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Statistical Inference: Confidence Intervals and Hypothesis Testing, Slides of Structural Analysis

An introduction to statistical inference, focusing on the concepts of confidence intervals and hypothesis testing. It covers the estimation of population means with unknown variances, the use of student's t-distribution, and the calculation of confidence intervals. The document also discusses the process of making decisions on properties of a population based on observed samples.

Typology: Slides

2012/2013

Uploaded on 04/24/2013

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Download Statistical Inference: Confidence Intervals and Hypothesis Testing and more Slides Structural Analysis in PDF only on Docsity!

Monte Carlo simulation approach-

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(^50) -

f^ t^^  

x t ^ 

Ensemble of inputs

Ensemble of outputs

System

Generate ensemble ofinputs obeying prescribedmodel for

( ) f t

Process ensemble of outputsusing statistical tools and arriveat probabilistic model for

( ) x^ t

f^ t^^  

x t ^ 

t^

t

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3

^ ^

^

1

Method of moments

1

Mean, Variance, Skewness, Kurtosis,...

Method of maximum likelihoodEstimate directly the parameters of the pdfquantities like mode, median, range

n^

k i

k

i

X n^

 ^

 

Estimation of paramters

, ...

can be estimated using this method.

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44

^ 

^ 

^

^

^ 

^

^

^

^ 

^ 

1

2

Let

be a random variable with PDF

, pdf

,

mean

, and standard deviation

.

Let

be an iid sequence with common pdf

.

That is,

1,^

,

,^

,^

1

i

X^

X

n i

X

i i^

j

i^

i^

X^

X

X^

P^

x^

p^

x

X^

p^

x

X^

X^

i^

j^

n

X^

Var

X

p^

x^

p^

x^

i

 

 

^

^

^

 

Estimation of mean

^

^

1

2

,^

.

1

is an unbiased estimator

of^

with minimum

variance and the lowest variance is

.

n

i i

n

X^

n

n

n

  

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5

1

2

2

1

Consider the estimator

is an unbiased estimator of

with variance

Let us consider the case in which

is known.

If^

is Gaussian, it w

n

i i

X

n

n

X

  

Sampling distribution for the estimator of mean

^

^1

ould mean that

is an iid

sequence of Gaussian random variables and consequently

would also be Gaussian distributed. If^

is not Gaussian, by virtue of central limit theorem, for large

, we may s

n i X

X

n

^

till consider

to be Gaussian.

It may be inferred that

~^
,^

or,

~^
N^
N

n^

n

^

^
^
^
^

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Confidence interval estimation

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^

^

^

^

Consider

~^

0,1. Consider

to be

a specified probability

e.g.,

N

 n

^
^

-^ -^ -^ -^

0

1

2

3

4

0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05^0

x

pdf

k^ /2^ ^

^

 1

/ k



Area in thisregion=1-

/^

1 / /

P k^^1

k n

   

  ^

^

^

^

^ 

^

^

^

^

1 /

1

1 /

1

/ 2 1

/ 2

k^  k

     

  

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^

^

^

^

/^

1

/

/^

1

/

/^

1

/

observed value of

from the sample.

,^

confidence

interval for the population mean

. Suppose, 1-

P k We sa

k n

P^

k^

k

n^

n

k^

k n^

n

 

^

^

 

 

^

^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^

y with 95% confidence that the true mean is contained in the interval

k^

k n^

n

^
^
^
^

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^

^

/^

1

/2 /

1

/ 1

This should be interpreted as the probability thatthe random interval

contains the population mean

is 1

Remember

is a determinist

P^

k^

k

n^

n

k^

k n^

n

 

^

^

^
^
^
^
^
^
^
^
^
^
^
^

Remark

^

(^1) /

1

/

ic quantity.

is a point estimate &,^

is a confidence interval

estimate.

n

i i

x n k^

k n^

n

^
^
^
^
^

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10

= x -0.4326-1.66560.12530.2877-1.14651.19091.1892-0.03760.32730.

^

10 1 /^

1 /

1

10 0.

1.96 &

95% confidence interval

=^

, 0. 10

10

0.6068, 0.

i i

x

k^

k

 

^

  

^

^

^

^

^

Example

^

The point estimate of mean is 0.0013.With 95% confidence we say that the populationmean is contained in the interval

0.6068, 0.. ^

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0

100

200

300

400

500

600

700

800

900

1000

(^3210) x -1 -2 -

n

^

1000 1 /^

1 /

1

-0.

1000 0.

1.96 &

95% confidence interval

=^

,^ 0. 1000

10

0.1065, 0.0.

i x i

k^

k

 

^

  

^

^

^

^

^

^

^ 

n =

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0 1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

(^43210) -1 -2 -3 -

^

 10000 1 /^

1 /

1

10000 0.

