This question is from the below book, Chapter 15 (Comparing means of more than two groups) , AP 27 Book-Analysis of biological data2nd EditionMichael C. Whitlock, Dolph SchluterPublisher: W. H. FreemanISBN: 978193622148 Does adding math to a scientific paper make readers think that it has more value? Erikkson (2012) sent two abstracts of scientific papers to 200 people with postgraduate degrees. For each participant, one of the abstracts was randomly chosen and had a meaningless sentence inserted describing an unrelated. mathematics model, while the other had no mathematical addition. The sentence had no conceptual connection to the subject matter of the abstract; it was just meaningless mathematics in that context. Participants were asked to rate the quality of the research in each abstract on a scale from 1 to 100 and the differences between the scores of their two abstracts -score of the abstract with math minus score of abstract without math-were recorded. Participants were also asked for the subject matter of their postgraduate degree : math, science, technology (MST); medicine (M), humanities, social sciences (HS) or other (O). A box plot of the data and summaries were of their result for each group are given below; the full data set can be found out at whitlockschluter.zoology.ubc.ca. Graph attached in the image.Examine the graph and judge by eye how well the data likely fit the assumptions of ANOVA.Test whether the subject background of the participants affected how much the added math changed their views of the abstracts on average.Is the relationship between degrees subject and score difference strong? Answer using R2. ### Analysis of Score Differences by Degree Subject#### Box Plot ExplanationThe box plot above illustrates the score differences across four different degree subjects: MST, M, HS, and O. Each box plot displays:- **Median**: The line inside each box represents the median score difference for each group.- **Interquartile Range (IQR)**: The box itself covers the middle 50% of the data, stretching from the 25th to the 75th percentile.- **Whiskers**: The lines extending from the boxes indicate variability outside the upper and lower quartiles. They typically represent 1.5 times the IQR.- **Outliers**: Dots beyond the whiskers indicate outliers or data points that fall significantly outside the range of the rest of the data.#### Statistical SummaryThe table below the box plot provides a summary of mean score differences, standard deviation (SD), and sample size (n) for each degree subject:- **MST** - Mean Score Difference: -1.28 - SD: 19.24 - n: 69- **M** - Mean Score Difference: 3.06 - SD: 15.99 - n: 16- **HS** - Mean Score Difference: 6.60 - SD: 21.15 - n: 84- **O** - Mean Score Difference: 13.90 - SD: 23.31 - n: 31This data provides an overview of how score differences vary among students in different degree subjects, highlighting variations in both central tendency and dispersion.

Name: This question is from the below book, Chapter 15 (Comparing means of m
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Description: This question is from the below book, Chapter 15 (Comparing means of more than two groups) , AP 27 Book-Analysis of biological data2nd EditionMichael C. Whitlock, Dolph SchluterPublisher: W. H. FreemanISBN: 978193622148 Does adding math to a scientif

(1). Comment on the assumptions of ANOVA: The Boxplot for the score difference of the groups MST, M, HS and O. Assumptions of ANOVA: The necessary conditions to use one way ANOVA are given below: Each group sample must be drawn from a normally distributed population. All population must have common variance. All samples must be independent of each other. Within each sample, the observations must be randomly and independently drawn from each other. The populations must be symmetrical and uni-modal. The sample sizes of the groups must be greater than or equal to 10. Here, we can assume about the randomness and independence of the data. The sample sizes of all the four treatment groups are greater than 10. From the graph it is seen that, except the sample M, remaining all 3 samples have outliers. The distribution of first 3 samples seems to be roughly symmetric. The spread of the data of all the 4 samples roughly seems to be even. Even-though, the data indicate violation from normality condition due to outliers, the data seems to be likely to fit the assumptions of ANOVA as the ANOVA is robust.(2). Test whether the subject background of the participants affected the score: There are 4 groups of degree subject. Denote the true mean score difference of four groups as μ1, μ2, μ3, and μ4 respectively. The hypotheses to be tested are: Null hypothesis: H0: μ1 = μ2 = μ3 = μ4 That is, the mean score difference for all the 4 groups are equal. Alternative hypothesis: Ha: At least one of the mean score difference of three treatment groups differs from the other. Evidently, the test is for equality of 4 means. The means take numerical values. Now, a one-way ANOVA is the most suitable for testing the equality of several means, provided the assumptions of independence of observations, normality of residuals and homogeneity of variances are satisfied. Calculate sum of squares within: The sum of squares within (SSW) is obtained as 82435.73 from the calculations given below: Calculate between sum of squares: The between sum of squares (SSB) is obtained as 5578.626 from the calculations given below: Calculate the F-statistic for ANOVA: Here, the total number of observations is N = 200 and the number of treatments is k = 4. The F–statistic for ANOVA is obtained as 4.4213 from the calculation given below: The F–statistic for ANOVA is 4.4213. Degrees of freedom: Here, the total number of observations is N = 200 and the number of treatments is k = 4. The degrees of freedom is obtained as given below: Decision rule: Here, the level of significance is α = 0.05 Decision rule based on critical-value approach: If F ≥ Fα, then reject the null hypothesis H0. If F < Fα, then fail to reject the null hypothesis H0. Critical-value: The critical value for the test is Fα,k–1,N –k., where, N is the sample size and k is the number of treatment groups. Here, k = 3 and N = 196. Step-by-step software procedure to obtain critical-value using EXCEL software is as follows: Open an EXCEL file. In cell A1, enter the formula “=F.INV.RT(0.05,3,196)” Thus, the critical-value using EXCEL is 2.6507. Conclusion: Conclusion based on critical value approach: The test statistic-value is 4.4213 and critical value is 2.6507. Here, test statistic value is greater than the critical value. That is, 4.4213 (=test statistic-value) > 2.6507 (= critical value). By the rejection rule, reject the null hypothesis. Hence, at least one of the mean score difference of three treatment groups differs from the other. Thus, there is enough evidence to conclude that there are significant differences in mobility among the 3 treatment groups. Therefore, the subject background of the participants affected the math changed views of the abstract on average. (3). Comment on the strength of the relationship between degree subject and score difference using R2: The value of coefficient of determination is obtained as given below: Thus, the coefficient of determination is 0.9966. Since, the coefficient of determination is high, the relationship between the degree subject and score difference is strong.