Abid ResearchMethods Vadodra

8/11/2019 Abid ResearchMethods Vadodra

1/71

1

Research Methods(Use of Statistical tools in Data Analysis)

Abid Ali Khan PhD (UL, Ireland)

Associate Professor,

Ergonomics Research Division

Department of Mechanical Engineering, Aligarh Muslim University, Aligarh

Email: [email protected]


2/71

Processing & Analysis of Data

Measurement of central tendency

Mean

Median

Measurement of dispersion

Range

Variance

Standard deviation

Measurement of skewness

2


3/71

Central Tendency

3


4/71

4


5/71

Variance

Need to know the

variability of a data set

How much each number in set

varies from central point Types of variability

Range

Variance Standard deviation

5


6/71

Skewness

6


7/71

Measurement of relationship

Correlation (strength of relationship between DV and IV)

Karl Pearsons coefficient of correlation (r)

7


8/71

Normal Distribution /SND

65% of scores fall within

1 st.dev. of mean

95% of scores fall within

2 st.dev. of mean Only 5% of scores fall in

extreme portions

8


9/719


10/71

Simple Regression Analysis

10


11/71

Least square estimates of regression line

11


12/7112


13/71

Testing of Hypothesis(Parametric or Standard Tests of Hypotheses)

What is Hypothesis?

Null Hypothesis

Alternative Hypothesis

The level of significance

Decision rule

Type I and Type II errors Two tailed and One tailed tests

Power of Hypothesis test

13


14/7114


15/71

Important parametric tests

t- test

Z-test

Chi-square test F-test

ANOVA

ANCOVA

15

T t i th f


16/71

Test concerning the mean of

Normal population

Case for known variance

16


17/7117


18/71

Case for unknown variance

t-test

18


19/71

Test of equality of means of two

normal populations

19


20/71

Hypothesis Testing

Tests Concerning MEANS


21/71

- known

H 0 : = 0 = ( 0)( )

H1 Critical Zone

< 0 Z 0 Z>-Z

0 ZZ/2


22/71

- unknown

H 0 : = 0

H1 Critical Zone

< 0 t 0 t>-t

0 tt/2

= ( 0)( ) ; = 1


23/71

1 & 2 - known

H 0 : 1-2 = d 0

H1 Critical Zone

1-2 < d 0 Z d 0 Z>-Z

1-2 d 0 ZZ/2

= (1 2) 0

12

1

22

2


24/71

1 = 2 - unknown

H 0 : 1-2 = d 0

H1 Critical Zone

1-2 < d 0 t d 0 t>-t

1-2 d 0 tt/2

=(

1 2)

01 1 + 1 2; =1 + 2 2

=(

1

1)

12 + (

2

1)

22

(1 +2 2)


25/71

1 2 - unknown

H 0 : 1-2 = d 0

H1 Critical Zone

1-2 < d 0 t d 0 t>-t

1-2 d 0 tt/2

= (1 2) 012 1 +22 2

=

(1

2

1 +2

2

2 )2

(12 1 )

2

1 1 +

(22 2 )

2

2 1


26/71

Pairwise t-test

H 0 : d = d 0

H1 Critical Zone

1-2 < d 0 t d 0 t>-t

1-2 d 0 tt/2

= ( 0) ; = 1


27/71

ExampleAn experiment was performed to compare the abrasive wear of two

different laminated materials. Twelve pieces of material 1 were

tested by exposing each piece to a machine measuring wear. Ten

pieces of material 2 were similarly tested. In each case, the depth of

wear was observed. The samples of material 1 gave an average

wear of 85 units with a sample standard deviation of 4, while the

samples of material 2 gave an average of 81 and a sample standarddeviation of 5. can we conclude at the 0.05 significance that the

abrasive wear of material 1 exceeds that of material 2 by more than

2 units?

Assume he populations to be approximately normal with equal

variances.


