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4th Semester Statistics notes
Bsc psychology (University of Calicut)
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4thSEMESTER
5thSEMESTER
PSYCHOLOGICAL STATISTICS
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PSYCHOLOGICAL STATISTICS
MODULE 1: ANALYSIS OF VARIANCE
⚫ One-way & Two-way Classification with Single Observation per Cell
⚫ Critical Difference
MODULE 2: NON-PARAMETRIC TESTS
⚫ Chi-square Test of Goodness of Fit
⚫ Test of Independence of Attributes
⚫ Test of Homogeneity of Proportions
MODULE 3: SIGN TEST
⚫ Sign Test
⚫ Wilcoxon’s Signed Rank Test
⚫ Wilcoxon’s Rank Sum Test
⚫ Run Test
⚫ Kruskal-Wallis Test
MODULE 4: FACTORIAL DESIGN
⚫ Basics of factorial Design
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⚫ Factorial experiments and their uses in Psychological studies
⚫ Concepts of 22, 23 factorial experiments (without derivation)
MODULE 5: PREPARATION OF QUESTIONNAIRE
⚫ Scores and Scales of Measurement
⚫ Reliability and Validity of Test Scores
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MODULE 1
ANALYSIS OF VARIANCE
Analysis of variance may be defined as a technique which analyses the variance
of 2 or more samples for determining the significance of difference in their
arithmetic means with the help of statistical technique called F-test.
ANOVA is used when simultaneous comparisons are made of measurements
from more than 2 samples.
Assumptions in ANOVA
It is a parametric test and has to satisfy certain assumptions in order to be
usedforstatisticalinference.
• Normality: Populations from which samples have been drawn are
normally distributed.
• Homogeneity: Populations from which samples are drawn have same
variance.
• The observations in the sample are randomly selected from the
population.
• The observations are non-correlated random variables.
• Any observation is the sum of the effects of the factors influencing it.
A. ONE WAY CLASSIFICATION – ONE WAY ANOVA
The term one-factor analysis of variance refers to the fact that a single variable
or factor of interest is controlled and its effect on the elementary units is observed.
In other words, in one-way classification the data are classified according to only
one criterion. Suppose, we have independent samples of n1, n2, n3…..nk observations,
from ‘k’ populations. This population means are denoted by μ1, μ2, μ3…. μ k.
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The one way analysis of variance is designed to test the null hypothesis,
HO: μ1 = μ2 = μ3 =…. μ k
ie., the arithmetic means of the population from which the ‘k’ samples are randomly
drawn are equal to one another.
The steps involved in carrying out the analysis are:
1. Null hypothesis: All sample means are equal.
2. Calculate the variance between the samples: Sum of squares is a measure of
variation. The sum of squares between samples is denoted by SSC. For
calculating variance between sample, we take the total of square of the
variations of the means of the various samples from the grand average and
divide this total by the degrees of freedom. Thus, the steps in calculating
variance between samples will be
a. Calculate the mean of each sample, ie., ̅̅̅,
𝜒1 𝜒
̅̅̅,
̅̅̅𝑘
2 …..𝜒
b. Calculate the grand average 𝜒̿.
̿=
𝝌
̅̅̅̅
̅̅̅̅
𝝌
𝝌𝟐 + …..+ 𝝌
𝟏 + ̅̅̅̅
𝒌
𝑵𝟏 + 𝑵𝟐 +⋯….+ 𝑵𝒌
c. Take the difference between the means of various samples and the
grand average.
d. Square these deviations and obtain the total which will give sum of squares
between the samples and divide the total by degrees of freedom.
The degrees of freedom will be one less than the number of samples.
3. Calculate the variance within the sample: The variance (sum of squares)
within samples measure those inter-sample difference that arise due to
chance only. It is denoted as SSE. For calculating the variance within the
samples, we take the total of the sum of the squares of the deviation of
various items from the mean values of the respective samples and divide this
total by the degrees of freedom. Thus, the step in calculating the variance
within the sample will be:
̅̅̅,
𝜒2 … . . ̅̅̅.
𝜒𝑘
(a) Calculate the mean value of each sample ie., 𝜒
1 ̅̅̅,
(b) Take the deviations of the various items in a sample from the mean
values of the respective samples.
(c) Square these deviations and obtain the total which gives the sum of
squares within the samples.
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(d) Divide this total by degrees of freedom.
4. Calculate the F-ratio
𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒔𝒂𝒎𝒑𝒍𝒆𝒔/𝒈𝒓𝒐𝒖𝒑𝒔
𝑺 𝟐
F = 𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 𝒘𝒊𝒕𝒉𝒊𝒏 𝒔𝒂𝒎𝒑𝒍𝒆𝒔/𝒈𝒓𝒐𝒖𝒑𝒔 = 𝟏 𝟐
𝑺𝟐
5. Compare the calculated value of F-ratio
Compare the calculated value of F with the table value of F for the given
degrees of freedom at a certain critical level. If the calculated value of F is
greater than the table value of F, we reject the null hypothesis and it indicates
that the difference in sample mean is significant or in other words the sample
do not come from the same population.
Short-cut method
The steps are given below:
1. Null hypothesis (Ho): All sample means are equal.
2. Compute Means squares between the sample (MSC) and Means square within
the sample (MSE).
For computing MSC & MSE, following calculations are made.
a) T = Sum of all observations in rows and columns.
b) Total sum of squares of variations,
SST= Sum of squares of all observations – T2/ N
Where, T2/ N = Correction factor
N = Total number of observations.
c) Sum of squares between samples,
SSC= (Σx1 )2 /n1 + (Σx2)2 /n2..….- T2/N.
