KB7044 Assignment

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KB7044 Assignment Support 2019/2020 semester 2
Tasks 1-3
KB7044 Assessment is an assignment. Unlike closed book examinations and MSc dissertation for
which past exam papers or reports can be made available for students to access (with examinations
there will be no repeat of questions, and with MSc Dissertations the project titles are normally
different from year to year.
For the academic year of 2019/2020, due to the technical extenuating circumstance with the
withdrawal of face to face sessions between academic staff and students, additional support for the
assignment is provided.
The support is in the form of examples from past student reports for some selected tasks of the
assignment. Hence, you will “have a feel” of the report that you are submitting for this module. It is
important to note that these are by no means “ideal solutions” or “exemplar” for the assignment, they
are merely as stated, examples from past reports.
You should be aware of the marks allocated to the various sections of your report:
From the assignment brief:
1. Introduction 10%
2. Computations Case 1 20%
3. Computations Case 2 20%
4. Independent Learning Project 35%
5. Discussion, recommendations, further work and conclusion(s) (including marks awarded to
referencing and citations 15%
Case 1 – Tasks 1 – 6.
Examples from past reports:
Introduction Section
Marks allocated is 10%.
Please provide a few short paragraphs for introduction. This section “sets the scene” for your report
and should provide context (Company ABC and its product, problem) and some elaboration. (With
this alone up to 50% of the 10% can be awarded).
You may include some citations with references on use of regression, probabilistic modelling etc. that
you think are relevant. Additional marks to be awarded based on the contents.
Example 1 (A little on the lengthy side):
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Introduction
The decision making plays an important role in every parts of project, especially in engineering design process.
The decision making can affect on job security, cost and opportunities for advancement that impact to
profitability of the company also the performance of employer (Henderson, 2010). Every manufacturing
organisation is aim to produce the product in the best quality in the shortest time and lowest cost. The
engineering design process involves the application of technical knowledge (Davis and Pecar, 2010). In order
to investigate the feasibility of incorporating a reliability and risk, we can examine safety factor in the face of
uncertainty use many variable factors for example mean load, design stress and actual stress (Albright, 2009).
So it is important to apply engineering data analysis to improve the process by applying statistical and logical
technique (Northern Illinois University, 2014). According to Albright (2009) suggests that engineering design
processes involve the application of technical knowledge in a structure way:
• The definition of a product that meets specified needs
• The best design strategy, methods and tools to adopt
• ……….
• The development and innovation
The case study of this project is ABC Company who designs and builds supporting structure for machinery
clients. ABC has got many complaints of exceeding yielding stress which can result in serious failure of
structure. There is high risk to loss of lives and damage to properties. So it is important for the ABC Company
to investigate the problems, ……….. The risk based method is adopted in this project to support decision
making under uncertainty. As a consultant investigates the problem and to look into the feasibility of
incorporating reliability, the company calculate the additional material cost and how does it affect the
company’s budget.
The analysis tool used in this assessment is Microsoft Excel by applying add-ins to analyse two cases of data;
Numerical case 1 for task 1 to 4 and Numerical case 2 for task 5 to 8. In this case the 22 samples are given with
specification such as mean load, load variation factor, design stress, actual stress and the ration of actual
stress over the design stress.
The objective of Numerical case 1 is;
Task1: To perform correlation and regression analysis. Then see and analyses their relationship.
Task2: To prove that there is any sufficient evidence to accept manufacturer’s claim with the variability
in the yield stress of material is 130MPa.
Task3: …………
Task4: To consider the trade-off of additional material cost, penalty cost and total cost by providing a
level of excess capacity and incurring by failed components that will be affected on expenses of company.
For objectives of case 2 were done by the following;
( …………….)
At first, we have to sort the data analyse variability in the yield stress of material is 130 MPa for making the
decision, Is there the sufficient evidence to accept the manufacturer’s claim? And analyse the data by plotting
histogram and determine an acceptable level of failure probability and the optimum excess capacity values.
Future more, the trade-off of additional material cost, penalty cost and total cost by providing a level of excess
capacity and incurring by failed components is considered also determine an acceptable level of failure
probability on the risk-based model.
………………..
In the development part of this report, a thorough discussion will be taking place. An analytical model will be
evaluated for comparing result of safety factor, failure probability and excess capacity. A thorough discussion
will be done on this case study and the findings will be related to the statistical value of life with
recommendations. Lastly, the conclusion and further work will be highlighted respectively
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Example 2 (Very short intro – too brief)
Task 1
Example 1: (Correlation should include all variables, Regression is incorrect! – this is a negative
example)
Introduction
In this analysis report it is expected to carryout research related data analysis and decision making
and based on the data’s given, have to do different tests like regression, correlation, and to plot
appropriate graphs based on the result got and to do the interpretation of the results. An in the
end an independent project is done based on the data’s collected from the recent events that
occurred in uk. And cost estimation is done because of the limitation of the available data’s some
data’s are assumed.
The first task is to investigate the reasonable degree of correlation between uncertainty and
actual stress in the section and if there is a correlation exist between certain factors a regression
test also must be done
Table 1 inserted

