Question

The following is sample information. Test the hypothesis that the treatment means are equal. Use the 0.05 significance level.

Treatment 1	Treatment 2	Treatment 3
4	5	5
8	9	7
6	10	6
8	5	9

a. State the null hypothesis and the alternate hypothesis.

H₀:                (Click to select)  The treatment means are not all the same.  The treatment means are the same.

H₁:                (Click to select)  The treatment means are not all the same.  The treatment means are the same.

b. What is the decision rule? (Round the final answer to 2 decimal places.)

Reject H₀ if the test statistic is greater than                 .

c. Compute SST, SSE, and SS total. (Round the final answers to 3 decimal places.)

SST =                

SSE =                

SS total =                

d. Complete the ANOVA table. (Round the SS, MS, and F values to 3 decimal places.)

Source	SS	DF	MS	F
Treatment
Error
Total

e. State your decision regarding the null hypothesis.

(Click to select)  Reject  Do not reject  H₀.

Answer 1

a. H₀: The treatment means are the same. H₁: The treatment means are not all the same

b. decision rule:

k=no of treatments=3,n=total no of observation, n-k=(3*4)-3=9

F_k-1,n-k=F_2,9=4.26

reject H₀ if the test statistics is greater than 3.86

c. SST= 1.167

SSE=40.500

SS total= 41.667

d.

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	1.166667	2	0.583333	0.12963	0.880031
Within Groups	40.5	9	4.5
Total	41.66667	11

as 0.12963<4.26.

so, We can not reject H₀

R codes and results used for this is:

t1<-c(4,8,6,8)
> t2<-c(5,9,10,5)
> t3<-c(5,7,6,9)
> treatment<-data.frame(cbind(t1,t2,t3))
> stacked<-stack(treatment)
> stacked
values ind
1 4 t1
2 8 t1
3 6 t1
4 8 t1
5 5 t2
6 9 t2
7 10 t2
8 5 t2
9 5 t3
10 7 t3
11 6 t3
12 9 t3
> anova_results<-aov(values~ind,data=stacked)
> anova_results
Call:
aov(formula = values ~ ind, data = stacked)

Terms:
ind Residuals
Sum of Squares 1.16667 40.50000
Deg. of Freedom 2 9

Residual standard error: 2.12132
Estimated effects may be unbalanced
> summary(anova_results)
Df Sum Sq Mean Sq F value Pr(>F)
ind 2 1.17 0.583 0.13 0.88
Residuals 9 40.50 4.500   
> anova(anova_results)
Analysis of Variance Table

Response: values
Df Sum Sq Mean Sq F value Pr(>F)
ind 2 1.167 0.5833 0.1296 0.88
Residuals 9 40.500 4.5000   

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