Where,
Σx1, Σx2… - column total.
n1, n2 …. – Number of observations in each column.
d) Sum of squares within samples, SSE= SST – SSC
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𝑆𝑆𝐶
e) Mean sum of squares between samples, MSC = 𝐶−1
C- Number of columns
𝑆𝑆𝐸
f) Mean sum of squares within samples, MSE= 𝑁−𝐶
𝑀𝑆𝐶
3. Thus, calculate F ratio. F = 𝑀𝑆𝐸
4. Obtain the table value of F for (C-1) and (N-C) degrees of freedom.
5. If the calculated value of F is less than table value, accept null hypothesis.
Otherwise, reject it.
ANOVA Table
Source of
variation
Between
samples
Sum of
squares
SSC
Degrees of
freedom (df)
C-1
Mean sum of
squares (MSS)
𝑆𝑆𝐶
MSC = 𝐶−1
Variance
ratio
SSE
N-C
MSE = 𝑁−𝐶
F = 𝑀𝑆𝐸
SST
N-1
Within
samples
Total
•
•
•
•
•
𝑆𝑆𝐸
𝑀𝑆𝐶
SST- Total sum of squares of variations
SSC- Sum of squares between samples
SSE- Sum of squares within samples
MSC- Mean sum of squares between samples
MSE- Mean sum of squares within samples
Question: Below are given the yield (in kg) per acre for 5 plots of 4
varieties oftreatment. Carryoutananalysisofvarianceandstateyour
conclusion.
Treatment
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Plot
1
2
3
4
5
Total
T1
42
50
62
34
52
240
T2
48
66
68
78
70
330
T3
68
52
76
64
70
330
T4
80
94
78
82
66
400
Answer
Steps:
a) Ho: The sample means are equal.
b) T= 1300
c) SST= Sum of squares of all observations- T2 / N
= 88736- (1300)2/20 = 4236
d) SSC= (Σx1 )2 /n1 + (Σx2)2 /n2...….- T2/N
= 11520 + 21780 + 21780 + 32000 – 84500
= 2580
e) SSE= SST- SSC
= 4236- 2580
= 1656
𝑆𝑆𝐶
f) MSC = 𝐶−1 =
𝑆𝑆𝐸
g) MSE = 𝑁−𝐶 =
h) F =
𝑀𝑆𝐶
𝑀𝑆𝐸
=
860
𝑆𝑆2580
4−1
1656
20−4
= 860
= 103.5
= 8.309
103.5
Source
ofvarian
ce
Between
Samples
Sum
ofsquar
es
2580
Degrees
offreedom(d
f)
4-1=3
Mean
square
Within
1656
20-4=16
103.5
Variance
ratio
860
F =8.309
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Samples
Total
4236
20-1=19
At 0.05 level, table value is 3.24. Thus, table value< calculated
value.Therefore,Horejected.i.e.Thetreatmenthavevariouseffectsonplots
.
B. TWO-WAY CLASSIFICATION – TWO WAY ANOVA
In the two-way classification, the observation are classified into groups on
thebasisof2criteria.
Example:Supposewewanttostudytheyieldofacrop.Thisstudycanbemadebased
ontesting2variables, thetypesoffertilizersanddifferencesinseed.
Here, we apply different kinds of seed on different field and try to find out
thedifferenceoneffectofthesedifferentfertilizeranddifferentseedsonthefield.
Thus, analysis of variance used to test the effect of two factors
simultaneouslyiscalled2-factor ANOVA.
ProcedureforcarryingoutANOVAintwo-wayclassification:
1. Null hypothesis- 2 null hypothesis
Ho: All column means are equal.
Ho: All row means are equal.
2. Compute mean square for column factor, row factor and within samples. i.e.
MSC, MSR and MSE respectively.
a) Find T= Sum of all observations.
b) Calculate Total sum of square,
SST=Sum of squares of all observations-T2/ N
c) Find SSC= (Σx1)2 /n1 + (Σx2)2 /n2..….- T2/N.
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Where, Σx1, Σx2… - column total.
n1, n2 …. – Number of observations in each column
d) Find SSR= (Σx1)2 /n1 + (Σx2)2 /n2..….- T2/N.
Σx1, Σx2… - Row total.
n1, n2 …. – Number of observations in each rows.
e) SSE= SST- SSC- SSR
𝑆𝑆𝐶
f) MSC = 𝐶−1 , C- Number of columns
𝑆𝑆𝑅
g) MSR= 𝑅−1 , R- Number of rows
𝑆𝑆𝐸
h) MSE= (𝐶−1)(𝑅−1)
3. Calculate F ratio
𝑀𝑆𝐶
FC= 𝑀𝑆𝐸
𝑀𝑆𝑅
FR= 𝑀𝑆𝐸
4. Obtain the table value of FC for ((C-1), (C-1)(R-1)) degrees of freedom and
tablevalue ofFR for((R-1),(C-1)(R-1))degreesoffreedom.
5. If FC is less than table value, then all column means are equal. If F Ris less
thantable value,then allrowmeans are equal.
2-WAY ANOVA TABLE
Source of
variance
Between
columns
Sum of
squares
(SSS)
SSC
Degrees of
freedom (df)
Means of squares
(MS)
C-1
MSC = 𝐶−1
𝑆𝑆𝐶
Variance
ratio (F)
𝑀𝑆𝐶
FC= 𝑀𝑆𝐸
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Between
rows
SSR
R-1
Within
samples
(residual)
SSE
(C-1)(R-1)
Total
SST
𝑆𝑆𝑅
MSR= 𝑅−1
𝑀𝑆𝑅
FR= 𝑀𝑆𝐸
𝑆𝑆𝐸
MSE= (𝐶−1)(𝑅−1)
•
•
•
•
•
•
N-1
SST- Total sum of squares
SSC- Sum of squares between columns
SSR- Sum of squares between rows
SSE- Sum of squares due to error
MSC- Mean sum of squares between samples
MSE- Mean sum of squares within samples
Question: 3 varieties of crops A, B, C are tested in a randomized block
designwith4replications.The yields aregivenbelow.
