Hi Rudra,

I am having trouble getting the double Anova to translate into the group write up in a clear manner. Please Help!

SPSS Worksheet 3: (Two-Way ANOVA) Name: Kelli Caras Instructions: Lesson 26 Exercise File 2 is located at the end of the chapter under the heading Exercises in your Green and Salkind textbook. Complete the exercise and then complete the worksheet below by filling in the blanks and answering the questions. Note: The two-way ANOVA looks at three null hypotheses at one time. H01: There is no significant difference among the amount of time fathers play with their children with no disability, a physical disability, or an intellectual disability. H02: There is no significant difference between the amount of time fathers play with their male or female children. H03: There is no significant interaction among the amount of time fathers play with their male or female children with no disability, a physical disability, or an intellectual disability. Assumptions Outliers: Create a Box and Whisker plot for each group. Hint: Go to Graph > Legacy Dialog > Boxplot and use the Cluster function. See page 184 in the Salkind and Green textbook for more information on how to display results. Fill in the blanks: Group Outliers (Item #) Are there any outliers? Male Typically No Male Physical No Male Mental No Female Typically No Female Physical No Female Mental No Assumption of Normality: Run a normality test each group. Hint: Begin by going to Data > Split File > Organize output by groups (see lesson 15, p. 64), then run Analyze > Descriptive > Explore (see lesson 40, p. 327). Insert six Tests of Normality tables below: Tests of Normalitya Kolmogorov-Smirnovb Gender of Child play Statistic Male df .161 Shapiro-Wilk Sig. 10 .200* Statistic df .954 Sig. 10 .713 *. This is a lower bound of the true significance. a. Disability status of the child = Typically Developing, Gender of Child = Male b. Lilliefors Significance Correction Tests of Normalitya Kolmogorov-Smirnovb Gender of Child play Statistic Female .242 df Shapiro-Wilk Sig. 10 .100 Statistic .819 a. Disability status of the child = Typically Developing, Gender of Child = Female b. Lilliefors Significance Correction df Sig. 10 .025 Tests of Normalitya Kolmogorov-Smirnovb Gender of Child play Statistic Male df .161 Shapiro-Wilk Sig. 10 Statistic .200* df .933 Sig. 10 .475 *. This is a lower bound of the true significance. a. Disability status of the child = Physical Disability, Gender of Child = Male b. Lilliefors Significance Correction Tests of Normalitya Kolmogorov-Smirnovb Gender of Child play Statistic Female df .224 Shapiro-Wilk Sig. 10 .168 Statistic df .942 Sig. 10 .573 a. Disability status of the child = Physical Disability, Gender of Child = Female b. Lilliefors Significance Correction Tests of Normalitya Kolmogorov-Smirnovb Gender of Child play Statistic Male df .183 Shapiro-Wilk Sig. 9 .200* Statistic df .901 Sig. 9 .255 *. This is a lower bound of the true significance. a. Disability status of the child = Mental Retardation, Gender of Child = Male b. Lilliefors Significance Correction Tests of Normalitya Kolmogorov-Smirnovb Gender of Child play Statistic Female df .187 Shapiro-Wilk Sig. 11 .200* Statistic .937 *. This is a lower bound of the true significance. a. Disability status of the child = Mental Retardation, Gender of Child = Female b. Lilliefors Significance Correction Fill in the blanks: Should you use a Shapiro-Wilks or Kolmogorov-Smirnov test? Why? Answer: Shapiro-Wilks because the sample size is smaller than 50. df Sig. 11 .480 Groups Significance Male Typically Male Physical Male Mental Female Typically Female Physical Female Mental .713 .475 .255 .025 .573 .480 Is the assumption of normality met? Yes Yes Yes No Yes Yes Assumption of Equal Variance: Insert Levene's Test of Equality of Error Variancesa table(s) below. Hint: Begin by going to Data > Split File > RESET > then run the Analyze. Levene's Test of Equality of Error Variancesa