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AOV Assumption Checking and
Transformations (§8.4-8.5)
• How do we check the Normality of residuals assumption in AOV?
• How do we check the Homogeneity of variances assumption in AOV? (§7.4)
• What to do if these assumptions are not met?
Model Assumptions
• Homoscedasticity (common group variances).
• Normality of residuals. • Independence of residuals. (Hopefully
achieved through randomization.)• Effect additivity. (Only an issue in multi-way
AOV; later).
Checking the Equal Variance Assumption
2t
22
210 :H
Hartley’s Test: A logical extension of the F test for t=2.
2min
2max
max s
sF
Reject if Fmax > F,t,n-1, tabulated in Table 12.
HA: some of the variances are different from each other
Little work but little power
Requires equal replication, n, among groups. Requires normality.
Bartlett’s Test
Bartlett’s Test: Allows unequal replication, but requires normality.
t
1i
2iei
2e
t
1ii slog)1n(slog)1n(C
t
1i
2i2
t
ssIf C > 2
(t-1),then apply the correction term
t
1ii
t
1i i )1n(
1
)1n(
1
)1t(3
11CF
Reject if C/CF > 2(t-1),
More work but better power
T.S.
R.R.
Levene’s Test
t
iTiij
n
ji
i
t
ii
tnzzn
tzznL
i
1
2
1
2
1
)/()(
)1/()(
More work but powerful result.
T.S.
Let ij ij iz y y iy = sample median of i-th group
R.R. Reject H0 if 1 2,df ,dfL F df1 = t -1df2 = nT - t
Use Table 8.Essentially an AOV on the zij
t
iiT nn
1
MinitabTest for Equal Variances
Response Resist
Factors Sand
ConfLvl 95.0000
Bonferroni confidence intervals for standard deviations
Lower Sigma Upper N Factor Levels
1.70502 3.28634 14.4467 5 15
1.89209 3.64692 16.0318 5 20
1.07585 2.07364 9.1157 5 25
1.07585 2.07364 9.1157 5 30
1.48567 2.86356 12.5882 5 35
Bartlett's Test (normal distribution)
Test Statistic: 1.890
P-Value : 0.756
Levene's Test (any continuous distribution)
Test Statistic: 0.463
P-Value : 0.762Do not reject H0 since p-value >
0.05 (traditional )
Minitab Help
Use Bartlett’s test when the data come from normal distributions; Bartlett’s test is not robust to departures from normality. Use Levene’s test when the data come from continuous, but not necessarily normal, distributions.
The computational method for Levene’s Test is a modification of Levene’s procedure [10] developed by [2]. This method considers the distances of the observations from their sample median rather than their sample mean. Using the sample median rather than the sample mean makes the test more robust for smaller samples.
Stat > ANOVA > Test for Equal Variances
SAS Programproc glm data=stress;
class sand;
model resistance = sand / solution;
means sand / hovtest=bartlett;
means sand / hovtest=levene(type=abs);
means sand / hovtest=levene(type=square);
means sand / hovtest=bf; /* Brown and Forsythe mod of Levene */
title1 'Compression resistance in concrete beams as';
title2 ' a function of percent sand in the mix';
run;
Hovtest only works when one factor in (right hand side) model.
SASBartlett's Test for Homogeneity of resistance Variance
Source DF Chi-Square Pr > ChiSq
sand 4 1.8901 0.7560
Levene's Test for Homogeneity of resistance Variance
ANOVA of Absolute Deviations from Group Means
Sum of Mean
Source DF Squares Square F Value Pr > F
sand 4 8.8320 2.2080 0.95 0.4573
Error 20 46.6080 2.3304
Levene's Test for Homogeneity of resistance Variance
ANOVA of Squared Deviations from Group Means
Sum of Mean
Source DF Squares Square F Value Pr > F
sand 4 202.2 50.5504 0.85 0.5076
Error 20 1182.8 59.1400
Brown and Forsythe's Test for Homogeneity of resistance Variance
ANOVA of Absolute Deviations from Group Medians
Sum of Mean
Source DF Squares Square F Value Pr > F
sand 4 7.4400 1.8600 0.46 0.7623
Error 20 80.4000 4.0200
hovtest=bf;
hovtest=bartlett;
hovtest=levene(type=abs);
hovtest=levene(type=square);
SPSS
Test of Homogeneity of VariancesRESIST
.947 4 20 .457
LeveneStatistic df1 df2 Sig.
