For the preceding problem you should find that there are significant differences
ID: 2956497 • Letter: F
Question
For the preceding problem you should find that there are significant differences among the three treatments. The primary reason for the significance is that the mean for the treatment I is substantially smaller than the means for the older two treatments. To create the following data we started with the values from problem 7 and added 3 points to each score in treatment I. Recall that adding a constant causes the mean to change but has no influence on the variability of the sample. In the resulting data the mean differences are much smaller. Treatments: I II III n= 6 n= 6 n= 6 M= 4 M= 5 M= 6 N= 18 T= 24 T= 30 T= 36 G= 90 b)Use an ANOVA with a=.05 to determine whether there are many significant differences among the three treatment means. ( Does your answer agree with your prediction in part a?)Explanation / Answer
I II III n=6 n=6 n=6 M=4 M=5 M=6 N=18 T=24 T=30 T=36 G=90 SS=30 SS=35 SS=40 SX^2=567 The main reason for the significant difference between the means in problem 7 was that the mean of the first treatment was substantially smaller than the means of the other treatments. By increasing the scores under the first treatment by 3, the mean of the first treatment increases by 3. This makes this mean more near to the combined mean. So SS within treatments will be substantially reduced. Since the variability within the first group is not affected by increasing the scores by 3 points, the error mean square remains the same as in problem 7. So when we calculate F value, the numerator decreases and the denominator is unaltered. So the value of F will much smaller and it is likely to be insignificant. The reduction in the treatment SS, makes the effect size measure smaller. Ho: The means are the same for all the three treatments. Ha: At least one treatment has a mean different from the rest. Total SS = SX^2 –G^2/N =567 - 90^2/18 =567 – 450 = 117 Treatment SS = T1^2/n1 + T2^2/n2 + T3^2/n3 - G^2/N = 24^2/6 + 30^2/6 + 36^2/6 - 90^2/18 = 96+150+216-450=12 Error SS = Total SS – Treatment SS = 117 -12 = 105 Treatment MS = Treatment SS/DF = 12/(3-1) = 12/2 =6 Error MS = Error SS /DF = 105/(18-3)=105/15 = 7 F = Treatment MS/Error MS = 6/7 = 0.8571 The p-value is 0.4441 The p value is not less than the significance level 0.05. We fail to reject H0 and conclude that there is no sufficient evidence that the treatment means are significantly different. Effect size r^2 = C.O.D = treatment SS/Total SS = 12/117 = 0.1026 Population estimate of effect size = ?^2 = (Treatment SS –(k-1)*Error MS)/(total SS + Error MS) =(12 –(3-1)*7)/(12+7) =(12-14)/19 =-2/19 =-0.1053 Our answer agree with the prediction in part(a)