Chapter Ten The Analysis Of Variance

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Chapter TenThe Analysis Of Variance
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ANOVA Definitionsgt Factor The characteristic
that differentiates the treatment or
populations from one another.gt Level
(Treatments) The number of different
treatments or populations.
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Randomized Experiment Randomizing the order of
sample observations will balance out any known or
unknown nuisance variable that may influence the
observed response.
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Mean Square for TreatmentsMSTr J (X1
X)2 (XI X)2 I 1For I number
of levelsFor J number of samples
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Mean Square for Error MSE S21 S22
S2I I
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Test Statistic for Single Factor ANOVA F
MSTr MSE With ?1 I 1 ?2
I(J-1)
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ANOVA on Single-Factor Experiment Several (I)
Means All Normal (Same ?2)Null Hypothesis
H0 u1 u2 uITest Statistic ? MSTr /
MSE Alternative Hypothesis
Ha (at least two means are not equal)Reject
Region (upper tailed)? ? F?, I-1, I
(J-1) Exp. (IJ 1) DOF
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ANOVA on Single-Factor ExperimentExperiments
were conducted to study whether commercial
processing of various foods changes the
concentration of essential elements for human
consumption. One such experiment was to study the
concentration of zinc in green beans. A batch of
green beans was divided into 4 groups. The 4
groups were then randomly assigned to be measured
(10) times each for zinc as follows group 1
measured Raw group 2 measured before Blanching
group 3 measured after Blanching and group 4
measured after the final processing step. Ten
independent measurements were taken from the 4
groups (treatments), yielding the following
data Zinc ConcentrationGroup 1 Group 2
Group 3 Group 4u1 2.01 u2 2.58 u3
2.10 u4 3.05S1 0.25 S2 0.50
S3 0.30 S4 1.00Test this hypothesis for
significance at the 5 level.Measurements of
this type are known to be Normal.
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ANOVA on Single-Factor Experiment ExampleThe
coded values for the measure of elasticity
(nt/m2) in plastic, prepared by two different
processes A B, for samples of (6) drawn
randomly from each of the two processes are as
follows Group A Group B u1 7.28
u2 8.02S12 0.48 S22 0.71 Do the
data present sufficient evidence to indicate a
difference in mean elasticity for the two
processes at a level of significance of a .05?
Measurements of this type are found to follow a
Normal pdf.
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ANOVA on Single-Factor Experiment Several (I)
Variances (Equal Samples J) All Normal Null
Hypothesis H0 ?21 ?22 ?2ITest
Statistic ?2 (2.3026) Q / h Bartletts
Test Alternative Hypothesis
Ha(at least 2 variances are not equal)Reject
Region (upper tailed test)?2 ? ?2 ?, I - 1
Q I(J1)log(MSE) (J-1)log(S21)log(S2I)
h 1 1 I 1
3(I-1) (J-1) I(J1)
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ANOVA on Single-Factor Experiment Several (I)
Variances (Equal Samples J) ExampleA study is
designed to investigate the sulfur content of (5)
major coal seams. Eight core samples are taken at
randomly selected points within each seam. The
measured response is the S content. Before
performing a Hypothesis Test on the data to
detect any differences that might exist in the
average sulfur content for these (5) seams, you
are required to test the condition that the (5)
seams all have the same population variance at a
level of significance of .05. The summary
statistics on the sulfur content of the (5) major
coal seams followsSeam 1 Seam 2 Seam 3 Seam
4 Seam 5?1 1.66 ?2 1.17 ?3 1.46 ?4
0.88 ?5 1.189S21.175 S22.144 S23.115 S24
.123 S25.074
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ANOVA on Single-Factor Experiment Several (I)
Variances (Equal Samples J) ExampleUse
Bartletts Hypothesis Test to determine whether
it is reasonable to assume homogeneity of
variances for the (4) treatment groups in the
study whether commercial processing of various
foods changes the concentration of essential
elements for human consumption. Use ?
.05.Rough rule of thumb If the largest s is
not much more than two times the smallest, it is
reasonable to assume equal variances.
