Please explain and fill table with steps by step results Comparison GroupsGroups(ascending order) Group AGroup BGroup CGroup DGroup A = 3Group B = 4Group C = 2Group D = 1 A professor gives an exam to a class with four recitation sections. Each section has 7 students. The professorhas no reason to assume that the exam scores would not be independent and normally distributed with equalvariance. However, it is unknown whether or not the section choice (different TAs and different days of theweek) has any relationship with how students performed on the test. The professor conducts a one-wayANOVA, and finds that Ho should be rejected at a significance level of a =0.01.SourceSSdfMSF-ratio P-valueBetween1889.293 3 629.7645.2570.0062Within2875.271 24 119.803As shown above, the professor finds that F(3,24)5.257, which has a probability of0.0062 ofoccurring randomly. Additionally, the error variance was estimated as MSwmeans for each group were obtained as shown in the table below.119.803, and the sampleLevelMeanGroup-188.214Group-2Group-3Group-487.65769.31474.557The professor decides to use the Tukey HSD method to test all possible pairwise contrasts. To do so, thefollowing table of q; statistics needs to be completed. For this presentation, the groups will be ordered fromsmallest to largest means. (In other words, Group A should be the group with the smallest mean.) First,indicate the appropriate order for the groups (use the group number). Then fill in the table with valuesrounded to 3 decimal places. (Furthermore, even though the sign does not matter, report all comparisonstatistics as positive values.)(Note: we often create a tablestatistics, which we will compare to qcritical)differences ancompare those to HSD. In thcase, we need a table of a

Post hoc tests: When there is significant difference between the means of several treatments, the…

A professor gives an exam to a class with four recitation sections. Each section has 7 students. The professor has no reason to assume that the exam scores would not be independent and normally distributed with equal variance. However, it is unknown whether or not the section choice (different TAs and different days of the week) has any relationship with how students performed on the test. The professor conducts a one-way ANOVA, and finds that Ho should be rejected at a significance level of a = 0.01.