Stat6305 — Unit6:PartialSolutions 1 of 11
6.1.1.This Latin Square design has 42 = 16 observations. As noted just above not all 3-way combinations of the three effects are present. How many observations would have been required in order to include all 3-way combinations of Blend, Model, and Driver?
If all combinations of Blend, Model, and Driver were included, 4 Blend * 4 Model * 4 Driver = 64 observations would be required.
6.1.2.Suppose that there were 5 Blends, 5 Drivers, and 5 Models. A full "factorial" design with one observation on each 3-way combination would require 125 observations. Make a table showing how you could assign Blends A-E to make a Latin Square design for this situation.
One possible Latin square of order 5 would be as follows:
ModelDriver / I / II / III / IV / V
1 / A / B / C / D / E
2 / B / C / D / E / A
3 / C / D / E / A / B
4 / D / E / A / B / C
5 / E / A / B / C / D
Note that each letter (fuel blend) appears exactly once in each row and exactly once in each column.
6.1.3.For your convenience, the 16 observations in the fuel efficiency study are listed below ....
15.5 33.9 13.2 29.1 16.3 26.6 19.4 22.8 10.8 31.1 17.1 30.3 14.7 34.0 19.7 21.6.
Put these observations into c1 of a Minitab worksheet. Use the patterned data procedure to put subscripts for Driver into c2 and Model into c3 (use 1 for I, 2 for II, etc.). The subscripts for Blend (use 1 for A, 2 for B, etc.) will have to be entered directly into the worksheet one at a time because the Latin Square pattern cannot be entered automatically by Minitab.... Label the four columns appropriately: MPG, Driver, Model, Blend.
When all the data have been entered in, they can be printed as follows:
MTB > print c1 - c4
Data Display
Row MPG Driver Model Blend
1 15.5 1 1 4
2 33.9 1 2 2
3 13.2 1 3 3
4 29.1 1 4 1
5 16.3 2 1 2
6 26.6 2 2 3
7 19.4 2 3 1
8 22.8 2 4 4
9 10.8 3 1 3
10 31.1 3 2 1
11 17.1 3 3 4
12 30.3 3 4 2
13 14.7 4 1 1
14 34.0 4 2 4
15 19.7 4 3 2
16 21.6 4 4 3
6.1.4.Because this experiment was done by an oil company it is reasonable to assume that the main issue is whether there are differences among the four Blends, and that Blend is a fixed effect because four particular blends are currently under study. Also, it may be reasonable to assume that Models of car and Drivers were chosen at random from among available models and drivers. Subject to these assumptions, write the model for this experiment. In specifying the range of the subscripts for Driver you can use "i = 1, 2, 3, 4." and for Model you can use "j = 1, 2, 3, 4." However, for Blend you need to say something like "the 4 values of the Blend subscript k are assigned to the 16 observations according to a Latin Square design," in order to make it clear that there are only 16 observations.
Here is the model for this experiment (with a slightly different approach to specifysubscripts of the Latin square design):
yijk = + Bi + Cj +k + eijk, where i, j, k = 1, 2, 3, 4; with observations in 16 cells, assigned according to a Latin square design of order 4. Restriction: kk = 0. Distributions: Bi ~ N(0, B2), Cj ~ N(0, C2), eijk ~ N(0, 2), where
all of the random variables Bi, Cj and eijk are mutually independent.
Also, denotes the grand mean, denotes the (fixed) blend effect, B denotes the (random) Driver effect,
C denotes the (random) Model effect, and eijkdenote random error, independent and identically distributed N(0, ).
6.1.5.The design space of this Latin Square is really a very carefully chosen subset of a cube. There are 43 = 64 possible combinations of the levels of the three factors, of which only 16 are observed. The table at the beginning of this unit can be viewed as a two-dimensional representation of the cube with dimensions z = Driver (height), y = Model (width), and x = Blend (depth). Minitab makes three-dimensional scatterplots; in Minitab 14 they can be conveniently rotated.
GRAPH 3D Scatterplot . Groups, z='Driver", y='Model', x='Blend', Group=Blend; Data view: symbols and lines. TOOLS Toolbars 3D Graph, rotate about appropriate axes.
In Minitab 14, use the tools to rotate such a plot so that, first, the y-axis is perpendicular to your view, and then the x axis. Also, with some fussing, you should be able to orient the cube to match the data table.