Variety
A
B
C
Total
I
6
7
8
21
Replicati
on
II
4
6
5
15
III
8
6
10
24
IV
6
9
9
24
Total
24
28
32
84
Answer:
1. HO = All column means are
equal.HO=Allrowmeansare
equal.
2. ComputeMSC,MSR,&MSE.
a) T=84
b) SST=Sumofsquares ofallobservations-T2/N
=624- 842/ 12 =36
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c) SSC=(Σx1)2 /n1+(Σx2)2/n2..….- T2/N
=147+75+192+192- 588
=18
d) SSR=(Σx1)2 /n1+(Σx2)2/n2..….- T2/N
=144+196+256-588
=8
e) SSE=SST- SSC–SSR
=36- 18-8 = 10
f)
𝑆𝑆𝐶
MSC = 𝐶−1 =
𝑆𝑆𝑅
18
4−1
8
=6
g)
MSR= 𝑅−1 = 3−1 = 4
h)
MSE= (𝐶−1)(𝑅−1) =
𝑆𝑆𝐸
10
(4−1)(3−1)
= 1.667
3. Compute FC& FR.
𝑀𝑆𝐶
6
FC= 𝑀𝑆𝐸 = 1.667 = 3. 599 ≈ 3.6
𝑀𝑆𝑅
4
FR= 𝑀𝑆𝐸 = 1.667 = 2.3995 ≈ 2.4
4. At0.05levelofsignificance,
ForFC,calculatedvalue(3.6)<tablevalue(4.76)→HOacceptedFor
FR, calculated value (2.4) < table value (5.14) →HO
acceptedTherefore, allcolumn meansandrowmeansare equal.
Source of
Sum of
Degrees of
Means of squares (MS)
Variance
variance
squares
freedom (df)
ratio (F)
(SSS)
𝑆𝑆𝐶
𝑀𝑆𝐶
Between
SSC = 18
C-1 = 3
MSC = 𝐶−1 = 6
FC= 𝑀𝑆𝐸 =
columns
3.6
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Between
rows
SSR = 8
R-1 = 2
Within
samples
(residual)
SSE = 10
(C-1)(R-1) =
6
Total
SST = 36
𝑆𝑆𝑅
MSR= 𝑅−1 = 4
𝑀𝑆𝑅
FR= 𝑀𝑆𝐸 =
2.4
𝑆𝑆𝐸
MSE= (𝐶−1)(𝑅−1) =
1.667
N-1 = 12
C. CRITICAL DIFFERENCE
In ANOVA, if we want to show the significance effect, we would be interested
to find out which pairs of treatment differ significantly. For this, instead of
calculating student‘s t-test for different pairs of treatment, we calculate the
least significant difference at the given level of size. This least difference is
known as the critical difference (CD)
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MODULE 2
NON-PARAMETRIC TESTS
Population: A population is a whole, it’s every member of a group.
Sample: A sample is a fraction or percentage of a group.
Parameter: It is a characteristic of a population. It is a measure based
upon a population.
Statistics: It is a characteristic of a sample. It is a measure based upon a
sample.
Parametric test: A parametric test is one which specifies certain
conditions about the parameter of the population from which a sample is
taken. It makes assumptions about the parameter of the population.
eg: ANOVA, z-test, Student t-test.
Non-parametric test: A non-parametric test is one which does not
specify any conditions about the parameter of the population from which
the sample is drawn. It is also called distribution free statistics.
Eg: Chi-square test, Mann Whitney U test, Wilcoxon Signed Rank test.
Parametric test
Non-parametric test
Information about population
completely known.
No information about population is
available.
Specific assumptions are made
regarding population.
No assumptions are made
regarding population.
Null hypothesis is made upon
parameters of the population
distribution.
Null hypothesis is free from
parameters.
Parametric test statistic is based on
the distribution.
Test statistic is arbitrary
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Applicable only for variables.
Applied for both variable and
attributes.
Powerful if it exist.
Not so powerful like parametric
tests.
CHI-SQUARE TEST (χ2 test)
The chi-square test is the most widely used non-parametric statistical
technique to determine whether there is a significant difference between
the expected/ theoretical frequencies and the observed frequencies. It is
used when the data are expressed in terms of frequencies of proportions or
percentages.
Assumptions of Chi-square test
•
•
•
•
Samples are randomly drawn from the population.
Each observation is independent of all others.
The data are in terms of frequency.
Observed frequencies should not be too small and the sample size ‘n’,
must be sufficiently large.
Application/ Uses of Chi-square test
• Chi-square test may be used as a test of equal probability hypothesis:
By equal probability hypothesis we means the probability of having
the frequencies in all the given categories are equal.
Example: 100 students answer an item in an attitude scale. The items
have 5 categories of response options: Strongly agree, Agree, Neutral,
Disagree and Strongly disagree. According to equal probability
hypothesis, the expected frequency of response given by 100 students
would be 20 in each.
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• Use of Chi-square test as a test for goodness of fit: It is used to
determine whether the distribution of cases in a single categorical
variable follows a known/ hypothesized distribution.
• Chi-square test in testing the significance of independence
hypothesis: The chi-square test is used to determine if there is a
significant relationship between 2 categorical variables.
