Consider the data in the table collected from four
independent populations. The conclusion of a one-way ANOVA test
using
alphaαequals=0.05 is that the population means are not all the same. Determine which means are different usingalphaαequals=0.05 |
Sample 1 |
Sample 2 |
Sample 3 |
Sample 4 |
||
---|---|---|---|---|---|---|
6 |
13 |
23 |
9 |
|||
7 |
16 |
13 |
8 |
|||
8 |
19 |
19 |
10 |
|||
9 |
22 |
Click here to view the ANOVA summary table.
Find the Tukey-Kramer critical range CR for each of these differences.
CR1,2 |
equals= |
? |
CR2,3 |
equals= |
? |
CR1,3 |
equals= |
? |
CR2,4 |
equals= |
? |
CR1,4 |
equals= |
? |
CR3,4 |
equals= |
? |
(Round to two decimal places as needed.)
Applying One way ANOVA from excel: data-data analysis: ANOVA(one factor):
Source of Variation | SS | df | MS | F |
Between Groups | 352.25 | 3 | 117.4167 | 13.69291 |
Within Groups | 85.75 | 10 | 8.575 | |
Total | 438 | 13 |
from above:
MSE= | 8.575 | ||
df(error)= | 10 | ||
number of treatments = | 4 | ||
pooled standard deviation=Sp =√MSE= | 2.928 |
critical q with 0.05 level and k=4, N-k=10 df= | 4.33 | ||
Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj) = |
CR1-2 =(4.33/sqrt(2))*(2.928)*sqrt(1/4+1/3)=6.85
CR1-3 =6.34
CR1-4 =6.85
CR2-3 =6.85
CR2-4 =7.32
CR3-4 =6.85
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