Can you write it in text please so I can copy and paste it!! Thank you There is some evidence that high school studentsjustify cheating in class on the basis of poor teacherskills or low levels of teacher caring (Murdock, Miller,& Kohlhardt, 2004). Students appear to rationalizetheir illicit behavior based on perceptions of how theirteachers view cheating. Poor teachers are thought notto know or care whether students cheat, so cheating intheir classes is okay. Good teachers, on the other hand,do care and are alert to cheating, so students tend notto cheat in their classes. Following are hypotheticaldata similar to the actual research results. The scoresrepresent judgments of the acceptability of cheating forthe students in each sample.GoodTeacherPoorAverageTeacherTeachern = 6n = 8n = 10N = 24M = 6М — 2М — 2G = 72SS = 30SS = 33SS = 42EX? = 393%3Da. Use an ANOVA with a = .05 to determine whetherthere are significant differences in student judg-ments depending on how they see their teachers.

a. The give study shows the scores of judgements of the acceptability of cheating for the students in each sample.The hypothesis is, Null hypothesis: H0:μ1=μ2=μ3 Alternative hypothesis: H1: At least one mean μiis different from the other means. The total sum of squares is, TSS=∑X2-G2N=393-72224=393-216=177 The error sum of squares is, SSE=∑SSi=SS1+SS2+SS3=30+33+42=105 The treatment sum of squares is, SST=TSS-SSE=177-105=72 The treatment degrees of freedom is, dftreat=k-1=3-1=2 The error degrees of freedom is, dferror=N-k=24-3=21 The total degrees of freedom is, dftotal=N-1=24-1=23 The mean squares treatment is, MST=SSTdftreat=722=36 The mean square for error is, MSE=SSEdferror=10521=5 The F-statistic is, F=MSTMSE=365=7.2 The complete ANOVA table is, Source SS df MS F Treatments 72 2 36 7.2 Error 105 21 5 Total 177 23 The critical value of chi-square distribution for 0.05 level of significance at (2, 21) degrees of freedom can be obtained using the excel formula “=F.INV.RT(0.05,2,21)”. The critical value is 3.467. Decision rule: If test statistic > critical value, then reject the null hypothesis. Conclusion: The test statistic (=7.2) is greater than the critical value (=3.467). Based on decision rule, reject the null hypothesis. Thus, there are significant differences in student judgments depending on how they see their teachers at 0.05 level of significance.