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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]
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Processing & Analysis of Data
Measurement of central tendency
Mean
Median
Measurement of dispersion
Range
Variance
Standard deviation
Measurement of skewness
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Central Tendency
3
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4
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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
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Skewness
6
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Measurement of relationship
Correlation (strength of relationship between DV and IV)
Karl Pearsons coefficient of correlation (r)
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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
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Simple Regression Analysis
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Least square estimates of regression line
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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
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Important parametric tests
t- test
Z-test
Chi-square test F-test
ANOVA
ANCOVA
15
T t i th f
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Test concerning the mean of
Normal population
Case for known variance
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Case for unknown variance
t-test
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Test of equality of means of two
normal populations
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Hypothesis Testing
Tests Concerning MEANS
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- known
H 0 : = 0 = ( 0)( )
H1 Critical Zone
< 0 Z 0 Z>-Z
0 ZZ/2
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- unknown
H 0 : = 0
H1 Critical Zone
< 0 t 0 t>-t
0 tt/2
= ( 0)( ) ; = 1
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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
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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)
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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
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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
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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.
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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
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Hypothesis concerning
variances of Normal Population
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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
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Hypothesis concerning the equality of
variances of two normal populations
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How to chose a Statistical Test?
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Design of Experiments
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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)
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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
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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
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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
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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
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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)
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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)]
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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
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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.
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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
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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 ..
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Model for RCB Design
Hypothesis
=+++
0:1 =2 =3 = 0
1:
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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
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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
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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
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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
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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 &
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Interaction between Blocks &
Treatments
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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
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Model:
Hypothesis:
=++
++
=+ 0: 1 = 2 = = 0
1:
S f S
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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
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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
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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
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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
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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
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Example ??????????????
B (???)
A(??) 1 2 3
1
2
3
4
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Two-factor ANOVA
Factor A a input levels
Factor B b input levels
n measurements for each input
combination
abn total measurements
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Two Factors, n Replications
Factor A
1 2 j a
FactorB
1
2
i yijk
b
n replications
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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
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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
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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
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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
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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-
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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
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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
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Thank you