chapter 24 single-factor (one-way) analysis of variance

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Chapter 24

Single-Factor (One-Way)

Analysis of Variance (ANOVA)

Analysis of Means (ANOM)

Introduction

• This chapter describes single-factor analysis of

variance (ANOVA) experiments with 2 or more levels

(or treatments).

• The method is based on a fixed effects model (as

opposed to a random effects model, or components of

variance model). It tests the 𝐻0 that the different

processes give an equal response.

• 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

• 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.

ANOVA and ANOM

• 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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ANOVA and ANOM

• 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 (𝜀2).

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

• 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.

• For a single-factor analysis of variance, a linear statistical

model can describe the observations of a level with 𝑗 observations taken under level 𝑖 (𝑖 = 1,2,… , 𝑎; 𝑗 = 1,2,… , 𝑛):

𝑦𝑖𝑗 = 𝜇 + 𝜏𝑖 + 𝜀𝑖𝑗 where 𝑦𝑖𝑗 is the (𝑖𝑗)th observation, 𝜇 is the overall mean, 𝜏𝑖 is

the 𝑖th level effect, and 𝜀𝑖𝑗 is random error.

24.3 Single-factor ANOVA

Hypothesis Test

• In ANOVA test, model errors are assumed to be normally

and independently distributed random variables with mean

0 and variance 𝜎2. The variance is assumed constant for

all factor levels.

• An expression for the hypothesis test of means is:

𝐻0: 𝜇1 = 𝜇2 = ⋯ = 𝜇𝑎

𝐻𝑎: 𝜇𝑖 ≠ 𝜇𝑗 for at least one pair (𝑖, 𝑗)

• When 𝐻0 is true, all levels have a common mean 𝜇, which

leads to an equivalent expression in terms of 𝜏𝑖.

𝐻0: 𝜏1 = 𝜏2 = ⋯ = 𝜏𝑎 = 0

𝐻𝑎: 𝜏𝑖 ≠ 0 (for at least one 𝑖)

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24.4 Single-factor ANOVA Table

Calculations

• The total sum of squares of deviations about the grand

average 𝑦 (also referred as the total corrected sum of

squares) represents the overall variability of the data:

𝑆𝑆total = (𝑦𝑖𝑗 − 𝑦 )2𝑛

𝑗=1

𝑖=1

• A division of 𝑆𝑆total by the number of degrees of freedom

would yield a sample variance of 𝑦’s. For this situation, the

overall number of degrees of freedom is 𝑎𝑛 − 1 = 𝑁 − 1.

Calculations

• Total variability in data, as measured by 𝑆𝑆total (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.

𝑆𝑆total = 𝑆𝑆factor levels + 𝑆𝑆error

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Calculations

• 𝑆𝑆factor levels is called the sum of squares due to factor

levels (i.e., between factor levels or treatments)

𝑆𝑆factor levels = 𝑛 (𝑦 𝑖 − 𝑦 )2𝑎

𝑖=1

• 𝑆𝑆error is called the sum of squares due to error (i.e., within

factor levels or treatments)

𝑆𝑆error = (𝑦𝑖𝑗 − 𝑦 𝑖)2

𝑗=1

𝑖=1

• When divided by the appropriate number of degrees of

freedom, these SSs give good estimates

Calculations

• 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).

• Expressions for the mean square are

𝑀𝑆factor levels =𝑀𝑆factor levels

𝑎 − 1

𝑀𝑆error =𝑆𝑆error

𝑎(𝑛 − 1)=

𝑆𝑆error

𝑁 − 𝑎

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Calculations

• 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.

• F-test statistic tests 𝐻0: there is no difference in factor

levels.

𝐹0 =𝑀𝑆factor levels

𝑀𝑆error

• 𝐻0 should be rejected if 𝐹0 > 𝐹𝛼,𝑎−1,𝑁−𝑎.

• Alternatively, a 𝑝-value could be calculated for 𝐹0. 𝐻0 should

be rejected if 𝑝 − value > 𝛼.

Source of

Variation

Sum of

Squares

Degrees of

Freedom

Square 𝑭𝟎

Between-

factor levels 𝑆𝑆factor levels 𝑎 − 1 𝑀𝑆factor levels 𝐹0 =

𝑀𝑆factor levels

𝑀𝑆error

(within-

factor levels)

𝑆𝑆error 𝑁 − 𝑎 𝑀𝑆error

Total 𝑆𝑆total 𝑁 − 1

Calculations

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24.5 Estimation of Model

Parameters

• 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.

