Anova with Spss – Post Hoc Comparisons
Autor: nihashukla • May 9, 2016 • Coursework • 1,241 Words (5 Pages) • 580 Views
Tasks Assignment 09: Introduction to Analysis of Variance (ANOVA)
Task 01: ANOVA with SPSS – Post hoc Comparisons
Data set: Data 09a.sav
An experiment was conducted in which a particular material was treated using 5 different methods (treatment). The experiment measured an important parameter of the material (parameter) in order to analyse the impact of the treatment. Do the methods vary in their impact on the material? If yes, are all methods different or are there any groups of methods?
Hint: First examine the averages of the 5 methods as a table and as a boxplot. (Analyze>Descriptive Statistics>Explore...).
Syntax:
EXAMINE VARIABLES=parameter BY treatment /PLOT BOXPLOT STEMLEAF /COMPARE GROUPS /STATISTICS DESCRIPTIVES /CINTERVAL 95 /MISSING LISTWISE /NOTOTAL.
UNIANOVA parameter BY treatment /METHOD=SSTYPE(3) /INTERCEPT=INCLUDE /POSTHOC=treatment(BONFERRONI) /PLOT=PROFILE(treatment) /PRINT=PARAMETER DESCRIPTIVE /CRITERIA=ALPHA(0.05) /DESIGN=treatment.
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ANOVA – Post hoc Comparisons: Results and Comments:
Examining the averages of the 5 methods (treatment) as a table and as a boxplot.
[pic 1][pic 2][pic 3][pic 4][pic 5][pic 6][pic 7]
[pic 8][pic 9]
Verification of the model:
[pic 10][pic 11][pic 12]
The overall model is significant (Corrected model: F(4,45) = 45.121, p = .000)
The factor treatment is also sig. p = .000, there is a main effect of treatment (methods- 1,2,3,4,5) on parameter, F(4,45) = 45.121, p = .000).
The value of adjusted R square = .783 which indicates that 78.3% of the variance in parameter around the grand mean can be predicted by the model (here by treatment)
Thus, we have sig. model and sig. factor and the null hypothesis (no model) can be rejected.
Parameter estimates:
[pic 13][pic 14][pic 15]
Linear model with parameter as the dependent variable (ygk = parameter value k in group g)
Ygk = Ӯ + αg + ε gk
Ӯ = grand mean
αg = effect of group g
ε gk =random term
Y1k = 36.10 – 9.10 + ε 1k
Y2k = 36.10 – 14.50 + ε 2k
Y3k = 36.10 + 10.50 + ε 3k
Y4k = 36.10 – 1.60 + ε 4k
Y5k = 36.10 + 0.0 + ε 5k
Post hoc comparisons:
(See table below)
Group 1 & 2 does not show significant difference (p = .099 > .05)
Group 1 & 3 show a significant difference (p=.000)
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