An article gives adaptation scores for three different treatments: (1) no shock,
ID: 3324908 • Letter: A
Question
An article gives adaptation scores for three different treatments: (1) no shock, (2) shock following each stuttered word, (3) shock during each moment of stuttering. These treatments were used on each of 19 stutterers, resulting in SST = 3462.00, SSTr = 25.61, and SSB1 = 2965.98.
(a) Construct the ANOVA table and test at level .05 to see whether true average adaptation score depends on the treatment given. (Give answers accurate to 2 decimal places.)
Conclusion
Reject the null hypothesis, there is no dependence in true average adaptation and the treatment given.
Reject the null hypothesis, there is dependence in true average adaptation and the treatment given.
Fail to reject the null hypothesis, there is no dependence in true average adaptation and the treatment given.
Fail to reject the null hypothesis, there is dependence in true average adaptation and the treatment given.
(b) Judging from the F ratio for subjects (factor B), do you think that blocking on subjects was effective in this experiment? Explain.
Yes, fB is quite large, suggesting great variability between subjects.
Yes, fB is quite small, suggesting great variability between subjects.
No, fB is quite large, suggesting little variability between subjects.
No, fB is quite small, suggesting little variability between subjects.
Source Df SS MS F Treatments Subjects Error TotalExplanation / Answer
(a)
SSE = SST - SSTr = 3462.00 - 25.61 = 3436.39
Df for Treatments = Number of treatments - 1 = 3-1 = 2
Df for Subjects = Number of Subjects - 1 = 19-1 = 18
Total number of observations = 19 * 3 = 57
Df for total = Number of observations - 1 = 57-1 = 56
Df for error = Number of observations - Number of treatments = 57 - 3 = 54
MS for each row = SS / Df
F for treatment = MS for treatment / MS for error = 12.805 / 63.63685 = 0.2012
F for subjects = MS for subjects / MS for error = 164.7767 / 63.63685 = 2.5893
Df for treatment is 2, 54
Critical value of F for 0.05 significance level is 3.168
As, observed F (0.2012) is less than the critical value,
Fail to reject the null hypothesis, there is no dependence in true average adaptation and the treatment given.
(b)
Df for treatment is 18, 54
Critical value of F for 0.05 significance level is 1.8
As, observed F (2.5893) is greater than the critical value, the correct option is,
Yes, fB is quite large, suggesting great variability between subjects.
Source Df SS MS F Treatments 2 25.61 12.805 0.2012 Subjects 18 2965.98 164.7767 2.5893 Error 54 3436.39 63.63685 Total 56 3462.00Related Questions
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