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 precise low Low Average 62 67 70 70 78 81 65 68 70 80 81 89 69 70 74 346 364 384 Total 69.2 72.8 76.8 257 297 263 318 277 64.25 74.25 65.75 79.5 69.25 Average Total Total meanspirited 2.Assumptions. Normal distribution freelancer random samples represented for each group lug and preaching is additive effect 3.Hypotheses. H0: Tj =0 for j = 1,2,3.k HA: Tj does non equal 0 for all sets of j 4.Test statistic. V.R. = MST MSE 5.Distribution of testify statistic. df 4 in numerator and 12 in denominator following(a) F distribution 6.Decision rule. a =0.05 Fvalue =3.26 7.Calculation of test statistic. ANOVA denotation Treatmen Block Error Total SS SSTr SSBi SSE 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 null surmise because too huge of a difference so the SS atomic number 18 not calculating like 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 are not calculating homogeneous quantities 9.Conclusion. All treatment effects are not 0 and are not equal either 10.Determination of p value. 0.00000349If you penury to get a full essay, order it on our website: OrderEssay.net

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