Tom Calabrese Final Assignment Multivariate Analysis EPSY346 page 1 10/18/2018
Tom CalabreseMultivariate Final AssignmentEPSY3464/22/07
Problem 1:
The following is the graphic display of the means of the four groups on the 3 scales. The design is a repeated measures design one factor within with 3 levels and one between group factor.
a.) Testing the hypothesis that the profiles are parallel is the same as testing for significant interaction of the within-subjects factor with the between-subjects factor (e.g. groups and scales). A non-significant finding means the profiles are not significantly different in shape and we would conclude they are parallel (from a statistical point of view). In this case we see from the multivariate analysis that the interaction effects are non-significant (ranging from .081 to .215) at the .05 level. Therefore I would conclude in this case that the profiles are parallel.
Multivariate Tests(c)
Effect / Value / F / Hypothesis df / Error df / Sig.Sfactor / Pillai's Trace / .465 / 5.225(a) / 2.000 / 12.000 / .023
Wilks' Lambda / .535 / 5.225(a) / 2.000 / 12.000 / .023
Hotelling's Trace / .871 / 5.225(a) / 2.000 / 12.000 / .023
Roy's Largest Root / .871 / 5.225(a) / 2.000 / 12.000 / .023
Sfactor * SOCG / Pillai's Trace / .550 / 1.643 / 6.000 / 26.000 / .175
Wilks' Lambda / .512 / 1.592(a) / 6.000 / 24.000 / .193
Hotelling's Trace / .834 / 1.529 / 6.000 / 22.000 / .215
Roy's Largest Root / .648 / 2.810(b) / 3.000 / 13.000 / .081
a Exact statistic
b The statistic is an upper bound on F that yields a lower bound on the significance level.
c Design: Intercept+SOCG
Within Subjects Design: Sfactor
b.) Given the results in (a) YES it is meaningful to test the main effect due to groups and scales for significant differences. It is only not meaningful when the interaction effects are significant. In this case they are not so we should test the main effects.
(i) First I tested the hypothesis that the means for S1, S2 and S3 are equal. Formally stated as. After running the repeated measures analysis I see from the within-subject effects table that there are significant differences between the three with significance levels of Sfactor being reported as .032 for the Sphericity test, .038 for the Greenhouse-Geisser test, and .032 for the Huynh-Feldt test all significant at the .05 alpha. (Note: the result for the lower bound test was .069 which is just above the .05 alpha.)
(ii) Next I would test the group main effect formally stated as
. Itested this hypothesis by running the repeated measures analysis. I see from the between-subjects effects table that the group effect SOCG is significant at the .000 level. I can conclude from this that there are significant differences between the groups.
c.) First I construct a new variable reflecting the average of the 3 scales and call it AVES. This variable is formed as. This formula is entered into SPSS via the TRANSFORM function and values for each subject are then automatically calculated. I then ran a GLM ANOVA on this variable. This provides the following information necessary to construct the confidence interval for contrast 1, formally stated as:
Mean of=19.833
Mean of =14.4
Mean of =14.0
Mean of =10.0
From these I can calculate the estimate as (19.833+14.4)/2 – (14 + 10)/2
34.233/2 - 24/2
17.117 - 12
5.117
The AVES Mean Square due to Error of 1.092 as provided from the Error row and Mean Square column of the Tests of Between-Subjects Effects table. From this I can construct the standard error as:
= .509
The Bonferroni adjustment will be used to correct for Type I error with an M=20, df=(4+4+5+4)-2 = 15, and an alpha of .05 which yields a value of 3.6245.
Now the confidence interval is constructed as:
CI = 5.117 .509 * 3.6254
CI = 5.1171.845
(3.272, 6.952)
Since 0 is not in the interval I would conclude this is a significant finding.
d.) First I construct a new variable reflecting the difference between the average of the first two scales and the third scale. This variable is formed as. This formula is entered into SPSS via the TRANSFORM function and values for each subject are then automatically calculated. I then ran a GLM ANOVA on this variable. This provides the following information necessary to construct the confidence interval for contrast 2 or (CI2) only within SOC1.
Mean of CI2 for group 1 = -.25. This will be used as the estimate for group 1.
The CI2 Mean Square due to Error of 2.388 as provided from the Error row and Mean Square column of the Tests of Between-Subjects Effects table. From this I can construct the standard error as:
= .773
The Bonferroni adjustment will be used to correct for Type I error with an M=20,
df=4-1=3, and an alpha of .05 which yields a value of 5.61.
