15.Consider the resulting table for a Two-Factor ANOVA with replication regardin

Question

15.Consider the resulting table for a Two-Factor ANOVA with replication regarding standardized final exam scores for twelve students from three universities: six day students (two from each university), and six night students (two from each university) ANOVA MS 588 P-value F crit 0.2022 5.14 Source of Variation Sample (Day/Night) Columns (University) Interaction Within (Error) SS 588 2328 4392 1720 2.05 2 1164 06 0.0767 5.99 2 2196 7.66 0.0223 5.14 6 286.67 Total 902811 a. At = 0.05, is there a difference in average levels of performance between day and night students based on these scores? Circle one: YES NO b. At a = 0.05, do the universities prepare students at different average levels of capability based on these scores? Circle one: YES NO C. At = 0.05, do the universities prepare day and night students at different average levels of capability based on these scores? Circle one: YES NO 16. Consider the resulting table for a Two-Factor ANOVA without replication regarding engine tune up times using two different types of electronic analyzer and 3 different types of automobile engine: ANOVA Source of VariationSS df MS P-value F crit 30 5 6 3.157894737 0.0573992 3.325835 Rows (engine type) Error (Random Variation) 19 10 1.9 Total a. At a 0.05, do the engine types generate different average tune up times? Columns (analyzer type) 21 2 10.5 5.526315789 0.0241807 4.102821 70 17 Circle one: YES NO b. At = 0.05, does the type of analyzer yield different average tune up times? Circle one: YES NO

Explanation / Answer

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15) p value = 0.2

alpha = 0.05

as p > alpha, null is accepted

So, there is no difference

NO

b) p = 0.077

alpha = 0.05

as p > alpha, null is accepted

So, there is no difference.

NO

c) p = 0.02

alpha = 0.05

as p > alpha, null is accepted

So, there is no difference.

NO

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