a) Fill in the correct values for the missing quantities in the ANOVA table abov
ID: 3062596 • Letter: A
Question
a) Fill in the correct values for the missing quantities in the ANOVA table above. Show your computations reporting a maximum of two decimal places.
b) Carry out a hypothesis test to determine whether the mean amounts of e-mail spam are not all equal across the different university communities. Please provide:
-Hypotheses
-Value of test statistic, to two decimal places
-The critical value for the corresponding test statistic at = .05 (include degrees of freedom)
-Exact p-value using software
-Conclusion; don’t simply say “reject” or “do not reject”; include a statement with practical meaning
c) What assumption of the ANOVA model can be examined with the information provided in the summary table above? Explain briefly. Assuming the assumption is warranted, show how the MSE is calculated using the sample variances.
d) Suppose one of the students received 20 spam e-mails. Explain how this would enable you to question another key assumption of the ANOVA model. What would you conclude?
e) What is the margin of error associated with the Bonferroni method for multiple paired comparisons between the population means? Assume a family confidence level of 95%. Using the Bonferroni method, determine which pairwise differences are or are not statistically significant. What do you conclude? Show your work.
f) If instead of considering only three university communities, we consider five. For example, part- time instructors, full-time instructors, administrators, undergraduate students and graduate students. How many pairwise comparisons are possible?
g) Suppose we now ask each group of participants to count the number of spam e-mail messages every day for a seven-day period. Identify this experimental design. Explain briefly.
SUMMARY Count Average Variance 14.3 12.1 6.6 Groups Professors Administrators Students 10 10 10 63.34 38.10 12.04 ANOVA Source of Variation df MS Between Groups Within Groups Total 37.83 1336.0 29Explanation / Answer
a)Since there are 3 groups, DF(Between groups) = 3-1 = 2.
Thus, DF(Within groups) = 29-2 = 27
SS(Within groups) = DF * MS = 27*37.83 = 1021.41
SS(Between groups) = Total SS – SSW = 1336 – 1021.41 = 314.59
MS(Between groups) = SSB/DF = 314.59/2 = 157.295
Thus, F = MSB/MSW = 157.295/37.83 = 4.157943
b) Null hypothesis = H0 : mean(Professors) = mean(Administrators) = mean(Students)
Alternative hypothesis = H1 : not H0
Value of test statistic : F = 4.16
Critical value : F 0.05; 2,27 = 3.354
P-value : P[ F2,27 > 3.354] = 0.0499
Since p-value < 0.05, we reject H0 and conclude that the mean amounts of email spam are significantly different across the different university communities.
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