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A one-way ANOVA was conducted at a = 0.05. The study involved three (3) different groups with 12 subjects in each. The means for each
A one-way ANOVA was conducted at a = 0.05. The study involved three (3) different groups with 12 subjects in each. The means for each group were as follows: M1 = 10.92, M2=13.83, and M3 = 14.92. The sum of squares values is shown in the table below. Source SS df MS F p Between-treatments 102.72 <.05 Within-treatments 229.5 Total 332.22 a. Assuming = 0.05, complete the rest of the table M1 M2 M3 Total 10.92 13.83 14.92 39.67 n 12 12 12 36 Mean 0.91 1.15 1.24 1.10 df total = 3*12 - 1 = 35 DfC = C - 1 = 2 Dfe= N - C = 33 F critical value= 3.28 MS(between) = SS(between)/df(between) 102.72/2 51.36 MS(within) = SS(within)/df(within) 229.5/33 6.95 F = MS(between)/MS(within) 51.36/6.95 7.39 Source SS df MS F p Between-treatments 102.72 2 51.36 7.39 <.05 Within-treatments 229.5 33 6.95 Total 332.22 The decision to be made is to reject the null hypothesis because the observed value of 7.39 is greater than the critical F-value which is 3.28. It means that there is a significant difference between the means of three groups. b. Conduct a Tukey's HSD and indicate which means differ significantly. (I did the answering on A I just need the B.) thank you .
A one-way ANOVA was conducted at a = 0.05. The study involved three (3) different groups with 12 subjects in each. The means for each group were as follows: M1 = 10.92, M2=13.83, and M3 = 14.92. The sum of squares values is shown in the table below. | |||||||
Source | SS | df | MS | F | p | ||
Between-treatments | 102.72 | <.05 | |||||
Within-treatments | 229.5 | ||||||
Total | 332.22 | ||||||
a. Assuming = 0.05, complete the rest of the table | |||||||
M1 | M2 | M3 | Total | ||||
10.92 | 13.83 | 14.92 | 39.67 | ||||
n | 12 | 12 | 12 | 36 | |||
Mean | 0.91 | 1.15 | 1.24 | 1.10 | |||
df total = 3*12 - 1 = | 35 | ||||||
DfC = C - 1 = | 2 | ||||||
Dfe= N - C = | 33 | ||||||
F critical value= | 3.28 | ||||||
MS(between) = | SS(between)/df(between) | ||||||
102.72/2 | |||||||
51.36 | |||||||
MS(within) = | SS(within)/df(within) | ||||||
229.5/33 | |||||||
6.95 | |||||||
F = | MS(between)/MS(within) | ||||||
51.36/6.95 | |||||||
7.39 | |||||||
Source | SS | df | MS | F | p | ||
Between-treatments | 102.72 | 2 | 51.36 | 7.39 | <.05 | ||
Within-treatments | 229.5 | 33 | 6.95 | ||||
Total | 332.22 | ||||||
The decision to be made is to reject the null hypothesis because the observed value of 7.39 is greater than the critical F-value which is 3.28. It means that there is a significant difference between the means of three groups. | |||||||
b. Conduct a Tukey's HSD and indicate which means differ significantly. | (I did the answering on A I just need the B.) thank you . |
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