• For the single-factor model

𝑦𝑖𝑗 = 𝜇 + 𝜏𝑖 + 𝜀𝑖𝑗 Estimates for the overall mean and factor-level effects are

𝜇 = 𝑦

𝜏 𝑖 = 𝑦 𝑖 − 𝑦 , 𝑖 = 1,2,… , 𝑎

• A 100(1-)% CI on the 𝑖th factor level is

𝑦 𝑖 ± 𝑡𝛼,𝑁−𝑎

𝑀𝑆error

24.6 Unbalanced Data

• A design is considered unbalanced when the number of

observations in the factor level is different. For this

situation, ANOVA equations need slight modification

𝑆𝑆factor levels = 𝑛𝑖(𝑦 𝑖 − 𝑦 )2𝑎

𝑖=1

• 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

• As in the regression model, valid ANOVA requires that

certain assumptions be satisfied.

• One typical assumption is that errors are normally and

independently distributed with mean 0 and constant but

unknown variance 𝑁𝐼𝐷(0, 𝜎2). • To help with meeting the independence and normal

distribution requirement, an experimenter needs to select

an adequate sample size and randomly conduct the trials.

• After data are collected, computer programs offer routines

to test the assumptions.

• Generally, in a fixed effects ANOVA, moderate departures

from normality of the residuals are of little concern.

24.7 Model Adequacy

• In addition to an analysis of residuals, there is also a direct

statistical test for equality of variance.

𝐻0: 𝜎12 = 𝜎2

2 = ⋯ = 𝜎𝑎2

𝐻𝑎: above not true for at least one 𝜎𝑖2

• Bartlett’s test is frequently used to test this hypothesis

when the normality assumption is valid. Levene’s 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

• 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.

• One phenomenon that may occur is inconsistent variance.

Fortunately, a balanced fixed effects model is robust to

variance not being homogeneous.

• A data transformation may then be used to reduce this

phenomenon in the residuals, which would yield a more

precise significance test.

• 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)

• 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

• With transformations, the conclusions of the analysis apply to

the transformed populations.

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Data Characteristics Data (𝒙𝒊 𝒐𝒓 𝒑𝒊) Transformation

𝜎 ∝ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 None

𝜎 ∝ 𝜇2 1 𝑥𝑖

𝜎 ∝ 𝜇3/2 1 𝑥𝑖

𝜎 ∝ 𝜇 log 𝑥𝑖

𝜎 ∝ 𝜇, Poisson (count) data 𝑥𝑖 or 𝑥𝑖 + 1

Binomial proportions 𝑠𝑖𝑛−1( 𝑝𝑖)

Upper- and lower-bound data

(e.g., 0~1 probability of failure)

Logit transformation:

𝑙𝑜𝑔𝑥𝑖 − 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡

𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 − 𝑥𝑖

24.9 Comparing Pairs of

Treatment Means

• 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.

• 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.

• Methods include those of Tukey and Fisher. Montgomery

(I997) describes several methods of making comparisons.

• 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

• 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:

One-Way

Graphs

Boxplot

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24.10 Example 24.1:

Single-Factor ANOVA

Minitab:

One-Way

Graphs

Ind. values

24.10 Example 24.1:

Single-Factor ANOVA

Minitab:

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:

Test of equal variances

24.10 Example 24.1:

Single-Factor ANOVA

Minitab:

One-Way

Graphs

Four in One

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24.11 Analysis of Means (ANOM)

• Analysis of means (ANOM) is a statistical test procedure in

a graphical format.

Group 1 2 3 ⋯ 𝒌

𝑥11 𝑥21 𝑥31 ⋯ 𝑥𝑘1

𝑥12 𝑥22 𝑥32 ⋯ 𝑥𝑘2

𝑥13 𝑥23 𝑥33 ⋯ 𝑥𝑘3

⋮ ⋮ ⋮ ⋮ ⋮

𝑥1𝑗 𝑥2𝑗 𝑥3𝑗 ⋯ 𝑥𝑘𝑗

𝑥 1 𝑥 2 𝑥 3 ⋯ 𝑥 𝑘

𝑠1 𝑠2 𝑠3 ⋯ 𝑠𝑘

• The grand mean 𝑥 is simply the average of the group

means (𝑥 𝑖).

