hw9 - STAT 350 – Spring 2009 Homework #9 - Solution...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: STAT 350 – Spring 2009 Homework #9 - Solution covers through Lecture N Text Exercises Chapter 9 problems #361, 38(a,b)1,2 Chapter 10 problems #81,2 1 be sure to give a full ANOVA table (including p-value) to accompany your answer! 2 you may do the analysis in SAS (if so, include the code and output immediately after that problem) 9.36 Source Assessor House (block) Error Total df 2 4 8 14 SS 11.7 113.5 25.6 150.8 MS 5.85 28.375 3.2 F 1.82813 8.86719 p 0.221882 0.004892 a. There is no significant difference among the assessors (p-value = 0.221882) b. The house effect was significant, thus it was warranted as a blocking factor. Recall that a Completely Randomized Design (CRD) and a 1-way ANOVA can be used when the experimental units are HOMOGENEOUS (not variable), the more variable the experimental units are, the more important it is to use blocks. Had the effect of "house" not been significant (p > 0.05), that would not mean blocking is definitely NOT warranted. Homework #9 - Solution Page 1 of 1 9.38 Source Grade Species Error Total df 2 2 4 8 SS MS F p 90754.66667 45377.33333 393.45 <.0001 2216.00000 1108.00000 9.61 0.0297 461.33333 115.33333 93432.00000 a. Yes, there is a significant difference in the three species of wood (p = 0.0297) b. Yes, there is a significant difference in the three grades of wood (p < 0.0001) SAS data prob938; input grade $ species $ bend; cards; ss doug 370 g2 doug 381 g3 doug 165 ss hem 386 g2 hem 390 g3 hem 182 ss spf 430 g2 spf 405 g3 spf 195 ; run; proc anova data=prob938; class grade species; model bend= grade species; run; The ANOVA Procedure Dependent Variable: bend DF Sum of Squares Mean Square F Value Pr > F Model 4 92970.66667 23242.66667 201.53 <.0001 Error 4 461.33333 115.33333 Corrected Total 8 93432.00000 Source R-Square Coeff Var Root MSE bend Mean 0.995062 3.328307 10.73934 322.6667 Source grade species Homework #9 - Solution DF Anova SS Mean Square F Value Pr > F 2 2 90754.66667 2216.00000 45377.33333 1108.00000 393.45 9.61 <.0001 0.0297 Page 2 of 2 10.8 Source A B A*B Error Total df 3 3 9 16 31 SS 635592.3438 79607.0938 19336.7813 1037.5000 735573.7188 MS 211864.1146 26535.6979 2148.5313 F 3267.30 409.23 33.13 p <0.0001 <0.0001 <0.0001 a. Gas flow rate (factor A) does have a significant effect on heat transfer coefficient b. liquid flow rate (factor B) does have a significant effect on heat transfer coefficient c. Yes, there is a significant interaction between the two factors SAS data prob108; input factA factB response; cards; 1 1 200 1 1 211 1 2 226 1 2 219 1 3 240 1 3 249 1 4 261 1 4 250 2 1 278 2 1 267 2 2 312 2 2 324 2 3 330 2 3 337 2 4 381 2 4 375 3 1 369 3 1 355 3 3 3 3 3 3 4 4 4 4 4 4 4 4 ; run; 2 2 3 3 4 4 1 1 2 2 3 3 4 4 416 402 462 457 517 524 500 487 575 593 645 632 733 718 proc anova data=prob108; class factA factB; model response = factA factB factA*factB; run; The ANOVA Procedure Dependent Variable: response Source DF Sum of Squares Mean Square F Value Pr > F Model 15 734536.2188 48969.0813 755.19 <.0001 Error 16 1037.5000 64.8438 Corrected Total 31 735573.7188 R-Square Coeff Var Root MSE response Mean 0.998590 2.006088 8.052562 401.4063 Source factA factB factA*factB Homework #9 - Solution DF Anova SS Mean Square F Value Pr > F 3 3 9 635592.3438 79607.0938 19336.7813 211864.1146 26535.6979 2148.5313 3267.30 409.23 33.13 <.0001 <.0001 <.0001 Page 3 of 3 Then answer the following problem: 1. For this problem, use the data from problem 9.46 in the textbook (page 430-431). Treat this like a RCBD and conduct the analysis of variance using “Drying Method” as the treatment, and “Fabric Type” as the block. You may do the analysis in SAS (if so, include the code and output immediately after that problem). a. Give the ANOVA table (including p-values). Source fabric drying Error Total DF 8 4 32 44 Sum of Squares 9.69600000 14.96222222 3.26177778 27.92000000 Mean Square 1.21200000 3.74055556 0.10193056 F Value 11.89 36.70 p-value <.0001 <.0001 b. Is the effect of drying method significant (α = 0.05)? Just answer “yes” or “no”. Yes c. Is the effect of fabric type significant (α = 0.05)? Just answer “yes” or “no”. Yes d. Conduct Tukey’s Multiple Comparison procedure to determine which drying methods result in significantly different smoothness from each other. Present your results using the lines method as discussed in class. Explain these results (that is, prove to me that you know how to correctly interpret the output). Drying Method: Mean Smoothness: 1 3.36 3 2.96 sheeting cord 2 denim 2 crepe 3 dbl_knit twill 3 twill_mix terry 3 broadcloth sheeting cord 3 denim 3 crepe 4 dbl_knit twill 4 twill_mix terry 4 broadcloth sheeting 2 1.3 1.4 2.8 3 3.8 3 2.8 3 3 2.8 2.4 2.5 4 3.1 4 2 4 4 4 2.02 5 2.01 2 1.99 SAS data cloth; input fabric $ drying smoothness; cards; crepe 1 3.3 dbl_knit 1 twill 1 4.2 twill_mix 1 terry 1 3.8 broadcloth 1 sheeting 1 cord 1 3.6 denim 1 2.6 crepe 2 2.5 dbl_knit 2 twill 2 3.4 twill_mix 2 terry 2 1.3 broadcloth 2 Homework #9 - Solution 3.6 3.4 2.2 3.5 2 2.4 1.5 2.1 3.6 2.9 2.7 2.8 2.4 1.6 1.5 2.1 cord 4 denim 4 crepe 5 dbl_knit twill 5 twill_mix terry 5 broadcloth sheeting cord 5 denim 5 ; 1.7 1.3 1.9 5 3.1 5 1.6 5 5 1.8 1.6 2.3 1.7 1.9 2.2 proc anova data=cloth; class fabric drying; model smoothness = fabric drying; means drying/tukey; run; Page 4 of 4 The ANOVA Procedure Class Level Information Class Levels Values fabric 9 broadclo cord crepe dbl_knit denim sheeting terry twill twill_mi drying 5 12345 Number of Observations Read Number of Observations Used 45 45 The ANOVA Procedure Dependent Variable: smoothness Source DF Sum of Squares Mean Square F Value Pr > F Model 12 24.65822222 2.05485185 20.16 <.0001 Error 32 3.26177778 0.10193056 Corrected Total 44 27.92000000 R-Square Coeff Var Root MSE smoothness Mean 0.883174 12.94320 0.319266 2.466667 Source DF Anova SS Mean Square F Value Pr > F fabric drying 8 4 9.69600000 14.96222222 1.21200000 3.74055556 11.89 36.70 <.0001 <.0001 The ANOVA Procedure Tukey's Studentized Range (HSD) Test for smoothness NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 32 Error Mean Square 0.101931 Critical Value of Studentized Range 4.08622 Minimum Significant Difference 0.4349 Means with the same letter are not significantly different. Tukey Grouping Mean N drying A A A 3.3556 9 1 2.9556 9 3 B B B B B 2.0222 9 4 2.0111 9 5 1.9889 9 2 Homework #9 - Solution Page 5 of 5 ...
View Full Document

This note was uploaded on 02/16/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online