Download ANALYSIS OF VARIANCE

2.08.14
One-way ANOVA
 The concept of an ANOVA
 The one-way ANOVA model
 Assumptions
 The F test
 CI estimate of the ith mean
 CI estimate of the difference of 2 means
 Bonferoni and Tukey estimates
Two-way ANOVA
 The two-way ANOVA model
 Assumptions
 The F test
 CI estimates
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ANALYSIS OF VARIANCE: ONE-WAY
Purpose
Compares multiple means: Tests for differences in the means for more than 2 groups
Examples of Applications
 Compare mean strength of concrete developed using 6 different forms
 Determine the variety of beans producing the greatest mean yield
How about multiple t-tests?
 Ease in interpretation: The number of t-tests increases rapidly as the number of
groups to be compared increases, and thus, analysis becomes difficult.

Error Reduction: In completing many analyses, the probability of committing at least
one type I error somewhere in the process increases with the number of tests that are
completed. The probability of committing at least one type I error in an analysis is
called the experiment-wise error rate.
Initial Investigation - Descriptive Comparisons
 Look at the means of each process or group (treatment); compare the variances (are
they approximately the same?).
 Examine side-by-side box plots by treatment. Consider the spread and center points.
Model
Suppose r samples of size n1, n2, …nr are independent samples from normal distributions
with a common variance 2.
Notation
yij is the jth observation in the ith sample
j is the observation number
i is the sample number
Sample 1
x11, x12…x1j
Sample 2
x21, x22…x2j
…
…
Sample i
xi1, xi2…xij
The one-way model is:
yij = i + ij where I is the mean by group
[An observation = population mean + random error]
Residuals: The error terms are independent random variables with mean 0 and
variance 2. They look like a random sample from a normal distribution. To
check, use a normal plot of the residuals
6360ho5_7
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^
eij = yij - I and is found by y ij - yi
eij : residual corresponding to yij
^
y ij is the jth observation from the ith sample (observation number is j and
sample number is i)
y ij = the fitted value corresponding to yij. It corresponds to the sample mean.
Assumptions
1. r samples of size n1, n2,..,nr samples are independently and randomly selected
2. Normality: the values in each group are normally distributed (note that the test is
robust relative to this assumption i.e. modest departures will not adversely affect the
results.)
3. Homogeniety of Variance: the variances within each population are equal (if sample
sizes are equal, modest departures are not serious.) Look at side-by-side box plots and
values of si.
4. Independence of Error: Residuals (e: IN(0,2)) are independent for each value.
[The residuals for one observation are not related to the residual for another.)
Construct a normal plot of the residuals or plot the residuals vs. the fits.
Example for Discussion
Consider the following times to complete a task using 3 different machines.
I
II
III
25
23
21
26
22
23
24
24
20
24
24
21
26
22
20
Compare the time to complete a task using three different machines.
The Hypothesis
Ho: 1 =  2 = … r
(All group means are equal)
Ha:
not all mean are equal
(At least one group mean is not equal to another)
Source of Variation (why do means vary?)
Between Group Variation (treatment) - The values may vary because the treatments are
different. The bigger the effect of the treatment, the more variation in-group means we
will find.
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Within Group (error) - People and specimens are different; they differ whether we treat
them the same or not. There are differences in means because there is variability in
samples.
NOTE:
 Even if Ho is true (i.e. all group means are equal), there will be differences in the
sample means; variability in the samples will create the differences.
 If Ho is true, the between group variation will estimate the variability as well as
the within group variation.
 If Ho is false the between group variation will be larger than the within group...
The F-Test
The variation between or within groups is the basis for the F-test. Consider the following
sum of squares.
Total Variation - includes both between group and within group variation
(variation without regard to treatment)
SST - Total sum of squares (difference between cell mean and grand
means)
SSB = SSTr = Sum of Squares Between groups - Variation from Treatment
SSW = SSE = Random variation-random error - Sum of Squares Within groups
SST = SSB + SSW = SSTr + SSE
MSTr = SSTr/(r-1)
MSE = SSE/(n-r)
F=
MSTr
MSE
df (numerator) = r -1
df(denominator) = n-r
If Ho is false, the Between-Group variation will be larger than the Within-Group
variation.
If Ha is false, the treatment variation and random variation are about the same.
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ANOVA Table
Source
Treatment
(Between)
Error
(Within)
Total
SS
 n (Y
i
i
 (Y
ij
j
 Y )2
 Y i )2
df
r-1
n-r
i
SST
MS
SSTr
= MSTr
r 1
SSE
= MSE
nr
F
F=
MSTr
MSE
n-1
Pooled Sample Variance: Estimates the baseline variation present in a response variable,
sp2 may be thought of as an “average” variance for each groups or the weighted
average of the sample variances.
(Note that the ANOVA assumes equal variances and equal variances are required to
find the pooled sd.)
r = number of samples
ni = sample sizes
si2 = sample variance
sp2 = MSE2
(n1  1) s1  (n2  1) s 22  ...  (nr  1) s 2p
2
2
sp =
(n1  1)  (n2  1)  ...  (nr  1)
r = number of samples
ni = sample sizes
si2 = sample variances
R2: fraction of the raw variability in y accounted for by fitting the model to the data
SSTr
R2 =
SST
Standardized Residuals – see page 458
Confidence Interval Estimate for the ith Population Mean – estimates a cell mean
sp
with df = n - r (total sample size-number of groups)
Yi t
ni
Two-Sided Confidence Interval Estimate of the Difference in PopulationMeans
1 1
with df = n – r
(Yi  Yi ' )  t s p

