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Chapter 24
Single-Factor (One-Way)
Analysis of Variance (ANOVA)
and
Analysis of Means (ANOM)
Introduction
โข This chapter describes single-factor analysis of
variance (ANOVA) experiments with 2 or more levels
(or treatments).
โข The method is based on a fixed effects model (as
opposed to a random effects model, or components of
variance model). It tests the ๐ป0 that the different
processes give an equal response.
โข With fixed effects model, the levels are specifically
chosen. The test hypothesis is about the mean
responses due to factor levels. Conclusions apply only
to the factor levels considered.
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24.1 S4/IEE Application Examples:
ANOVA and ANOM
โข Transactional 30,000-foot-level metric: DSO reduction was
chosen as an S4/IEE project. A cause-and-effect matrix
ranked company as an important input that could affect the
DSO response (i.e., the team thought that some
companies were more delinquent in payments than other
companies). From randomly sampled data, a statistical
assessment was conducted to test the hypothesis of
equality of means for the DSOs of these companies.
24.1 S4/IEE Application Examples:
ANOVA and ANOM
โข Manufacturing 30,000-foot-level metric (KPOV): An S4/IEE
project was to improve the capability/performance of the
diameter of a manufactured product (i.e., reduce the
number of parts beyond the specification limits). A cause-
and-effect matrix ranked cavity of the four-cavity mold as
an important input that could be yielding different part
diameters. From randomly sampled data, statistical tests
were conducted to test the hypotheses of mean diameter
equality and equality of variances for the cavities.
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24.1 S4/IEE Application Examples:
ANOVA and ANOM
โข Transactional and manufacturing 30.000-foot-level cycle
time metric (a lean metric): An S4/IEE project was to
improve the time from order entry to fulfillment. The WIP
at each process step was collected at the end of the day
for a random number of days. Statistical tests were
conducted to test the hypothesis that the mean and
variance of WIP at each step was equal.
24.2 Application Steps
1. Describe the problem using a response variable that
corresponds to the KPOV or measured quality characteristic.
2. Describe the analysis (e.g., determine if there is a difference)
3. State the null and alternative hypotheses.
4. Choose a large enough sample and conduct the experiment
randomly.
5. Generate an ANOVA table.
6. Test the data normality and equality of variance assumptions.
7. Make hypothesis decisions about factors from ANOVA table.
8. Calculate (if desired) epsilon squared (๐2).
9. Conduct an analysis of means (ANOM).
10. Translate conclusions into terms relevant to the problem or
process in question.
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24.3 Single-factor ANOVA
Hypothesis Test
โข ANOVA assesses the differences between samples taken
at different factor levels to determine if these differences
are large enough relative to error to conclude that the
factor level causes a statistically significant difference in
response.
โข For a single-factor analysis of variance, a linear statistical
model can describe the observations of a level with ๐ observations taken under level ๐ (๐ = 1,2,โฆ , ๐; ๐ = 1,2,โฆ , ๐):
๐ฆ๐๐ = ๐ + ๐๐ + ๐๐๐ where ๐ฆ๐๐ is the (๐๐)th observation, ๐ is the overall mean, ๐๐ is
the ๐th level effect, and ๐๐๐ is random error.
24.3 Single-factor ANOVA
Hypothesis Test
โข In ANOVA test, model errors are assumed to be normally
and independently distributed random variables with mean
0 and variance ๐2. The variance is assumed constant for
all factor levels.
โข An expression for the hypothesis test of means is:
๐ป0: ๐1 = ๐2 = โฏ = ๐๐
๐ป๐: ๐๐ โ ๐๐ for at least one pair (๐, ๐)
โข When ๐ป0 is true, all levels have a common mean ๐, which
leads to an equivalent expression in terms of ๐๐.
๐ป0: ๐1 = ๐2 = โฏ = ๐๐ = 0
๐ป๐: ๐๐ โ 0 (for at least one ๐)
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24.4 Single-factor ANOVA Table
Calculations
โข The total sum of squares of deviations about the grand
average ๐ฆ (also referred as the total corrected sum of
squares) represents the overall variability of the data:
๐๐total = (๐ฆ๐๐ โ ๐ฆ )2๐
๐=1
๐
๐=1
โข A division of ๐๐total by the number of degrees of freedom
would yield a sample variance of ๐ฆโs. For this situation, the
overall number of degrees of freedom is ๐๐ โ 1 = ๐ โ 1.