28/71

Solution

=

(1 2) 01 1 + 1 2

;

=

1 +

2

2

=(

1

1)

12 + (

2

1)

22

(1 +2 2)

= (85 81) 24.4781 12 + 1 10

; = 12+ 10 2

t= 1.04 >1.725 (Critical region)Decision: Do Not Reject H0

=(12

1)42 + (10

1)52

(12+10 2) = 4.478


29/71

Hypothesis concerning

variances of Normal Population

29


30/71

Example Chi Square-test

A manufacturer of car batteries claims that the life of his batteries is

approximately normally distributed with a standard deviation equal

to 0.9 year. If a random sample of 10 of these batteries has a

standard deviation of 1.2 years, do you think that >0.9 year? Use

0.05 level of significance.

30Decision:

2

>0.81


31/71

Hypothesis concerning the equality of

variances of two normal populations

31


32/71


33/71

How to chose a Statistical Test?

33


34/71

34

Design of Experiments


35/71


36/71

36

Terminology

Response variable

Measured output value

E.g. total execution time

Factors

Input variables that can be changed

E.g. cache size, clock rate, bytes transmitted

Levels Specific values of factors (inputs)

Continuous (~bytes) or discrete (type of system)


37/71

37

Terminology

Replication

Completely re-run experiment with same input

levels

Used to determine impact of measurementerror

Interaction

Effect of one input factor depends on level ofanother input factor


38/71

One Way ANOVA (Complete

Randomised Design)

Assumptions & Hypothesis

H0: 1= 2=.= k

H1: atleast two of means are not equalYij = i+ij = + i+ ijH0: 1=2=..=k=0

H1: atleast one of the is is not equal to zero


39/71

k Random Samples

MODEL (One Way ANOVA yij = + i + ij)Treatment: 1 2 3 ------

---

i ----- k

Y11 Y21 --- ----- Yi1 ----- Yk1

Y12 Y22 ---- ------ Yi2 ----- Yk2

::

::

::

::

Y1n Y2n ---- ---- Yin ------ Ykn

Total Y1. Y2. ---- ---- Yi. ---- Yk. y..

Mean: 1. 2. i. k. ..

T l


40/71

Total

variabilitySum of Squares

SST = SSA + SSETotal sum of squares Treatment sum of squares Error Sum of Squares

Degrees of Freedom

(nk-1) (k-1) k(n-1)

ANOVA T bl O W


41/71

ANOVA Table: One Way

ANOVASource of

Variation

Sum of

Squares

Degrees

of

freedom

Mean

Squares

F-value p-value

Treatments SSA k-1 MSA=(SSA

/(k-1))

MSA/SSA - level of

significance

Error SSE k(n-1) MSE=(SSE

/k(n-1))

Total SST nk-1 SST/(nk-1)


42/71

Conclusions-OneWay ANOVA

The null hypothesis H0 is rejected at the -level of significance when

Fcalculated > F [1=(k-1), 2=k(n-1)]

Another approach (p-value)

p- value = at F[1=(k-1), 2=k(n-1)]


43/71

Example- One Way ANOVA

Suppose in an industrial experiment that an engineer is interested in

how the mean absorption of moisture in concrete varies among 5

different concrete aggregates. The samples are exposed to moisture

for 48 hours. It is decided that 6 samples are to be tested for each

aggregate, requiring a total of 30 samples to be tested.

We are interested to make comparisons among 5 populations.

The data are recorded as follows:

Aggregate: 1 2 3 4 5

551 595 639 417 563

457 580 615 449 631

450 508 511 517 522

731 583 573 438 613

499 633 648 415 656

632 517 677 555 679


44/71

Results

Source of

Variation

Sum of

Squares

Degrees

of

freedom

Mean

Squares

F-value p-value

Treatments 85356.47 4 21339.12 4.30 0.0088

(i.e.


45/71

Randomised Complete Block

Design (RCBD)Block 1t2

t1

t3

Block 2t1

t3

t2

Block 3t3

t2

t1

Block 4t2

t1

t3

A typical layout for the randomised complete block design using 3

measurements in 4 blocks

Treatments Blocks: 1 2 3 4

1 Y11 Y12 Y13 Y14

2 Y21 Y22 Y23 Y24

3 Y31 Y32 Y33 y34


46/71

k x b Array for the RCB DesignTreatm

ents

Blocks

:

1 2 --- j ---- b Total Mean

1 Y11 Y12 --- Y1j ---- Y1b Y1. 1.

2 Y21 Y22 -- Y2j ---- Y2b Y2. 2.

: : : : : : : : :

i Yi1 Yi2 ---- Yij ---- Yib Yi. i.: : : : : : : : ; :

k Yk1 Yk2 --- Ykj ---- Ykb Yk. k.