ni ni '
df = n - r
where n is the total sample size and r is the number of groups or cells
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Prediction Interval of a Single Observation
This interval predicts or locates a single additional observation from a particular
one of the r distributions.
Tolerance Interval
The tolerance interval locates most of a large number of values within one of the r
distributions.
Simultaneous Confidence Levels
 Bonferroni Inequality
 Tukey’s Comparison
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EXAMPLE OF ONE WAY ANOVA (Completely Randomized Design)
An experiment was conducted to compare the wearing qualities of three types of paint when subjected to the
abrasive action of a slowly rotating cloth-surfaced wheel. Ten paint specimens were tested for each paint
type and the number of hours until visible abrasion was apparent was recorded for each specimen.
WEAR DATA FOR 3 TYPES OF PAINT (in hours)
Type
Hours of Wear
148
76 393 520 236 134
55
1
166
415
153
Totals
2296
2
513
264
433
94
535
327
214
165
280
304
3129
3
335
643
216
536
128
723
258
380
594
465
4278
Note that the numbers of the following items are used on the attached Minitab printout. Items in italics are
my notes regarding the interpretation of the data. For XL, see:
http://www.uh.edu/~tech132/6360ho5_7phs.xls.xlsx
Preliminaries
1. Run the descriptive statistics on the data.
2.
Explore the test assumptions

Three (3) samples of size 10 are independently and randomly selected.
This assumption is satisfied with a proper experimental design.

Normality: values in each group are normally distributed (the model is robust relative to this
assumption i.e.... modest departures will not adversely effect the results.)

Homogeneity of Variance: the variance within each group should be equal for all populations ( =
2
1

2
2
=
2
3
).

Look at the variances found in calculating the descriptive statistics.

Do a dot plot or side-by-side box plots.
My print-out does not provide one of these plots, but you should run one to check the spread of the
values.


There is a test called Hartley’s test for Homogeneity of Population Variances that can be run to
check this assumptions. It is not routinely run; it is extremely sensitive to departures from
normality of the populations. A second test is Bartlett’s Test, also sensitive to departures from
normality.