24.4 Single-factor ANOVA Table
Calculations
โข Total variability in data, as measured by ๐๐total (the total
corrected sum of squares) can be partitioned into a sum of
two elements. The first element is the sum of squares for
differences between factor level averages and the grand
average. The second element is the sum of squares of the
differences of observations within factor levels from the
average of factorial levels. The first element is a measure
of the difference between the means of the levels, whereas
the second one is due to random error.
๐๐total = ๐๐factor levels + ๐๐error
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24.4 Single-factor ANOVA Table
Calculations
โข ๐๐factor levels is called the sum of squares due to factor
levels (i.e., between factor levels or treatments)
๐๐factor levels = ๐ (๐ฆ ๐ โ ๐ฆ )2๐
๐=1
โข ๐๐error is called the sum of squares due to error (i.e., within
factor levels or treatments)
๐๐error = (๐ฆ๐๐ โ ๐ฆ ๐)2
๐
๐=1
๐
๐=1
โข When divided by the appropriate number of degrees of
freedom, these SSs give good estimates
24.4 Single-factor ANOVA Table
Calculations
โข When divided by the appropriate number of degrees of
freedom, these SSs give good estimates of the total
variability, the variability between factor levels, and the
variability within factor levels (or error).
โข Expressions for the mean square are
๐๐factor levels =๐๐factor levels
๐ โ 1
๐๐error =๐๐error
๐(๐ โ 1)=
๐๐error
๐ โ ๐
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24.4 Single-factor ANOVA Table
Calculations
โข If there is no difference in treatment means, the two
estimates are presumed to be similar. If there is a
difference, we suspect that the observed difference is
caused by differences in the treatment (factor) levels.
โข F-test statistic tests ๐ป0: there is no difference in factor
levels.
๐น0 =๐๐factor levels
๐๐error
โข ๐ป0 should be rejected if ๐น0 > ๐น๐ผ,๐โ1,๐โ๐.
โข Alternatively, a ๐-value could be calculated for ๐น0. ๐ป0 should
be rejected if ๐ โ value > ๐ผ.
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square ๐ญ๐
Between-
factor levels ๐๐factor levels ๐ โ 1 ๐๐factor levels ๐น0 =
๐๐factor levels
๐๐error
Error
(within-
factor levels)
๐๐error ๐ โ ๐ ๐๐error
Total ๐๐total ๐ โ 1
24.4 Single-factor ANOVA Table
Calculations
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24.5 Estimation of Model
Parameters
โข In addition to factor-level significance, it is useful to
estimate the parameters of the single-factor model and the
CI on the factor-level means.
โข For the single-factor model
๐ฆ๐๐ = ๐ + ๐๐ + ๐๐๐ Estimates for the overall mean and factor-level effects are
๐ = ๐ฆ
๐ ๐ = ๐ฆ ๐ โ ๐ฆ , ๐ = 1,2,โฆ , ๐
โข A 100(1-)% CI on the ๐th factor level is
๐ฆ ๐ ยฑ ๐ก๐ผ,๐โ๐
๐๐error
๐
24.6 Unbalanced Data
โข A design is considered unbalanced when the number of
observations in the factor level is different. For this
situation, ANOVA equations need slight modification
๐๐factor levels = ๐๐(๐ฆ ๐ โ ๐ฆ )2๐
๐=1
โข A balanced design is preferable to an unbalanced design.
With a balanced design, the power of the test in maximized
and the test statistic is robust to small departure from the
assumption of equal variances.
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24.7 Model Adequacy
โข As in the regression model, valid ANOVA requires that
certain assumptions be satisfied.
โข One typical assumption is that errors are normally and
independently distributed with mean 0 and constant but
unknown variance ๐๐ผ๐ท(0, ๐2). โข To help with meeting the independence and normal
distribution requirement, an experimenter needs to select
an adequate sample size and randomly conduct the trials.
โข After data are collected, computer programs offer routines
to test the assumptions.
โข Generally, in a fixed effects ANOVA, moderate departures
from normality of the residuals are of little concern.
24.7 Model Adequacy
โข In addition to an analysis of residuals, there is also a direct
statistical test for equality of variance.
