Bioe339 Hw 8 #2 Ans

HW8-Problem II 1.Description of data. Nil A B C D E 58 68 60 68 64 318 63.6 1412 70.6 very(prenominal) low Low bonny 62 67 70 70 78 81 65 68 70 80 81 89 69 70 74 346 364 384 count 69.2 72.8 76.8 257 297 263 318 277 64.25 74.25 65.75 79.5 69.25 Average Total Total mean 2.Assumptions. Normal distribution freelancer random samples represented for each multitude lug and preaching is analog effect 3.Hypotheses. H0: Tj =0 for j = 1,2,3.k HA: Tj does non live 0 for all mensurates of j 4.Test statistic. V.R. = MST MSE 5.Distribution of testify statistic. df 4 in numerator and 12 in denominator interest F distribution 6.Decision rule. a =0.05 Fvalue =3.26 7.Calculation of test statistic. ANOVA beginning Treatmen Block Error Total SS SSTr SSBi sou-sou-east SST df 4 3 12 19 SS MS VR P-Value 632.8 158.2 30.2293 3.49719E-06 471.2 157.0667 MSTr/MSE 62.8 5.233333 1166.

8 8.Statistical decision. V.R. > F value 30.229 > 3.26 hence reject unserviceable surmise because too long of a difference so the SS atomic number 18 not calculating alike(p) quantities 8.Statistical decision. V.R. > F value 30.229 > 3.26 accordingly reject null hypothesis because too great of a difference so the SS atomic number 18 not calculating confusable quantities 9.Conclusion. All treatment make are not 0 and are not equal either 10.Determination of p value. 0.00000349If you motive to get a fully essay, order it on our website: Ordercustompaper.com

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