If population variances are extremely different, it may be desirable to do a transformation of the
data. For example transform the data by finding the square root (often useful for count or
percentage data) or the logarithm of each value of x. The type of transformation depends upon
the type of data. Once the data are transformed, perform the ANOVA on the transformed data.
Test results for differences in transformed treatment means and apply to the original treatment
means. It is often difficult to assign a practical interpretation to the transformed variable and
the various treatment means, and some applied statisticians do not use transformations unless
there is evidence to suggest sizable differences among the treatment population variances
(Mendenhall and Beavers)
Independence of Error: Residual are independent for each value.
Check by doing a normal probability plot of the residuals or plot the residuals vs. fits. If the
residuals are normally distributed then the resulting plot will be linear. There is a simple correlation
test that can be used to test the hypothesis that the values are correlated; you need a table that I
will distribute in class.
The normal probability plot is linear and indicates that this assumption is satisfied.
Note: The ANOVA has been show n to be ‘robust’ to violations of these assumptions.
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3. Test the hypotheses:
 1 =  2 =  3 (Population means are equal)
Ha: At least two of the population means are not the same.
The test is an F-test. It looks at the ratio of the mean square between samples (MSB) to the mean square
error, random error within samples (MSE). If the null hypothesis is true, we would expect these values
to be approximately the same and the ratio to be one; if Ho is not true, then MSB should be larger than
MSE.
In this problem the F ratio is 3.51. With 2 and 27 degrees of freedom, the F-table can be used to find a
critical value of 3.35 when  is 0.05. (Recall,  is the probability of rejecting a true Ho.) Since the
calculated F exceeds the table value of F, reject Ho and conclude that at least two of the population
means are not equal. This conclusion is also verified by p = 0.04.
4. If the null hypothesis is rejected then we need to determine which means are different. This is typically determined
by calculating confidence intervals for the difference in population means (1 -  2  2 -  3 ,1 -  3).
We have discussed two procedures for determining simultaneous confidence intervals for multiple
confidence intervals (there are others).

Bonferroni Inequality
( y i  y i ' )  ts p
1
1

n1
n2
This process involves calculating a t-interval for the difference in means where the value of t is
adjusted to allow for an acceptable family confidence level. When many intervals are
constructed, this procedure can be inefficient because the confidence level for each individual
interval would be extremely large.

Tukey’s Method involves the use of Studentized range distribution.
D-10-B)
( yi  yi' ) 
(See Tables D-10-A and
q*
1 1
sp

n1 n2
2
These intervals can be most easily computed using computer packages such as Minitab.

If there is only one difference, a simple t confidence interval could be calculated for the
difference between (usually the largest and smallest) means.

