applied statistical inference [modo de compa

8/11/2019 Applied Statistical Inference [Modo de Compa

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Hypothesis Testing

Can you convert your data into meaningful information?

o you know how to conduct experiments?

- D r. O. -


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Population Sample

Statistics

Inference

Parameters

Presumption: belief dictated by probability


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IndependentRelatedIndependentRelated

Intervaland

Ratio

Ordinal

Nominal

Two Samples N-SamplesSingle

Sample

Hypothesis Testing Map

Eyes Color, Sex, Race

Level Pain, Preference,.. #of Children,..{Integers}

{Any Real #}


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:

:

1

0

:

:

1

0

::

1

0

Two-Tailed

Test

One-Tailed

Test

One-Tailed

Test


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LetX1,, Xnbe alarge(e.g.,n 30) sample from a

population with mean and standard deviation . To test a

null hypothesis of the form

H0: = 0

H0: 0

H0: 0

Compute thez-score:

If is unknown it may be approximated by .

nz

/

0

nSt

/

0


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H0:D= 3%H1:D 3%

Single Sample CaseEXERCI SE #1


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n

stn

2

0

1

H0: D= 3%

H1: D 3% =0.494

df = n-1

=10-1= 9


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EXERCI SE #


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EXERCISE #


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df = 13


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Many studies have shown that direct eye contact and even patterns that look like eyes are

avoided by many animals. Some insects, such as moths, have even evolved large eye-spot patterns on their wings to help ward off predators. This research example, modeled

after Scaife's (1976) study, examines how eye-spot patterns affect the behavior of moth-

eating birds.

A sample ofn = 16 insectivorous birds is selected. The animals are tested in a box that

has two separate chambers (see Figure). The birds are free to roam from one chamber to

another through a doorway in a partition. On the wall of one chamber, two large eye-spotpatterns have been painted. The other chamber has plain walls. The birds are tested one

at a time by placing them in the doorway in the center of the apparatus. Each animal is

left in the box for 60 minutes, and the amount of time spent in the plain chamber is

recorded. Suppose that the sample of n= 16 birds spent an average of = 35 minutes in

the plain side with SS = 1215. Can we conclude that eye-spot patterns have an effect on

behavior? Note that while it is possible to predict a value for, we have no information

about the population standard deviation.

EXE RCISE #4


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H0:D= 30

H1:D 30

97.5% confidence


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H0:D 30

H1:D> 30

95% confidence


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More Applications


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H0:D= 0

H1:D 0

Two Related Samples Case

nSt D

/

Cluster Before After

X2-X1


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Site X2X1

nS

Dt D

/


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Site X2X1

df=4

-2.13

0:H

0:H

1

0


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Cluster Before

H0:D 0H1:D< 0

After

The environmental policy must be well known across all the company levels. To certify

ISO 14001 a company wide training effort has been made to make sure that every

cluster in the company need less overtime dedication to this matter.

Note that weested the

mean = 0

already


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H0:D 0

H1:D< 0


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In-Class Work


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Compare the mean numbers of days

spent in hibernation by hedgehogs in

two areas of the country using

randomly selected samples of each.

Suppose 10 hedgehogs in each area are

captured, radio-tagged and released.Their movements are monitored

throughout the winter to determine the

number of days when they did not

leave their nests.

Two Independent SamplesTest ( comparable)


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Two Independent Samples Test


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Two Independent Samples Test ( comparable)


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Two Independent Samples Test ( comparable)

df =


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Exercise


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More Applications


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Goodness of fit test (Chi2 test)It is suggested that dispersal of seeds from the edge of a plantation of trees

follows an inverse square law (i.e. the number of seeds travelling m is

proportional to 1/x2). We want to test this theory using an isolated stand of

trees on an area of moorland. Seeds will be collected in four 0.25 m2 plots at

each of the five distances, on the downwind side of the plantation. The total

numbers of seeds counted at each distance were:

Using the equation (theoretical or empirical)0.04 0.0564

X 294


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168.923

If P-value (=0.05) >P-score; Then, H0is true

If P-value (=0.05)

P=score; Then, we have

significant evidence that

the results are not

consistent w/H0 , so H1is true.

One tailDist.

Multiply by

2

=0.054


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If P-value (=0.05) >P-score; Then, H0is true

If P-value (=0.05)

P=score; Then, we have

significant evidence that

the results are not

consistent w/H0 , so H1is true.

7.814

168.923


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Highschool College Graduate

67 25 8

Education

A pool of 100 individual have been surveyed with respect to level of

education (see table below). The population is expected to show the

following trend: High school 60%, College degree 25%, and Master and PhD15%. Is the sample representative of the population?

EXE RCISE #1


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Poor Average Rich

A 8.19 -1.35 -6.84

B -1.64 5.93 -4.29

C -6.55 -4.58 11.13

Reactant

cata

lyst

Poor Average RichA 25 12 7 44

B 6 12 2 20

C 3 3 19 25

34 27 28 89

catalyst

Reactant

Poor Average Rich

A 16.81 13.35 13.84 44

B 7.64 6.07 6.29 20C 9.55 7.58 7.87 25

34 27 28 89

Reactant

cata

lyst

Difference (O-E)

Observed

Expected


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Poor Average Rich

A 3.99 0.14 3.38 7.51

B 0.35 5.80 2.93 9.08

C 4.49 2.77 15.76 23.03

8.84 8.71 22.07 39.62

Reactant

catalyst

Difference (O-E)2

E

If P-score> 0.05 Then,there is insufficient

evidence to conclude that

there is association

between two factors

If P-score 0.05 Then,

there is significantevidence of an

association between two

factors.