1.96 &

95% confidence interval

=^

,^ 0. 10000

10000

-0.0163, 0.

i i

x

k^

k

 

^

  

^

^

^

^

^

^

n =

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^

 ^

^

^

^

^

^

^

^

^

^

^

^

^

2

2

1

1

2

2

1

2

1

2

2

1

2 2

2

2

2

2 1

1

with X=

1 1 1 1 1 1

2

1 Note:

&

Similarly we can show that Var

n^

n

i^

j

i^

j

n

i i n

i i n

i^

i

i

i

S^

X^

X^

X

n^

n

S^

X^

X

n

X^

X

n

X^

X^

X^

X

n

X^

X^

n

S

S

^

^

^ 

   ^

 

 

^

^

^

^

 ^

^

^

^

^

^

^

^

^

^

 

  

Sampling distribution for variance

4

4

2

4

(^31) n

n^

n

^

^  

 

^

 

 ^

^

^

^

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^

^

2

2

1

2

2

2

2 2

1 1

is an unbiased and

1

consistent estimator of

.

1

/

If population is Gaussian,

and

are Gaussian.

RHS: sum of squares of Gaussian random variable

n

i i n

i i

i

S^

X^

X

n n^

S^

X^

X

n

X^

X

 

 

^

^

 ^

^

^

^

^

^

 ^

^

^

^

^

 

^

^

^

^ 

2 2

2

2

(^1 )

(^12)

s

such sums have

distributions.

1

is^

distributed with

-^

dof.

The pdf of such a random variable is given by

1

exp

; 2

1

2

2

n

n n^

S^

n u

p u

u^

u

  n



 ^

 ^

 ^

^

^

^

^

 

^

 ^

^

^

 

 ^

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15

^

 2  

1 2

2

~^
~^

distribution with

-dofs

X^
Y

T^

n

X
T^
Y^

n

X^

N
Y^

n

n

p^

t^

t

n n^

t n

^

^
^
^
^
^
^
^
^
^
^

Student's t - distribution

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1

2

Consider the estimator

is an unbiased estimator of

with variance

~ Student t-distribution with

-1 dofs.

n

i i

X

n

n

n

s^

n s

  

Sampling distribution for the estimator of meanwith variance not known

^

^

^

 

^

 1

1 2

2

estimate of standard deviation from the sample.

dof

f

f

p^

f

f^

f

f

^

^
^
^
^
^
^
^
^
^
^

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^

^

^

 

^

1

,^1

,^1

2

2

1

1 2

2

Consider the statement

From this one can obtain the confidence intervalfor the population mean.

n^

n

f

f

p^

f

f^

f

P^

s^

n

 

^

 ^ ^ 

^

^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^

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0 10

20 30

40

50 60

70

80 90

100

(^3210) -1 -2 -

sample

x

Point estimatesˆ^

ˆ

=-0.0677;

=0.

^

^

^

^

^

100

1

,^

,

-0.

-0.

-0.

-0.

-0.

-0.

-0.

-0.

-0.

-0.

-0.

 

 

^

^

^

 

^

-0.

Example

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  • -1.
  • -1.
  • -0.
  • -0.
  • -0.
    • -0.
    • -0.
    • -1.
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                    • -1. - 0. - 0. - -0. - -0. - -0. - 0. - -0. - -0. - -1. - 0. - 2. - 0. - -1. - 0. - -1. - 0. - -1. - 0. - n = Data used in the example

20

0

10

20

30

40

50

60

70

80

90

100

0.06 0.05 0.04 0.03 0.02 0.01^0 -0.01-0.02-0.03-0.

100*(1-alpha)

mu

lower conf-limitupper conf-limitpoint estimate

5000 n^ 

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10

20

30

40

50

60

70

80

90

100

1.06 1.05 1.04 1.03 1.02 1.01^1 0.99 0.98

100*(1-alpha)

sigma

lower conf-limitupper conf-limitpoint estimate

5000 n^ 

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Factors influencing confidence interval^ The statistic used as the estimate  The observations made Confidence level Sampling distribution Sample size

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^

^

1 /2^

1

/2

Consider the estimator for population meanwith known variance.

~^

,^

or,

~^

n

i i

X^

N^

N

n^

n^

n

P^

k^

k

n^

n

 

^

^

^

^

^

^

^

^

^

^

^

^

^

Number of samples needed for a given widthof confidence interval

^

 

 1

/2^1

/2

2

Let

half width of confidence interval

be specified. Minimum number of samples required

w k^1

n

n^

k

w

  

^

^

^

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0

0.1^

0.2^

0.3^

0.4^

0.5^

0.6^

0.7^

0.8^

0.9^

1

(^43210) -1 -2 -3 -4

w

Mean^0

0.1^

0.2^

0.3^

0.4^

0.5^

0.6^

0.7^

0.8^

0.9^

5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 1

MeanLower Conf limitUpper Conf limit

10 log

n

^

1

0.95 

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0.01^

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0.11

(^510410310210)

half conf width

n

MU-Mw

n

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