The conclusion of a one-way ANOVA procedure for the data shown in the table is t
ID: 2932421 • Letter: T
Question
The conclusion of a one-way ANOVA procedure for the data shown in the table is to reject the null hypothesis that the means are all equal. Determine which means are ditterent using =0.05. Sample 1 12 17 Sample 2 10 Sample 3 20 21 23 Click the icon to view the ANOVA summary table. Click the icon to view a studentized range table for 0.05 Let x1, X2, and x3 be the means for samples 1, 2, and 3, respectively. Find the absolute values of the differences between the means Find the Tukey-Kramer critical range CR. Note that since the samples are all the same size, CRI ,2 = CRI ,3-CR2.3-CR. CR(Round to two decimal places as needed.) ANOVA Summary Table Source Between Within Total Sum of Squares 253.5 92.75 346.25 Degrees of Freedom 2 9 Mean Sum of Squares 126.75 10.306 12.299Explanation / Answer
Answer:
ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
s1
4
58
14.5
12.3333
s2
4
34
8.5
7.0000
s3
4
79
19.75
11.5833
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
253.5000
2
126.7500
12.2992
0.0027
4.2565
Within Groups
92.7500
9
10.3056
Total
346.2500
11
Level of significance
0.05
Tukey-Kramer critical range CR = 3.95
Absolute
Std. Error
Critical
Comparison
Difference
of Difference
Range
Results
Group 1 to Group 2
6
1.605113357
6.34
Means are not different
Group 1 to Group 3
5.25
1.605113357
6.34
Means are not different
Group 2 to Group 3
11.25
1.605113357
6.34
Means are different
ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
s1
4
58
14.5
12.3333
s2
4
34
8.5
7.0000
s3
4
79
19.75
11.5833
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
253.5000
2
126.7500
12.2992
0.0027
4.2565
Within Groups
92.7500
9
10.3056
Total
346.2500
11
Level of significance
0.05
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