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# HW 9 - Azfar Khandoker Dr Knapp Stat 350 4:30 HW#9 36 SST =...

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Azfar Khandoker Dr. Knapp Stat 350 4:30 HW #9 36. SST = 11.7 + 113.5 + 25.6 = 150.8 MSA = 11.7 / 2 = 5.85 MSB = 113.5 / 4 = 28.375 MSE = 25.6 / (2 x 4) = 3.2 F A = 5.85 / 3.2 = 1.828125 F B = 28.375 / 3.2 = 8.8671875 Source df SS MS F p-value A 2 11.7 5.85 1.828125 0.221883 B 4 113.5 28.375 8.8671875 0.004892 Error 8 25.6 3.2 Total 14 150.8 a. With α = 0.05, and p-value = 0.221883, we fail to reject H o that there are no differences between the average values reported by the three assessors. b. Yes, the results show us that there is a significant block effect. Therefore, the use of houses as the blocking factor shows that they influence the value of the property.

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38. Source df SS MS F p-value Species 2 2216 1108 9.61 0.0297 Wood Grade 2 90754.67 45377.33 393.45 <0.0001 Error 4 461.33 115.33 Total 8 93432 a. Yes, since α = 0.05 and the p-value = 0.0297, we conclude that there is a significant difference between the mean bending parameters of the three types of wood. b. Yes, since α = 0.05 and the p-value is less than 0.0001, we conclude that there is a significant difference between mean bending parameters for the three grades of wood. SAS CODE/OUTPUT: data q38; input tree \$ grade \$ bend; cards ; DouglasFir SS 370 DouglasFir G2 381 DouglasFir G3 165 Hem-Fir SS 386 Hem-Fir G2 390 Hem-Fir G3 182 Spruce-Pine-Fir SS 430 Spruce-Pine-Fir G2 405 Spruce-Pine-Fir G3 195 ; proc anova data = q38; class tree grade; model bend = tree grade; run ;
The SAS System 11:43 Monday, March 30, 2009 5 The ANOVA Procedure Class Level Information Class Levels Values tree 3 DouglasF Hem-Fir Spruce-P grade 3 G2 G3 SS

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