Homework
ANNOVA
Systems Engineering and Analysis (MRSL 1113)
Semester 2 2020/2021
Lecturer:
NURUL FATHIA MOHAMAND NOOR
MRS201040
TS DR. SITI ARMIZA MOHD ARIS
TS. DR. SITI ZURA A.JALIL
OUTLINE
❏ Exercise 1
❏ Exercise 2
❏ Exercise 3
EXERCISE 1
ANSWER EX 1: STEP 1 to 5
1. Parameter of interest:
•
•
Treatment
Data
2. H0 = ԏ1 + ԏ2 + ԏ3 + ԏ4 =0
3. Alternative hypothesis, H1 ≠ 0 for at least one i.
4. Test statistic: F0
5. Reject H0 if P value is less than 0.05.
Descriptive Analysis
ANSWER EX 1: STEP 6 & 7
Step 6 Analysis of Variance
Step 7 Conclusion
From Table obtained, the computed value of the test
statistic is f0= 1.14 and the computed value of the P
value is higher than 0.05 thus we failed to reject H0
We could bound the P value by using F table find that f
0.05,3,40 = 2.84, and because f0 = 1.14 does not
exceeds this value, we know that P value is higher
than 0.05.
ANSWER EXERCISE 1: Pairwise Comparison
The Tukey pairwise comparisons suggest
that all the means are different. Therefore,
treatment 1 is the highest treatment 3 is the
lowest
EXERCISE 2
ANSWER EX 2: STEP 1 to 5
1. Parameter of interest:
•
•
Cutting Speed
The mean tool life of 4 different cutting speed
2. H0 = ԏ1 + ԏ2 + ԏ3 + ԏ4 =0
3. Alternative hypothesis, H1 ≠ 0 for at least one i.
4. Test statistic: F0
5. Reject H0 if P value is less than 0.05.
Descriptive Analysis
ANSWER EX 2: STEP 6 & 7
Step 6 Analysis of Variance
Step 7 Conclusion
From Table obtained, the computed value of the test
statistic is f0= 3.17 and the P value is less than
0.05 thus we conclude that H0 is rejected.
We could bound the P value by using F table find that f
0.05,3,20 = 3.10, and because f0 = 3.17 exceeds this
value, we know that P value is less than 0.05.
ANSWER EXERCISE 2: Pairwise Comparison
The Tukey pairwise comparisons suggest that all
the means are different. Therefore, cutting speed 2
have the highest tool life and cutting speed 3 have
the lowest
EXERCISE 3
ANSWER EX 3: STEP 1 to 5
1. Parameter of interest:
•
•
Five machines
The average machine performance in hours
2. H0 = ԏ1 + ԏ2 + ԏ3 + ԏ4 + ԏ5 =0
3. Alternative hypothesis, H1 ≠ 0 for at least one i.
4. Test statistic: F0
5. Reject H0 if P value is less than 0.05.
Descriptive Analysis
ANSWER EX 3: STEP 6 & 7
Step 6 Analysis of Variance
Step 7 Conclusion
From Table obtained, the computed value of the test
statistic is f0= 3.96 and the P value is less than
0.05 thus we conclude that H0 is rejected.
We could bound the P value by using F table find that f
0.05,4,30 = 2.69, and because f0 = 3.96 exceeds this
value, we know that P value is less than 0.05.
ANSWER EXERCISE 3: Pairwise Comparison
The Tukey pairwise comparisons suggest that all
the means are different. Therefore, machine 4
have the highest performance and machine 2 have
the lowest performance