In: Statistics and Probability
Three classes in elementary statistics are taught by three different persons : a regular faculty member, a graduate teaching assistant, and an adjunct from outside the university. At the end of the semester, each student is given a standardized test. Five students are randomly picked from each of these classes, and their scores are as shown in Table
(a) Construct an ANOVA table and interpret your results.
(b) Test at the 0.05 level whether there is a difference between the mean scores for the three persons teaching. Assume that the ANOVA assumptions are met.
Solution
(a) Construct an ANOVA table and interpret your results.
The ANOVA table is :
we have k = 3, n_1 = 5, n_2 = 5, n_3 = 5, and N = n_1 + n_2 + n_3 = 15
then we have , SSTr = 339.7333, SSE = 2785.2, SST = 3124.9333
Since MSTr =[SSTr /(k − 1)] = 169.8667, MSE =[SSE/(n − k)]= 232.1
Then, F = [MSTr/MSE]= 0.73187
Therefore, ANOVA Table
(b) Test at the 0.05 level whether there is a difference between the mean scores for the three persons teaching. Assume that the ANOVA assumptions are met.
For significance level 0.05
we have p-value = P(F ≥ f) = 0.501295.
Since the p-value is too high (0.501295 >> 0.05) we can say that, there is a difference between the mean scores for the three persons teaching.