Since the p-value (0.457) is greater than our (typical) =0.05 Type I error risk level, we do not reject the null hypothesis.
This is Levene’s original test in which the zij are centered on group means and not medians.
R
Tests of Homogeneity of Variances
bartlett.test(): Bartlett’s Test.
fligner.test(): Fligner-Killeen Test (nonparametric).
Checking for Normality
TOOLS
1. Histogram of all residuals (ij).2. Normal probability (Q-Q) plot.3. Formal test for normality.
Reminder: Normality of the RESIDUALS is assumed. The original data are assumed normal also, but each group may have a different mean if HA is true. Practice is to first fit the model, THEN output the residuals, then test for normality of the residuals. This APPROACH is always correct.
Histogram of Residualsproc glm data=stress;
class sand;
model resistance = sand / solution;
output out=resid r=r_resis p=p_resis ;
title1 'Compression resistance in concrete beams as';
title2 ' a function of percent sand in the mix';
run;
proc capability data=resid;
histogram r_resis / normal;
ppplot r_resis / normal square ;
run;
Probability Plots
A scatter plot of the percentiles on the residuals versus the percentiles of a standard normal distribution. The basic idea is that if the residuals are truly normally distributed, values for these percentiles should lie on a straight line.
• Compute and sort the residuals (1), (2),…, (n).
• Associate to each residual a standard normal
percentile. [ z(i) = normsinv((i-.5)/n)].
• Plot z(i) versus (i). Compare to straight line (don’t care
so much about which line).
Spreadsheet & R
Percentile pi = (i-0.5)/n
Normal percentile
=NORMSINV(pi)
Percentile Residual Normal Percentile0.02 -4.4 -2.05370.06 -3.8 -1.55480.10 -3.6 -1.28160.14 -3.4 -1.08030.18 -2.6 -0.91540.22 -2.6 -0.77220.26 -2.6 -0.64330.30 -1.6 -0.52440.34 -0.8 -0.41250.38 -0.6 -0.30550.42 0.2 -0.20190.46 0.2 -0.10040.50 0.4 0.00000.54 0.4 0.10040.58 0.4 0.20190.62 0.4 0.30550.66 1.4 0.41250.70 1.4 0.52440.74 1.4 0.64330.78 1.6 0.77220.82 2.4 0.91540.86 2.6 1.08030.90 3.6 1.28160.94 4.2 1.55480.98 5.4 2.0537
In EXCEL: scatterplot of percentile versus Normal percentile. Use AddLine option.
In R with residuals in “y”:
qqnorm(y)
qqline(y)
Excel Probability PlotProbability Plot - Percent Sand Data
-2.500
-2.000
-1.500
-1.000
-0.500
0.000
0.500
1.000
1.500
2.000
2.500
-6 -4 -2 0 2 4 6
Data Percentiles
No
rmal
Per
cen
tile
s
Probability Plot
SAS (note axes changed)
Minitab
These look normal!
Formal Normality Tests
• Kolmogorov-Smirnov test.
• Shapiro-Wilks test (n < 50).
• D’Agostino’s test (n>=50)
Many, many tests (a favorite pass-time of statisticians is developing new tests for normality.)
All quite conservative – they fail to reject the null hypothesis of normality more often than they should.
Shapiro-Wilk’s W test
2k
1i n j 1 jd
j 1
W a ( )
1, 2, …, n represent data ranked from smallest to largest.
H0: The population has a normal distribution.HA: The population does not have a normal distribution.
T.S.n
2i
i 1
d ( )
nk
2(n 1)
k2
If n is even
If n is odd.R.R. Reject H0 if W < W0.05
Coefficients ai come from a table.
Critical values of W come from a table.
Shapiro-Wilk Coefficients
Shapiro-Wilk Coefficients
Shapiro-Wilk W Table
D’Agostino’s Test
(D 0.28209479) nY
0.02998598
12n
21jn
j 1
n1
j2j 1
2
s ( )
[ j (n 1)]
Dn s
1, 2, …, n represent data ranked from smallest to largest.
