Hypothesis Testing
DOE PRACTICAL TEAM MEMBERS :
1. Nander
2. Tzer
3. Ashwati
4. Daryl
5. Nikkisha
Data collected for FULL factorial design using CATAPULT A
Response Table (Data)
Main
Effects Replicates used to normalize
the data
![]()
A B C R1 R2 R3 R4 R5 R6 R7 R8 Ave. Std.Dev.
|
1 |
1 |
- |
- |
- |
144.2 |
139.8 |
140.6 |
152.2 |
149.6 |
142.5 |
144.3 |
153.6 |
145.9 |
5.28 |
|
2 |
3 |
+ |
- |
- |
123.0 |
126.5 |
123.6 |
123.5 |
123.6 |
123.5 |
126.0 |
129.6 |
124.9 |
2.29 |
|
3 |
2 |
- |
+ |
- |
140.0 |
138.0 |
134.5 |
131.2 |
135.0 |
132.2 |
138.6 |
130.7 |
135.0 |
3.55 |
|
4 |
6 |
+ |
+ |
- |
98.6 |
98.8 |
98.4 |
106.0 |
98.5 |
101.3 |
104.0 |
99.0 |
100.6 |
2.93 |
|
5 |
4 |
- |
- |
+ |
82.0 |
81.7 |
85.0 |
85.0 |
86.0 |
82.0 |
82.5 |
85.0 |
83.7 |
1.75 |
|
6 |
5 |
+ |
- |
+ |
85.0 |
90.5 |
89.5 |
87.5 |
85.0 |
85.0 |
88.7 |
85.0 |
87.0 |
2.32 |
|
7 |
8 |
- |
+ |
+ |
81.5 |
82.0 |
93.2 |
80.0 |
81.0 |
83.0 |
80.0 |
82.3 |
82.9 |
4.30 |
|
8 |
7 |
+ |
+ |
+ |
83.2 |
90.0 |
86.5 |
84.0 |
86.0 |
85.1 |
86.5 |
84.2 |
85.7 |
2.12 |
FiguResponse Table Collected for Full Factorial Design
Reference Key to Factor Selection:
Factor A: Arm length
Factor B: Start angle
Factor C: Stop angle
Data collected for FRACTIONAL factorial design using CATAPULT B
Response Table (Data)
Main
Effects Replicates used to normalize
the data
![]()
A B C R1 R2 R3 R4 R5 R6 R7 R8 Ave. Std.Dev.
|
2 4 5 7 |
1 |
+ + - - |
- + - |
- - + + |
133.5 |
147.0 |
150.5 |
158.0 |
138.5 |
144.5 |
151.3 |
133.0 |
144.5 |
8.94 |
|
2 |
119.8 |
119.8 |
117.7 |
121.0 |
117.3 |
124.0 |
113.3 |
117.4 |
118.8 |
3.15 |
||||
|
3 |
103.0 |
100.0 |
103.0 |
100.0 |
103.5 |
103.5 |
103.5 |
96.6 |
100.8 |
2.89 |
||||
|
4 |
88.0 |
96.0 |
88.0 |
88.4 |
88.5 |
94.0 |
89.0 |
91.2 |
90.4 |
3.07 |
FiResponse Table Collected for Fractional Factorial Design
Nander will use Run #2 from FRACTIONAL
factorial and Run#2 from FULL factorial.
Daryl will use Run #4 from FRACTIONAL
factorial and Run#4 from FULL factorial.
Tzer will use Run #4 from FRACTIONAL factorial and
Run#4 from FULL factorial.
Ashwati(me)
will use Run #5 from FRACTIONAL factorial and Run#5 from FULL factorial.
Nikkisha will use Run #7 from FRACTIONAL
factorial and Run#7 from FULL factorial.
|
The QUESTION |
The
catapult (the ones that were used in the DOE practical) manufacturer needs to
determine the consistency of the products they have manufactured. Therefore,
they want to determine whether CATAPULT A produces the same flying distance
of projectile as that of CATAPULT B.
|
|
Scope of the test |
The human factor is assumed to be
negligible. Therefore, the different users will not have any effect on the flying
distance of the projectile.
Flying distance for catapult A and
catapult B is collected using the factors below: Arm length = _28___cm Start angle = __5___ degree Stop angle = __90___ degree
|
|
Step 1: State the statistical Hypotheses: |
State the null hypothesis (H0): Ho :m1 =m2 Catapult A and Catapult B produce the same flying distance of projectile(no significant difference)
State the alternative hypothesis (H1): H1: m1¹m2 Catapult A and Catapult B have a different flying distances of projectile(there is a significant difference)
|
|
Step 2: Formulate an analysis plan. |
The sample size is __8__ Therefore t-test
will be used.
Since the sign of H1 is __¹__, a two-tailed test is used.
The significance level (α) used in this
test is _0.05___
|
|
Step 3: Calculate the test statist |
State the mean and standard deviation of sample catapult A:
State the mean and standard deviation of
sample catapult B:
Compute the value of the test statistic (t):
|
|
Step 4: Make a decision based on result |
Type of test (check one only) 1.
Left-tailed
test: [ __ ] Critical value tα = - ______ 2.
Right-tailed
test: [ __ ] Critical value tα = ______ 3.
Two-tailed
test: [ _/_ ] Critical value tα/2 = ± _2.145_____
At significant level of 0.05, since its
two-tail distribution, a/2 =0.05/2 = 0.025 percentile From appendix A, at V=14, t(1-0.025)
= ± 2.145
Use the t-distribution table to determine
the critical value of tα or tα/2 Compare the values of test statistics, t, and critical value(s), tα or ± tα/2 Therefore, Ho is _rejected__________.
|
|
Conclusion that answers the initial
question |
Since t = -17.1 lies in the rejection
region, the null hypothesis is rejected. Hence, the alternative hypothesis(H1)
where the flying distance the projectile was launched by catapult A and B is
different is correct. At 0.05 level of significance, the catapults produced by
the manufacturer are not consistent.
|
|
Compare your conclusion with the
conclusion from the other team members.
What inferences can you make from these
comparisons? |
By comparing my conclusion with most of
my teammates, the results support the alternative hypothesis because of the
critical values which lies in the rejection zone of the test statistic value.
Hence, I could infer that the flying distance the projectile was launched is
different for both catapults. My conclusion is acceptable and reliable as most
of my teammates also achieved the same conclusion. |
Comments
Post a Comment