Dependent Variable: Play F df1 .427 df2 5 Sig. 54 .828 Tests the null hypothesis that the error variance of the dependent variable is equal across groups. a. Design: Intercept + gender + disable + gender * disable Fill in the blanks: Significance Is the assumption of equal variance met? .828 Yes Results Insert Tests of Between-Subjects Effects table(s) below: Tests of Between-Subjects Effects Dependent Variable: play Type III Sum Source Corrected of Squares Mean df Square 182.278a 5 1276.571 1 gender .763 1 .763 disable 178.579 2 4.294 Error Total Model Intercept Sig. Noncent. Observed Squared Parameter Powerb 11.081 .000 .506 55.405 1.000 1276.571 388.025 .000 .878 388.025 1.000 .232 .632 .004 .232 .076 89.289 27.140 .000 .501 54.281 1.000 2 2.147 .653 .525 .024 1.305 .154 177.656 54 3.290 1648.000 60 359.933 59 gender * disable Corrected Total 36.456 F Partial Eta a. R Squared = .506 (Adjusted R Squared = .461) b. Computed using alpha = .05 Differences among disabilities Fill in the blanks: Results for Disability: d.f. between Groups Value 2 d.f. within Groups 54 F-statistic 27.140 F-critical (See Appendix C in Warner) 3.168 p- value 0.000 Partial Eta Squared .501 Is the F- statistic greater than F-critical? Answer: Yes Is the p- value less than .05? Answer: Yes Should you reject or fail to reject the null? Answer: Reject the null What is the effect size small, medium, or large (See Table 5.2 in Warner, p. 208)? Answer: Medium Should you run post hoc analysis? Answer: Yes If so, between which groups do the differences exist? Answer: Typically Developing and Physical Disability, Typically Developing and Mental Retardation Differences between genders Fill in the blanks: Results for Gender: d.f. between Groups Value 1 d.f. within Groups 54 F-statistic F-critical (See Appendix C in Warner) .232 4.020 p- value 0.632 Partial Eta Squared 0.004 Is the F- statistic greater than F-critical? Answer: No Is the p- value less than .05? Answer: No Should you reject or fail to reject the null? Answer: Fail to reject the null What is the effect size small, medium, or large (See Table 5.2 in Warner, p. 208)? Answer: Small Should you run post hoc analysis? Hint: There are only two groups (Male and Females). Answer: No Interaction among groups Fill in the blanks: Results for Interaction: d.f. between Groups Value 2 d.f. within Groups F-statistic 54 .653 F-critical (See Appendix C in Warner) 3.168 p- value .525 Partial Eta Squared .024 Is the F- statistic greater than F-critical? Answer: No Is the p- value less than .05? Answer: No Should you reject or fail to reject the null? Answer: Fail to reject the null What is the effect size small, medium, or large (See Table 5.2 in Warner, p. 208)? Answer: Small Descriptive Statistics Fill in the blanks: Groups Male Typically Mean 7.30 S.D. 1.829 Male Physical 3.00 1.563 Male Mental 3.22 1.716 Female Typically 6.80 2.201 Female Physical 3.40 1.897 Female Mental 4.00 1.612 Liberty University SOE 12-19-16 Note: This is an example only. You will need to expand on each section. This example should only be used as a guide and for correct format. Refer to the end of each chapter in your textbooks and the Dissertation Handbook on how to write up the findings section. Remember to put the information in your own words in order to avoid plagiarism. GROUP # WRITE-UP: ASSIGNMENT NAME by Group Members Liberty University Partial Fulfillment Of the Requirements for EDUC 812 Liberty University Year Liberty University SOE 12-19-16 FINDINGS Research Question The research question for this study was: RQ1: Is there a difference in learning attitude among traditional, adult, and senior vocational learners at a Northwestern public college? Null Hypothesis The null hypothesis for this study was: H01: There is no significant difference in learning attitude as measure by the Learning Attitude Learning Inventory among traditional, adult, and senior vocational learners at a Northwestern public college. Descriptive