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ANOVA Multiple ComparisonsProcedures for
identifying which uis significantly differ when
H0 is rejected gt Tukey gt Bonferroni gt Duncan
gt Fisher LSD gt Newman-Keuls
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Tukeys T Method (Equal Samples)1. Select ?
find Q?, I, I(J-1) from Studentized Range
Distribution Table A.10 on pg. 736. (m
I)2. Determine w Q?, I, I(J-1)?MSE/J3. List
uis in increasing order underline those
pairs that differ by less than w. Any pair
of uis not underscored by the same line
corresponds to a pair of population or
treatment means that are judged significantly
different.
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Examples of Tukeys MethodSummary Results w
5.37 x1 x5 x2 x3
x4 9.8 10.8 15.4 17.6 21.6Summary
Results w 0.40 x5 x3 x2
x4 x1 6.1 6.3 6.8 7.3
7.5Summary Results w 0.40 x5 x3
x2 x4 x1 6.1 6.3 7.15
7.3 7.5
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ANOVA Multiple Comparison Tukeys MethodA
product development engineer is interested in
maximizing the tensile strength of a new
synthetic fiber. Previous experience indicates
that the strength is affected by the of cotton
in the fiber. The engineer suspects that
increasing the cotton content will increase the
strength, at least initially. He decides to test
(5) specimens at (5) levels of cotton content.
Summary data followsCotton 15
20 25 30 35 Mean 9.8 15.4
17.6 21.6 10.8 (psi) s
3.35 3.13 2.07 2.61
2.86The Null Hypothesis H0 is rejected because
the F statistic falls in the Reject Region. The
of cotton in the fiber significantly affects the
mean tensile strength. Now use Tukeys T method
to find significant differences among the means.
Use ? .05.
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ANOVA Multiple Comparison Tukeys MethodAn
experiment is developed to measure the effect
that teaching methods have on a students
performance. The following table lists the
numerical grades on a standard arithmetic test
given to 45 students divided randomly into (5)
equal-sized groups. Groups 1 2 were taught by
the current method. Groups 3, 4, 5 were taught
together for a number of days on each day group
3 students were praised publicly for their
previous work while group 4 students were
criticized publicly. Group 5 students while
hearing the praise and criticism of groups 3 4,
were ignored. Group 1 2 3 4
5Mean 19.67 18.33 27.44 23.44
16.11 s2 17.72 12.75 6.05
9.55 13.104 Test the null hypothesis that
there is no difference in the mean grades
produced by these teaching methods using ? at
.05. Then use Tukeys T method to compare
illustrate the difference in the teaching
methods.
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Least Significant Difference Method (Equal
Samples1. Select ? find t?/2, I(J-1) 2.
Determine w t?/2, I(J-1) ?2MSE/J3. Compare
the observed difference between each pair of
averages to the corresponding LSD. If
ui uJ gt LSD, we conclude that the
population mean ui and uJ differ.
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Example Least Significant Difference MethodA
manufacturer of paper used for making grocery
bags is interested in improving the tensile
strength of the product. Product engineering
thinks that the tensile strength is a function of
the hardwood concentration in the pulp and that
the range of hardwood concentrations of practical
interest is between 5 and 20. You decide to
investigate (4) levels of hardwood concentration.
Six specimens at each of the (4) concentration
levels are prepared and tested on a tensile
tester in random order. The summary data from
this experiment are shown in the following
tableHardwood (psi)Concentration gt 5
10 15 20 Mean 10.00
15.67 17.00 21.17 S2 8.00
7.87 3.20 6.97Test the null
hypothesis that there is no difference in the
mean tensile strength produced by these (4)
concentration levels using ? at .01. Then use the
LSD method at ? .05 to compare illustrate the
difference at each level of concentration.
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Example Least Significant Difference MethodThe
effective life of insulating fluids at an
accelerated load of 50 m/sec2 is being studied.
Test data have been obtained for (4) types of
fluids. The summary results for (7) trials on
each fluid are as follows Life (in hours) at
50 m/sec2 Fluid Type gtgt 1 2 3
4 Mean 18.65 17.95 20.95
18.82 S2 3.81
3.44 3.53 2.42Is there any
indication that the fluids differ at a
significance level of .05?Which fluid or fluids
would you select if the objective is long life?