𝑥 = 𝑥 𝑖

𝑘𝑖=1

• The pooled estimate for the standard deviation is as follows

𝑠 = 𝑠𝑖

2𝑘𝑖=1

• The lower and upper decision lines (LDL and UDL) are

𝐿𝐷𝐿 = 𝑥 − ℎ𝛼𝑠𝑘 − 1

𝑘𝑛 𝑈𝐷𝐿 = 𝑥 + ℎ𝛼𝑠

𝑘 − 1

𝑘𝑛

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𝑘𝑛 𝑈𝐷𝐿 = 𝑥 + ℎ𝛼𝑠

𝑘 − 1

𝑘𝑛

• ℎ𝛼 is from Table I for risk level 𝛼, number of means 𝑘, and

degrees of freedom 𝑛 − 1 𝑘 .

• 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.

• If normality can be assumed (𝑛𝑝 > 5 𝑎𝑛𝑑 𝑛 1 − 𝑝 > 5),

ANOM is also directly applicable to attribute data.

24.12 Example 24.2:

• 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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• The grand mean 𝑥 is

𝑥 = 𝑥 𝑖

𝑘𝑖=1

7= 64.6

• The pooled standard deviation is

𝑠 = 𝑠𝑖2

𝑘𝑖=1

7= 3.00

𝑘𝑛= 64.6 − 2.94 3.0

7 − 1

7 4= 60.49

𝑈𝐷𝐿 = 68.65

24.12 Example 24.2:

ANOM of Injection-Molding Data

24.12 Example 24.2:

ANOM of Injection-Molding Data

Minitab:

Analysis of Means

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24.13 Example 24.3:

Example 15.1

Example 22.4

• 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:

𝜀factor level2 = 100 ×

𝑆𝑆factor

𝑆𝑆total

𝜀error2 = 100 ×

𝑆𝑆error

𝑆𝑆total

• 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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• For this type of data, the sums of squares from an ANOVA

table can be used to break down total variability into 2 parts.

• 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).

• The estimator of total variability gives an estimate for LT

capability, while the estimator of within group variability

gives an estimate for ST capability.

• These concepts of variability can be used to represent the

influence of time on a process.

• The ST and LT standard deviation estimates from an

ANOVA table are

𝜎 𝐿𝑇 = (𝑦𝑖𝑗 − 𝑦 )2𝑛

𝑗=1𝑎𝑖=1

𝑛𝑎 − 1

𝜎 𝑆𝑇 = (𝑦𝑖𝑗 − 𝑦 𝑖)2

𝑛𝑗=1

𝑎𝑖=1

𝑎(𝑛 − 1)

• These two estimators are useful in calculating the ST and

LT capability / performance of the process.

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• The variable used to measure this capability / performance

is Z. Short-term Z values for the process are

𝑍LSL,ST =𝐿𝑆𝐿 − 𝑇

𝜎 𝑆𝑇 𝑍USL,ST =

𝑈𝑆𝐿 − 𝑇

𝜎 𝑆𝑇

• The nominal specification 𝑇 value is used because it

represents the potential capability of the process.

• Long-term Z values for the process are

𝑍LSL,LT =𝐿𝑆𝐿 − 𝜇

𝜎 𝐿𝑇 𝑍USL,ST =

𝑈𝑆𝐿 − 𝜇

𝜎 𝐿𝑇

• Probability values can then be obtained from the normal

distribution for the different values of 𝑍.

• These probabilities correspond to the frequency of

occurrence beyond specification limits.

• Multiplication of these probabilities by one million gives

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24.15 Example 24.4:

