report writing
DESCRIPTION
Report Writing. A report should be self-explanatory. It should be capable of being read and understood without reference to the original project description. Thus, for each question, it should contain all of the following:. a statement of the problem; - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Report Writing](https://reader035.vdocument.in/reader035/viewer/2022062723/56813db2550346895da77a0e/html5/thumbnails/1.jpg)
Report Writing
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A report should be self-explanatory. It should be capable of being read and understood without reference to the original project description. Thus, for each question, it should contain all of the following:
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(a)a statement of the problem;
(b) a full and careful description of how it is investigated;
(c) All relevant results, including graphical and numerical analyses; variables should be carefully defined, and figures and tables should be properly labelled, described and referenced;
(d) relevant analysis, discussion, and conclusions.
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It should be written in the third person.
NOT: I think the Central Limit Theorem is true for this example because I see that the graph is normal.
INSTEAD: It can be clearly seen that the graph displays a normal distribution confirming that the Central Limit Theorem holds.
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The Central Limit
Theorem
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Let X1, X2………. Xn be independent identically distributed random variables with mean µ and variance σ 2.
Let S = X1,+ X2+ ………. +Xn
Then elementary probability theory tells us that E(S) = nµ and var(S) = nσ 2 .
The Central Limit Theorem (CLT) further states that, provided n is not too small, S has an approximately normal distribution with the above mean nµ, and variance nσ 2.
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In other words,
S approx ~ N(nµ, nσ 2)
The approximation improves as n increases.
We will use R to demonstrate the CLT.
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Let X1,X2……X6 come from the Uniform distribution, U(0,1)
0 1
1
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For any uniform distribution on [A,B],
µ is equal to
and variance, σ2, is equal to
2
A B
2( )
12
B A
So for our distribution, µ= 1/2 and
σ2 = 1/12
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The Central Limit Theorem therefore states that S should have an approximately normal distribution with mean nµ (i.e. 6 x 0.5 = 3)
and var nσ2 (i.e. 6 x 1/12 = 0.5)
This gives standard deviation 0.7071
In other words,
S approx ~ N(3, 0.70712)
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Generate 10 000 results in each of six vectors for the uniform distribution on [0,1] in R.
> x1=runif(10000)> x2=runif(10000)> x3=runif(10000)> x4=runif(10000)> x5=runif(10000)> x6=runif(10000)>
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Let S = X1,+ X2+ ………. +X6
> s=x1+x2+x3+x4+x5+x6> hist(s,nclass=20)>
![Page 15: Report Writing](https://reader035.vdocument.in/reader035/viewer/2022062723/56813db2550346895da77a0e/html5/thumbnails/15.jpg)
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Consider the mean and standard deviation of S
> mean(s)[1] 3.002503> sd(s)[1] 0.7070773>
This agrees with our earlier calculations
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A method of examining whether the distribution is approximately normal is by producing a normal Q-Q plot.
This is a plot of the sorted values of the vector S (the “data”) against what is in effect a idealised sample of the same size from the N(0,1) distribution.
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If the CLT holds good, i.e. if S is approximately normal, then the plot should show an approximate straight line with intercept equal to the mean of S (here 3) and slope equal to the standard deviation of S (here 0.707).
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> qqnorm(s)>
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> qqnorm(s)>
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> qqnorm(s)>
4.4 – 1.8
4
= 0.7 to 1 DP
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From these plots it seems that agreement with the normal distribution is very good, despite the fact that we have only taken n = 6, i.e. the convergence is very rapid!