SPSS Worksheet 3: (Two-Way ANOVA) Name: Demario Preston 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 < Note: Remove any outliers from the dataset before continuing.> 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 INDEPENDENT WRITE-UP 1 Independent Write up: ANOVA Demario Preston Liberty University Partial Fulfillment Of the Requirements for EDUC 812 Liberty University ANOVA WRITEUP 2 FINDINGS Research Question The research question for this study was: RQ1: Is there a difference in extrovertedness among blondes, brunettes, and redheads? Null Hypothesis The null hypothesis for this study is: H01: There is no significant in extrovertedness among blondes, brunettes, and redheads. Descriptive Statistics Data obtained for the dependent variable extrovertedness among blondes, brunettes, and redheads can be found in Table 1. The mean extrovertedness for blondes is 5.17 contrasting to the mean extrovertedness of 2.33 for a redhead; all three hair categories mean is 3.72. Table 1 Descriptive Statistics Dependent Variable: Social Extroversion Hair Mean Std. N Color Deviation Blond 5.17 2.787 6 Brunet 3.67 1.211 6 Redhe 2.33 1.033 6 ad Total 3.72 2.109 18 Results Data screening Data screening was conducted on each group's dependent variables (Blond, Brunette and Redhead) regarding data inconsistencies, outliers and normality. The researcher sorted the data on each variable and scanned for inconsistencies. No data errors or inconsistencies were ANOVA WRITEUP 3 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. Figure 1. Box and Whisker Plot for Blond, Brunet, and Redhead Assumptions An Analysis of Variance (ANOVA) was used to test the null hypothesis that examined the differences among type of hair color and their extrovertedness. The ANOVA required that the assumptions of normality and homogeneity of variance are met. Normality was examined using a Shapiro-Wilk test; a Shapiro-Wilk test was used because the sample size was less than 50. No ANOVA WRITEUP violations of normality were found. The assumption of normality was met each group due to their significance level exceeding .05 with blond at .525, brunette at .415 and redhead at .473. See Table 2 for Shapiro-Wilk test. Table 2 The assumption of homogeneity of variance was examined using the Levene's test of equality of error variances. The F test returned a ratio of 1.52. No violations were found (p = . 250), so the assumption of homogeneity was met. For this reason, the researcher continued with the analysis. See Table 3 for Levene's Test. Table 3 Results for Null Hypothesis One An ANOVA was used to test the null hypothesis: the differences in extrovertedness among blonde, brunette, and redheads. The null hypothesis was failed to be rejected at a 95% 4 ANOVA WRITEUP confidence level where F(2, 15) = 3.511, p = .056, 2 = .319. See Table 4 Tests of BetweenSubjects Effects. Table 4 Because the null was rejected and the F test was significant, follow up tests were ran to using a Turkey HSD test was ran to control for Type I errors. There was a substantial disparity between the extrovertedness (M = 5.17, S.D. = 2.787) and Blond haired subjects (P = .056). See Table 5 for Multiple Comparisons Groups. Table 5 5