• Name the variables used in this analysis and whether they are categorical or continuous.
• State a research question, null hypothesis, and alternate hypothesis for the analysis of variance (ANOVA).

VERSITY Col Default* Testing Assumptions Style Tests of Normality Text Color Kolmogorov-Smirnova Shapiro-Wilk its Statistic df Sig. Statistic df Sig. Alignment quiz3 .140 105 .000 .935 105 .000 on Announcement a. Lilliefors Significance Correction E E 021 to Dec 17 2 T ght 6PMCST-8PM ent Walkthrough T There is sufficient evidence to prove that the data are not normally distributed therefore the ght @ 7pm CST normality assumption has been violated. The Shapiro-.Wilk test determines if a given dataset is Spacing normally distributed. At Least Null hypothesis: The data are normally distributed Before Paragraph Alternative hypothesis: The data are not normally distributed. At the 5% level of significance, we reject the null hypothesis if the p-value is less than or equal to 0.05, from the output, the test After Paragraph statistic is 0.935 while the p-value is 0.000 which is less than 0.05 therefore we reject the null hypothesis and conclude that there is sufficient evidence to prove that the data are not normally Bullets & Lists distributed. The normality assumption has been violated. No Bullets Indent:Descriptives quiz 3 95% Confidence Interval for Std. Mean N Mean Deviation Std. Error Lower Bound Upper Bound Minimum Maximum 1 33 7.24 1.324 230 6.77 7.71 5 10 2 39 6.72 1.538 .246 6.22 7.22 4 10 3 33 7.82 1.976 .344 7.12 8.52 4 10 Total 105 7.23 1.677 .164 6.90 7.55 A 10 Post Hoc Tests Multiple Comparisons Dependent Variable: quiz3 Tukey HSD Mean 95% Confidence Interval Difference (1- (1) section () section J) Std. Error Sig. Lower Bound Upper Bound 1 2 524 385 .365 -.39 1.44 3 -.576 401 .327 -1.53 .38 2 1 -.524 385 .365 -1.44 .39 3 -1.100' .385 014 -2.02 -.18 3 1 576 401 .327 -.38 1.53 2 1.100 385 014 .18 2.02 *. The mean difference is significant at the 0.05 level. ANOVA quiz 3 Sum of Squares df Mean Square F Sig. Between Groups 21.647 2 10.824 4.076 020 Within Groups 270.867 102 2.656 Total 292.514 104 926 wordsMea Standard n Deviation Group 7.24 1.324 Group 6.72 1.538 2 Group 7.82 1.976 3 To Test :- Ho : MI = My =M3 [ there is no difference in means quiz3 scores for three groups } H : : 7=M; , for at least on pair [ at least one groups differ in mean ] From provided output of ANOVA table, we can observe that Test Statistics value is : F =4.076 . . The p-value is : P-Value = 0.020 We reject the null hypothesis at 5% of significance level. (as P-Value for F-Test is 0.020