9 - STAT 350 Fall 2008 Homework#9 Solution through...

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STAT 350 – Fall 2008 Homework #9 – Solution through Lecture #13 Text Exercises Chapter 9, Problems #6(a-f), 10, 12 1,2 , 14, 16 1 , 22 2 , 24 3,4 1 You may conduct the analysis using SAS or Minitab, but must present the ANOVA table by hand. DO not paste in extraneous computer output. State your conclusion in a clear and complete sentence! 2 You must also give the ANOVA table for this analysis 3 You may use Minitab to construct the probability plot 4 Be sure to construct the probability plot using the deviations , not on the original observations. To get the deviations, take each observation and subtract the average for that group. That is, if x i,j is the j th observation in the i th group, then the deviation is ( ) , ij i x x . 9.6 a. 3.69 b. 4.82 c. 6.63 d. 10.29 e. 3.10 f. 0.95 9.10 Source df SS MS F Brand 4 39.2192 9.8048 15.32 Error 45 28.8 0.64 Total 49 68.0192 9.12 Source DF SS MS F P Factor 3 456.50 152.17 17.11 0.000 Error 14 124.50 8.89 Total 17 581.00 The p -value is clearly < 0.05, so we can conclude that there is a statistically significant difference in the yield of tomatoes among the soil salinity levels Homework #9 – Solution Page 1 of 5
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9.14 The overall mean, () ( ) ( ) ( ) ( ) 6 2.58 6 2.63 6 2.13 6 2.41 6 2.49 2.448 30 x ++++ == SSTr = 6(2.58-2.448) 2 + 6(2.63-2.448) 2 + 6(2.13-2.448) 2 + 6(2.41-2.448) 2 + 6(2.49-2.448) 2 = 0.92928 SSE = SST – SSTr = 3.619 – 0.92928 = 2.68972 a. Source df SS MS F Sugar 4 0.92928 0.23232 2.15933 Error 25 2.68972 0.107589 Total 29 3.619 b. For df 1 = 4, and df 2 = 25, the critical value of F for an α =0.05 test is 2.76 The F
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This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue.

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9 - STAT 350 Fall 2008 Homework#9 Solution through...

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