Total Y.1 Y.2 Y.j Y.b Y..

Mean .1 .2 .j .b ..


47/71

Model for RCB Design

Hypothesis

=+++

0:1 =2 =3 = 0

1:


48/71

Sum of Squares

SST = SSA + SSB + SSETotal Treatment Block Error

Sum of Squares Sum of Squares Sum of Squares Sum of Squares

Degrees of Freedom

(bk-1) = (k-1) + (b-1) + (k-1)(b-1)

( ..)2

=1

=1=( . ..)2 + (. ..)2 +

=1

=1 ( . . +..)2

=1

=1


49/71

ANOVA Table RCB Design

Source of

Variation

Sum of

Squares

Degrees

of

freedom

Mean

Squares

F-value p-value

Treatments SSA k-1 MSA=(SSA

/(k-1))

MSA/MSE - level ofsignifican

ce

Blocks SSB b-1 MSB=(SSB

/(b-1))

MSB/MSE

Error SSE (k-1)(n-1) MSE=(SSE

/k(n-1))

Total SST kb-1


50/71

Conclusions

The null hypothesis H0 is rejected at the

- level of significance whenF

calculated> F

[1=(k-1), 2=k(n-1)]

Another approach (p-value)

p- value = at F[1=(k-1), 2=k(n-1)]

E l RCB D i


51/71

Example-RCB DesignFour different machines, M1, M2, M3, and M4 are being considered for the

assembling of a particular product. It is decided that 6 different operators

are to be used in a randomised block experiment to compare the machines.

The machines are assigned in a random order to each operator. The

operation of the machines requires physical dexterity, and it is anticipated

that there will be a difference among the operators in the speed with which

they operate the machine. The amount of time (in seconds) were recorded

for assembling the product:

Test the hypothesis H0, at the 0.05 level of significance, that the machines

perform at the same mean rate of speed.

Machine Operator: 1 2 3 4 5 6

1 42.5 39.3 39.6 39.9 42.9 43.6

2 39.8 40.1 40.5 42.3 42.5 43.1

3 40.2 40.5 41.3 43.4 44.9 45.1

4 41.3 42.2 43.5 44.2 45.9 42.3


52/71

Results- RCB Design (Example)

Source of

Variation

Sum of

Square

s

Degrees

of

freedom

Mean

Squares

F-value p-value

Machines 15.93 3 5.31 3.34 - level ofsignifican

ce

Operators 42.09 5 8.42 8.42

Error 23.84 15 1.59

Total 81.86 23

I t ti b t Bl k &


53/71

Interaction between Blocks &

Treatments


54/71

Latin Square Design (LSD) The randomised block design is very effective for

reducing experimental error by removing one source of

variation

Another design useful in controlling two sources of

variation, while reducing the required number of

treatment combinations is called the LATIN SQUARE

Row Column: 1 2 3 4

1 A B C D

2 B C D A

3 C D A B

4 D A B C

A, B, C, & D represents Treatments


55/71

Model:

Hypothesis:

=++

++

=+ 0: 1 = 2 = = 0

1:

S f S


56/71

Sum of Squares

SST = SSR + SSC + SSTr + SSE

Degrees of Freedom

(r2-1) = (r-1) + (r-1) + (r-1) + (r-1)(r-2)

( )2 = (.. )2

+ (. . )2 + (.. )2 +

( .. . . ..+ 2)2

ANOVA Table LATIN SQUARE


57/71

ANOVA Table LATIN SQUARE

Design

Source of

Variation

Sum of

Squares

Degrees of

freedom

Mean

Squares

F-value p-value

ROW SSR (r-1) MSR=(SSR/(r-1))

COLUMN SSC (r-1) MSC=(SSC/(r-1))

TREATMENTS SSTr (r-1) MSTr=SSTr/(r-1) F=MSTr/MSE - level ofsignificance

Error SSE (r-1)(r-2) MSE=SSE/((r-1)(r-2))

Total SST (r 2-1)

E l L i S D i


58/71

Example Latin Square Design

To illustrate the analysis of a Latin Square design let us return to the

experiment where the letters A, B, C, & D represent 4 varieties ofwheat; the rows represent 4 different fertilizers; and the columns

account for 4 different years. The date in the table are the yields for

the 4 varieties of wheat, measured in kg per plot. It is assumed that

the various sources of variation do not interact. Using 0.05 level of

significance, test the hypothesis H0: there is no difference in the

average yields of the 4 varieties of wheat.