Limitations of Chi-square test
• Its use is restricted by certain conditions like total of the frequencies
should not be less than 50.
• It is not as reliable as a parametric test hence it should be used only
when parametric test cannot be used.
• It is not possible to compute chi-square value when the given values
are in proportion or in percentage.
A. CHI-SQUARE TEST FOR GOODNESS OF FIT
Suppose we are given a set of observed frequencies obtained under some
experiments and we want to test whether the experimental result support a
particular hypothesis/ theory. Karl Pearson in 1900 developed a test for
testing the significance of discrepancy between experimental values and the
theoretical values obtained under some theory of hypothesis. This test is
known as chi-square test of goodness of fit and is used to determine
whether there is a significant difference between the frequencies of
occurrence of observation in a given sample and expected frequencies
obtained from this specified distribution.
Consider a set of possible events 1, 2, 3…..n arranged in ‘n’ classes or cells.
Let these events occur with frequencies O 1, O2, O3….On called the observed
frequencies. Let them be expected to occur using the concept of probability
with frequencies E1, E2, E3…..En called the expected frequencies.
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A measure of discrepancy existing between the observed and expected
frequencies can be found by using the test statistics χ2 and is given by:
χ2 =
𝟐
∑𝒏
𝒊=𝟏(𝑶𝒊 − 𝑬𝒊 )
𝑬𝒊
n = Total number of frequencies
𝑛
𝑛
𝑖=1
𝑖=1
∑ 𝑂𝑖 = ∑ 𝐸𝑖
Due to this, the distribution losses 1 degrees of freedom; df = n-1
The expected frequencies are calculated on the supposition that the data
obey certain law of distribution, such as, for example, normal, binomial,
poison.
If chi-square is equal to 0, the observed frequencies and expected
frequencies will coincide and this shows that there is perfect agreement
between theory and observation.
The following are some points to be considered while using chi-square test:
1. In a set of observation which are subject to linear constraints
∑𝑛𝑖=1 𝑂𝑖 ∑𝑛𝑖=1 𝐸𝑖 , 1 degrees of freedom is lost. Therefore, the application
of chi-square with ‘N’ cell of frequencies, the df is n-1.
2. If you have to estimate the parameter (P for Binomial distribution, λ
for poison, μ and σ2 for normal distribution), then one more degrees
of freedom has to be subtracted from each parameter estimated.
3. Sample size should be large, say more than 50.
4. If any of the expected cell frequencies is less than 5, we combine or
pool it with proceeding succeeding frequency, so that the result in
frequency is greater than 5. In this case also 1 df has to be adjusted.
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5. If the classification is in the form of a contingency table (2 way table)
and if there are ‘r’ rows ‘c’ columns and no parameters are estimated
the df is (r-1) × (c-1).
6. Chi-square formula can also be written as
χ2 =
∑𝒏
𝒊=𝟏 𝑶𝒊
𝑬𝒊
−𝑵
Question: 4 coins are tossed 80 times. The distribution of number of
heads are given below:
Heads: 0, 1, 2, 3, 4. Total = 10
Frequencies: 4, 20, 32, 18, 6. Total = 80
Apply chi-square test at 1% level of significance if the coin is unbiased.
Answer:
1
1
n=4, N=80, p=𝑞, q=𝑝
HO: There is no significance difference between 𝑂𝑖 𝑎𝑛𝑑 𝐸𝑖 .
f (x) = nCx px qn-x; x=0,1,2,3,4
E(O) = 4C0 (½)0 (1/2)4-0
= 0.0625 × N
= 0.0625 × 80 = 5
E(1) = 80 × 4C1 (½)1 (1/2)3 = 20
E(2) = 80 × 4C2 (½)2 (1/2)2 = 30
E(3) = 80 × 4C3 (½)3 (1/2)1 = 20
E(4) = 80 × 4C4 (½)4 (1/2)0 = 5
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𝑶𝒊
(𝑶𝒊 − 𝑬𝒊 )𝟐
𝑬𝒊
4
20
32
18
6
Total - 80
5
20
30
20
5
Total - 80
χ2 =
1
0
4
4
1
2
∑𝑛
𝑖=1(𝑂𝑖 − 𝐸𝑖 )
𝐸𝑖
(𝑶𝒊 − 𝑬𝒊 )𝟐
𝑬𝒊
1/5 = 0.2
0/20 = 0
4/30 = 0.133
4/20 = 0.2
1/5 = 0.2
Total – 0.733
= 0.733
Df = 4 at 1% level of significance (table) = 13. 277
𝑋 2 (∝) = 𝑋 2 (4) = 13.277 (1% = 1/100 = 0.01)
Here, Critical Value > Table Value. Therefore, accept H 0.
Therefore, there is no significant difference between 𝑂𝑖 𝑎𝑛𝑑 𝐸𝑖 .
Question: Fit a poison distribution to the following data test the goodness
of fit.
X : 0, 1, 2, 3, 4, 5, 6
Frequency : 275, 72, 30, 7, 5, 2, 1
Answer:
H0: There is no signicant difference between 𝑂𝑖 𝑎𝑛𝑑 𝐸𝑖 .
N = Σf = 392
∑ 𝑥𝑓
λ= 𝑁
x
f
xf
0
275
0
1
72
72
2
30
60
3
7
21
4
5
20
5
2
10
6
1
6
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∑ 𝑥𝑓
189
E(X) = N
𝑒 −𝜆 𝜆𝑥
λ = 𝑁 = 392 = 0.48
𝑥!
