026.3126.31 21.8021.80 22.2022.2024.1024.10 23.5023.50 19.7519.7523.7423.74 22.7522.75 20.6020.6025.1025.10 21.6021.60 20.4020.40
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One-Way ANOVA F-Test Solution
One-Way ANOVA F-Test Solution
H0: 1 = 2 = 3
H1: Not all equal
= .05
df1 = 2 df2 = 12
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
Reject at Reject at = .05 = .05
There is evidence pop. There is evidence pop. means are differentmeans are differentF0 3.89 F0 3.89
= .05= .05 = .05= .05
FMSAMSW
23 5820
921125 6
..
.FMSAMSW
23 5820
921125 6
..
.
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Summary TableSolution
Summary TableSolution
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquare
(Variance)
F
Among(Machines)
3 - 1 = 2 47.1640 23.5820 25.60
Within(Error)
15 - 3 = 12 11.0532 .9211
Total 15 - 1 = 14 58.2172
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquare
(Variance)
F
Among(Machines)
3 - 1 = 2 47.1640 23.5820 25.60
Within(Error)
15 - 3 = 12 11.0532 .9211
Total 15 - 1 = 14 58.2172
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Summary TableExcel Output
Summary TableExcel Output
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One-Way ANOVA Thinking Challenge
One-Way ANOVA Thinking Challenge
You’re a trainer for Microsoft Corp. Is there a differencein mean learning times of 12 people using 4 different training methods ( =.05)?
M1 M2 M3 M410 11 13 18
9 16 8 235 9 9 25
© 1984-1994 T/Maker Co.
AloneAlone GroupGroup Class Class
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One-Way ANOVA Solution*One-Way ANOVA Solution*H0: 1 = 2 = 3 = 4
H1: Not all equal
= .05
df1 = 3 df2 = 8
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
Reject at Reject at = .05 = .05
There is evidence pop. There is evidence pop. means are differentmeans are differentF0 4.07 F0 4.07
= .05= .05
FMSAMSW
11610
116.FMSAMSW
11610
116.
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Summary Table Solution*
Summary Table Solution*
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquare
(Variance)
F
Among(Methods)
4 - 1 = 3 348 116 11.6
Within(Error)
12 - 4 = 8 80 10
Total 12 - 1 = 11 428
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquare
(Variance)
F
Among(Methods)
4 - 1 = 3 348 116 11.6
Within(Error)
12 - 4 = 8 80 10
Total 12 - 1 = 11 428