The above test shows that there is a reasonable correlation between design stress and actual
stress and can be noted that there is a negative correlation. A negative correlation will result in
decrease in dependant variable. Even though the correlation between design stress and actual
stress looks similar big difference can be seen in varying mean load. And a scattered graph is
plotted.
0
50
100
150
200
250
300
118 120 122 124 126 128 130 132
Actual stress
Design stress
Actual stress(MPa)
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Example 2
Insert Table 1
Table 1 presented the data that will be used in correlation and regression analysis. From this table the data is
sorted by its load variation that shown in table 2. There are 22 data with different three load deviation factors, 1,
2 and 3. The mean load has varied range between 83 and 102. The fourth and fifth column shows the design
stress (σd) and actual stress (σa ) respectively with MPa unit. For the last column it illustrates the ration of actual
stress over the design stress (σa/σd).
The question one is to inspect the correlation between safety factor (σa / σd) and load variation factor to identify
the relationships among the variances. The correlation coefficient provides a satisfactory measure for bivariate
data. In this case, given 5 variable: F (Mean Load), load variation factor (LVF), design stress (σd), actual stress (σa)
and actual stress/design stress ratio (σa / σd). At first, we sorted data by their load variation factor to observe its
pattern which shown in the table as below

A strong relationship can be noticed between actual stress/design stress (σa / σd) and actual stress both with
variation factor which scored 0.9044 and 0.9055 respectively. For the weak relationship will happen when the
relationship close to zero. From this figure, the weak relationship can be seen between mean load and design stress
(σd) with scored 0.029 same as the relationship between mean load and actual stress/design stress (σa / σd) with
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Example 2 (continue)
R =0.915 and R2 = 0.838. It can been seen that correlation coefficient is almost 1 and also a measure of value that is
fit for linear regression.
•Standard error: It measures the accuracy which a sample represents a population. Generally, smaller the standard
error, the more representative the sample will be of the overall population
•Lower σa / σd = 0.403-0.006-3.240
• σa / σd = 0.520+0.011-0.081
•Upper σa / σd = 0.638+0.027+3.079
•P-value is 0.539 for load variation factor. It is more than50%. It is not significant for this project.
(More texts to explain and describe this second attempt)_ ……
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Task 2
Example 1:
Descriptive test

The descriptive test shows that the mean value is 124.32MPa and the hypothesis t test below shows
that there is not enough evidence to reject hypothesis. After doing the hypothesis t-test it shows that
the alpha value is not above 0.058 and we would get enough evidence to reject the hypothesis. And
hence it proves that there is no need to reject the hypothesis, but the suppliers claim about the yield
stress of material to be 130MPa is not true.
One tailed t-test
It’s given that the material yields at 130MPa, and this will be the null hypothesis
Null hypothesis (Ho): µ= 130
Test hypothesis (H1): µ= ˃/˂ 130
Tcrit = -1.83
α = 0.05
tstat = (124.32-130)/10.2968
= -1.74
After doing the t test the critical t value is –1.83 which is less than -1.744,
tcrit ˂ tstat. The critical t exceeds t when the alpha value is 0.05
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The above graph shows the cumulative percentage and frequency. The safe option to claim
the yield stress and to reject the hypothesis is when the mean value is 125MPa. The t test shows that
the the critical t value is 1.833 and to reject this the alpha value should be increased.
Example 2:
At first, the histogram was plotted to analyse the data. The histogram of frequency in the different
bin ranges are shown in Figure 6
Figure 6 Histogram
0.00%
20.00%
40.00%
60.00%
80.00%
100.00%
120.00%
76543210
Frequency
Bin
Histogram
Frequency
Cumulative %
Frequency, 100,

Frequency, 110,
1
Frequency, 120,
2
Frequency, 130,
4
Frequency, 140,
2
Frequency, 150,
1
Frequency,
More, 0
Cumulative %,
120, 30.00%
Cumulative %,
130, 70.00%
Cumulative %,
140, 90.00%
Cumulative %,
150, 100.00%
Cumulative %,
More, 100.00%
Frequency
Range
Histogram
Frequency
Cumulative %
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The second task is to contemplate the variability in the yield stress of the material used. According the
supplier of the material, the yield stress of the material is 130 MPa. A sample comprising 10 specimens
was tested. We use Hypothesis test to figure out this problem. Prior the test, two hypotheses must be
introduced, Ho and H1, Ho: the “null hypothesis”. The underlying principle of H0 is that observations in
the sample result from chance and there is no meaningful pattern or relationship. And H1: the
“alternative hypothesis”. The opposite of the null hypothesis, there is some relationship, pattern or
influence in the data.
Note: The concept of one-tail (left or right) and two-tail tests can be applied pretty much in the same
manner as described before. The Excel functions TDIST and TINV can be used as before. Note that the
Excel t distributions take alpha values as the two-tailed values, hence this need to be taken into
consideration when you are performing a one-tail test (the alpha values need to be multiplied by two
for a one-tailed test).
Our intention is to prove that the population mean value of variable is 130 MPa on the basis of this
sample. Our Hypothesis should be as below.
Hypothesis

The t statistic:
t =

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