6.2.1.Use the following procedure and the stacked-subscripted data to make a table that resembles the original data table of Section 1. Now, by hand, make a table similar to the original data table, except that the rows are Drivers, the columns are Blends, and the labels within cells are Roman numerals I-IV designating Models. Verify your result in Minitab with a procedure similar to the one shown just above. Repeat (both by hand and in Minitab) for a table in which Drivers are designated within the cells.
Using the table command in Minitab to display the data in a format similar to the original table results in the following:
MTB > table 'Driver' 'Model';
SUBC> data 'MPG' 'Blend'.
Tabulated statistics: Driver, Model
Rows: Driver Columns: Model
1 2 3 4
1 15.5 33.9 13.2 29.1
4 2 3 1
2 16.3 26.6 19.4 22.8
2 3 1 4
3 10.8 31.1 17.1 30.3
3 1 4 2
4 14.7 34.0 19.7 21.6
1 4 2 3
Cell Contents: MPG : DATA
Blend : DATA
Another arrangement of the data, with roman numerals designations for Model, is as follows:
BlendsDriver / A / B / C / D
1 / IV 29.1 / II 33.9 / III 13.2 / I 15.5
2 / III 19.4 / I 16.3 / II 26.6 / IV 22.8
3 / II 31.1 / IV 30.3 / I 10.8 / III 17.1
4 / I 14.7 / III 19.7 / IV 21.6 / II 34.0
The Minitab representation of this table is as follows:
Tabulated statistics: Driver, Blend
Rows: Driver Columns: Blend
1 2 3 4
1 4 2 3 1
29.1 33.9 13.2 15.5
2 3 1 2 4
19.4 16.3 26.6 22.8
3 2 4 1 3
31.1 30.3 10.8 17.1
4 1 3 4 2
14.7 19.7 21.6 34.0
Cell Contents: Model : DATA
MPG : DATA
The Minitab representation of yet another arrangement of the data, with Drivers designated within the cells, is as follows:
Tabulated statistics: Blend, Model
Rows: Blend Columns: Model
1 2 3 4
1 4 3 2 1
14.7 31.1 19.4 29.1
2 2 1 4 3
16.3 33.9 19.7 30.3
3 3 2 1 4
10.8 26.6 13.2 21.6
4 1 4 3 2
15.5 34.0 17.1 22.8
Cell Contents: Driver : DATA
MPG : DATA
6.2.2.Try to use Minitab's balanced ANOVA procedure to analyze this block design. What happens? (A Latin Square is "balanced" in the sense that the four subspaces corresponding to the three factors and error are orthogonal. In Minitab's "Balanced ANOVA" all combinations of treatment levels must have the same number of observations; here 16 frequencies are 1 and 64 – 16 = 48 are 0.)
Attempting to use the balanced ANOVA to analyze this design results in the following error message from Minitab: (Proper English would be 'imbalance' rather than 'unbalance', but one properly speaks of 'unbalanced designs.')
* ERROR * Unbalanced design. A cross tabulation of your factors will show
* where the unbalance exists.
6.2.3.Repeat the GLM procedure shown in this section to obtain the EMS table and components of variance, and make a column of residuals. (Notice that we have not used the restrict subcommand because it is not available with GLM; for a Latin Square design this causes no difficulty.)
The output of the GLM procedure is as follows:
General Linear Model: MPG versus Driver, Model, Blend
Factor Type Levels Values
Driver random 4 1, 2, 3, 4
Model random 4 1, 2, 3, 4
Blend fixed 4 1, 2, 3, 4
Analysis of Variance for Fuel Eff, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P Orthogonal design so Seq SS = Adj SS
Driver 3 5.897 5.897 1.966 0.50 0.699
Model 3 736.912 736.912 245.637 61.90 0.000
Blend 3 108.982 108.982 36.327 9.15 0.012
Error 6 23.809 23.809 3.968
Total 15 875.599
S = 1.99202 R-Sq = 97.28% R-Sq(adj) = 93.20%
Expected Mean Squares, using Adjusted SS
Expected Mean
Square for Each
Source TermMathematical Expressions
1 Driver (4) + 4.0000 (1)2 + 4B2
2 Model (4) + 4.0000 (2)2 + 4C2
3 Blend (4) + Q[3]2 + 4
4 Error (4)2
(a)What mean square is in the denominator of each F-ratio. How do the EMSs lead you to the conclusion that this is correct?