E(0) = 392 ×
𝑒 −(0.48) (0.48)0
= 242.5 ≈ 243
E(1) = 392 ×
𝑒 −(0.48) (0.48)1
= 116.4
E(2) = 392 ×
𝑒 −(0.48) (0.48)2
= 27.94 ≈ 28
E(3) = 392 ×
𝑒 −(0.48) (0.48)3
= 4.47 ≈ 5
E(4) = 392 ×
𝑒 −(0.48) (0.48)4
= 0.53 ≈ 1
E(5) = 392 ×
𝑒 −(0.48) (0.48)5
=0
E(6) = 392 ×
𝑒 −(0.48) (0.48)6
=0
0!
1!
2!
3!
4!
5!
6!
𝑶𝒊
𝑬𝒊
275
72
30
7
5
2 15(pooled)
1
243
116
28
5
1
0 5
0
𝑋 2 (∝) = 𝑋 2 (0.05) (2) = 5.991
(𝑶𝒊 − 𝑬𝒊 )𝟐
𝑬𝒊
4.213
16.689
0.142
20
41. 044
∑ 𝐸𝑖
Df = 7-1-1-3 = 2
𝑋 2 >𝑋 2 (𝛼)
(𝑶𝒊
− 𝑬𝒊 )𝟐
1024
1936
4
100
= 392
Therefore, we reject Ho
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Therefore, there is significant difference between 𝑂𝑖 𝑎𝑛𝑑 𝐸𝑖 .
Therefore, Poison distribution is not a good fit for given data.
B. TEST OF INDEPENDENCE OF ATTRIBUTES
We shall now apply chi-square test for testing some hypothesis which can
be formulated in the case of contingency table. In general, number of
classification of variable may be larger, but we consider only the case of 2
way classification. Such pairs of classification system may be gender versus
educational level, smoking habit versus cause of death and so on. And the
data will be given in 2 × 2 contingency.
Let there be 2 attributes A and B. A is divided into ‘m’ classes and B is
divided into ‘n’ classes. The various cell frequency can be represented in the
following table having ‘m’ rows and ‘n’ columns.
A/B
1
2
3
:
:
I
:
:
m
Total
1
𝑓11
2
𝑓12
𝑓31
𝑓32
𝑓21
𝑓𝑖1
𝑓𝑚1
𝑓.1
3
𝑓13
4……
𝑓14
j…..
𝑓1𝑗
𝑓33
𝑓34
𝑓3𝑗
𝑓22
𝑓23
𝑓𝑖2
𝑓𝑖3
𝑓𝑚2
𝑓𝑚3
𝑓.2
𝑓.3
𝑓24
𝑓𝑖4
𝑓𝑚4
𝑓.4
n
𝑓1𝑛
Total
𝑓1.
𝑓3𝑛
𝑓3.
𝑓2𝑗
𝑓2𝑛
𝑓𝑖𝑗
𝑓𝑖𝑛
𝑓𝑚𝑗
𝑓.𝑗
𝑓𝑚𝑛
𝑓.𝑛
𝑓2.
𝑓𝑖.
𝑓𝑚.
Grand
total
Such a table is called m × n contingency table. We test the hypothesis if that
A and B are independent i.e., there is no association or relationship
between A and B.
The test statistics is given by:
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χ2 = ∑ 𝒎
𝒊=𝟏
∑𝒏𝒋=𝟏
(𝑶𝒊𝒋 − 𝑬𝒊𝒋 )𝟐
𝑬𝒊𝒋
The value of χ2 is obtained by referring the χ2 table for (m-1) (n-1) df. If the
calculated value of χ2 is greater than the table value, we reject HO.
Otherwise accept it.
Question: From the following table shown, the number of plans having
certain characters, test the hypothesis that the flower color is independent
of flatness of leaves.
Flower color/ Leaves
Flat leaves
Curved leaves
Total
99
20
119
36
5
41
135
25
160
White flower
Red flower
Total
Answer:
HO: The flower color and flatness of leaves are independent.
Expected frequency =
𝐸(99) =
135 ×119
135 ×41
= 34.5937 ≈ 35
𝐸(20) =
25 ×119
= 18.5937 ≈ 19
𝐸(36) =
𝐸(5) =
160
160
160
25 ×41
160
𝑹𝒐𝒘 𝒕𝒐𝒂𝒕𝒂𝒍 ×𝑪𝒐𝒍𝒖𝒎𝒏 𝒕𝒐𝒕𝒂𝒍
𝑮𝒓𝒂𝒏𝒅 𝒕𝒐𝒕𝒂𝒍
= 100.4062 ≈ 100
= 6.4062 ≈ 6
𝑶𝒊𝒋
99
36
20
5
𝑬𝒊𝒋
100
35
19
6
(𝑶𝒊𝒋 − 𝑬𝒊𝒋 )𝟐
1
1
1
1
(𝑶𝒊𝒋 − 𝑬𝒊𝒋 )𝟐
𝑬𝒊𝒋
1/100 = 0.01
1/35 = 0.02
1/19 = 0.05
1/6 = 0.16
0.24
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Df = (m-1) × (n-1) = (2-1) (2-1) = 1
𝑋 2 (∝) = 𝑋 2 (0.24) (1) = 3.841 (table value)
𝑋 2 = 0.24
𝑋 2 (𝛼) >𝑋 2
Therefore, we accept HO
Therefore, the flower color is independent of flatness of leaves.
YATES CORRECTION
When we have a small sample of cases such that the least expected
frequency in any cell of the 2×2 contingency table with df=1 is less than 5,
computation in the usual way of 𝑋 2 gives an over estimate of the true value.
As a result, we may reject some hypothesis which in fact should not be
rejected. This problem arising particularly in 2×2 table with df=1 can be
avoided by using a correction called Yate’s correction.
