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STAT Homework #7a - Dor STAT 427 Assignment 7 Spring 2005...

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Dor STAT 427 Assignment 7 Spring 2005 ANOVA Table for Model (1) Source of Variation Degrees of freedom Sum of Squares Mean Square Among the 5 Groups 4 165113.8303 41278.4576 Within the 5 Groups 54 168449.3900 3119.4331 Total 58 333563.2203 ANOVA Table for Model (2) Source of Variation Degrees of freedom Sum of Squares Mean Square Among the 4 Groups 3 160009.1407 53336.3802 Within the 4 Groups 55 173554.0796 3155.5287 Total 58 333563.2203 The ANOVA table for comparing model (1) and model (2), which is provided below, can be constructed from the information in the preceding ANOVA tables. The only computation required is to subtract the reduced model among groups sum of squares from the full model among groups sum of squares to get SS(full model / reduced model) = 165113.8303- 160009.1407 = 5104.689560, with 4-3 = 1 degrees of freedom. ANOVA Table for Comparing Model (1) and Model (2). Source of Variation Degrees of freedom Sum of Squares Mean Square Reduced Model 3 160009.1407 53336.3802 Full Model after Reduced Model 1 5104.689560 5104.689560 Within the 5 Groups 54 168449.3900 3119.4331 Total 58 333563.2203 The calculated F-statistic for comparing model (1) and model (2) is Fcalc = 5104.689560/3119.4331 = 1.64 with a P-value of .2063. This P-value is not small enough to allow us to reject the hypothesis that uL = uS so we conclude that the four means (uH, uV, uM, and uC) of the reduced model (2) will suffice and we do not need the five means of the full model (1). We now proceed to compare the current full model (2) to the reduced model (3). The ANOVA table for model (3) is given below. ANOVA Table for Model (3). Source of Variation Degrees of freedom Sum of Squares Mean Square Among the 3 Groups 2 147506.5317 73753.2659 Within the 3 Groups 56 186056.6886 3322.4409 Total 58 333563.2203
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We can produce the ANOVA table for comparing model (2) and model (3) as before. In this case we find that SS (full model / reduced model) = 160009.1407 – 147506.5317 = 12502.60903, with 3 – 2 = 1 degrees of freedom.
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