It is sometimes desirable to calculate a confidence interval for one of the means; this CI is
calculated using the t-interval. The Bonferroni Inequality can be applied in determining t.
y
i
 t
s
p
ni
On the print-out , the graph that is shown following the ANOVA summary table indicates that there is a
difference between Paint 1 and Paint 3. This conclusion is verified by the Tukey pairwise comparisons.
(Tukey’s method has a 95% CI family rate with individual comparisons at 98%) With 95% confidence, it
is asserted that the wear time for Paint 3 is from 12 to 385 hours longer than for Paint 1; thus, Paint 3
appears to possess slightly greater wearing quality than Paint 1. There is no statistically significant
difference in the wear time between Paints 1 and 2 or between Paints 2 and 3.
5.
sp2 is needed for the calculations described in item 4. It is the pooled standard deviation and may be
thought of as an “average” variance for each groups or the weighted average of the sample variances.
sp2 estimates the baseline variation that is present in a response variable (hours until apparent
abrasion).
(Note that the ANOVA assumes equal variances and equal variances are required to find the
pooled sd.)
From the ANOVA summary table, find the MSE and take the square root to get sp= 167.88.This value is
the pooled standard deviation which can be thought of as the average (or base) variation of the hours of
wear for any of the three treatments i.e. paint types.
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C1
T1
C2
T2
C3
T3
C6
Hrs
148
76
393
520
236
134
55
166
415
153
513
264
433
94
535
327
214
165
280
304
335
643
216
536
128
723
258
380
594
465
148
76
393
520
236
134
55
166
415
153
513
264
433
94
535
327
214
165
280
304
335
643
216
536
128
723
258
380
594
465
6360ho5_7
C7
Paint
Type
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
9
Worksheet size: 100000 cells
MTB >
describe c1-c3
Descriptive Statistics
Variable
T1
T2
T3
N
10
10
10
Mean
229.6
312.9
427.8
Variable
T1
T2
T3
Minimum
55.0
94.0
128.0
Median
159.5
292.0
422.5
Maximum
520.0
535.0
723.0
TrMean
215.1
312.5
428.4
Q1
119.5
201.7
247.5
StDev
158.2
144.2
196.8
SE Mean
50.0
45.6
62.2
Q3
398.5
453.0
606.2
MTB > Boxplot c1 c2 c3;
SUBC>
Box;
SUBC>
Symbol;
SUBC>
Outlier;
SUBC>
Overlay;
SUBC>
ScFrame;
SUBC>
ScAnnotation.
MTB >
Note: The Following is obtained from the
menu using:
Stat/Anova/One-Way (Unstacked)
Responses in separate columns: C1 C2 C3
MTB > AOVOneway c1 c2 c3.
One-way Analysis of Variance
Analysis of Variance
Source
DF
SS
Factor
2
198080
Error
27
760987
Total
29
959067
MS
99040
28185
F
3.51
P
0.044
Individual 95% CIs For Mean
Based on Pooled StDev
Level
T1
T2
T3
6360ho5_7
N
10
10
10
Mean
229.6
312.9
427.8
StDev
158.2
144.2
196.8
----------+---------+---------+-----(--------*--------)
(--------*--------)
(--------*--------)
----------+---------+---------+------
10
Pooled StDev =
MTB >
Note:
167.9
240
360
480
The Following is obtained from the menu using:
Stat/Anova/One-Way
Responses is stacked in C6; I have factors in C7.
(See p2)
I used the following from this dialog box.
Comparisons: Tukey's Family Error Rate
Graphs: Dotplot, Boxplot, Normal Plot of Residuals,
Residuals vs Fits
MTB > Name c11 = 'RESI1' c12 = 'FITS1'
MTB > Oneway c6 c7 'RESI1' 'FITS1';
SUBC>
Tukey 5;
SUBC> GDotplot;
SUBC> GBoxplot;
SUBC> GNormalplot;
SUBC> GFits.
One-way Analysis of Variance
Analysis of Variance for Hrs
Source
DF
SS
MS
Paint Ty
2
198080
99040
Error
27
760987
28185
Total
29
959067
F
3.51
P
0.044
Individual 95% CIs For Mean Based on Pooled StDev
Level
1
2
3
N
10
10
10
Mean
229.6
312.9
427.8
Pooled StDev =
StDev
158.2
144.2
196.8
167.9
----------+---------+---------+-----(--------*--------)
(--------*--------)
(--------*--------)
----------+---------+---------+------
240
360
480
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0196
Critical value = 3.51
Intervals for (column level mean) - (row level mean)
6360ho5_7
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1
2
-270
103
3
-385
-12
2
-301
71
MTB >
6360ho5_7
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EXAMPLE OF TWO- WAY ANOVA (a Factorial Experiment)
An experiment was conducted to determine the effect of sintering time (two levels) on
the compressive strength of two different metals. Five test specimens were sintered for
each metal at each of two sintering times. The data (in thousands of pounds per square
inch) are shown in the following table.
Metal 1
Metal 2
100 minutes
17.1 15.2 16.5 16.7
14.9
12.3 13.8 10.8 11.6
12.1
200 minutes
19.4 17.2 18.9 20.7
20.1
15.6 17.2 16.7 16.1
18.3
This experiment represents a factorial experiment; every level of every variable is paired
with every level of every other variable. To collect this data, we could process a fixed
quantity of metal 1 for 100 minutes and the same quantity of the metal for 200 minutes.