That is, if any of the cell frequency is less than 5, then a correction is
applied in computing chi-square, which is known as Yates correction. It is
done by subtracting 0.5 from the difference of observed and expected
frequencies for each category or cell. Thus the formula for computing chisquare after Yates correction becomes:
χ2 =
𝟏
𝑵 ×[|𝑨𝑫−𝑩𝑪|−𝟐 𝑵]𝟐
(𝑨+𝑩)(𝑪+𝑫)(𝑨+𝑪) (𝑩+𝑫)
Question: Consider the 2 by 2 contingency table. Apply 𝑋 2 at 5% level of
significance.
𝐵1
𝐵2
Total
𝐴1
7
6
13
𝐴2
1
8
9
Total
8
14
22
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Answer:
HO: A and B are independent.
Here, second self frequency is less than 5. So we apply Yales correction.
1
𝑁 ×[|𝐴𝐷−𝐵𝐶|−2 𝑁]2
χ2 = (𝐴+𝐵)(𝐶+𝐷)(𝐴+𝐶)
1
(𝐵+𝐷)
χ2 =
22 ×[|7×8−1×6|−2 × 22]2
χ2 =
22 ×[|50|−11]2
(7+1)(6+8)(7+6) (1+8)
13.104
= 2.55
df = (2-1) (2-1) = 1
χ2 = 2.55
𝑋 2 (∝) = 3.841
𝑋 2 (𝛼) >𝑋 2
Therefore, we accept HOie., A and B are independent
C. TEST OF HOMOGENEITY OF PROPORTIONS
The chi-square test of homogeneity is made to determine whether several
populations are similar or equal or homogeneous in some characteristics. It
is used to determine whether frequency counts are distributed identically
across different population.
Question: In a study of television viewing habits of children, a
developmental psychologist selects a random sample of 300 first graders
and was asked which of the following TV programs they like best. Use 0.05
level of significance and find both boys and girls preferences differ
significantly.
Boys
Girls
Viewing preferences
A
B
C
50
30
20
50
80
70
Row total
100
200
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Column total
100
110
90
300
Answer:
HO: There is a significant difference between preferences of boys and girls.
Expected frequency =
Boys A =
100 ×100
= 33.33
Boys B =
110 ×100
= 36.667
Boys C =
90 ×100
Girls A =
100 ×200
= 66.667
Girls B =
110 ×200
= 73.33
Girls C =
90 ×200
300
300
300
300
300
300
𝑶𝒊𝒋
50
30
20
50
80
70
𝑅𝑜𝑤 𝑡𝑜𝑎𝑡𝑎𝑙 ×𝐶𝑜𝑙𝑢𝑚𝑛 𝑡𝑜𝑡𝑎𝑙
𝐺𝑟𝑎𝑛𝑑 𝑡𝑜𝑡𝑎𝑙
= 30
= 60
𝑬𝒊𝒋
33.33
36.667
30
66.667
73.33
60
(𝑶𝒊𝒋 − 𝑬𝒊𝒋 )𝟐
277.7
44.4
100
277.7
44.4
100
(𝑶𝒊𝒋 − 𝑬𝒊𝒋 )𝟐
𝑬𝒊𝒋
8.33
1.212
3.33
4.16
0.606
1.66
19.312
Df = (3-1) (2-1) = 2
𝑋 2 (∝) =5.991
Since, calculated value > critical value, HO may be rejected.
ie., There is a significant difference between preferences of boys and girls.
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MODULE : 3
SIGN TEST
A. SIGN TEST
It is the simplest of non parametric test. It's name comes from the fact that
it is based on the direction (or signs). For plus or minus of a pair of
observation and not on their numerical magnitude. The sign test can be of
two types
1. One sample sign test
2. Two sample sign test (paired sampling test)
(sign test is a statistical method to test for consistence differences between
pairs of observations)
1.ONE SAMPLE SIGN TEST
One sample sign test : on this test the null hypothesis that Ho:μ=μo
against the alternate hypothesis on the basis of random sample size 'n', we
replace each sample value greater than μo with a + sign and each sample
value less than μo with a - sign. If the sample value happens to be equal to
μo, we do not assign any sign. After doing this we find the proportion of +
signs out of the total number of + and - signs are values of a random
variable having the binomial distribution with P=½
Ho:P=½ against H1:P≠1/2
The test statistic is P' - P
√pq/n
Where P' is the proportion of + signs out of total signs and P=½ and q=½.
If the test statistic is less than table value we accept Ho, otherwise reject it.
When Ho is accepted we conclude that the sample belongs to the
population so that the given population mean is true.
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Q. A 4 round golf play scores of 11 professional are
202,210,200,203,193, 203,204, 195,199, 202,201apply sign test
at 5% level of significance to test the Hothat professional golfers
average is 204?
Ans: We have To Test, Ho:P = 1/2 V/S H1: P= 1/2
X
SIGNS
202
-
210
+
200
-
203
-
193
-
203
-
204
No sign
195
-
199
-
202
-
201
-
The test statistic is
Z= P' - P
√pq/n
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Where p'=1/10 α=0.05
Zα=1.645
Zα/2=1.96
Z=1/10 - ½
√1/4 ÷10
Z=-2.5
Given α=0.05 then Zα/2=1.96
|Z|>Zα/2
2.5 >1.96
We reject Ho , Golfer's average is not 204
Q. The following are the measurement of breaking the strength of a certain
kind of 2 inch cotton ribbon.