๐ป0: ๐12 = ๐2
2 = โฏ = ๐๐2
๐ป๐: above not true for at least one ๐๐2
โข Bartlettโs test is frequently used to test this hypothesis
when the normality assumption is valid. Leveneโs test can
be used when the normality assumption is questionable.
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24.8 Analysis of Residuals: Fitted Value Plots and Data Transformations
โข Residual plots should show no structure relative to any
factor included in the fitted response; however, trends in
the data may occur for various reasons.
โข One phenomenon that may occur is inconsistent variance.
Fortunately, a balanced fixed effects model is robust to
variance not being homogeneous.
โข A data transformation may then be used to reduce this
phenomenon in the residuals, which would yield a more
precise significance test.
24.8 Analysis of Residuals: Fitted Value Plots and Data Transformations
โข Another situation occurs when the output is count data,
where a square root transformation may be appropriate,
while a lognormal transformation is often appropriate if the
trial outputs are standard deviation values and a logit might
be helpful when there are upper and lower limits.(Table 24.2)
โข As an alternative to the transformations included in the table,
Box (I988) describes a method for eliminating unnecessary
coupling of dispersion effects and location effects by
determining an approximate transformation using a lambda
plot.
โข With transformations, the conclusions of the analysis apply to
the transformed populations.
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24.8 Analysis of Residuals: Fitted Value Plots and Data Transformations
Data Characteristics Data (๐๐ ๐๐ ๐๐) Transformation
๐ โ ๐๐๐๐ ๐ก๐๐๐ก None
๐ โ ๐2 1 ๐ฅ๐
๐ โ ๐3/2 1 ๐ฅ๐
๐ โ ๐ log ๐ฅ๐
๐ โ ๐, Poisson (count) data ๐ฅ๐ or ๐ฅ๐ + 1
Binomial proportions ๐ ๐๐โ1( ๐๐)
Upper- and lower-bound data
(e.g., 0~1 probability of failure)
Logit transformation:
๐๐๐๐ฅ๐ โ ๐๐๐ค๐๐ ๐๐๐๐๐ก
๐ข๐๐๐๐ ๐๐๐๐๐ก โ ๐ฅ๐
24.9 Comparing Pairs of
Treatment Means
โข The rejection of the null hypothesis in an ANOVA indicates
that there is a difference between the factor levels
(treatments). However, no information is given to determine
which means are different.
โข Sometimes it is useful to make further comparisons and
analysis among groups of factor level means. Multiple
comparison methods assess differences between treatment
means in either the factor level totals or the factor level
averages.
โข Methods include those of Tukey and Fisher. Montgomery
(I997) describes several methods of making comparisons.
โข The analysis of means (ANOM) approach is to compare
individual means to a grand mean.
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24.10 Example 24.1:
Single-Factor ANOVA
โข The bursting strengths of diaphragms were determined in an
experiment. Use analysis of variance techniques to determine
if there is a statistically significant difference at a level of 0.05.
Type 1 Type 2 Type 3 Type 4 Type 5 Type 6 Type 7
59.0 65.7 65.3 67.9 60.6 73.1 59.4
62.3 62.8 63.7 67.4 65.0 71.9 61.6
65.2 59.1 68.9 62.9 68.2 67.8 56.3
65.5 60.2 70.0 61.7 66.0 67.4 62.7
24.10 Example 24.1:
Single-Factor ANOVA
Minitab:
Stat
ANOVA
One-Way
Graphs
Boxplot
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24.10 Example 24.1:
Single-Factor ANOVA
Minitab:
Stat
ANOVA
One-Way
Graphs
Ind. values
24.10 Example 24.1:
Single-Factor ANOVA
Minitab:
Stat
ANOVA
One-Way
One-way ANOVA: Strength versus Type
Source DF SS MS F P
Type 6 265.34 44.22 4.92 0.003
Error 21 188.71 8.99
Total 27 454.05
S = 2.998 R-Sq = 58.44% R-Sq(adj) = 46.57%
Individual 95% CIs For Mean Based on Pooled StDev
Level N Mean StDev ------+---------+---------+---------+---
1 4 63.000 3.032 (-----*-----)
2 4 61.950 2.942 (-----*-----)
3 4 66.975 2.966 (-----*-----)
4 4 64.975 3.134 (-----*-----)
5 4 64.950 3.193 (-----*-----)
6 4 70.050 2.876 (-----*-----)
7 4 60.000 2.823 (-----*-----)
------+---------+---------+---------+---
60.0 65.0 70.0 75.0
Pooled StDev = 2.998
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24.10 Example 24.1:
Single-Factor ANOVA
Minitab:
ANOVA
Test of equal variances
24.10 Example 24.1:
Single-Factor ANOVA
Minitab:
Stat
ANOVA
One-Way
Graphs
Four in One
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24.11 Analysis of Means (ANOM)
โข Analysis of means (ANOM) is a statistical test procedure in
a graphical format.
