Running Head: TWO-WAY ANOVA WRITE UP Group 2 Write-Up: Two-Way ANOVA Partial Fulfillment Of the Requirements for EDUC 812 2017 1 TWO-WAY ANOVA WRITE UP 2 FINDINGS Research Question The research question for this study was: RQ1: Is there a difference in the amount of time fathers play with their male and female children based on ability level? Null Hypothesis The null hypotheses for this study were: 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. Descriptive Statistics A Two-Way ANOVA was conducted to compare the main effects of types of disabilities and gender and the interaction effect between types of disabilities and gender on amount of play with fathers. Data obtained for the dependent variable amount of play by groupings can be found in Table 1. TWO-WAY ANOVA WRITE UP 3 Table 1 Descriptive Statistics Dependent Variable: play Disability status of the child Typically Developing Physical Disability Intellectual Disability Total Gender of Child Male Mean 7.30 Std. Deviation 1.829 N Female 6.80 2.201 10 Total 7.05 1.986 20 Male 3.00 1.563 10 Female 3.40 1.897 10 Total 3.20 1.704 20 Male 3.22 1.716 9 Female 4.00 1.612 11 Total 3.65 1.663 20 Male 4.55 2.613 29 Female 4.71 2.369 31 Total 4.63 2.470 60 10 Results Data screening Data screening regarding inconsistencies and outliers was conducted on the influence of two independent variables (children's disability type and gender) for the amount of time fathers spend playing with their children. Disability type included three levels (typically developing, physical, and intellectual) and gender consisted of two levels (male and female). No data errors or inconsistencies were identified. Box and Whisker plots were used to detect outliers on each variable. No outliers were identified (see Figure 1). TWO-WAY ANOVA WRITE UP 4 Figure 1. Box and Whisker plot for typically developing, physical disability, and mental retardation. Assumptions A two-way Analysis of Variance (ANOVA) was used to test the null hypotheses. The twoway ANOVA requires that the assumptions of normality and homogeneity of variance are met. Normality was examined using a Kolmogorov-Smirnov test because the sample size was greater than 50 (N = 60) (Warner, 2013). No violations of normality were found as p > 0.05 for all areas (see Table 2). Tests of Normality Kolmogorov-Smirnova Disability status of the child Typically Developing Physical Disability Gender of Child Male Residual for play Female Residual for play Male Residual for play Female Residual for play Statistic 0.161 0.242 0.161 0.224 df 10 10 10 10 Sig. 0.200* 0.100 0.200* 0.168 Shapiro-Wilk Statistic 0.954 0.819 0.933 0.942 df 10 10 10 10 Sig. 0.713 0.025 0.475 0.573 TWO-WAY ANOVA WRITE UP 5 Mental Retardation Male Residual for play Female Residual for play *. This is a lower bound of the true significance. a. Lilliefors Significance Correction b. Alpha = .05. 0.183 0.187 9 11 0.200* 0.200* 0.901 0.937 9 11 0.255 0.480 The assumption of homogeneity of variance was examined using the Levene's test. No violation was found (p = 0.828) so the assumption was tenable (see Table 3). Table 3 Levene's Test of Equality of Error Variancesa Dependent Variable: play F df1 df2 Sig. 0.427 5 54 0.828 Tests the null hypothesis that the error variance of the dependent variable is equal across groups. a. Design: Intercept + disable + gender + disable * gender Results for Null Hypothesis One A 32 factorial ANOVA test was utilized to examine the null hypothesis that there is no difference in the amount of 10-minute play segments between fathers and their children who have no disability, a physical disability, and an intellectual disability. F(2, 54) = 27.140, p = 0.001, 2 = 0.004 (see Table 4). Because the results were significant, (p < 0.05), the null hypothesis failed to reject. Table 4 Tests of Between-Subjects Effects Dependent Variable: play Type III Sum of Source Corrected Model Intercept Squares 182.278a 1276.571 Partial Eta df 5 1 Mean Square 36.456 1276.571 F 11.081 388.025 Sig. 0.001 0.001 Squared 0.506 0.878 TWO-WAY ANOVA WRITE UP disable gender disable * gender Error Total Corrected Total 178.579 0.763 4.294 177.656 1648.000 359.933 6 2 1 2 54 60 59 89.289 0.763 2.147 3.290 27.140 0.232 0.653 0.001 0.632 0.525 0.501 0.004 0.024 a. R Squared = .506 (Adjusted R Squared = .461) Significances were found in the disability variable, therefore post hoc tests were run to determine where there were significances. The results show that scores were significant for the categories Typically Developing Physical Disability (p = 0.001), Typically Developing Intellectual Disability (p = 0.001), Physical Disability Typically Developing (p = 0.001),and Intellectual Disability Typically Developing (p = 0.001) (see Table 5). TWO-WAY ANOVA WRITE UP 7 Table 5 Multiple Comparisons Dependent Variable: Score Tukey HSD (I) Disability (J) Disability status of the status of the child child 95% Confidence Interval Mean Difference (I-J) Physical Typically Developing Disability Intellectual Disability Typically Physical Disability Developing Intellectual Disability Typically Intellectual Disability Developing Physical Std. Error Sig. Lower Bound Upper Bound 3.85* 0.574 0.001 2.47 5.23 3.40* 0.574 0.001 2.02 4.78 -3.85* 0.574 0.001 -5.23 -2.47 -0.45 0.574 0.714 -1.83 0.93 -3.40* 0.574 0.001 -4.78 -2.02 0.45 0.574 0.714 -0.93 1.83 Disability Based on observed means. The error term is Mean Square(Error) = 3.290. Results for Null Hypothesis Two A 3 x 2 factorial ANOVA was used to examine the second null hypothesis that: there is no significant difference between the amount of time fathers play with their male or female children. The test was not significant in that F(1, 54) = 0.232, p = 0.632, 2 = 0.501. Therefore, we fail to reject the null hypothesis at = 0.05. Because the null failed to be rejected, no post hoc testing was needed (see Table 4). Results for Null Hypothesis Three A 3 x 2 factorial ANOVA was used to test the third null hypothesis that there is no significant difference in the amount of time fathers play with their male or female children with TWO-WAY ANOVA WRITE UP 8 no disability, a physical disability, or an intellectual disability. The test was not significant in that F(2, 54) = 0.653, p= 0.525, 2 = 0.024. Therefore, we fail to reject the null hypothesis at = 0.05. Since the null failed to be rejected, no post hoc testing was needed (see Table 4). TWO-WAY ANOVA WRITE UP 9 Reference Warner, R. (2013). Applied statistics: From bivariate through multivariate techniques. Thousand Oaks, CA: Sage Publishing