Brainstorming A researcher investigated whether...

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Statistics 106 Solutions for Homework 6 Due : Nov. 14, 2016, In Class 20.5 Brainstorming. A researcher investigated whether brainstorming is more e ff ective for larger groups than for smaller ones by setting up four groups of agribusiness executives, the group sizes being two, three, four, and five, respectively. He also set up four groups of agribusiness scientists, the group sizes being the same as for the agribusiness executives. The researcher gave each group the same problem: ”How can Canada increase the value of its agricultural exports?” Each group was allowed 30 minutes to generate ideas. The variable of interest was the number or di ff erent ideas proposed by the group. The results, classified by type of group (factor A ) and size of group (factor B ), were: Factor B (size of group) Factor A j = 1 j = 2 j = 3 j = 4 (type of group) Two Three Four Five i = 1 Agribusiness executives 18 22 31 32 i = 2 Agribusiness scientists 15 23 29 33 Assume that no-interaction ANOVA model is appropriate. a. Plot the data in the format of Figure 20.1 in the textbook (plot of the observations Y i j ). Does it appear that interaction e ff ects are present? Does it appear that factor A and factor B main e ff ects are present? Discuss. b. Conduct separate tests for type of group and size of group main e ff ects. In each test, use level of significance α = . 01 and state the alternatives, decision rule, and conclusion. What is the P -value for each test? c. Obtain confidence intervals for D 1 = μ . 2 - μ . 1 , D 2 = μ . 3 - μ . 2 , and D 3 = μ . 4 - μ . 3 ; use the Bonferroni procedure with a 95 percent family confidence coe ffi cient. State your findings. d. Is the Bonferroni procedure used in pan (c) the most e ffi cient one here? Explain. Solution:
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It appears that interaction e ff ects are very little, since the two curves are nearly parallel. Factor A main e ff ects are also very little, since the two curves are nearly overlapped. Factor B main e ff ects are present, since the two curves have large departure from horizontal. b. We first construct the ANOVA table. Note that the error sum of squares is given by S S AB , which can be obtained by S S AB = S S TO - S S A - S S B in R. Source S S d f MS Type of group 1.125 1 1.125 Size of group 318.375 3 106.125 Error 6.375 3 2.125 Total 325.875 7 F test for type of group main e ff ects: H 0 : α 1 = α 2 = 0, H a : not both α 1 and α 2 are 0. F * = MS A / MS AB = 1 . 125 / 2 . 125 = . 53, F (1 - α ; a - 1 , ( a - 1)( b - 1)) = F ( . 99; 1 , 3) = 34 . 1. If F * < 34 . 1, conclude H 0

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