Statistics and Mathematics for Economics

1
Statistics and Mathematics for Economics

• Statistics Component Lecture Seven

2
Objectives of the Lecture

• To present diagrammatically and mathematically a
  normal distribution
• To show how to calculate the probability that the
  value of a normally-distributed random variable
  lies within a specified range of values
• To indicate how to produce a point estimate and
  an interval estimate of the value of the
  population mean of a normally-distributed random
  variable

3
The Normal Distribution

• In the previous lecture, consideration was given
  to continuous random variables
• Examples were provided of the probability density
  function of a continuous random variable
• In Econometrics and Statistics, one of the most
  frequently encountered probability density
  functions is the normal distribution
• Diagrammatically, a normal distribution has the
  appearance of a symmetrical, bell-shaped curve
  which is centred over the expected value of the
  random variable

4
Diagrammatic Presentation of the Normal
Distribution
f(x)
EX ?x
X
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Mathematical Form of the Normal Distribution

• f(x) (2??x2)-1/2.exp.(-1/(2?x2)).(x - ?x)2,
• -? lt x lt ?
• ?x denotes the population mean of X
• ?x2 denotes the population variance of X
• 3.14159
• exp. exponential 2.71828

6
A Property of a Normal Distribution

• A normal distribution is completely characterised
  by its first moment and its second central moment
• Consequently, given knowledge of the values of
  the population mean and the variance of a
  normally-distributed random variable, it is
  possible to calculate the probability that its
  value lies within any specified range
• Short-hand notation X N(?x, ?x2)

7
The Calculation of a Probability
Assume that X N(10, 4). In theory, through
integration, it is possible to calculate P(X gt
12). ? ?(2??x2)-1/2.exp.(-1/(2?x2)).(x -
?x)2dx 12 ? ?(2?(4))-1/2.exp.(-1/(2(4))
.(x - 10)2dx 12
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The Table of the Cumulative Standardised Normal
Distribution

• Fortunately, there is a more convenient method
  which is available for the purpose of calculating
  the probability
• Typically, there is to be found, towards the back
  of a Statistics or an Econometrics textbook, a
  table of probabilities relating to a normal
  distribution
• One such table has been produced by Christopher
  Dougherty, and accompanies his textbook,
  Introduction to Econometrics, Second Edition, 2002

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An Apparent Problem

• The table of probabilities relates specifically
  to a particular type of normally-distributed
  random variable, Z N(0, 1)
• However, any normally-distributed random variable
  can be easily transformed to create a variable
  which has a standardised normal distribution
• The strategy is to subtract from the variable its
  expected value, and to divide the result by its
  standard deviation

10
Using the Table of the Standardised Normal
Distribution
Hence, if X N(10, 4) then Z (X 10) N(0,
1)
----------
?4 And so, P(X gt 12)
P(Z gt (12 10) /2) P(Z gt 1)

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Interpreting the Probabilities in the Table

• A figure in the table indicates the probability
  of obtaining a value of Z which is less than the
  specified value
• Thus, 0.8413 represents P(Z lt 1)
• Upon recognising that Z lt 1 and Z gt 1 cover all
  possible values of Z then P(Z lt 1) P(Z gt 1) 1
• Hence, P(Z gt 1) 1 P(Z lt 1)
• So, P(Z gt 1) P(X gt 12) 1 0.8413 0.1587

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P(Z gt 1)
Prob.(Z lt 1.00)
f(z)
Prob.(Z gt 1.00)
0.8413
0.1587
0
1.00
Z
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A Second Example of the Use of the Table
Let us suppose that we are seeking to
calculate P(X lt 6). X N(10, 4) Hence, Z (X
10) N(0, 1) ----------
?4 P(X lt 6) P(Z lt (6
10)/2) P(Z lt -2)
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Calculation of P(X lt 6)
On the basis of the symmetrical nature of the
graph of the standardised normal distribution
about Z 0 on the horizontal axis P(Z lt -2)
P(Z gt 2). But, P(Z gt 2) 1 P(Z lt 2) Upon
consulting the table, P(Z gt 2) 1
0.9772 0.0228 P(X
lt 6)
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P(Z lt -2)
f(z)
-2
0
Z
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P(Z gt 2)
f(z)
2
0
Z
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P(Z gt 2) 1 P(Z lt 2)
Prob.(Z lt 2.00)
f(z)
Prob.(Z gt 2.00)
0.9772
0.0228
2
0
Z
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Point and Interval Estimates of the Value of the
Population Mean of a Random Variable

• Assume that there is a population of female
  students
• The concern is with a particular characteristic
  of a female student, namely, her height
• X denotes the height of a female student
• The population of values of X is described by a
  normal distribution
• More specifically, X N(?x, ?x2)
• The value of ?x is unknown

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Point Estimate of the Population Mean

• Suppose that there is a desire to know the
  average height of a female student
• It is impractical to approach every female
  student in the population
• Consequently, an estimate of the average height
  is formed, having taken a random sample from the
  population of values of X
• Random sample X1, X2, , Xn
• The estimate is achieved by calculating the
  average of the sample values

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The Sample Mean
The sample mean, -_
n X (1/n)?Xi i
1 The value of the sample mean has the
interpretation of the best guess of the value of
the population mean.
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An Interval Estimate

• It may be more useful to have access to an
  interval estimate, rather than a point estimate,
  of the value of the population mean
• An interval estimate constitutes a range of
  values within which the value of the population
  parameter falls with a specified probability
• In order to be able to produce an interval
  estimate of ?x, it is necessary to know the
  statistical properties of the sample mean of X

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The Probability Density Function of the Sample
Mean
- X (1/n) (X1 X2 Xn)
(1/n)X1 (1/n)X2 (1/n)Xn On the basis
of the nature of the population from which the
sample has been drawn, Xi N(?x, ?x2), i 1,
2, , n. Any linear combination of
normally-distributed random variables, itself,
has a normal distribution.
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The Expected Value and the Variance of the Sample
Mean
It can be demonstrated that - EX
?x and - Var.(X) ?x2/n.
- Consequently, Z (X -
?x)/(?x2/n)1/2 N(0, 1)
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P(Z lt 1.96) 0.975
Prob.(Z lt 1.96)
f(z)
Prob.(Z gt 1.96)
0.9750
0.0250
1.96
0
Z
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P(-1.96 lt Z lt 1.96)
If P(Z gt 1.96) 0.025 then, on the basis of
the symmetrical nature of the standardised
normal distribution about zero, P(Z lt -1.96)
0.025. It follows that P(-1.96 lt Z lt 1.96) 1
0.025 0.025 0.95
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Ninety-five per cent confidence interval for ?x
On substitution, then
-
Prob.(-1.96 lt (X - ?x)/?(?x2/n) lt 1.96)
0.95 Through a sequence of manipulations, it is
possible to obtain the interval estimate
-
- Prob.X 1.96?(?x2/n) lt ?x lt X
1.96?(?x2/n) 0.95