Homework Answers
Solution
From given formation
n for each cell is 9
Explnation:
Since for number of factor A = 3
number of factor B = 2
And interaction A*B = 3*2 = 6
Error degrres of freedom = 48
So total number of units = 54
divide into 6 equal parts so 54/6 = 9
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Question 21 1 pts Results from a 2 Factor ANOVA produced the following FA(2, 48) =...
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Results from a 2 Factor ANOVA produced the following FA(2, 48) = 2.25, p >.05, FB(1,48) = 12.07, p<.05 and FaxB(2, 48) = 7.48, p <.05. Using this information determine the n of each cell, if there is an equal number of participants in each condition.
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Please show your work The following table summarizes the results of a two-factor ANOVA evaluating an...
Please show your work
The following table summarizes the results of a two-factor ANOVA evaluating an independent- measures experiment with two levels of factor A three levels of factor B, and n= 5 subjects in each separate sample. a. Fill in all missing values in the table. (Hint: Start with the df column.) Source SS df MS Between Treatments 60 Fbtw = Factor A FA= 5 Factor B FB= Ах В 30 FAB= I Within Treatments Total
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I need help with questions 11 & 12 please.
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I need help with this question please
Results from a 2 Factor ANOVA produced the following FA(2, 48) = 2.25, p >.05, FB(1,48) = 12.07, p<.05 and FaxB(2, 48) = 7.48, p <.05. Using this information determine the n of each cell, if there is an equal number of participants in each condition.
Question 12 15 pts The following table summarizes the results of a two-factor ANOVA evaluating an independent-measures experiment with two levels of factor A, three levels of factor B, and n = 5 subjects in each separate sample. Fill in all missing values in the table. (Hint: Start with the df column.) Source SS df MS 60 Between Treatments Factor A FA = 5 Factor B _ — FB = AXB 30 FAXB = N Within Treatments Total HTML Editor...
Please show your work
The following table summarizes the results of a two-factor ANOVA evaluating an independent- measures experiment with two levels of factor A three levels of factor B, and n= 5 subjects in each separate sample. a. Fill in all missing values in the table. (Hint: Start with the df column.) Source SS df MS Between Treatments 60 Fbtw = Factor A FA= 5 Factor B FB= Ах В 30 FAB= I Within Treatments Total
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The following results are from an independent-measures, two-factor study with n = 5 participants in each treatment condition Factor A: Factor B: 3 M=5 M=8 M=14 T=25 T=40 T=70 SS 30 SS 38 SS46 n=5 n=5 n=5 2 T= 15 T-20 T=40 SS 22 SS 26 SS 30 ZX2 = 2,062 Use a two-factor ANOVA with α = .05 to evaluate the main effects and interaction. Source df MS Between treatments A x B Within treatments Total F Distribution Numerator...
I need help with questions 11 & 12 please.
D Question 11 1 pts A study reports the following final notation: F(2, 37) = 9.00,p<.05; how many total participants were involved in the study? O 37 O 38 O 39 о 40 Question 12 1 pts A study reports the following final notation: F(2, 37) - 9.00,p<.05; if MSwithin is 5, what is the MSbetween? 40 45
The following results are from an independent-measures, two-factor study with n condition. 10 participants in each treatment Factor B Factor A 2 T 40 M=4.00 SS = 50 T=50 M = 5.00 SS = 60 T= 10 M 1.00 SS 30 T=20 M 2.00 SS 40 N = 40; G = 120; Σ? = 640 Use a two-factor ANOVA with α =。05 to evaluate the main effects and the interaction Source df MS Between treatments AxB Within treatments Total For...
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