163,165, 160,189, 161,171, 158, 151,169, 162,163,139, 172,165, 148,166,
172,163, 187,173.
use signtest to test Ho:μ=160 v/s H1:μ>160 at 5% level of significance
Ans: We have to test Ho:P=½ v/sH1:P≠1/2
X
SIGN
163
+
165
+
160
No sign
189
+
161
+
171
+
158
-
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151
-
169
+
162
+
163
+
139
-
172
+
165
+
148
-
166
+
172
+
163
+
187
+
173
+
The test statistic is
Z=P' - P
√pq/n
where P'=15/19
Z= 15/19 - ½
√14/19÷19
=2.5
Given α=0.05 then Zα = 1.645
Z>Zα
2.5>1.645
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We reject Ho, The average is not 160
2.TWO SAMPLE SIGN TEST (PAIRED SIGN TEST)
Suppose x and y are 2 variables and their 'n' values are known, then we get
'n' pairs of values. 1st value of each pair being a value x and 2nd is that of y.
That is (x1, y1) is a pair were x1 belongs to x and y1 belong to y. In such
problems each pair each pair can be replaced by + or - sign. If both value
are equal to the concerning pair is disregarded. Then we proceed in some
manner as in one sample sign test.
Q. The following are number of tickets issued by two salesman on 11 days.
1st salesman: 7,10,14,12,6,9,11,13,7,6,10
2nd salesman: 10,13,14,11,10,7,15, 11,10,9,8
Use the sign at 1% level of significance to test the hypothesis that the
average the 2 salesman use. Issue equal number of tickets
Ans: Ho:P=1/2 v/s H1:P≠1/2 Now P'=2/10 = ⅖
The test statistic
Z = P' - P
|z|=0.632
At 1% level of significance Zα/2=2.58 |Z| = Zα/2
We accept Ho , The α salesman issued equal tickets
Q.To determine the effectiveness of traffic control system the number of
accident that occurred 12 dangers in the selection during 4 weeks before
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and 4 weeks after, the installation of the new system were observed and
following data were obtained.
X = 3,5,2,3,3,3,0,4,1,6,4,1
Y=1,2,0,2,2,0,2,3,3,4,1,0
Ans: Ho:P=½ v/s H1:P≠1/2
Now P'=10/12 = ⅚
The test statistic z=P'- P
√pq/n
z = 5/6 - 1/2 = 2.2
√1/4÷12
α=0.05Zα=1.96|z|>Zα/
2
2.2>1.96
we rejectHo.
B. WILCOXEN'S SIGNED RANK TEST / WILCOXON'SMATCHED
PAIR TEST
In the case of 2 related samples one significance of difference we can use
the Np test known as wilcoxen's signed rank test.
When this technique is employed we first find the difference between each
pair of values and assigned ranks to the difference from the smallest to
largest without regarding the sign. The actual sign of each different are then
put to the corresponding ranks and test the statistics. Then test statistic is
calculated.
T is smaller of 2, namely the sum of - ve ranks and the sum of +ve ranks
Case 1 : When the given number of matched pairs is less than or equal to
25, we use the table (wilcoxen's sign rank test table). When T is greater than
or equal to the table value we accept that there is difference (we reject Ho).
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Note 1: By applying this test we may have situation where some matched
pairs are equal. That is difference between values zero in this case we broke
out these pairs
Note 2: Sometimes two or more pairs have some difference, in such case
we assign average of ranks to each of these pairs.
Case 2: when N>25, that is given number of matched pairs exceeds 25,
we apply 'z' test. The test statistic is
Z=T-μ σ
Where μ=n ( n+1 )
4
Q. Given below is 16 pairs of values showing the performance of 2 machines
A and B. Test whether there is difference between performance at 5% level
of significance.
A
73 43 47 53 58 47 52 58 38 61 56 56 34 55 65 75
B
51 41 43 41 47 32 24 58 43 53 52 57 44 57 40 68
A
B
A-B
Rank
+
73
51
22
13
13
43
41
2
2.5
2.5
47
43
4
4.5
4.5
53
41
12
11
11
58
47
11
10
10
47
32
15
12
12
-
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52
24
28
15
15
58
58
0
-
-
38
43
-5
6
61
53
8
8
8
56
52
4
4.5
4.5
56
57
-1
1
-1
34
44
-10
9
-10
55
57
-2
2.5
-2
65
40
25
14
25
75
68
7
7
7
-5
T+ = 101.5 (sum of +ve ranks)
T - = 18.5 (absolute value of sum of -veranks)
T = smaller value of T+ and T=18.5 .
The table value at 5%level significanceis 25
T ≯ Tα 18.5≯25
We accept Ho, There is no difference between A
and B
C. WILCOXEN'S RANK SUM TEST / MANN WHITNEY U TEST
This is a two sample sign test as an alternative to the T test when
assumptions in test above the population are not made.
Here we want to to test whether the population are identical. i.e we want to
test the two samples have come from two identical population. The test
statistic is
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Z=μ-U
SE
Where μ=n1n2
2
U=n1n2+ n1 ×(n1+1) - R1
2
Where R1 denotes the sum of ranks of values of first sample in a combined
order sample
SE=
Q. There are two sample first containing the observation.
Sample 1: 54,39,70, 58,47,40, 74,75,61,79 and second contains
Sample 2: 45,41,62,53,33,45,71,42,68, 73,54,73
Apply rank sum test to test at 5% level of significance that they came
population will be the same reason
Here we have to test , Ho:samples come from the population within the
same mean
Values in ascending Rank
Sample
I
II
33
1
II
1
2
39
I
2
3
40
I
3
4
41
II
4
5
42
II
5
6.5
45
II
6.5
6.5
45
II
6.5
8
47
I
8
9
49
I
9
10
53
II
10
11.5
54
I
11.5
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54
58
61
62
68
70
71
73
73
74
74
75
79
11.5
13
14
15
16
17
18
19.5
19.5
21.5
21.5
23
24
II
I
I
II
II
I
II
II
II
I
I
I
I
11.5
13
14
15
16
17
18
19.5
19.5
21.5
21.5
23
24
167.5
132.5
Z=μ-U SE
μ=n1n2 = 12 ×12 = 72
22
U=n1n2+ n1 ×(n1+1) - R1
2
= 12 × 12 +
12×(12+1)
=54.5
2
- 167.5
SE= √ n1n2(n1+n2+1)
SE=
= 17.32
Z=μ-U
SE
= 72—54. 5 =1.01
17.32
Table value = 1.96 , CV table value , We accept Ho
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D.
