N = 8*5 = 40
n = 8
Degress of Freedom:
dfBetween = a – 1 = 5-1 =4
dfWithin = N-a = 40-5 = 35
dfTotal = N-1 = 40-1 = 39
MS = Sum of Squares / degrees of Freedom
F = MSTreatments / MSError
Completed ANOVA Table
1. /16.66 points ASWSBE13 13.E.002.MI My Notes Ask Your Teacher You may need to use the...
1, + -/16.66 points ASWSBE13 13E002.MI. My Notes You may need to use the appropriate technology to answer this question. In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Complete the following ANOvA table. (Round your values for MSE and Fto two decimal places, and your p-value to fou r decimal places.) Source of Variation Sum of Squares Degrees of Freedom Mean Square p-value Treatments 340 Total 480 Need Help?...
1. + -/16.66 points ASWSBE13 13.Е.023.MI My Notes Ask Your Teacher You may need to use the appropriate technology to answer this question An experiment has been conducted for four treatments with seven blocks. Complete the following analysis of variance table (Round your values for mean squares and F to two decimal places, and your p-value to three decimal places.) Source of Variation Sum Degrees of Freedom Mean p-value of Squares Square Treatments 900 Blocks 500 Error Total 1,800 Use...
5, + -/16.66 points ASWSBE13 13.Е.031.MI My Notes Ask Your Teacher You may need to use the appropriate technology to answer this question An amusement park studied methods for decreasing the waiting time (minutes) for rides by loading and unloading riders more efficiently. Two alternaive loading/unloading methods have been proposed. To account for potential differences due to the type of ride and the possible interaction between the method of loading and unloading and the type of ride, a factorial experiment...
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