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In a completely randomized design (one factor ANOVA), 12 individuals were assigned to the first treatment (sample), 15 individuals to the second treatment, and 18 individuals to the third treatment. A partial ANOVA table is shown below: Source of Variation SS df MS F Treatments * * * 9 Error * * 35 Total * * Fill in the blanks (identified by asterisks) in the above ANOVA table. Test at the 5% significance level to determine if differences exist among the three treatment means. Show all 5 steps (step 4 is filling in the table in part a.)

          In a completely randomized design (one factor ANOVA), 12 individuals were assigned to the first treatment (sample), 15 individuals to the second treatment, and 18 individuals to the third treatment. A partial ANOVA table is shown below:
Source of Variation	SS	df	MS	F
Treatments	*	*	*	9
Error	*	*	35	
Total	*	*		
Fill in the blanks (identified by asterisks) in the above ANOVA table.
Test at the 5% significance level to determine if differences exist among the three treatment means. Show all 5 steps (step 4 is filling in the table in part a.)

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in a completely randomized design one factor anova 12 individuals were assigned to the first treatment sample 15 individuals to the second treatment and 18 individuals to the third treatment 54768

Added by Mark G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Ivan Kochetkov

09/24/2024

Step 1

The total number of individuals is $12 + 15 + 18 = 45$. The total degrees of freedom is $45 - 1 = 44$. The degrees of freedom for treatments is the number of treatments minus 1, which is $3 - 1 = 2$. The degrees of freedom for error is the total degrees of freedom Show more…

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In a completely randomized design (one factor ANOVA), 12 individuals were assigned to the first treatment (sample), 15 individuals to the second treatment, and 18 individuals to the third treatment. A partial ANOVA table is shown below: Source of Variation SS df MS F Treatments * * * 9 Error * * 35 Total * * Fill in the blanks (identified by asterisks) in the above ANOVA table. Test at the 5% significance level to determine if differences exist among the three treatment means. Show all 5 steps (step 4 is filling in the table in part a.)

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