See the mathematical expressions added to the EMS table just above. If there is no Blend () effect so that , then EMS(Blend) = 2 + 42 = EMS(Error), so MS(Blend) and MS(Error) have the same expected value. This indicates that they can be used as numerator and denominator of an F-statistic. Similar arguments can be made concerning the F-ratios to test for the random Driver and Model effects. The Error Terms table just below summarizes the result that MS(Error) is used in the denominator of all three F-tests done in the ANOVA table.
Error Terms for Tests, using Adjusted SS
Synthesis
of Error
Source Error DF Error MS MS
1 Driver 6.00 3.968 (4)
2 Model 6.00 3.968 (4)
3 Blend 6.00 3.968 (4)
Variance Components, using Adjusted SS
Estimated
Source Value
Driver -0.5006
Model 60.4173
Error 3.9681
(b)What is strange about the component of variance for Driver? How do you interpret this result in practice?
The estimated value of B2 (component of variance for Driver) is negative. We know that a variance cannot actually be negative. We conclude that the variance component for Driver is negligibly small. (These drivers must have been professionals carefully trained to use similar driving styles during the test. In ordinary driving, differences in driving styles can have a major impact on MPG.)
(c)What is the P-value of the Anderson-Darling test for normality of the residuals, and how do you interpret it?
The Anderson-Darling statistic has a large P-value = .970. This indicates that the null hypothesis of normality of the data cannot be rejected.
6.2.4.Multiple comparison procedures for a Latin Square design. We have established that there are significant differences among the Blends. By hand, establish the pattern of significant differences using Fisher's LSD procedure and Tukey's HSD procedure (both at the 5% level). The formulas are similar to the ones for a one-way ANOVA, except that you must use MS(Error) from the ANOVA table as the variance estimate (i.e., instead of sw2) and n = t. (The exact formula for LSD is given in O/L 6e, page 1146.)
The Tukey comparisons using Minitab GLM are as follows:
Tukey 95.0% Simultaneous Confidence Intervals
Response Variable MPG
All Pairwise Comparisons among Levels of Blend
Blend = 1 subtracted from:
Blend Lower Center Upper +------+------+------+------
2 -3.41 1.475 6.3554 (------*------)
3 -10.41 -5.525 -0.6446 (------*------)
4 -6.11 -1.225 3.6554 (------*------)
+------+------+------+------
-12.0 -6.0 0.0 6.0
Blend = 2 subtracted from:
Blend Lower Center Upper +------+------+------+------
3 -11.88 -7.000 -2.120 (------*------)
4 -7.58 -2.700 2.180 (------*------)
+------+------+------+------
-12.0 -6.0 0.0 6.0
Blend = 3 subtracted from:
Blend Lower Center Upper +------+------+------+------
4 -0.5804 4.300 9.180 (------*------)
+------+------+------+------
-12.0 -6.0 0.0 6.0
Tukey Simultaneous Tests
Response Variable MPG
All Pairwise Comparisons among Levels of Blend
Blend = 1 subtracted from:
Difference SE of Adjusted
Blend of Means Difference T-Value P-Value
2 1.475 1.409 1.047 0.7307
3 -5.525 1.409 -3.922 0.0297
4 -1.225 1.409 -0.870 0.8204
Blend = 2 subtracted from:
Difference SE of Adjusted
Blend of Means Difference T-Value P-Value
3 -7.000 1.409 -4.970 0.0100
4 -2.700 1.409 -1.917 0.3136
Blend = 3 subtracted from:
Difference SE of Adjusted
Blend of Means Difference T-Value P-Value
4 4.300 1.409 3.053 0.0808
Descriptive Statistics: MPG
Variable Blend Mean
FuelEff 1 23.58
2 25.05
3 18.05
4 22.35
Significant differences according to Tukey's HSD method are printed in red. The order of the means from smallest to largest is 3, 4, 1, 2. Linking codes for means that do not differ, we get the following underline diagram:
3412. Blends 1 and 2 have better MPG than Blend 3, but Blend 4 cannot be distinguished from other blends.
–––
———
Hand calculations for Tukey HSD and Fisher LSD methods of multiple comparisons.