Group 1 2 3 โฏ ๐
๐ฅ11 ๐ฅ21 ๐ฅ31 โฏ ๐ฅ๐1
๐ฅ12 ๐ฅ22 ๐ฅ32 โฏ ๐ฅ๐2
๐ฅ13 ๐ฅ23 ๐ฅ33 โฏ ๐ฅ๐3
โฎ โฎ โฎ โฎ โฎ
๐ฅ1๐ ๐ฅ2๐ ๐ฅ3๐ โฏ ๐ฅ๐๐
๐ฅ 1 ๐ฅ 2 ๐ฅ 3 โฏ ๐ฅ ๐
๐ 1 ๐ 2 ๐ 3 โฏ ๐ ๐
24.11 Analysis of Means (ANOM)
โข The grand mean ๐ฅ is simply the average of the group
means (๐ฅ ๐).
๐ฅ = ๐ฅ ๐
๐๐=1
๐
โข The pooled estimate for the standard deviation is as follows
๐ = ๐ ๐
2๐๐=1
๐
โข The lower and upper decision lines (LDL and UDL) are
๐ฟ๐ท๐ฟ = ๐ฅ โ โ๐ผ๐ ๐ โ 1
๐๐ ๐๐ท๐ฟ = ๐ฅ + โ๐ผ๐
๐ โ 1
๐๐
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24.11 Analysis of Means (ANOM)
โข The lower and upper decision lines (LDL and UDL) are
๐ฟ๐ท๐ฟ = ๐ฅ โ โ๐ผ๐ ๐ โ 1
๐๐ ๐๐ท๐ฟ = ๐ฅ + โ๐ผ๐
๐ โ 1
๐๐
โข โ๐ผ is from Table I for risk level ๐ผ, number of means ๐, and
degrees of freedom ๐ โ 1 ๐ .
โข The means are then plotted against the decision lines. If
any mean fall outside the decision lines, there is a
statistically significant difference for this mean from the
grand mean.
โข If normality can be assumed (๐๐ > 5 ๐๐๐ ๐ 1 โ ๐ > 5),
ANOM is also directly applicable to attribute data.
24.12 Example 24.2:
Analysis of Means (ANOM)
โข The bursting strengths of diaphragms were determined in an
experiment. Use analysis of variance techniques to determine
if there is a statistically significant difference at a level of 0.05.
Type 1 Type 2 Type 3 Type 4 Type 5 Type 6 Type 7 Sum 59.0 65.7 65.3 67.9 60.6 73.1 59.4
62.3 62.8 63.7 67.4 65.0 71.9 61.6
65.2 59.1 68.9 62.9 68.2 67.8 56.3
65.5 60.2 70.0 61.7 66.0 67.4 62.7
Mean 63.0 62.0 67.0 65.0 65.0 70.1 60.0 451.9
Var. 9.19 8.66 8.80 9.82 10.20 8.27 7.97 62.9
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โข The grand mean ๐ฅ is
๐ฅ = ๐ฅ ๐
๐๐=1
๐=
451.9
7= 64.6
โข The pooled standard deviation is
๐ = ๐ ๐2
๐๐=1
๐=
62.90
7= 3.00
โข The lower and upper decision lines (LDL and UDL) are
๐ฟ๐ท๐ฟ = ๐ฅ โ โ๐ผ๐ ๐ โ 1
๐๐= 64.6 โ 2.94 3.0
7 โ 1
7 4= 60.49
๐๐ท๐ฟ = 68.65
24.12 Example 24.2:
ANOM of Injection-Molding Data
24.12 Example 24.2:
ANOM of Injection-Molding Data
Minitab:
Stat
ANOVA
Analysis of Means
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24.13 Example 24.3:
Analysis of Means (ANOM)
Example 15.1
Example 22.4
โข A comparison of the proportion of total variability of the
factor levels to the error term could be made in % units
using a controversial epsilon square relationship:
๐factor level2 = 100 ร
๐๐factor
๐๐total
๐error2 = 100 ร
๐๐error
๐๐total
โข Consider the situation in which a process was randomly
sampled using conventional, rational sampling practices.