Use the Least Significance Difference method with
an alpha of .05 to support you conclusion.
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?-Error for Single Factor ANOVA
F-TestNon-centrality parameter ? J ?
(?I - ?)2 ?2For Non-central F
distribution.With Degrees of Freedom?1
I-1?2 I(J-1)
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?-Error for Single Factor ANOVA1) Find the value
of ?2 (Experience)2) Find the
values of ?(?i - ?) 3) Compute ?2 using
(Replaces ?) ?2 J ?
(?i - ?)2 I ?24) Use Power
Curves (pg. 422) to look-up power value ?
1 Power gt Use appropriate set curves for
?1 gt ? (with ?) is on the horizontal axis
gt Move up to the curve associated with ?2
gt Find value of power value on vertical
axis
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?-Error ANOVA ExampleA product development
engineer is interested in maximizing the tensile
strength of a new synthetic fiber. Previous
experience indicates that the strength is
affected by the of cotton in the fiber. The
engineer suspects that increasing the cotton
content will increase the strength, at least
initially. He decides to test (5) specimens at
(5) levels of cotton content. Summary data
followsCotton 15 20 25
30 35 Mean 9.8 15.4 17.6 21.6
10.8 (psi) s2 11.22
9.80 4.28 6.81 8.18What is the
?-error if the engineer is interested in
rejecting the null hypothesis if the five
treatment means are as follows ?15 11 ?20 12
?25 15 ?30 18 ?35 19 Historically, the
standard deviation of tensile strength is usually
equal to 3 psi. Assume ? .01 for this test.
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?-Error ANOVA ExampleSuppose that (5) means are
being compared in a completely randomized
experiment with ? .01. The design engineer
would like to know how many samples to take if it
is important to reject the Null Hypothesis with
probability at least 0.90 if ? (?i - ?)2 25
the population variance is known to be 5.0.
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?-Error ANOVA ExampleSuppose that (4) Normal
populations have common variance ?2 25 and
means ?1 50, ?2 60, ?3 50, and ?4 60. How
many observations should be taken on each
population so that the probability of rejecting
the hypothesis of equality of means is at least
0.90? Use ? 0.05.
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Single-Factor ANOVA (Unequal Sample Sizes Ji)
F MSTr MSE With ?1
I 1 ?2 N-IWhere MSTr SSTr I
1And MSE SSE N -I
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ANOVA DefinitionsSum of Squares
Treatment SSTr ? Ji (?i - ?)2
iSum of Square Error SSE ?? (xij- ?i)2
i jSum of Square Total SST SSTr SSE
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Example of Unbalanced DesignTwenty-seven coins
discovered in Cyprus were grouped into (4)
classes, corresponding to (4) different coinages
during the reign of King Manuel I Comnenus
(1143-1180). Archaeologists are interested in
whether there were significant differences in the
Ag content of coins minted early and late in King
Manuels reign. Test the H0 at ? .01. Summary
data for testing the Ag content of early coins
(group 1) to later coins (group 4)
followsGroup Ji Mean SSE SSTr 1 9
6.74 11.02 37.75 2 7 8.24 3 4
4.88 4 7 5.61
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Multiple Comparisons (Unequal Samples)Tukeys
method modified1. Select ? find Q?, I, N-I
from Studentized Range Distribution Table
A.10 on pg. 736. (m I)2. Determine wij Q?,
I, N-I?MSE x ( 1 1 )
2 Ji Jj
Uses averages of pairs 1/Jis instead of
1/J.3. List uis in increasing order
underline those pairs that differ by less than
wij.
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Example of Multiple Comparison(Unequal Sample
Sizes)Use Tukeys modified T method at ? .01
to compare illustrate the difference in the
means of Ag percentage in coins found on Cyprus.
Group Ji Mean SSE SSTr 1 9 6.74
11.02 37.75 2 7 8.24 3 4 4.88
4 7 5.61