ANOVA Analysis of Grades and Response Variables: Key Findings

PSYC3000D Courtney Orr Assignment 5 April 22, 2023 1A. Questions (i) For this ANOVA, report the results given in the ANOVA table in the output (i.e., the effect of ___explanatory variable__ on ___response variable__ was significant/not significant, F(_df1_, _df2_) = ___, p = ___ [with all the blanks filled in with the appropriate values or variable names]) using α = .05 as the level of significance. State what conclusion (in words) can be made from those ANOVA table For the effect of grades on wmscores response variables were significant. F=4.800 df1= 2 df2 =21 P=3 Reject the null hypothesis and accept the alternative hypothesis, at least 2 means differ. (ii) The assumption of constancy of variance across the groups is violated (i.e., does not hold) if the “Levene” test of homogeneity of variances is significant. Based on the results of this test for these data (see the output), can it be assumed here that the variances are constant or not constant? Based on the results of this test for these data it can be assumed that the variances are constant and equal, as well as not significant. We accept null hypothesis. The alternative hypothesis has non-constant variances that are not equal Null hypotheses are constant and equal

(iii) For this ANOVA, report the follow-up 95% Tukey confidence intervals for each possible pair of group means. Use those confidence intervals to determine which pairs of group means, if any, differ significantly (and report the result of that determination). [Note that SPSS gives results for each pair twice - once for each order of comparison.] Grade level Lower bound 95% CI Upper bound 95% CI Grade 4 -5.2540 1.2540 Grade 5 -7.2540 -.7460 Grade 6 -1.2540 5.2540 1B. For the ANOVA analysis, using the “Test of Within Subjects Effects” results in the report the results in the ANOVA table (i.e., the effect of explanatory variable on response variable was significant/not significant, F(_df1_, _df2_) = ___, p = ___ [with all the blanks filled in with the appropriate values or variable names]) using α = .05 as the level of significance. State what conclusion (in words) can be made from those ANOVA table results. For the effect of grade456 on grade response variable was significant because it was <.001 F= 13.714 df1= 2 df2= 14 P= 3 Reject the null hypothesis and accept the alternative hypothesis because at least 2 means differ. (ii) The assumption of sphericity across the groups is violated (i.e., does not hold) if the “Mauchly’s” test is significant. Based on the results of this test for these data (see the output), can it be assumed here that sphericity holds or does not hold? Based on the results of this test for the data it can be assumed that sphericity does not hold across the groups, mauchlys test is significant.

(iii) For this ANOVA, report the follow-up 95% Bonferroni confidence intervals for each possible pair of group means. Use those confidence intervals to determine which pairs of group means, if any, differ significantly for this analysis (and report the result of that determination) Grade 1(2) = (-3.448, -.552) Grade 1 (3) = (-6.896, -1.104) Grade 2(1) = (.552, 3.448) Grade 2(3) = (-4.576, .576) Grade 3 (1) = (1.104, 6.896) Grade 3 (2) = (-.576, 4.576) 2. Questions (i) For each of these two factors, what factor levels do the numbers in the data file refer to? For the gender factor, value label 1 represents male (N=30) For the gender factor, value label 2 represents female (N=20) For academic level factor, value label 1 represents undergraduate (N=20) For academic level factor, value label 2 represents MA (N=20) For academic level factor, value label 3 represents PhD (N=20) (ii) Report the results of the three F-tests for the two main effects and the interaction found in the ANOVA table for this analysis using α = .05 as the level of significance for each test (e.g., the main effect of Factor /interaction effect of Factor 1 and Factor 2 on response variablewas significant/not significant, F(_df1_, _df2_) = ___, p = ___ [with all the blanks filled in for each test with the appropriate values or variable names]).

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1. Gender -the interaction effect of factor 1 and factor 2 on response variable was significant because it does not exceed 0.50 F= 4.992 Df1= 1 Df2=1 P= .030 2. Academic level -the interaction effect of factor 1 and factor 2 on response variable was significant because it does not exceed 0.50 F= 6.492 Df1= 2 Df2=2 P= .003 3. Gender X academic level -the interaction effect of factor 1 and factor 2 on response variable was significant because it does not exceed 0.50 F= 4.164 Df1= 2 Df2=2 P= .021 (iii) Using the means and/or the profile plot of means provided by SPSS, describe how the effect of academic level on extroversion differs for males and females. Does this difference represent a main effect or an interaction?

-males had a higher mean academic level for extraversion than females for undergraduate (35.099 vs 33.500) -females had a higher mean academic level for extraversion than males for both MA and PhD (40.200 & 45.400 vs 35.800 & 37.300)

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