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The following data are from a completely randomized design. In the following calculations, use a = 0.05. Treatment TreatmentUse Fishers LSD procedure to determine which means are different. Difference Absolute Value Conclusion Significant differenc

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Answer #1

a)applying one way ANOVA on above data:

Source SS df MS F P value
Between 1448.0 2 724.0 7.24 0.013
Within 900.0 9 100.0
Total 2348.0 11

p value : between 0.01 and 0.025

b)

critical value of t with 0.05 level and N-k=9 degree of freedom= tN-k= 2.262
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   = 15.99
Difference Absolute Value Conclusion
x2-x1 26.00 significant difference
x3-x1 7.00 not significant difference
x3-x2 19.00 significant difference
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