H0: The population has a normal distribution.HA: The population does not have a normal distribution.
T.S.
R.R. (two sided test) Reject H0 if
0.025 0.975Y Y or Y Y Y0.025 and Y0.975 come from a table of percentiles of the Y statistic.
Transformations to Achieve Homoscedasticity
What can we do if the homoscedasticity (equal variances) assumption is rejected?
1. Declare that the AOV model is not an adequate model for the data. Look for alternative models. (Later.)
2. Try to “cheat” by forcing the data be homoscedastic through a transformation of the response variable Y. (Variance Stabilizing Transformations.)
ii yz
i ik2 This transformation works when we notice the variance changes as a linear function of the mean.
• Useful for count data (Poisson Distributed).• For small values of Y, use Y+.5.
Typical use: Counts of items when countsare between 0 and 10.
Square Root Transformation
k>0
Response is positive and continuous.
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
0 10 20 30 40
Sample Mean
Sam
ple
Var
ian
ce
This transformation tends to work when the variance is a linear function of the square of the mean
• Replace Y by Y+1 if zero occurs.• Useful if effects are multiplicative (later).• Useful If there is considerable
heterogeneity in the data.
Z Yln( )2 2ki i
Typical use: Growth over time.Concentrations.Counts of times when countsare greater than 10.
Logarithmic Transformation
k>0
Response is positive and continuous.
0
20
40
60
80
100
120
140
160
180
200
0 10 20 30 40
Sample Mean
Sam
ple
Var
ian
ce
With proportions, the variance is a linear function of the mean times (1-mean) where the sample mean is the expected proportion.
• Y is a proportion (decimal between 0 and 1).• Zero counts should be replaced by 1/4, and N by N-1/4 before converting to percentages
YarcsinYsinZ 1
i i ik2 1
Response is a proportion.
Typical use: Proportion of seeds germinating.Proportion responding.
ARCSINE SQUARE ROOT
0
0.05
0.1
0.15
0.2
0.25
0.3
0 0.2 0.4 0.6 0.8 1
Sample Mean
Response is positive and continuous.
This transformation works when the variance is a linear function of the fourth root of the mean.
• Use Y+1 if zero occurs.• Useful if the reciprocal of the original
scale has meaning.
ZY
1
i ik2 4
Typical use: Survival time.
Reciprocal Transformation
0
5000
10000
15000
20000
25000
30000
35000
0 10 20 30 40 50 60 70
Sample Mean
ii k
Suppose we apply the power transformation: pyz Suppose the true situation is that the variance is proportional to the th power of the mean.
* 1pi i
If p is taken as 1-, then the variance of Z will not depend on the mean, i.e. the variance will be constant. This is a Variance stabilizing transformation.
Power Family of Transformations (1)
In the transformed variable we will have:
With replicated data, can sometimes be found empirically by fitting:
Estimate:2
1
1ˆ ( )
1
ˆ
in
i ij iji
i i
y yn
y
can be estimated by least squares (regression – Next Unit).
ˆˆ 1p If is zero use the logarithmic transformation.
p̂
Power Family of Transformations (2)
0.0000
0.5000
1.0000
1.5000
2.0000
2.5000
3.0000
3.5000
4.0000
4.5000
5.0000
0.980 1.080 1.180 1.280 1.380 1.480 1.580 1.680 1.780 1.880
Sample Mean
)ˆln()ˆln( ii C
n
ii
i
i
i
yn
y
yz
1
1
ln1
exp
0ln
01
suggestedtransformation
geometric mean of the original data.
Exponent, , is unknown. Hence the model can be viewed as having an additional parameter which must be estimated (choose the value of that minimizes the residual sum of squares).
Box and Cox Transformations
Regression? ANOVAno
yes
Test forHomoscedasticity
reject
accept
Transform
Transformed Data
Box/Cox FamilyPower Family
Traditional
Fit linearmodel
Plotresiduals
Fit Effect Model
Handling Heterogeneity
Not OK
OK
OK
Regression? ANOVAno
yes
Fit linearmodel
Estimategroup means
Residuals Normal?
no
yes
Transform Different Model
Probability plotFormal Tests
Transformations to Achieve Normality
OK