Now the confidence interval is constructed as:
CI = -.25.773 * 5.61
CI = -.254.37
(-4.62, 4.12)
Since 0 is in the interval I would conclude this is not a significant finding.
Problem 2
The design of this problem is an Multivariate Analysis of Covariance (MANCOVA).
From my discipline a context for this design would be that there are four groups of students from different sections of Algebra I (groups 1 – 4), who each have an opportunity to use each of 4 computer based training tools (intelligent tutoring system, web based collaboration tool, a scripted computer based learning module, and an automated work book.) After using the tools, the students are given a test on a set of Algebra I word problems. These score are assigned to y1, y2, y3, y4 respectively. Prior to using the tools they are given a test of pre-existing algebra knowledge. That score is assigned to U. The researcher is interested in which tool best assists the student after accounting for previous skill in algebra.
The analysis was done in SPSS using the Multivariate Analysis of Covariance, where U is listed as the covariate, y1, y2, y3, y4 are the dependent variables and Group is the between subject factor. The following table shows the multivariate test results from the SPSS analysis.
Multivariate Tests(c)
Effect / Value / F / Hypothesis df / Error df / Sig.Intercept / Pillai's Trace / .899 / 17.808(a) / 4.000 / 8.000 / .000
Wilks' Lambda / .101 / 17.808(a) / 4.000 / 8.000 / .000
Hotelling's Trace / 8.904 / 17.808(a) / 4.000 / 8.000 / .000
Roy's Largest Root / 8.904 / 17.808(a) / 4.000 / 8.000 / .000
u / Pillai's Trace / .568 / 2.628(a) / 4.000 / 8.000 / .114
Wilks' Lambda / .432 / 2.628(a) / 4.000 / 8.000 / .114
Hotelling's Trace / 1.314 / 2.628(a) / 4.000 / 8.000 / .114
Roy's Largest Root / 1.314 / 2.628(a) / 4.000 / 8.000 / .114
group / Pillai's Trace / 1.464 / 2.382 / 12.000 / 30.000 / .027
Wilks' Lambda / .042 / 4.151 / 12.000 / 21.458 / .002
Hotelling's Trace / 11.482 / 6.379 / 12.000 / 20.000 / .000
Roy's Largest Root / 10.452 / 26.130(b) / 4.000 / 10.000 / .000
a Exact statistic
b The statistic is an upper bound on F that yields a lower bound on the significance level.
c Design: Intercept+u+group
To account for the differences on the variables the researcher needs to test 24 comparisons (for each group the researcher would compare y1 with y4, y1 with y2, y1 with y3, y2 with y3, y2 with y4, and y3 with y4 – or 6 tests across 4 groups = 24 tests).
Type I error rate will be controlled by making a Bonferroni adjustment with an M of 24 (plus any additional special contrasts that the researcher might want to evaluate – e.g. combinations of the tools) at an alpha of .05.
The research also would like to compare the average of the first two groups with the average of the last two groups on y1. Formally stated:
To compute the estimate and the SE we must execute the following steps:
- Run the MANCOVA (previously done)
- From the multivariate test table above we see that there are significant differences between the groups (sigs ranging from .000 to .027). We also see that the sig levels of U (the covariate) are not significant at the .05 level (all sigs .114).
- Create an estimate
To do this I make use of the Parameter estimates produced as part of the output of MANCOVA. The estimates produce by default the values of
Dependent Variable / Parameter / By1 / Intercept / 25.372
u / .217
[group=1.00] / 7.413
[group=2.00] / -2.109
[group=3.00] / -.630
[group=4.00] / 0(a)
This is interpreted as:
G1Y1 – G4Y1 = 7.413
G2Y1 – G4Y1 = -2.109
G3Y1 – G4Y1 = - .630
To use this information to calculate the estimate I must first rework the contrast (algebraic manipulation to get it to fit this information)
call this equation 1 (needed in step 4)
{At this time the equation has been worked such that direct substitution from the parameter estimate table is possible}
Estimate of ((7.413) – (-.630)) + (-2.109)
= 8.043 – 2.109
=5.934
- I first go to the Residual SSCP matrix
Covariance / y1 / 2.188 / .822 / 1.042 / 3.316
y2 / .822 / 2.766 / 2.230 / -1.057
y3 / 1.042 / 2.230 / 10.128 / -6.429
y4 / 3.316 / -1.057 / -6.429 / 32.825
and get the variance of Y1 which is 2.188. This is the MSW part of the SE calculation
- To compute the portion of the formula, I used the SYNTAX command and typed in
MANOVA y1 y2 y3y4 U BY GROUP(1,4)/
ANALYSIS = y1 y2 y3 y4/
CONTRASTS (GROUP) = SIMPLE/
Print = Design(overall) Param(estim, cov)/
Design = GroupU.
