The following table summarizes the results from a two-factor study with 2 levels of factor A...
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
The following table summarizes the results of a two-factor ANOVA evaluating an independent-measures experiment with 2 levels of factor A, 3 levels of factor B, and n = 6 participants in each treatment condition. A. Fill in all missing values in the table. Show your work (i.e., all computational steps for finding the missing values). Hint: start with the df values. B. Do these data indicate any significant effects (assume p < .05 for hypothesis testing of all three effects)?...
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...
The following table shows the results of a two-factor ANOVA evaluating an independent-measures experiment with three levels of factor A, three levels of factor B, and n = 10 participants in each treatment condition. a. What is the calculated F value for the interaction? Source SS df MS Between Treatments 124 Factor A 20 10 F A= Factor B 42 F B= A*B 20 F A*B=? Within Treatments 324 Total 1. F=13 2. F=5 3. F=2.5 4. F =...
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...
The following table summarizes the results of a study on SAT prep courses, comparing SAT scores of students in a private preparation class, a high school preparation class, and no preparation class. Use the information from the table to answer the remaining questions Number of Sum of Treatment Private prep class High school prep class No prep class Observations Sample Mean Squares (SS) 690 680 640 60 60 60 265,500.00 276,120.00 302,670.00 Using the data provided, complete the partial ANOVA...
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...
4. ANOVA calculations and rejection of the null hypothesis The following table summarizes the results of a study on SAT prep courses, comparing SAT scores of students in a private preparation dlass, a high school preparation dlass, and no preparation dass. Use the information from the table to answer the remaining questions. Number of Observations Treatment Sample Mean Sum of Squares (SS) Private prep dlass 40 610 97,500.00 High school prep class 40 600 101,400.00 No prep class 40 590...
The following table shows the results of an analysis of variance comparing two treatment conditions with a sample of n = 11 participants in each treatment. Note thatseveral values are missing in the table. What is the missing value for the F-ratio?Source SS df MSBetween xx xx 14 F = xxWithin xx xx xxTotal 154 xxa. 14b. 28c. 7d. 2
1) A researcher conducts an experiment comparing four treatment conditions with a separate sample of n 10 in each treatment. An ANOVA is used to evaluate the data, and the results of the ANOVA are presented in the following table Complete all missing values in the table. Hint: Begin with the values in the df column Source SS Between treatments 10 Within treatments Total df MS 164