The data below was collected from a two-factor ANOVA experiment, factor A has 4 levels and factor B has 3 levels: Factor B 1 2 3 1 60 48 50 Factor A 53 50 48 46 44 49 4 51 43 52
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- The following table summarizes the results from a two-factor study with two levels of factor A and four levels of factor B using a separate sample of n=5 participants in each treatment condition. Fill in the missing values. Determine if any of the F-statistics are significant at . Source SS df MS Between Treatments ____ ____ Factor A 20 ____ ____ F=____ Factor B __ ____ 12 F=____ AxB Interaction ____ ____ 6 F=____ Within Treatments 64 ____ ____ Total ____ ____The following data represent the results from an independent-measures study comparing two treatment conditions. TreatmentOne TreatmentTwo 4.52.45.44.04.73.03.26.0 3.93.62.83.74.43.74.85.0 The data can be download in serial format with this link: Download CSVUsing SPSS, run the independent-measures single-factor ANOVA for this data:F-ratio: p-value: Now, run the independent-measures t test on the same data:t-statistic: p-value:The following table summarizes the results from a two-factor study with two levels of factor A and three levels of factor B using a separate sample of n=5 participants in each treatment condition. Fill in the missing values.
- A study was performed on 200 elementary school students to investigate whether regular Vitamin A supplementation was effective in preventing colds during the month of March. 100 were randomized to receive daily Vitamin A supplements during the month of March, and 100 students were randomized to a placebo group (and did not receive Vitamin A) during the same month. The number of students getting at least one cold in March was computed in the two groups, and the results are given in the following 2 X 2 table. Using a 5% level of significance determine whether there is an association between Vitamin A supplementation and prevention of Common Cold ColdNo Cold Vitamin A1585100 Placebo2575100 40160200The following tables presents the results of a single factor ANOVA comparing 5 treatment means with n= 10 in each condition. Complete the missing values in the table Source SS df Ms F between treatment 15 Within treatment Total 190The following results are from an independent-measures, two-factor study with n=5 participants in each treatment condition. Use a two-factor ANOVA with α=.05 to evaluate the main effects and the interaction.
- An experiment was conducted to study the life (in hours) of two different brands of batteries in threedifferent devices (radio, camera, and portable DVD player). A completely randomized two-factorexperiment was conducted, and the following data resulted.Brand Deviceof Battery Radio Camera DVD PlayerA 8.6 7.9 5.48.2 8.4 5.7B 9.4 8.5 5.88.8 8.9 5.9(a) Analyze the data and draw conclusions, using α = 0.05. (b) Investigate model adequacy by plotting the residuals. (c) Which brand of batteries would you recommend?The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: SST = 280, SSA = 26, SSB = 23, SSAB = 175. Set up the ANOVA table and test for any significant main effects and any interaction effect. Use α=\alpha =α= 0.05.The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: SST = 271, SSA = 22, SSB = 24, SSAB = 170. Set up the ANOVA table and test for any significant main effect and may interaction effect. use a= .05.
- Which of the following is a component of a 2 x 3 Factorial ANOVA? Factor A main effect Factor B main effect Interaction effect All of the aboveComplete the following ANOVA summary table for a two-factor fixed-effects ANOVA, where there are four levels of factor A (school) and six levels of factor B (curriculum design). Each cell includes 8 students. Use a significance level of α=0.05α=0.05. Round values for SS and MS to 3 decimal places as needed. Round p-value to 4 decimal places. Source SSSS dfdf MSMS FF pp AA 2544.8 BB 4243.6 A×BA×B 9760.1 Error TOTAL 75180.5You sample 1500 dung beetles in a population for 2 alleles A1 and A2 and get the following data: A1A1 825 individuals A1A2 525 individuals A2A2 135 Individuals Is this population in Hardy Weinberg Euilibrium, use a chi square test to determine?