Statistics Data obtained for the dependent variable learning attitude for traditional, adult, and senior learners can be found in Table 1. Table 1 Descriptive Statistics Dependent Variable: Score Group Mean Std. Deviation N TL 18.7000 2.90784 10 AL 19.6667 1.73205 9 SL 22.8182 4.46807 11 Total 20.5000 3.70228 30 Results Data screening Data screening was conducted on each group's dependent variables (TL, AL, SL attitude) regarding data inconsistencies and outliers. The researcher sorted the data on each variable and Liberty University SOE 12-19-16 scanned for inconsistencies. No data errors or inconsistencies were identified. Box and whiskers plots were used to detect outliers on each dependent variable. No outliers were identified. See Figure 1 for box and whisker plot. Tests of Normality Group Kolmogorov-Smirnova Statistic Score df Shapiro-Wilk Sig. Statistic df Sig. TL .179 10 .200* .883 10 .140 AL .224 9 .200* .921 9 .399 SL .105 11 .200* .958 11 .751 *. This is a lower bound of the true significance. a. Lilliefors Significance Correction Figure 1. Box and Whisker Plot for Traditional, Adult, and Senior Learners. Assumptions An Analysis of Variance (ANOVA) was used to test the first null hypothesis that looked at the differences among type of learner and their learning attitudes. The ANOVA required that the assumptions of normality and homogeneity of variance are met. Normality was examined using a Shapiro-Wilk test. Shapiro-Wilk was used because the sample size was less than 50. No violations of normality were found. See Table 2 for Shapiro-Wilk test. Liberty University SOE 12-19-16 Table 2 Tests of Normality Kolmogorov-Smirnova Group Statistic Score df Shapiro-Wilk Sig. Statistic df Sig. TL .179 10 .200* .883 10 .140 AL .224 9 .200* .921 9 .399 SL .216 11 .158 .893 11 .151 *. This is a lower bound of the true significance. a. Lilliefors Significance Correction The assumption of homogeneity of variance was examined using the Levene's test. A violation was found (p = .009) so the assumption of homogeneity was not met. However, the ANOVA is considered a robust test against the homogeneity assumption (Warner, 2013, p. 474). For this reason, the researcher continued with the analysis. See Table 3 for Levene's Test. Table 3 Levene's Test of Equality of Error Variance Dependent Variable: Score F df1 5.647 df2 2 Sig. 27 .009 Tests the null hypothesis that the error variance of the dependent variable is equal across groups. a. Design: Intercept + Group Results for Null Hypothesis One An ANOVA was used to test the first null hypothesis; the differences in learning attitude among traditional, adult, and senior vocational learners. The first null hypothesis was rejected at a 95% confidence level were F(2, 27) = 4.40, p = .02, 2 = .25. The effect size was very large. See Table 4 Tests of Between-Subjects Effects. Liberty University SOE 12-19-16 Table 4 Tests of Between-Subjects Effects Dependent Variable: Score Source Type III Sum of df Mean Square F Sig. Partial Eta Squares Squared 97.764a 2 48.882 4.403 .022 .246 12395.150 1 12395.150 1116.545 .000 .976 Group 97.764 2 48.882 4.403 .022 .246 Error 299.736 27 11.101 Total 13005.000 30 397.500 29 Corrected Model Intercept Corrected Total a. R Squared = .246 (Adjusted R Squared = .190) Because the null was rejected, post hoc analysis was conducted using a Tukey Test HSD. There was a significant difference between the attitude scores of traditional (M = 18.7, S.D. = 2.9) and senior (M = 22.8, S.D. = 4.5) vocational learners (p = .02). See Table 5 for Multiple Comparisons Groups. Table 5 Multiple Comparisons Dependent Variable: Score Tukey HSD (I) Group (J) Group Mean Difference Std. Error Sig. (I-J) TL AL SL 95% Confidence Interval Lower Bound Upper Bound AL -.9667 1.53089 .804 -4.7624 2.8290 SL -4.1182* 1.45580 .023 -7.7277 -.5087 TL .9667 1.53089 .804 -2.8290 4.7624 SL -3.1515 1.49756 .108 -6.8646 .5616 TL * 4.1182 1.45580 .023 .5087 7.7277 AL 3.1515 1.49756 .108 -.5616 6.8646