Which returned the table:
Parameter Variance/Covariance Factors(upper) and Correlations(lower)
PARAMETER
PARAMETER 1 2 3 4 5
1 4.888 .634 .063 .222 -.127
2 .375 .583 .258 .279 -.017
3 .041 .478 .501 .253 -.002
4 .140 .512 .500 .510 -.006
5 -.994 -.378 -.041 -.141 .003
This matrix can be transformed to be symmetric as:
PARAMETER 1 2 3 4 5
1 4.888 .634 .063 .222 -.127
2 .634 .583 .258 .279 -.017
3 .063 .258 .501 .253 -.002
4 .222 .279 .253 .510 -.006
5 -.127 -.017 -.002 -.006 .003
Call this matrix X
Given equation 1 above it should be obvious that this can be transformed easily into
The reason for this is that each of these component transfers to the matrix diagonals above
Which can be represented by their coefficients in the matrix [011-10].
= [0 1 1 -1 0] X
= [.475 .56199 .50599 .02200 -.0130]
= 1.04598
(done using the Joemath matrix multiplier)
- To computer the Standard Error I now simply insert the computations above into the formula
=
=
= 1.5129
Problem 3
This is a repeated measures design with 2 within group factors (3 levels for A and 2 levels for B) and 1 between group factor. The effect for this design are:
Main:
A
B
Group
Interaction:
AxB
GroupxA
GroupxB
GroupxAxB
For this experiment the relevant hypothesis are:
across the average of the dependent variables (main effect group)
across the three groups (main effect A)
across the three groups (main effect B)
A context for this experiment from my field would be that three groups of students selected from three different classes were each given three different tutoring applications to try (A1, A2, A3) in combination with either the assistance of a tutor (B1) or with the assistance of another classmate (B2). The researcher is trying to find out how the scores differ between groups and which combination of tool and tutor works the best.
From the multivariate test results we see that there are significant interaction effects:
Multivariate Tests(c)
Effect / Value / F / Hypothesis df / Error df / Sig.A / Pillai's Trace / .001 / .007(a) / 2.000 / 15.000 / .993
Wilks' Lambda / .999 / .007(a) / 2.000 / 15.000 / .993
Hotelling's Trace / .001 / .007(a) / 2.000 / 15.000 / .993
Roy's Largest Root / .001 / .007(a) / 2.000 / 15.000 / .993
A * Group / Pillai's Trace / .752 / 4.816 / 4.000 / 32.000 / .004
Wilks' Lambda / .302 / 6.146(a) / 4.000 / 30.000 / .001
Hotelling's Trace / 2.133 / 7.465 / 4.000 / 28.000 / .000
Roy's Largest Root / 2.046 / 16.367(b) / 2.000 / 16.000 / .000
B / Pillai's Trace / .616 / 25.679(a) / 1.000 / 16.000 / .000
Wilks' Lambda / .384 / 25.679(a) / 1.000 / 16.000 / .000
Hotelling's Trace / 1.605 / 25.679(a) / 1.000 / 16.000 / .000
Roy's Largest Root / 1.605 / 25.679(a) / 1.000 / 16.000 / .000
B * Group / Pillai's Trace / .869 / 52.937(a) / 2.000 / 16.000 / .000
Wilks' Lambda / .131 / 52.937(a) / 2.000 / 16.000 / .000
Hotelling's Trace / 6.617 / 52.937(a) / 2.000 / 16.000 / .000
Roy's Largest Root / 6.617 / 52.937(a) / 2.000 / 16.000 / .000
A * B / Pillai's Trace / .753 / 22.839(a) / 2.000 / 15.000 / .000
Wilks' Lambda / .247 / 22.839(a) / 2.000 / 15.000 / .000
Hotelling's Trace / 3.045 / 22.839(a) / 2.000 / 15.000 / .000
Roy's Largest Root / 3.045 / 22.839(a) / 2.000 / 15.000 / .000
A * B * Group / Pillai's Trace / .979 / 7.667 / 4.000 / 32.000 / .000
Wilks' Lambda / .046 / 27.536(a) / 4.000 / 30.000 / .000
Hotelling's Trace / 20.286 / 71.002 / 4.000 / 28.000 / .000
Roy's Largest Root / 20.260 / 162.079(b) / 2.000 / 16.000 / .000
a Exact statistic
b The statistic is an upper bound on F that yields a lower bound on the significance level.
