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Comparison between-Group Variance and within-Group Variance - Assignment Example

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The paper "Comparison between-Group Variance and within-Group Variance" highlights that there is no difference determined between the three groups (those that lived at home, those that ate from campus dining halls, and those that cooked for themselves)…
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Comparison between-Group Variance and within-Group Variance
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Extract of sample "Comparison between-Group Variance and within-Group Variance"

Order 1092499 Assignment statistics: Modules 24, 25 and 26 Problem 4, Page 292: Analysis of variance (ANOVA) is used to compare between-group variance and within-group variance by partitioning the sums of squares (SS). The within group variability (SS within) is the deviation of scores within each group from the mean of the group, whereas the between group variability (SS between) is the deviation of each group from the total mean. The deviation of individual scores from the total mean (mean of all scores) constitutes total variability (SS total). The sum of squares represents the variation of the data. Therefore, if the between group deviation is 254.22 and the within group deviation is 106.33, the total deviation will be calculated as follows: SS total (SS between + SS within) = 254.22 + 106.33 = 360.55 Problem 8, Page 295 F= MS between / MS within; therefore F = 100/25 = 4 Problem 10, Page 314: The degrees of freedom (df) are defined by df 1 = k-1; df 2 = N-k, where, k is the number of comparison groups and N is the total number of observations in the study. From the ANOVA table, the df 1= 4; df 2 = 45 and df total =49. Substituting for N and k in the formulae: (4 = k-1; 45= N-k and 49 = N-1), will give N=50 and k=5.Therefore: a) There are 5 groups. b) Each group has 10 subjects (i.e. .50/5) c) The df between and df within are used to get the F critical by using the tabled values of F. Therefore, the F critical = 3.77 at 0.01(1 %). d) Given that the F ≥ 3.77, there is a 99 % level of confidence of rejecting the null hypothesis (H0) e) If the significance level is set at 0.01, there is a 0.01 probability, if the null hypothesis is true, that the researcher will make a Type I error ; any comparison that will be made will be wrongly assumed to be significant . Problem 5, Page 312 a) H0: Students’ weight gain has nothing to do with the type of food served in campus dining centres. H0: μ1= μ2= μ3. b) Table 1: ANOVA summary weight gained by student Sum of Squares df Mean Square F Sig. Between Groups 87.2 2 43.6 4.557 0.02 Within Groups 258.3 27 9.567 Total 345.5 29 c) Result: In this study, one-way ANOVA determined that weight gain of students was statistically significant, F (2, 27) = 4.557, p < 0.05. Problem 5, Page 323 The Tukey HSD post hoc test is used to identify particular independent variable groups or treatments, after F is determined to be significant. In this study, Tukey HSD test has been performed to determine differences between group pairs at α =0 .01 (Table 2 and 3) and α = 0.05 (Table 4 and 5). Table 2: Tukey HSD on the ANOVA at α = .01 Multiple Comparisons Dependent Variable: weight gained by student Tukey HSD (I) place where student eats from (J) place where student eats from Mean Difference (I-J) Std. Error Sig. 99% Confidence Interval Lower Bound Upper Bound home campus dining hall -3.4 1.383 0.052 -7.8 1 cook for themselves 0.4 1.383 0.955 -4 4.8 campus dining hall home 3.4 1.383 0.052 -1 7.8 cook for themselves 3.8 1.383 0.028 -0.6 8.2 cook for themselves home -0.4 1.383 0.955 -4.8 4 campus dining hall -3.8 1.383 0.028 -8.2 0.6 Mean is significant at the 0.01 level Table 3: Homogenous Subsets weight gained by student Tukey HSD,a place where student eats from N Subset for alpha = 0.01 1 cook for themselves 10 3.1 home 10 3.5 campus dining hall 10 6.9 Sig. 0.028 Means for groups in homogeneous subsets are displayed. a Uses Harmonic Mean Sample Size = 10.000. Table 4: Tukey HSD on the ANOVA at α =0 .05 Multiple Comparisons Dependent Variable: weight gained by student Tukey HSD (I) place where student eats from (J) place where student eats from Mean Difference (I-J) Std. Error Sig. 95% Confidence Interval Lower Bound Upper Bound home campus dining hall -3.4 1.383 0.052 -6.83 0.03 cook for themselves 0.4 1.383 0.955 -3.03 3.83 campus dining hall home 3.4 1.383 0.052 -0.03 6.83 cook for themselves 3.800* 1.383 0.028 0.37 7.23 cook for themselves home -0.4 1.383 0.955 -3.83 3.03 campus dining hall -3.800* 1.383 0.028 -7.23 -0.37 Mean is significant at the 0.05 level Table 5: Homogenous Subsets weight gained by student Tukey HSD,a place where student eats from N Subset for alpha = 0.05 1 2 cook for themselves 10 3.1 home 10 3.5 3.5 campus dining hall 10 6.9 Sig. 0.955 0.052 Means for groups in homogeneous subsets are displayed. a Uses Harmonic Mean Sample Size = 10.000. In order to determine which specific pairs were responsible for the overall significant difference in the study, the means of the three groups were computed and compared with the HSD Home Campus dining hall Cook for themselves X X2 X X2 X X2 2 4 5 25 3 9 4 16 9 81 2 4 0 0 12 144 5 25 3 9 0 0 6 36 0 0 8 64 0 0 5 25 10 100 2 4 2 4 7 49 0 0 4 16 5 25 3 9 10 100 4 16 1 1 5 25 9 81 9 81 Total 35 199 69 585 31 169 1 3.5 2 6.9 3 3.1 1 -2=3.4 1 -3= 0.4 2 -3= 3.8 HSD is calculated by the following formula: HSD= q*(SQRT (MS within)/n), where n is the number of subjects in a sample, and q is the studentized range statistic. The studentized range statistic (q) at α =0 .01 and α =0 .05 for the three groups and 27 df within, are found by interpolating from rows 24 and 30, because row 27 is not available. Since 27 is found halfway between 24 and 30, the q values interpolated for α =0 .01 and α =0 .05 are 3.51 and 4.50, respectively. Therefore, the HSD values for α =0 .01 and α =0 .05 are as listed below: i. α =0 .01: 3.51 * (SQRT (9.567)/10) = 3.433 ii. α =0 .05: 4.50 * (SQRT(9.567)/10) = 4.401 Where the difference between the mean computed of a particular group is larger than HSD, then it means there is a significant difference. Therefore, the results can be summarized as follows: At α =0 .01, 1 -2 =3.4 = 3.433; 1 -3= 0.4 3.433. There is a significant difference between group 2 (students that ate from campus dining halls) and 3 (students that cooked for themselves). At α =0 .05, 1 -2=3.4 < 4.401; 1 -3= 0.4 Read More

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