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
ic

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.



Reflection:  In this lesson, I got to learn about hypothesis testing. I learned that it is a set of formal procedures used by experimenters and researchers to accept or reject the statistical hypothesis. It is a calculated prediction or assumption about a population parameter based on limited evidence. The main idea behind this testing is to evaluate the calculated assumptions to know whether they are true or false. In this lesson, I learned that there is a null hypothesis and an alternative hypothesis. As the name suggests, a null hypothesis is formed when a researcher suspects that there is no relationship between the variables. In other words, there is no significant change. Whereas the latter alternative hypothesis is an assumption to disapprove the null hypothesis. It is the opposite of what the null hypothesis means. In terms of setting the significance level, researchers have created a 5% allowance for accepting the value of an alternative hypothesis, even if the value is untrue. It Is basically the default value. I also learned that the smaller the significance level, the greater the burden of proof needed to reject the null hypothesis and support the alternative hypothesis.


In terms of my experience learning this lesson, I was a little excited because this was the first lesson in CPDD that involves calculations and practices. It was fun for me to learn as I found it manageable to understand the content taught. Before learning this lesson, I always assumed that an ideal product is easy to be built or produced with no errors. I used to think that hypothesis testing doesn’t require any calculations, just a simple observation from the experiments would suffice. But I  was stand corrected. I learned that a product produced has its own acceptable and reject range that must be determined based on the parameters that we are observing so that the product can deliver its function. I am looking forward to making use of this knowledge that I have learned in the future for my capstone project as well as in the workforce.



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