Consider also that there were between 25 and 100 sets of
samples taken over time.
24.14 Six Sigma Considerations*
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โข For this type of data, the sums of squares from an ANOVA
table can be used to break down total variability into 2 parts.
โข The division of these sums of squares by the correct
number of degrees of freedom yields estimates for the
different source of variation (total, between subgroups, and
within subgroups).
โข The estimator of total variability gives an estimate for LT
capability, while the estimator of within group variability
gives an estimate for ST capability.
โข These concepts of variability can be used to represent the
influence of time on a process.
24.14 Six Sigma Considerations*
โข The ST and LT standard deviation estimates from an
ANOVA table are
๐ ๐ฟ๐ = (๐ฆ๐๐ โ ๐ฆ )2๐
๐=1๐๐=1
๐๐ โ 1
๐ ๐๐ = (๐ฆ๐๐ โ ๐ฆ ๐)2
๐๐=1
๐๐=1
๐(๐ โ 1)
โข These two estimators are useful in calculating the ST and
LT capability / performance of the process.
24.14 Six Sigma Considerations*
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โข The variable used to measure this capability / performance
is Z. Short-term Z values for the process are
๐LSL,ST =๐ฟ๐๐ฟ โ ๐
๐ ๐๐ ๐USL,ST =
๐๐๐ฟ โ ๐
๐ ๐๐
โข The nominal specification ๐ value is used because it
represents the potential capability of the process.
โข Long-term Z values for the process are
๐LSL,LT =๐ฟ๐๐ฟ โ ๐
๐ ๐ฟ๐ ๐USL,ST =
๐๐๐ฟ โ ๐
๐ ๐ฟ๐
24.14 Six Sigma Considerations*
โข Probability values can then be obtained from the normal
distribution for the different values of ๐.
โข These probabilities correspond to the frequency of
occurrence beyond specification limits.
โข Multiplication of these probabilities by one million gives
DPMO.
24.14 Six Sigma Considerations*
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24.15 Example 24.4:
Determining Process Capability Using
One-Factor ANOVA # ๐ฅ R
1 0.65 0.70 0.65 0.65 0.85 0.70 0.20
2 0.75 0.85 0.75 0.85 0.65 0.77 0.20
3 0.75 0.80 0.80 0.70 0.75 0.76 0.10
4 0.60 0.70 0.70 0.75 0.65 0.68 0.15
5 0.70 0.75 0.65 0.85 0.80 0.75 0.20
6 0.60 0.75 0.75 0.85 0.70 0.73 0.25
7 0.75 0.80 0.65 0.75 0.70 0.73 0.15
8 0.60 0.70 0.80 0.75 0.75 0.72 0.20
9 0.65 0.80 0.85 0.85 0.75 0.78 0.20
10 0.60 0.70 0.60 0.80 0.65 0.67 0.20
11 0.80 0.75 0.90 0.50 0.80 0.75 0.40
12 0.85 0.75 0.85 0.65 0.70 0.76 0.20
13 0.70 0.70 0.75 0.75 0.70 0.72 0.05
14 0.65 0.70 0.85 0.75 0.60 0.71 0.25
15 0.90 0.80 0.80 0.75 0.85 0.82 0.15
16 0.75 0.80 0.75 0.80 0.65 0.75 0.15
โข Example 11.2 Process
capability / performance
metrics
โข Example 22.3 Variance
components
24.15 Example 24.4:
Determining Process Capability Using
One-Factor ANOVA
One-way ANOVA: Data versus Subgroup
Source DF SS MS F P
Subgroup 15 0.10950 0.00730 1.12 0.360
Error 64 0.41800 0.00653
Total 79 0.52750
๐ ๐ฟ๐ = (๐ฆ๐๐ โ ๐ฆ )2๐
๐=1๐๐=1
๐๐ โ 1=
0.52750
5 16 โ 1= 0.081714
๐ ๐๐ = (๐ฆ๐๐ โ ๐ฆ ๐)2
๐๐=1
๐๐=1
๐(๐ โ 1)=
0.41800
16(5 โ 1)= 0.080816
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๐LSL,ST =๐ฟ๐๐ฟ โ ๐
๐ ๐๐=
.5 โ .7
.080816 = โ2.4747 ๐USL,ST=
๐๐๐ฟ โ ๐
๐ ๐๐= 2.4747
๐ ๐ > ๐USL,ST = ๐ ๐ > 2.4747 = 0.006667
๐ ๐ < ๐LSL,ST = ๐ ๐ < โ2.4747 = 0.006667
๐๐๐ = 0.006667 + 0.006667 = 0.013335
