Question

• Nine observations were selected from each of three populations:    
• (You may think of the rows as weeks)  

	Population 1	Population 2	Population 3	Ave.
1	13, 12, 14	17, 13, 15	9, 13,11	13
2	12, 10, 8	15, 13, 14	11, 7, 9	11
3	12, 13, 14	14, 18, 16	16, 16, 16	15
Means	12	15	12	13

a.   What is the value of the Treatment Sum of Squares:………………………..

b.   The value of the Sum of Squares Total:……….. c.   The value of the Error Sum of Squares:…………..

d.   Complete the Anova Table:

Source	Sum of Squares	D. F.	Mean Square	F
Treatment
Error
Total

e.   State the Null Hypothesis Ho, and the Alternate Ha:

Ho:                                                             Ha:

f.   Do you reject the Null Hypothesis?........................

Answer 1

Sol:

Rcode:

df1 =read.table(header = TRUE, text ="
Pop value
1 13
1 12
1 14
1 12
1 10
1 8
1 12
1 13
1 14
2 17
2 13
2 15
2 15
2 13
2 14
2 14
2 18
2 16
3 9
3 13
3 11
3 11
3 7
3 9
3 16
3 16
3 16
"
)
df1
summary(df1)

df1$Pop <- ordered(df1$Pop ,
levels = c("1", "2", "3"))

res.aov <- aov(value ~ Pop, data = df1)
summary(res.aov)

Output:

summary(res.aov)
Df Sum Sq Mean Sq F value Pr(>F)
Pop 2 54 27.000 4.378 0.0239 *
Residuals 24 148 6.167   

a.   What is the value of the Treatment Sum of Squares: 54

b.   The value of the Sum of Squares Total:202

C.The value of the Error Sum of Squares:148

d.   Complete the Anova Table:

Source	Sum of Squares	D. F.	Mean Square	F
Treatment	54	2	27.000	4.378
Error	148	24	6.167
Total	208

e.   State the Null Hypothesis Ho, and the Alternate Ha:

Ho:          

All the three popualtion means are equal   

Ha:atleast one of the popualtion means are different

f.   Do you reject the Null Hypothesis?

p=0.0239

p<0.05

Reject the null hypothesis