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A randomized block design yielded the ANOVA table to the right. Complete parts a through g Source Treatments Blocks Error Tot plz answer from a to g
d. What test statistic should be used to conduct the hypothesis test? OA. F= 14.857 OC. F=6.248 OB. F=18.743 OD. F=2.971 e. S
F. Conduct the test of parts ce, and state the proper conclusion O A. Reject H. There is insufficient evidence to indicate th
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Answer #1

a) We know that in randomized block design, df_treatment=no_of_teatment - 1 & df_block=no. of Block - 1

So, no. of Block=3+1=4 & no. of treatment = 5+1=6

b) No. of observation= (No. of treatment) * (No. of block) = 6*4 =24

c) Since the Null hypothesis assumes there is no significant difference among various treatment means & Alternative hypothesis is the complement of the Null Hypothesis.So

(B) H0: µ1=µ2=µ3=µ4=µ5=µ6, H1: At least two treatment differ.

d) F statistic for this hypothesis will be defined as is defined as the ratio of the mean sum of square due to treatment to the Mean sum of square due to error i.e. F=131.2/7=18.743 (B)

e) Tabulated value of F at (5,15) degree of freedom at 0.01 level of significance is equal to 4.56. Thus (D) the rejection region is F>4.56

f) Since calculate F=18.743 is much greater than than than tabulated F=4.56. It implies that we can reject and we can accept H1. H0.  (C) Reject H0. There is sufficient evidence to indicate that at least two treatment means differ.

g) Since in any design, The null hypothesis is usually that the variance of the means 0. but from (f) we can see that there is a significant difference among treatment means. It means there is variability among treatment means. So, (D) The distribution of observation corresponding to all the block-treatment combination is normal, and their variances are not equal.

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