Fertilizer

Treatment

1981 1982 1983 1984

t1 A70 B75 C68 D81

t2 D66 A59 B55 C63

t3 C59 D66 A39 B42

t4 B41 C57 D39 A55

R lt E l L ti S D i


59/71

Results Example Latin Square Design

Source of

Variation

Sum of

Square

s

Degrees

of

freedom

Mean

Squares

F-value p-value

Fertilizer 1557 3 519.00

Year 418 3 139.33

TREATMENTS 264 3 88.00 2.02 - level ofsignificance

Error 261 6 43.50

Total 2500 15


60/71

60

Two-factor Experiments

Two factors (inputs)

A, B

Separate total variation in output values

into:

Effect due to A

Effect due to B

Effect due to interaction of A and B (AB)

Experimental error


61/71

61

Example ??????????????

B (???)

A(??) 1 2 3

1

2

3

4


62/71

62

Two-factor ANOVA

Factor A a input levels

Factor B b input levels

n measurements for each input

combination

abn total measurements


63/71

63

Two Factors, n Replications

Factor A

1 2 j a

FactorB

1

2

i yijk

b

n replications


64/71

64

Two-factor ANOVA

Each individual

measurement is

composition of

Overall mean Effects

Interactions

Measurement errors

errortmeasuremen

BandAofninteractiotodueeffect

Btodueeffect

Atodueeffect

meanoverall...

...

=

=

=

=

=

++++=

ijk

ij

j

i

ijkijjiijk

e

y

eyy


65/71

65

Sum-of-Squares

As before, use sum-of-squares identity

SST = SSA + SSB + SSAB + SSE

Degrees of freedom

df(SSA) = a 1

df(SSB) = b 1

df(SSAB) = (a 1)(b 1)

df(SSE) = ab(n 1)

df(SST) = abn - 1


66/71

66

Two-Factor ANOVA

)]1(),1)(1(;1[)]1(),1(;1[)]1(),1(;1[

222222

2222

Tabulated

Computed

)]1([)]1)(1[()1()1(squareMean

)1()1)(1(11freedomDeg

squaresofSum

ErrorABBA

===

====

nabbanabbnaba

eababebbeaa

eabba

FFFF

ssFssFssFF

nabSSEsbaSSABsbSSBsaSSAs

nabbaba

SSESSABSSBSSA


67/71

67

Need for Replications

If n=1

Only one measurement of each configuration

Can then be shown that

SSAB = SST SSA SSB

Since

SSE = SST SSA SSB SSAB

We have

SSE = 0

Generalized m-factor


68/71

68

Generalized m-factor

Experiments

effectstotal12

nsinteractiofactor-1

nsinteractiofactor-three3

nsinteractiofactor-two

2

effectsmain

factors

m

=

m

m

m

m

m

m

m

Effects for 3

factors:

A

B

C

AB

AC

BCABC

Degrees of Freedom for m-


69/71

69

Degrees of Freedom for m-

factor Experiments df(SSA) = (a-1)

df(SSB) = (b-1)

df(SSC) = (c-1)

df(SSAB) = (a-1)(b-1)

df(SSAC) = (a-1)(c-1)

df(SSE) = abc(n-1) df(SSAB) = abcn-1

Procedure for Generalized


70/71

70

Procedure for Generalized

m-factor Experiments1. Calculate (2m-1) sum of squares terms (SSx)

and SSE

2. Determine degrees of freedom for each SSx

3. Calculate mean squares (variances)4. Calculate F statistics

5. Find critical F values from table

6. If F(computed) > F(table), (1-) confidence thateffect is statistically significant


71/71

Thank you

Date post:	02-Jun-2018
Category:	Documents
Upload:	vivek-sharma
View:	234 times
Download:	0 times

Abid ResearchMethods Vadodra

Documents