c Design: Intercept+Group
Within Subjects Design: A+B+A*B
Therefore there is no need to consider the results of the main effects and it is best to proceed with testing for contrasts of interaction effects.
a.) To construct a confidence interval I first create the new dependent variable
I first will use the transform command in SPSS to calculate the new variable stated above. I run a Univariate Analysis of Variance on the new variable.
Dependent Variable: W1
Group / Mean / Std. Deviation / N1.00 / 2.7778 / .58373 / 6
2.00 / 1.1429 / 1.57359 / 7
3.00 / -2.1111 / .77936 / 6
Total / .6316 / 2.28763 / 19
Now we want to look at the contrast of (G1W1 – G2W1). Then I use the means of the new variable to construct my estimate. (2.7778 – 1.1429) = 1.6349 as the estimate,
From the table of Between subjects effects
Tests of Between-Subjects Effects
Dependent Variable: W1
Source / Type III Sum of Squares / df / Mean Square / F / Sig.Corrected Model / 74.601(a) / 2 / 37.300 / 30.453 / .000
Intercept / 6.876 / 1 / 6.876 / 5.614 / .031
Group / 74.601 / 2 / 37.300 / 30.453 / .000
Error / 19.598 / 16 / 1.225
Total / 101.778 / 19
Corrected Total / 94.199 / 18
a R Squared = .792 (Adjusted R Squared = .766)
I get the MSE for the new variable of 1.225.
From this I can calculate the Standard Error as
.616
Given the Bonferroni adjusted T value for M= 20, df = (6+7)-2=11, and alpha=.05 of 4.0 (approximated), we can construct the interval as:
CI= 1.6349.616*4
= 1.63492.464
(-.8251, 4.0989)
Since 0 is in the interval this finding is not significant.
b.) To examine the levels of the A factor within group 3 I would construct the following confidence intervals:
i.)To examine the difference between A1 and A3 in group three I start with the contrast:
Next I used the transform command in SPSS to create this new variable and calculate new values for it for each of the subjects. I then ran a univariate ANOVA to obtain the means of the variable and the MSE as is seen from the following tables
Descriptive Statistics
Dependent Variable: W2
Group / Mean / Std. Deviation / N1.00 / -2.1667 / .98319 / 6
2.00 / -2.2143 / .90633 / 7
3.00 / 2.0000 / .94868 / 6
Total / -.8684 / 2.19116 / 19
From this we see that the mean of the new variable computed by SPSS for group 3 is 2.0000. This is the estimate that I will use to calculate the confidence interval.
Tests of Between-Subjects Effects
Dependent Variable: W2
Source / Type III Sum of Squares / df / Mean Square / F / Sig.Corrected Model / 72.159(a) / 2 / 36.080 / 40.477 / .000
Intercept / 11.905 / 1 / 11.905 / 13.356 / .002
Group / 72.159 / 2 / 36.080 / 40.477 / .000
Error / 14.262 / 16 / .891
Total / 100.750 / 19
Corrected Total / 86.421 / 18
a R Squared = .835 (Adjusted R Squared = .814)
From this table we see that the MSE is .891.
From this information I can now calculate the standard error as being:
Standard Error of the Estimate= .385
From the Bonferroni table I chose an M = 20 (assuming that I will run additional tests), df = (6-1)=5, alpha = .05 for an adjusted T of 5.6099.
