Question: (a) The accompanying table is an (improper) ANOVA table for the data in Exercise 11.6.8. This analysis does not account for the blocking that was
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(b) The proper ANOVA table for the data, which accounts for blocking, follows. Based on this proper analysis, is there evidence that fish affect the number of mayfly nymphs present in the channels? Use α = 0.05.
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(c) Compute and compare Spooled using the ANOVA table from parts (a) and (b). Why is one estimate larger than the other? What is Spooled measuring in part (a)? In part (b)?
df Sum sq Mean sq Fvalue Between 242.889 21.444 2.924 groups Within 6 44.000 groups Total 7.333 8 86.889 df Sum sq Mean sq value Between 42.889 21.444 16.783 groups Between 238.889 19.444 15.217 blocks Within groups Total 4 5.111 1278 8 86.889
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a H 0 The fish treatments do not affect the mean number of nymphs 1 2 3 H A The fish treatments affe... View full answer
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