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 post-hoc tests must be performed to identify which pairs of treatments are significantly different. For post-hoc tests, there may be various methods. Here, Tukey’s method is implemented. The general formula to obtain Tukey’s HSD post-hoc test is given below:Calculation: Given that the groups should be ordered from smallest to largest means. That is, the Group-A should be the group with smallest mean. Therefore, the appropriate order for the groups is given below: Group-A = Group-3 Group-B = Group-4 Group-C = Group-2 Group-D = Group-1 The differences between the means of each group are reported in the table given below:What is the critical value of the standardized range for the Tukey HSD test: Here, the error degrees of freedom is 24 and the number of groups is 4. The table value corresponding to k = 4, df for error term = 24 and α = 0.01 is q = 4.91. The value of MSWithin = 119.803. The Tukey’s HSD is obtained as given below: Thus, the value of Tukey’s HSD is HSD = 20.313.Comparison of mean difference with HSD: If mean difference > HSD, then the difference is said to be significant. Since, all the pairwise comparisons of means are less than the HSD value, none of the pairwise comparisons are statistically significant. Thus, None of the groups are statistically significantly different.

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Statistics

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.

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.

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Question

Please explain and fill table with steps by step results

Comparison Groups
Groups
(ascending order) Group A
Group B
Group C
Group D
Group A = 3
Group B = 4
Group C = 2
Group D = 1

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.
Source
SS
df
MS
F-ratio P-value
Between
1889.293 3 629.764
5.257
0.0062
Within
2875.271 24 119.803
As shown above, the professor finds that F(3,24)
5.257, which has a probability of
0.0062 of
occurring randomly. Additionally, the error variance was estimated as MSw
means for each group were obtained as shown in the table below.
119.803, and the sample
Level
Mean
Group-1
88.214
Group-2
Group-3
Group-4
87.657
69.314
74.557
The professor decides to use the Tukey HSD method to test all possible pairwise contrasts. To do so, the
following table of q; statistics needs to be completed. For this presentation, the groups will be ordered from
smallest 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 values
rounded to 3 decimal places. (Furthermore, even though the sign does not matter, report all comparison
statistics as positive values.)
(Note: we often create a table
statistics, which we will compare to qcritical)
differences an
compare those to HSD. In th
case, we need a table of a

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