The CI for this contrast is:
CI = 2.000 .385 * 5.6099
= 2.000 2.16
(-.16 , 4.16)
Since 0 is in the interval I would say this finding is not significant.
ii.)To examine the difference between A1 and A2 in group three I start with the contrast:
As in number i above I will use the transform command to create the new variable and to calculate the new values of the transformed variable for each subject in SPSS. I then ran a univariate ANOVA to obtain the mean of this variable for Group 3 and the MSE of the new variable. We see from the following table that:
Descriptive Statistics
Dependent Variable: W3
Group / Mean / Std. Deviation / N1.00 / -3.9167 / .97040 / 6
2.00 / -3.7857 / 1.28638 / 7
3.00 / 3.2500 / 1.12916 / 6
Total / -1.6053 / 3.55738 / 19
I used 3.25 as the estimate of the new variable for group three and from
Tests of Between-Subjects Effects
Dependent Variable: W3
Source / Type III Sum of Squares / df / Mean Square / F / Sig.Corrected Model / 206.778(a) / 2 / 103.389 / 78.728 / .000
Intercept / 41.630 / 1 / 41.630 / 31.700 / .000
Group / 206.778 / 2 / 103.389 / 78.728 / .000
Error / 21.012 / 16 / 1.313
Total / 276.750 / 19
Corrected Total / 227.789 / 18
I get the MSE of 1.313.
I can now calculate the Standard error as:
= .468
The Bonferroni adjusted T is the same as in the last problem: 5.6099
The CI for this contrast is: 3.25 .468 * 5.6099
3.25 2.625
(.625, 5.875)
Since 0 is in the interval I would say that this is a significant finding.
iii.)To examine the difference between A2 and A3 in group three I start with the contrast:
I again use the transform command in SPSS to calculate new values for each subject on this variable. Next I ran a univariate ANOVA on this variable and obtained the new mean for this variable from the descriptive statistics table as -1.25
Descriptive Statistics
Dependent Variable: W4
Group / Mean / Std. Deviation / N1.00 / 1.7500 / 1.08397 / 6
2.00 / 1.5714 / 1.20515 / 7
3.00 / -1.2500 / 1.57321 / 6
Total / .7368 / 1.85119 / 19
And I obtained the MSE of 1.685 from
Tests of Between-Subjects Effects
Dependent Variable: W4
Source / Type III Sum of Squares / df / Mean Square / F / Sig.Corrected Model / 34.720(a) / 2 / 17.360 / 10.301 / .001
Intercept / 9.011 / 1 / 9.011 / 5.347 / .034
Group / 34.720 / 2 / 17.360 / 10.301 / .001
Error / 26.964 / 16 / 1.685
Total / 72.000 / 19
Corrected Total / 61.684 / 18
a R Squared = .563 (Adjusted R Squared = .508)
I will use the -1.25 as the estimate. Next I calculate the standard error of the estimate as:
= .53
The Bonferroni adjusted T value will be the same as in the previous two problems at 5.6099.
The confidence interval for this contrast is:
CI = -1.25 .53 * 5.6099
= -1.25 2.97
(-4.22, 1.72)
Since 0 is in the interval I would say this finding is not significant.
Final Examination Part II
a.)The two discriminate functions are:
b.)Group 1 mean
= .217 + 2.9645 - 3.04 - 3.354 - 1.986 + .91 + 2.681
= -1.6075
= -8.256 + .792 + 4.545 + 1.182 + .162 + .44 - .308
= -1.443
Group 2 mean
= .217 + 4.312 - 4.256 - 4.472 - 2.648 + 1.092 + 3.064
= -2.691
= -8.256 + 1.152 + 6.363 +1.576 + .216 + .528 - .352
= 1.227
Group 3 mean
= .217 + .468 - 3.648 - 2.236 – 1.986 +1.638 + 3.83
= -1.717
= -8.256 + 1.728 + 5.454 + .788 + .162 +.792 - .44
= .228
D2
^
2-
|
|
1.5-
|
G2* |
1-
|
|
.5-
G3* New * |
______|______D1
-2.5 -2 -1.5 -1 -.5 | .5 1 1.5 2
|
- .5-
|
|
-1-
|
G1* |
-1.5-
|
|
-2-
c.)The new person’s D1* and D2* scores are
= .217 + 3.234 - 4.249 - 3.354 - 1.324 + .91 + 3.447
= -1.119
= -8.256 + .864 + 6.363 + 1.182 +.108 + .44 - .396
= .305
d.)Compute the Euclidean distance between the D1* and D2* point of the new person and the centroid of each group:
Group 1 and the new point
=
= .239+3.056
= 3.295
Distance =
=1.815
Group 2 and the new point
=
= 2.471+.85
= 3.321
Distance =
= 1.822
Group 3 and the new point
=
= .358 + .006
= .364
Distance =
= .603
Therefore the shortest distance to the new point from any centroid is the centroid for group 3. Therefore this new person is a member of group 3.