Fisher: LSD = t*[2 MS(Error) / t]1/2 = 2.447 [2(3.968)/4]1/2 = 3.467, where t* cuts off the top 2.5%
of Student's t distribution with df = df(Error) = 6.
|18.05 – 22.35| = 4.30 > 3.467, so Blends 3 and 4 differ;
|25.05 – 22.35| = 2.70 < 3.467, so Blends 1, 2 and 4 do not differ.
3412 .
Tukey: HSD = q*[MS(Error) / t]1/2 = 4.90 [3.968 / 4]1/2 = 4.880, where q* cuts off the top 5% of the Studentized range distribution for 4 treatment groups and df = 6.
So Blends 3 and 4 do not differ, and so on. The underline diagram is the same as we deduced from Minitab.
We leave it to you to verify all necessary pairwise comparisons.
Notice that the length of the MInitab CIs is 2(4.880) = 9.76.
For example, the CI comparing Blends 2 and 3 has length –2.120 – (–11.88)= 9.76.
Because LSD < HSD, the Fisher method is more aggressive in declaring differences. On the other hand the more conservative Tukey method carries a 5% error rate for the simultaneous comparisons of all C(4, 2) = 4!/(2! 2!) = 6 possible pairs of 4 Blends. For the Fisher method, a 5% error rate attaches to each individual paired comparison.
6.2.5.Suppose that the MPG value for Driver 3/ Model II is missing. Copy c1 'MPG' to c11 and name the copied column 'MPG_M'. Replace the appropriate observation with a * (the asterisk is Minitab's symbol for a missing observation), and run glm c11 = c2 c3 c4. How can you tell from the output that the Latin square design with a missing value is not an orthogonal design? Try running both glm c11 = c2 c4 c3and glm c11 = c4 c2 c3. What changes and what remains the same? Interpret Minitab's (approximate) F-ratios.
The output from the GLM procedure glm c7 = c2 c3 c4 is as follows:
General Linear Model: Fuel MPG_M versus Driver, Model, Blend
Factor Type Levels Values
Driver random 4 1, 2, 3, 4
Model random 4 1, 2, 3, 4
Blend fixed 4 1, 2, 3, 4
Analysis of Variance for Fuel Eff_M, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Driver 3 25.138 8.832 2.944 0.91 0.500
Model 3 634.486 645.669 215.223 66.37 0.000
Blend 3 116.334 116.334 38.778 11.96 0.010
Error 5 16.215 16.215 3.243
Total 14 792.173
S = 1.80083 R-Sq = 97.95% R-Sq(adj) = 94.27%
Notice that df(Error) = 5 now. A nonorthogonal design results in Sequential SS that are not the same as
Adjusted SS.
The output from the GLM procedure glm c7 = c2 c4 c3 is as follows:
General Linear Model: Fuel Eff_M versus Driver, Blend, Model
Factor Type Levels Values
Driver random 4 1, 2, 3, 4
Blend fixed 4 1, 2, 3, 4
Model random 4 1, 2, 3, 4
Analysis of Variance for Fuel Eff_M, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Driver 3 25.138 8.832 2.944 0.91 0.500
Blend 3 105.151 116.334 38.778 11.96 0.010
Model 3 645.669 645.669 215.223 66.37 0.000
Error 5 16.215 16.215 3.243
Total 14 792.173
S = 1.80083 R-Sq = 97.95% R-Sq(adj) = 94.27%
The output from the GLM procedure glm c7=c4 c2 c3 is as follows:
General Linear Model: Fuel Eff_M versus Blend, Driver, Model
Factor Type Levels Values
Blend fixed 4 1, 2, 3, 4
Driver random 4 1, 2, 3, 4
Model random 4 1, 2, 3, 4
Analysis of Variance for Fuel Eff_M, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Blend 3 101.057 116.334 38.778 11.96 0.010
Driver 3 29.233 8.832 2.944 0.91 0.500
Model 3 645.669 645.669 215.223 66.37 0.000
Error 5 16.215 16.215 3.243
Total 14 792.173
S = 1.80083 R-Sq = 97.95% R-Sq(adj) = 94.27%
Rearranging the terms in the GLM command changes the values of Sequential SS, but does not change the Adjusted SS and MS values, nor the resulting F-statistics and P-values. Therefore, the conclusions of the GLM analysis are the same, regardless of the order in which the factors are specified at the right of the equals sign in the GLM command.
The approximate F-ratios and p-values indicate that the blend is significant at the 5% level, that the drivers do not add significant variability (p-value = .500), but that the model does significantly affect the variability of the data (p-value = 0.000). These conclusions are identical to those reached for the GLM without the missing data.
