chapter 13

Name : Widiyati
NIM : 12030120410009
a. State the null and the alternate hypotheses.
Answer :
H0 : T1=T2=T3
H1 : T1 ≠ T2 ≠ T3
b. What is the decision rule?
Answer : The umuretor degrees of freedom, k-1, equal 3-1 or 2. The denominato degees of reedom, n-k, equal 12-2.
The value o F at 2 and 10 degree of freedom is 4.10. Thus, H0 is rejected I F>4.10 or p<0.05
SSE, and SS total.
c. Compute SST, SSE, and SS total.
Answer :
Compute of SSE
Treatment
1
8
6
10
9
Xk
SSE :
XG :
SS (T1)
0.0625
5.0625
3.0625
0.5625
8.75
8.25
12.75
5.083
Treatment
2
3
2
4
3
SS
(T2)
0
1
1
0
2
3
Treatment
3
3
4
5
4
SS
(T3)
1
0
1
0
2
4
Compute SST
Treatment
1
8
6
10
9
SST
XG :
SSE
Treatment
2
8.51
0.84
24.17
15.34
48.86
74.92
5.083
12.75
Treatmen
t3
3
2
4
3
Treatme
nt 4
4.34
9.51
1.17
4.34
19.36
Treatment
5
3
4
5
4
SS (T3)
4.34
1.17
0.01
1.17
6.69
Name : Widiyati
NIM : 12030120410009
Compute SS Total
Treatment
XT
1
8.25
2
3
3
4
SST
40.11
17.36
4.69
62.17
d. Complete an ANOVA table.
Mannually
Anova Table
Source o
Variation
Treatments
Error
Total
Sum of
Square
62.17
12.75
74.92
Degrees of
Freedom
3-1 = 2
12-2 = 10
12-1 = 11
Mean
Square
31.08
6.38
F
4.88
Using SPSS
Nilai_Obesrvasi
N
Treatment 1
Treatment 2
Treatment 2
Total
4
4
4
12
Mean
8.25
3.00
4.00
5.08
Std.
Deviation
Std. Error
1.708
.854
.816
.408
.816
.408
2.610
.753
Test of Homogeneity of Variances
Nilai_Obesrvasi
Levene
Statistic
1.500
df1
df2
2
9
Sig.
.274
95% Confidence Interval for
Mean
Lower Bound Upper Bound Minimum Maximum
5.53
10.97
6
10
1.70
4.30
2
4
2.70
5.30
3
5
3.43
6.74
2
10
Name : Widiyati
NIM : 12030120410009
Test of Homogeneity of Variances
Nilai_Obesrvasi
Levene
Statistic
df1
df2
Sig.
1.500
2
9
.274
e. State your decision regarding the null hypothesis.
Since we got value of F is 4.88 and it is more than 4.10 we can decide that we rejected H0 hypothesis. We can
conclude that there is difference among the treatments held in that observation.