Measure the compressive strength on the processed metal samples. This procedure is
duplicated for metal 2.
Note this problem involves two factors: metal type and sintering time. The different
settings for a given factor are call levels. In this problem, there are two levels of each
factor i.e. two types of metal and 2 different times. By completing the ANOVA, we can
determine 1) if there is a difference in strength based on metal type, 2) if there is a
difference in strength based on the sintering time, or 3) if there is a difference in strength
based on a particular combination of the metal type and the sintering time interaction.
The main effects of an independent variable is the effect of the variable averaging
over all levels of other variables in the experiment. For example, the main effect of for
metal type involves a comparison of the means of metal 1 and metal 2. If the effect of
metal type depends on sintering time, then there is an interaction effect.
Note that the numbers of the following items are used on the attached Minitab printout. Items in italics are
my notes regarding the interpretation of the data.
Preliminaries
1. Run the descriptive statistics on the data.
2.
Plot the data.
In this example, strength is plotted as the dependent variable on the vertical axis; time is plotted on the
horizontal axis. There should be a separate set of points based on the metal type, producing two ‘lines’.
It can be noted that as you increase the sintering time from 100 to 200 minutes, the compressive
strength increases for both metal 1 and metal 2. Since these lines are approximately ‘parallel,’ we can
speculate that there is no significant interaction, and if there are statistically significant differences in the
compressive strength of the metal, they will be differences resulting from metal type or sintering time.
(Note it would enhance the graph to note error bars that reflect CI estimates of each of the sample
means; this step is not done automatically by the software). There is an interaction when the magnitude
of an effect is greater at one level of a variable than at another level of the variable.
3.
Explore the test assumptions
Independence of Error: Residuals are independent for each value.
Check by doing a normal probability plot of the residuals; plot the residuals vs. fits.
The normal plot is approximately linear and thus, indicates no major problems with the normal
distribution assumption. Similarly, the plot of the residuals vs. the fits (sample means) does not
indicate a trend in  as a function of mean response.
Common Variance: Variances are the same for each population group.
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Check by doing a dot plot of the residuals vs. sample means and check the spread (variability).
3. Test the hypotheses. There are three tests of significance.
Test for Main Effects, Metal Type
 1 =  2 or i = 0 (There are no differences in population means for the metal types)
Ha: At least two of the population means are not the same.
In this problem the F ratio is 41.15, with 1 and 16 degrees of freedom, the F-table can be used to find a
critical value of F when  is 0.05. From the summary table p< 0.001, and thus, we can reject Ho and
conclude that population compressive strengths for metal 1 and metal 2 differ; metal 1 appears to have
a the compressive strength for metal 1 appears to be greater than the strength for metal 2.
Test for Main Effects, Sintering Time
  =  or j = 0 (There are no differences in population means for the two sintering times)
Ha: At least two of the population means are not the same.
In this problem the F ratio is 60.99, with 1 and 16 degrees of freedom, the F-table can be used to find a
critical value of F when  is 0.05. From the summary table p< 0.001, and thus, we can reject Ho and
conclude that population compressive strengths for sintering for 100 and 200 minutes differ; sintering for
200 minutes appears to result in a greater compressive strength.
Test for Interaction Effects
 or ij = 0 (Metal type and sintering time do not interact to produce different compressive strengths)
Ha: Metal type and sintering types interact.
In this problem the F ratio is 2.17, with 1 and 16 degrees of freedom, the F-table can be used to find a
critical value of F when  is 0.05. From the summary table = 0.16, and thus, we cannot reject Ho; we
conclude that population compressive strengths based on the interaction of metal type and sintering
time are not different i.e. all combinations of metal/time result in the same compressive strength.
4. If the null hypothesis is rejected then we need to determine which means are different. This is typically determined
by calculating confidence intervals for the difference in population means (1 -  2  2 -  3, 1 -  3).
We have discussed two procedures for determining simultaneous confidence intervals for multiple
confidence intervals (there are others).

