The one way Anova test results are listed above (See excel file). Before beginning the one way Anova, the main assumptions are: all groups are normally distributed and have equal variances. We have three different groups: type one, type two and type three. Within the groups degree of freedom is df1=37 and between group’s degree of freedom is df2=2. At 5% level of confidence the f-value calculated is greater than the tabulated (critical value) of F. That is, 5.5774>3.2516, hence the null hypothesis is rejected. The variances for the three types of groups are significantly different.
The Levene’s test also reveals that the variances are significantly different, F calculated is greater than F tabulated, that is, 14.403>3.2519. Hence an additional test is conducted. The Tukey test for pairwise comparison of group means is conducted. The Tukey test is the follow up test. Looking at the excel table below for the tukey test, one can infer on the significance between the means of the groups.
If “classes with teachers with a humanists approach with control will have the fewest number of behavior problems” is the null hypothesis, then considering the null hypothesis was rejected, one can say that a less humanistic or a humanistic teacher bears little behavioral problem in the students.
Type 1 and Type 3 groups have significantly different means same to type 2 and type 3 groups. On the hand, type 1 and type 2 groups don’t have significantly different means.
Ancova Receptive Vocab
The covariate and dependent variable table for type one and type two classes above.
For the two samples, the degree of freedom for the adjusted mean has one degree of freedom while the adjusted error has degree of freedom of 7. F calculated in this case is 34.93 and p-value is 0.0006. The p-value is lower than the critical p-value tabulated in statistical tables at 5% confidence level. We accept the null hypothesis and say that, students taught vocab in classes with windows perform better.
Test for Homogeneity of Regressions
The test of homogeneity of slope displayed above has an F-statistic value of 1.46 and a p value of 0.2666. With the inclusion of the covariate the one can see there is significant difference between the calculated values. F: 34.93>F: 1.46 and P-value: 0.0006<P-value: 0.2667. If the covariate is excluded, then there results obtained will be significantly different and null hypothesis will be rejected.