One tailDist.

Multiply by

2

=1.03E-07

62.3924


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Poor Average Rich

A 3.99 0.14 3.38 7.51

B 0.35 5.80 2.93 9.08

C 4.49 2.77 15.76 23.03

8.84 8.71 22.07 39.62

Reactant

catalyst

Difference (O-E)2

E

If P-score> 0.05 Then,there is insufficient

evidence to conclude that

there is association

between two factors

If P-score 0.05 Then,

there is significantevidence of an

association between two

factors.

One tailDist.

Multiply by

2

=1.03E-07

62.3924

9.487


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A scientist is concern that his new developed drug is affecting the proportion

of sheep having stillborn lambs in his flock of sheep. He has records from

before the drug was tested in the population so he is able to compare withcurrent data.

EXE RCISE #1


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Compare the mean numbers of

days spent in hibernation by

hedgehogs in two areas of the

country, using randomly

selected samples of each.

Suppose 10 hedgehogs in each

area are captured, radio-tagged

and released. Their movements

are monitored throughout the

winter to determine the number

of days when they did not leave

their nests

Two Independent Samples Test ( check ifcomparable)

33.344

32.233


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12 2

2

12 2

2

These are one-tailedprobabilities, so we need to

multiply them by 2.

Therefore P= 96%

4.0255

1.034


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2

2

2

11

2

2

2

10

:H

:H

2

2

2

11

2

2

2

10

:H

:H

One-Tailed

Test

One-Tailed

Test

)1N,1N,( 21FFscore

)1N,1N,1( 21FFscore

The hypothesis that the two standard deviations are equal is rejected if

1.034


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EXE RCI SE #1


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3.912

1.34


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ANOVA Test

A B C D

x x x x

x x x x

x x x x

x x x x

x x x x

ONE-FACTOR

FLUX

LEVELS

REPLICATES

SampleValues

1 2 3

A x,x,x,x x,x,x,x x,x,x

B x,x,x,x x,x,x,x x,x,x

C x,x,x,x x,x,x,x x,x,x

D x,x,x,x x,x,x,x x,x,x

LEVELS

REAGENT

LEVELS

CATALYST

One-Way ANOVA Two-Way ANOVA

1 2 3 c 1 2 3 c 1 2 3 c

1 x, x, x x, x, x x, x,x x, x,x x ,x, x x ,x, x x ,x, x x ,x, x x ,x ,x x ,x ,x x, x, x x, x, x2 x, x, x x, x, x x, x,x x, x,x x ,x, x x ,x, x x ,x, x x ,x, x x ,x ,x x ,x ,x x, x, x x, x, x

3 x, x, x x, x, x x, x,x x, x,x x ,x, x x ,x, x x ,x, x x ,x, x x ,x ,x x ,x ,x x, x, x x, x, x

a x, x, x x, x, x x, x,x x, x,x x ,x, x x ,x, x x ,x, x x ,x, x x ,x ,x x ,x ,x x, x, x x, x, x

FACTORB

FACTORC

LEVELS

1 2 b

LEVELS

FACTORC

LEVELS

FACTORA

LEVELS

FACTORC

LEVELS

Three-Way ANOVA


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One-Way ANOVA Test

A B C D

x x x x

x x x x

x x x x

x x x x

x x x x

ONE-FACTOR

FLUX

LEVELS

REPLICATES

SampleValues

ONE FACTOR


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One-Way ANOVA TestA B C Dx x x xx x x x

x x x x

x x x x

x x x x

ONE-FACTOR

FLUX

LEVELS

REPLICATES

SampleValues

ONE-FACTOR

FLUX

LEVELS

A B C

REPLICATES

SampleV

alues 16 38 19

13 21 14

36 36 17

29 39 15

18 26 12

22.4 32 15.4 23.267

ONE FACTOR


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One-Way ANOVA TestA B C Dx x x xx x x x

x x x x

x x x x

x x x x

ONE-FACTOR

FLUX

LEVELS

REPLICATES

SampleValues

5.098

6.272


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Two-Way ANOVA Test


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Two-Way ANOVA Test

20y

i

iB


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Two-Way ANOVA Test

20y

i

iB


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Two-Way ANOVA Test

20y

i

iB)05.0%(5


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Two-Way ANOVA Test

20y

i

iB)05.0%(5


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Two-Way ANOVA Test

Two Way ANOVA Test


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Two-Way ANOVA Test

Two Way ANOVA Test


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Two-Way ANOVA Test

T W ANOVA T t


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Two-Way ANOVA Test

Th W ANOVA T t


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Three-Way ANOVA Test

1 2 3 c 1 2 3 c 1 2 3 c

1 x ,x ,x x ,x ,x x, x, x x, x, x x, x,x x, x,x x ,x ,x x ,x, x x ,x, x x ,x, x x ,x ,x x ,x ,x



a x ,x ,x x ,x ,x x, x, x x, x, x x, x,x x, x,x x ,x ,x x ,x, x x ,x, x x ,x, x x ,x ,x x ,x ,x

FACTORB

FACTORC

LEVELS

1 2 b

LEVELS

FACTORC

LEVELS

FACTORA

LEVELS

FACTORC

LEVELS

Simplified Case

Th W ANOVA T t


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Th W ANOVA T t


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Th W ANOVA T t


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Computed F > F(5%)

Therefore, We reject Ho

for AB, AC, and BC. In

other words, these

interactions do have animpact in the dependent

variable.


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