Bonferroni Inequality
( y i  y i ' )  ts p

1
1

n1 n2
Tukey’s Method involves the use of Studentized range distribution
See equation 8-17 on page 456; use table D -10
Difference in strength based on metal type (rows)
( yi  yi' ) 
q * sp
JM
df = n - IJ = 16; I = 2; J = 2; M common cell size = 5; s p = 1.25
(17.65-14.45) + 1.6
[1.6,4.8] lbs. per sq. inch
With 99% confidence, it is asserted that the compressive strength of metal 1 is from 1.6 to 4.8
pounds per square inch greater than for Metal 2.
Difference in strength based on sintering time (columns)
( yi  yi' ) 
q * sp
IM
(17.65-14.45) + 1.6
[2.3,5.5] lbs. per sq. inch
With 99% confidence, it is asserted that the compressive strength of for a 100-minute sintering
time is from 2.3 to 5.5 pounds per square inch greater than for the 200 minute sintering time.
6360ho5_7
14

If there is only one difference, a simple t confidence interval could be calculated for the
difference between (usually the largest and smallest) means.

It is sometimes desirable to calculate a confidence interval for one of the cell means; this CI is
calculated using the t-interval. The Bonferroni Inequality can be applied in determining t.
y
6360ho5_7
ij
 t
s
p
n ij
15
Example of Two-Way ANOVA
Effect of the sintering time on the compressive strength of two different metals
1.
Run the descriptive statistics on the data.
MTB > print c1-c3
Data Display
Row Strength
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
MTB >
SUBC>
SUBC>
SUBC>
Metal
C1
17.1
15.2
16.5
16.7
14.9
12.3
13.8
10.8
11.6
12.1
19.4
17.2
18.9
20.7
20.1
15.6
17.2
16.7
16.1
18.3
C2
Time
C3
1
1
1
1
1
2
2
2
2
2
1
1
1
1
1
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
table c2 c3;
data c1;
means c1;
standard deviation c1.
Tabulated Statistics
Rows: Metal
1
1
17.100
15.200
16.500
16.700
14.900
6360ho5_7
Columns: Time
2
All
19.400
-17.200
18.900
20.700
20.100
16
16.080
0.971
2
19.260
1.339
17.670
2.006
12.300
15.600
-13.800
17.200
10.800
16.700
11.600
16.100
12.100
18.300
12.120
16.780
14.450
1.103
1.043
2.656
All
---14.100
18.020
16.060
2.306
1.729
2.824
Cell Contents -Strength:Data
Mean
StDev
6360ho5_7
17
2. Plot the data and test the hypotheses.
MTB > %Main c2 c3;
SUBC>
Response c1.
Minitab Plot [Use: Stat/ANOVA/Twoway/Interaction Plot]
Interaction Plot - Data Means for Strength
Metal
1
Mean Strength
18
2
16
14
12
1
2
Time
MTB > Name c4 = 'RESI1' c5 = 'FITS1'
MTB > Twoway c1 c3 c2 'RESI1' 'FITS1';
SUBC>
Means c3 c2;
SUBC> GNormalplot;
SUBC> GFits.
Two-way Analysis of Variance
Analysis of Variance for Strength
Source
DF
SS
MS
Time
1
76.83
76.83
Metal
1
51.84
51.84
Interaction
1
2.74
2.74
Error
16
20.16
1.26
Total
19
151.57
Time
1
2
6360ho5_7
Mean
14.10
18.02
F
60.99
41.15
2.17
P
0.000
0.000
0.160
Individual 95% CI
--+---------+---------+---------+--------(----*----)
(----*----)
--+---------+---------+---------+--------13.50
15.00
16.50
18.00
18
Metal
1
2
Mean
17.67
14.45
Individual 95% CI
------+---------+---------+---------+----(-----*------)
(-----*------)
------+---------+---------+---------+----14.40
15.60
16.80
18.00
MTB >
C1
Strength
17.1
15.2
16.5
16.7
14.9
12.3
13.8
10.8
11.6
12.1
19.4
17.2
18.9
20.7
20.1
15.6
17.2
16.7
16.1
18.3
6360ho5_7
C2
Metal
1
1
1
1
1
2
2
2
2
2
1
1
1
1
1
2
2
2
2
2
C3
Time
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
C4
RESI1
1.02
-0.88
0.42
0.62
-1.18
0.18
1.68
-1.32
-0.52
-0.02
0.14
-2.06
-0.36
1.44
0.84
-1.18
0.42
-0.08
-0.68
1.52
C5
FITS1
16.08
16.08
16.08
16.08
16.08
12.12
12.12
12.12
12.12
12.12
19.26
19.26
19.26
19.26
19.26
16.78
16.78
16.78
16.78
16.78
19
Normal Probability Plot of the Residuals
(response is Strength)
2
Normal Score
1
0
-1
-2
-2
-1
0
1
2
Residual
6360ho5_7
20