Question

The following are six observations collected from treatment 1, four observations collected from treatment 2, and five observations collected from treatment 3. Test the hypothesis at the 0.05 significance level that the treatment means are equal.

Treatment 1 ___ Treatment 2 ___ Treatment 3

9_____________ 13 __________ 10

7_____________ 20___________ 9

11___________    14___________ 15

9 ____________ 13 ____________14

12____________ 0 ___________ 15

10_____________0_____________0

a. State the null and the alternate hypothesis.

b. What is the decision rule

c. Compute SST, SSE and SS total

d. Complete the ANOVA table.

Answer 1

(a) Null hypothesis, H0: The three groups means are equal.

Alternative hypothesis, H1: At least one group mean is different.

(b) For significance level of 0.05, critical value, F(0.05,2,15) = 3.6823

Decision rule: We reject the null hypothesis, if the test statistic is more than the critical value of 3.6823.

(c) SST = 102.7778 (Sum of Square Treatments)

SSE = 115.6667 (Sum of Square Error)

SS Total = 218.4444 (Sum of Square Total)

(d) ANOVA Table:

Source of Variations	Degrees of Freedom	Sum of Squares	Mean Squares	F Value	P Value
Treatments	2	102.7778	51.3889	6,6642	0.00849
Error	15	115.6667	7.7111
Total	17	218.4444	12.8497

>> R-Studio Code & Outputs:

## One Way ANOVA:
T1<-c(9,7,11,9,12,10)
T2<-c(13,20,14,13)
T3<-c(10,9,15,14,15)
my_data <- data.frame(cbind(T1,T2,T3))
Stacked_Groups <- stack(my_data)
Anova_Results <- aov(values ~ ind, data = Stacked_Groups)
summary(Anova_Results)

Df Sum Sq Mean Sq F value Pr(>F)   
ind 2 102.8 51.39 6.664 0.00849 **
Residuals 15 115.7 7.71   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1