๐LSL,LT =๐ฟ๐๐ฟ โ ๐
๐ ๐ฟ๐=
.5 โ .7375
.081714= โ2.9065 ๐USL,ST =
๐๐๐ฟ โ ๐
๐ ๐ฟ๐= 1.9886
๐ ๐ > ๐USL,LT = ๐ ๐ > 1.9886 = 0.023373
๐ ๐ < ๐LSL,LT = ๐ ๐ < โ2.9065 = 0.001828
๐๐ฟ๐ = 0.023373 + 0.001828 = 0.02520
24.15 Example 24.4:
Determining Process Capability Using
One-Factor ANOVA
โข When conducting the 2-sample t-test to compare the
average of two groups, the data in both groups must be
sampled from a normally distributed population. If that
assumption does not hold, the nonparametric Mann-Whitney
test is a better safeguard against drawing wrong conclusions.
โข The Mann-Whitney test compares the medians from two
populations and works when the Y variable is continuous,
discrete-ordinal or discrete-count, and the X variable is
discrete with two attributes. Of course, the Mann-Whitney
test can also be used for normally distributed data, but in that
case it is less powerful than the 2-sample t-test.
24.16 Non-Parametric Estimate:
Mann-Whitney Test Procedure
http://www.isixsigma.com/tools-templates/hypothesis-
testing/making-sense-mann-whitney-test-median-comparison/
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Testing Hypothesis on the Difference
of Medians โ Mann Whitney Test
โข Null Hypothesis: ๐ป0: 1 = 2
โข Test statistic: ๐ = ๐1 โ๐1(๐1+1)
2
โข ๐ป๐: 1 < 2
โข Critical value ๐๐ (from Table, with n1, n2, )
โข Reject ๐ป0 if ๐ < ๐๐
โข Ha: 1 > 2
โข Find ๐โ (from Table, with n1, n2, ), ๐๐ = ๐1๐2 โ ๐โ
โข Reject ๐ป0 if ๐ > ๐๐
โข ๐ป๐: 1 โ 2
โข Find ๐๐๐๐ค๐๐ (from Table, with n1, n2, /2), ๐๐ข๐๐๐๐ = ๐1๐2 โ ๐๐๐๐ค๐๐
โข Reject ๐ป0 if ๐ < ๐๐๐๐ค๐๐ or ๐ > ๐๐ข๐๐๐๐
Where S1 is the sum of ranks of sample 1
Testing Hypothesis on the Difference
of Medians โ Mann Whitney Test
http://www.lesn.appstate.edu/olson/stat_directory/Statistical%20procedu
res/Mann_Whitney%20U%20Test/Mann-Whitney%20Table.pdf
Two-tailed Test
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24
Testing Hypothesis on the Difference
of Medians โ Mann Whitney Test
http://www.lesn.appstate.edu/olson/stat_directory/Statistical%20procedu
res/Mann_Whitney%20U%20Test/Mann-Whitney%20Table.pdf
One-tailed Test
Testing Hypothesis on the
Difference of Medians โ Example
โข =.05
โข Null Hypothesis: H0: 1 = 2
โข Test statistic:
๐ = ๐1 โ๐1 ๐1 + 1
2= 29.5 โ
5 5 + 1
2= 14.5
โข Ha: 1 โ 2
โข Ulower= 3 (Table VI, with n1=5, n2=6, .025), Uupper = n1n2 โ Ulower = (5)(6)-3 = 27
โข Reject H0 if U<Ulower or U>Uupper
โข Fail to reject H0
X Rank
4.6 B 1 1
8.9 B 2 2
9.5 A 3 3
9.7 A 4 4
10.8 A 5 5
11.1 B 6 6
12.9 A 7.5 7.5
12.9 B 7.5 7.5
13.0 B 9 9
16.0 A 10 10
16.4 B 11 11
29.5 36.5
Model A 9.5 10.8 12.9 16.0 9.7
Model B 12.9 11.1 13.0 16.4 8.9 4.6
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Testing Hypothesis on the
Difference of Medians โ Example
Mann-Whitney Test and CI: Model A, Model B
N Median
Model A 5 10.800
Model B 6 12.000
Point estimate for ETA1-ETA2 is -0.050
96.4 Percent CI for ETA1-ETA2 is (-3.502,6.202)
W = 29.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 1.0000
The test is significant at 1.0000 (adjusted for ties)
Minitab:
Stat
Nonparametrics
Mann-Whitney
โข A Kruskal-Wallis test provides an alternative to a one-way
ANOVA. This test is a generalization of Mann-Whitney test.
