Setup: In an experiment to compare different methods of teaching arithmetic, a group of students were randomly divided into three equal-sized groups. Group A was taught by the current method, while the other two groups were taught by one of two new methods. At the end, each student took a standardized test, with the following results: Group A B C Grand Score 17 14 24 20 24 23 16 15 14 18.55 19 28 26 26 19 24 24 23 22 23.44 21 14 13 19 15 15 10 18 20 16.11 19.37 Y. (Y.-Y..) 8² 0.664 18.03 16.60 9.53 10.62 13.11 1. What type of design method is used in the analysis? Write down appropriate statistical models for the analysis with necessary assumptions and constraints. 2. Based on the summary statistics, calculate the sum of squares (SSTR. SSE, SSTO), and construct an ANOVA table. 3. Perform a hypothesis test to determine if there is a difference in the group means at a 0.05 significance level. State the null and alternative hypotheses. Find the test statistic and calculate its p-value. State your conclusion.

Since you have posted a question with multiple subparts, we will provide solution to the first three sub-parts as per our Q&A guidelines. Please repost the remaining sub parts separately.The design model used in the study was completely randomized design (CRD) specifically a one-way analysis of variance (ANOVA) design, to compare the means of three or more groups to determine whether there are statistically significant differences among them. The model can be given as: where Yij is the jth observation in ith group, is the overall/grand mean, is the effect of ith group and eij is the error term. Assumptions and Constraints:We assume that the test scores of students in each group are independent of each other.We assume that the errors are normally distributed..We assume that the variances of the test scores within each group are approximately equal.SSTR (treatment sum of squares)= = 250. 956SSTO (total sum of squares) = = (17-19.37)2+(14-19.37)2+(24-19.37)2+.................................+(18-19.37)2+(20-19.37)2 = 576.296SSE (error sum of squares) = SSTO - SSTR = 576.296 - 250.956 = 325.340Treatment degrees of freedom (DF) = number of groups - 1 = 3-1 = 2Error degrees of freedom = Total sample size - number of groups = 9*3 - 3 = 27-3 = 24Total degrees of freedom = 24 + 2 = 26ANOVA table:SourceSSDFMSF-valueTreatment250.9562125.4789.256Error325.3402413.55583Total576.29626MS = SS/DF and F = MS(TR)/MS(E)Null hypothesis: There are no significant differences in mean test scores among the three groups.Alternative hypothesis: The mean test score of at least one group mean is significantly different from the others.Given 0.05 significance level.Test statistic, F = 9.256 (from ANOVA table).P-value = Pr(F>9.256) = 0.00092 (from R software).Since the p-value < 0.05, we reject the null hypothesis. We can conclude that the mean test score of at least one group mean is significantly different from the others.R-codes:1> pf(9.256, df1=2, df2=26, lower.tail = FALSE) #p-value 2[1] 0.00092Answers are given above.

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