Determining Process Capability Using

One-Factor ANOVA # 𝑥 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

• Example 11.2 Process

capability / performance

metrics

• Example 22.3 Variance

components

24.15 Example 24.4:

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

𝜎 𝐿𝑇 = (𝑦𝑖𝑗 − 𝑦 )2𝑛

𝑗=1𝑎𝑖=1

𝑛𝑎 − 1=

0.52750

5 16 − 1= 0.081714

𝜎 𝑆𝑇 = (𝑦𝑖𝑗 − 𝑦 𝑖)2

𝑛𝑗=1

𝑎𝑖=1

𝑎(𝑛 − 1)=

0.41800

16(5 − 1)= 0.080816

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𝑍LSL,ST =𝐿𝑆𝐿 − 𝑇

𝜎 𝑆𝑇=

.5 − .7

.080816 = −2.4747 𝑍USL,ST=

𝑈𝑆𝐿 − 𝑇

𝜎 𝑆𝑇= 2.4747

𝑃 𝑍 > 𝑍USL,ST = 𝑃 𝑍 > 2.4747 = 0.006667

𝑃 𝑍 < 𝑍LSL,ST = 𝑃 𝑍 < −2.4747 = 0.006667

𝑝𝑆𝑇 = 0.006667 + 0.006667 = 0.013335

𝑍LSL,LT =𝐿𝑆𝐿 − 𝜇

𝜎 𝐿𝑇=

.5 − .7375

.081714= −2.9065 𝑍USL,ST =

𝑈𝑆𝐿 − 𝜇

𝜎 𝐿𝑇= 1.9886

𝑃 𝑍 > 𝑍USL,LT = 𝑃 𝑍 > 1.9886 = 0.023373

𝑃 𝑍 < 𝑍LSL,LT = 𝑃 𝑍 < −2.9065 = 0.001828

𝑝𝐿𝑇 = 0.023373 + 0.001828 = 0.02520

24.15 Example 24.4:

One-Factor ANOVA

• 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.

• 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 – Mann Whitney Test

• Null Hypothesis: 𝐻0: 1 = 2

• Test statistic: 𝑈 = 𝑆1 −𝑛1(𝑛1+1)

• 𝐻𝑎: 1 < 2

• Critical value 𝑈𝑐 (from Table, with n1, n2, )

• Reject 𝐻0 if 𝑈 < 𝑈𝑐

• Ha: 1 > 2

• Find 𝑈’ (from Table, with n1, n2, ), 𝑈𝑐 = 𝑛1𝑛2 − 𝑈’

• Reject 𝐻0 if 𝑈 > 𝑈𝑐

• 𝐻𝑎: 1 ≠ 2

• Find 𝑈𝑙𝑜𝑤𝑒𝑟 (from Table, with n1, n2, /2), 𝑈𝑢𝑝𝑝𝑒𝑟 = 𝑛1𝑛2 − 𝑈𝑙𝑜𝑤𝑒𝑟

• Reject 𝐻0 if 𝑈 < 𝑈𝑙𝑜𝑤𝑒𝑟 or 𝑈 > 𝑈𝑢𝑝𝑝𝑒𝑟

Where S1 is the sum of ranks of sample 1

http://www.lesn.appstate.edu/olson/stat_directory/Statistical%20procedu

res/Mann_Whitney%20U%20Test/Mann-Whitney%20Table.pdf

Two-tailed Test

3/18/2013

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 – Example

• =.05

• Null Hypothesis: H0: 1 = 2

• Test statistic:

𝑈 = 𝑆1 −𝑛1 𝑛1 + 1

2= 29.5 −

5 5 + 1

2= 14.5

• Ha: 1 ≠ 2

• Ulower= 3 (Table VI, with n1=5, n2=6, .025), Uupper = n1n2 – Ulower = (5)(6)-3 = 27

• Reject H0 if U<Ulower or U>Uupper

• 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

3/18/2013

Testing Hypothesis on the

Difference of Medians – 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:

Nonparametrics

Mann-Whitney

• A Kruskal-Wallis test provides an alternative to a one-way

ANOVA. This test is a generalization of Mann-Whitney test.

• The null hypothesis is all medians are equal. The alternative

hypothesis is the medians are not all equal.

• For this test, it is assumed that independent random samples

taken from different populations have a continuous

distribution with the same shape.

• For many distributions, the Kruskal-Wallis test is more

powerful than Mood’s median test, but it is less robust

against outliers.

Kruskal-Wallis Test

3/18/2013

• 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

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:

Nonparametrics

Kruskal-Wallis

3/18/2013

• Like the Kruskal-Wallis test, a Mood’s median test (also

called a median test or sign scores test) is a nonparametric

alternative to ANOVA.

• In this chi-square test, the null hypothesis is the population

medians are equal. The alternative hypothesis is the

medians are not all equal.

• For this test, it is assumed that independent random samples

taken from different populations have a continuous

distribution with the same shape.

• Mood’s median test is more robust to outliers than the

Kruskal-Wallis test.

Mood’s Median Test

24.19 Example 24.6:

Nonparametric Mood’s Median Test

Minitab:

Nonparametrics

Mood’s 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

3/18/2013

• Variability in an experiment can be caused by nuisance

factors in which we have no interest.

• These nuisance factors are sometimes unknown and not

controlled. Randomization guards against this type of factor

affecting results.

• 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.

• 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

• Experiment results can often be improved dramatically

through the wise management of nuisance factors.

• Statistical software can offer blocking and covariance

analysis options.

24.20 Other Considerations

chapter 24 single-factor (one-way) analysis of variance

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