A professor has a class with four recitation sections. Each section has 16 students (rare, but...

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A professor has a class with four recitation sections. Eachsection has 16 students (rare, but there are exactly the samenumber in each class...how convenient for our purposes, yes?). Atfirst glance, the professor has no reason to assume that these examscores from the first test would not be independent and normallydistributed with equal variance. However, the question is whetheror not the section choice (different TAs and different days of theweek) has any relationship with how students performed on thetest.

Group-1Group-2Group-3Group-4
73.576.77565.7
8166.477.850.5
61.860.366.783
69.58170.381.4
77.457.977.774.9
91.259.268.182.9
70.667.983.585.4
6454.987.863.6
73.263.280.667.6
77.769.858.973.6
73.669.186.781.5
77.251.874.780.5
5460.566.971.8
65.455.476.768.1
77.868.276.355.8
81.664.869.570.4



First, run an ANOVA with this data and fill in the summary table.(Report P-values accurate to 4 decimal places and allother values accurate to 3 decimal places.

SourceSSdfMSF-ratioP-value
Between
Within



To follow-up, the professor decides to use the Tukey-Kramer methodto test all possible pairwise contrasts.

What is the Q critical value for the Tukey-Kramer critical range(alpha=0.01)?
Use the website link in your notes(http://davidmlane.com/hyperstat/sr_table.html) to locate the Qcritical value to 4 decimal places.
Q =

Using the critical value above, compute the critical range and thendetermine which pairwise comparisons are statisticallysignificant?

  • group 1 vs. group 2
  • group 1 vs. group 3
  • group 1 vs. group 4
  • group 2 vs. group 3
  • group 2 vs. group 4
  • group 3 vs. group 4
  • none of the groups are statistically significantlydifferent

Answer & Explanation Solved by verified expert
3.6 Ratings (272 Votes)
Group1 Group2 Group3 Group4 Total Sum 11695 10271 11972 11567 45505 Count 16 16 16 16 64 Mean sumn 7309375 6419375 74825 7229375 Sum of square xx 1195049 9410294 91731 1551929 Null and Alternative Hypothesis Ho 1 2 3 4 H1 At    See Answer
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