โข The null hypothesis is all medians are equal. The alternative
hypothesis is the medians are not all equal.
โข For this test, it is assumed that independent random samples
taken from different populations have a continuous
distribution with the same shape.
โข For many distributions, the Kruskal-Wallis test is more
powerful than Moodโs median test, but it is less robust
against outliers.
24.16 Non-Parametric Estimate:
Kruskal-Wallis Test
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โข The yield per acre for 4 methods of growing corn
24.17 Example 24.5:
Nonparametric Kruskal-Wallis Test
Method 1 Method 2 Method 3 Method 4
83 91 101 78
91 90 100 82
94 81 91 81
89 83 93 77
89 84 96 79
96 83 95 81
91 88 94 80
92 91 81
90 89
84
24.17 Example 24.5:
Nonparametric Kruskal-Wallis Test
Kruskal-Wallis Test: Yield versus Mothod
Kruskal-Wallis Test on Yield
Mothod N Median Ave Rank Z
1 9 91.00 21.8 1.52
2 10 86.00 15.3 -0.83
3 7 95.00 29.6 3.60
4 8 80.50 4.8 -4.12
Overall 34 17.5
H = 25.46 DF = 3 P = 0.000
H = 25.63 DF = 3 P = 0.000 (adjusted for ties)
Minitab:
Stat
Nonparametrics
Kruskal-Wallis
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โข Like the Kruskal-Wallis test, a Moodโs median test (also
called a median test or sign scores test) is a nonparametric
alternative to ANOVA.
โข In this chi-square test, the null hypothesis is the population
medians are equal. The alternative hypothesis is the
medians are not all equal.
โข For this test, it is assumed that independent random samples
taken from different populations have a continuous
distribution with the same shape.
โข Moodโs median test is more robust to outliers than the
Kruskal-Wallis test.
24.18 Non-Parametric Estimate:
Moodโs Median Test
24.19 Example 24.6:
Nonparametric Moodโs Median Test
Minitab:
Stat
Nonparametrics
Moodโs Median Test
Mood Median Test: Yield versus Mothod
Mood median test for Yield
Chi-Square = 17.54 DF = 3 P = 0.001
Individual 95.0% CIs
Mothod N<= N> Median Q3-Q1 ---------+---------+---------+-----
--
1 3 6 91.0 4.0 (--*---)
2 7 3 86.0 7.3 (---*-----)
3 0 7 95.0 7.0 (---*------)
4 8 0 80.5 2.8 (---*)
---------+---------+---------+-----
--
84.0 91.0 98.0
Overall median = 89.0
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โข Variability in an experiment can be caused by nuisance
factors in which we have no interest.
โข These nuisance factors are sometimes unknown and not
controlled. Randomization guards against this type of factor
affecting results.
โข In other situation, the nuisance factor is known but not
controlled. When we observe the value of a factor, it can be
compensated for by using analysis of covariance techniques.
โข In yet another situation, the nuisance factor is both known
and controllable. We can systematically eliminate the effect
on comparisons among factor level considerations (i.e.,
treatments) by using a randomized block design.
24.20 Other Considerations
โข Experiment results can often be improved dramatically
through the wise management of nuisance factors.
โข Statistical software can offer blocking and covariance
analysis options.
24.20 Other Considerations