6.3.1.Perform two additional incorrect one-way ANOVAs, using first Driver and then Model as the single factor.
If you were forced to use a spreadsheet program that will compute only one-way ANOVAs, how could you piece together results from this program to construct a correct ANOVA table for a Latin Square design? In other words, how can you combine information from three incorrect one-way ANOVAs to give the correct analysis of a Latin Square?"
Running three (incorrect) one way ANOVA's, with Blend, Driver, and Model as the single factors, respectively, yields the following tables.
One-way ANOVA: Fuel Eff versus Blend
Source DF SS MS F P
Blend 3 109.0 36.3 0.57 0.646
Error 12 766.6 63.9
Total 15 875.6
One-way ANOVA: Fuel Eff versus Driver
Source DF SS MS F P
Driver 3 5.9 2.0 0.03 0.994
Error 12 869.7 72.5
Total 15 875.6
One-way ANOVA: Fuel Eff versus Model
Source DF SS MS F P
Model 3 736.9 245.6 21.25 0.000
Error 12 138.7 11.6
Total 15 875.6
Note that the SS(Total) is the same and is correct in each case. SS(Error) in each case above incorrectly "absorbs" the variability caused by ignoring the other factors. Therefore, a correct ANOVA table can be constructed, using the three SS(Total) and the three treatment SSs above to calculate the correct SS(Error):
SS(Error)correct = SS(Total) – SSTBlend – SSTDriver - SSTModel = 875.6 – 109.0 – 5.9 – 736.9 = 23.8
This SS(Error) of 23.8 is identical to the original GLM ANOVA. The MS's, F-ratios, and P-values are then calculated as for a standard ANOVA. The correct MS(Error) is used as the denominator for each F-statistic:
SourceDF SS MS F P
Driver 3 5.9 2.0 .5.699
Model 3 736.9 245.661.4.000
Blend 3 109.0 36.39.08.012
Error 6 23.8 4.0
Total15 875.6
The P-values obtained in this way agree with those obtained with the original GLM (box above).
6.3.2.Perform an incorrect analysis on these data, treating them as if they came from a block design with Blends as the fixed effect and Models as the blocking effect (ignoring Drivers). Compare sums of squares and degrees of freedom in the resulting ANOVA table with the correct ones from the Latin Square analysis. If you were to use the F-ratios from this incorrect procedure to draw conclusions about the significance of Blend and Model effects, would your conclusions happen to be correct or incorrect? What about the conclusions drawn from an incorrect block design that ignores Models instead of Drivers? Comment.
The ANOVA table that Minitab generates for this incorrect analysis of Latin square data as a block design is as follows:
Source DF SS MS F P
Model 3 736.91 245.64 74.42 0.000
Blend 3 108.98 36.33 11.01 0.002
Error 9 29.71 3.30
Total 15 875.60
Using the P-values above, the conclusions happen to be correct – the low values of both P-values would lead to the conclusion that both Blend and Model were significant at the 5% level. Ignoring Drivers is not significant in this case because the variability due to Drivers is so low in the original data.
If Model were ignored instead, the resulting table would be:
Source DF SS MS F P
Driver 3 5.90 1.97 0.02 0.995
Blend 3 108.98 36.33 0.43 0.737
Error 9 760.72 84.52
Total 15 875.60
In this case, the large variability of Model causes the MSE term to be very high and, thus, all the F-ratios to be very low and the P-values to be very high. Thus, incorrect conclusions would be drawn for both Driver and Blend.
Also, notice that the correct ANOVA table for the Latin square design can be constructed from information in one incorrect one-factor ANOVA and in an incorrect block design using the remaining two factors. The correct SSs and DFs for each of the factors and Total can be obtained from these incorrect outputs and the correct MS(Error) can be obtained by difference as in Problem 6.3.1. Then correct F-statistics and P-values can be obtained.
Notes: We have seen that a Latin square design is a multifactor design with t levels of each factor in which
only t2 of the t3 possible factor combinations are observed. There are many other experimental designs in
which k factors have levels a1, a2, ..., ak, respectively, and in which some of the i ai factor combinations are strategically omitted in ways that save money or effort, yet make possible reasonable inferences about
some factors. Fractional factorial designs and partially balanced incomplete block designs are among them.