KRUSHKAL WALLIS H TEST(1952)
If several independent samples are involved, Analysis of variation(
ANOVA)is the usual procedure.when the assumptions of anovaare not met,
an alternative technique is developed called the krushkal wallis one way
anova or h test. This test helps in testing the null hypothesis that 'k'
independent random samples come from identical populations against the
alternative hypothesis that the mean of these samples are not equal. As
done in the 'Mann - Whitney U test' all data are ranked as ifthey were in
one sample, from lowest to highest,the rank some ofeach sample is
calculated from the formula.
𝟏𝟐
𝑹𝟏𝟐
H = 𝑵 (𝑵+𝟏) 𝒏𝟏 +
𝑹𝟐𝟐
𝑹𝒏𝟐
+ 𝒏𝒌 - 3(N+1)
𝒏𝟐
where n1,n2 …….nk are no.of in each of ksamples.
N= n1+n2+. ...+nk
R1,R2….Rk are the rank sum of each sample.if there are tiesusual
procedure as follows,so for small samples, H isapproximately distributed
as χ2 with (k-1) as degrees of freedom.If the null hypothesis is true each
sample has atleast 5observations.the sampling distribution of H can
beapproximately closely with χ2 distribution. If any sample has less than
5 items,the χ2 approximation cannot be used,in this case the test must be
based on special tables.
Q: A companies training are randomly assigned groups which taught a
certain industrial instruction procedure 3 diff methods. At the end of the
instruction period they are tested for inspecting performance equality.
the following are their scores:Method A: 80, 83, 79 ,85, 90 ,68
Method B: 82,84,60,72,86,67,91
Method C: 93,65,77,78,88 ,
use the H test to determine at 5% level of significance whether the 3
methods are equally effective.
ans:-here we defined
H0:There is no significance difference in performance.
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H1: There is significance difference in performance.
samples
Rank
60
1
65
2
67
3
68
4
72
5
77
6
78
7
79
8
80
9
82
10
83
11
84
12
85
13
86
14
88
15
90
16
91
17
93
18
Method A:9,11,8,13,16,4
Method B:10,12,1,5,14,3,17
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Method C:18,2,6,7,15
R1=61
R2=62
R3=48
n1=6
n2=7
n3=5
N=18.
𝟏𝟐
H = 𝑵 (𝑵+𝟏)
𝟏𝟐
𝑹𝟏𝟐
𝒏𝟏
H = 𝟏𝟖 (𝟏𝟖+𝟏)
𝟔𝟏𝟐
2
𝟔
𝑹𝟐𝟐
𝑹𝒏𝟐
𝟔𝟐𝟐
𝟒𝟖𝟐
+ 𝒏𝟐 + 𝒏𝒌 - 3(N+1)
+ 𝟕 + 𝟓 - 3(18+1)
2
2
=0.1968
Table value= 5.119
df=N-1 = 18-1=17
c.v ≯ 5.119 , so,we accept it.
such that there is significant difference between H0.
E.RUN TEST
Run test is a non parametric statistical test that checks a randomness
hypothesis for a 2 value data sequence.more precisely it can be used to test
the hypothesis that the elements of the sequence are mutually independent.
The run test can be used to find any diff exist in a simple sample or between
2 independent sample drawn from the population.
1.PROCEDURE FOR USING RUN TEST
●case -1
A single and small sample, a coin was tossed 20 times and the result
obtained were ;
HTHHTHHHTTHTTTHTHTHH
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the hypothesis that the win is not in order.
●step -1
H0: head and tail,the coin is erratic
HTH HTH H HT THT T THTHTH H
1 2 3
4
5
6 7
8
9 10 11 12 13
total no.of runs:13 r=13 here we show the procedure for smallsamples.
N1:no.of heads or first events=11
N2:no.of fails or second events=9
N= N1+N2= 11+9 =20
construct the table 01 and O2 given in the appendix to the find out the wave
and higher value of critical 'r' we significance at 0.05 (5%) level.
01
r13
02
6
accept
16
01=6 (lowest)
if 'r' falls between the critical value cannot be regarded as significant.so we
accept H0 and conclude that the coin was not erratic.
note:- when a sample size is termed small when the no of observations are
less than or equal to 20. ie 20- small sample.
≤20- small sample less than 20.
Q: 10 boys and 10 girls class +1 are selected from higher secondary school
respectively.where examining interns of their attitude scale are shown in
the below table.test the hypothesis of boys and girls intern of their attitude
towards the population of education.
Attitude of boys:15,6,7,19,12,4,20,5,18,10
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Attitude of girls:17,9,16,15,13,3,8,14,11,12
H0:there is no difference between the attitudes of boys and girls in
ascending order.
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
15
15
16
16
17
17
18
18
19
19
20
20
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GGBBBBGGBGBGGGBGGBBB
1
2345678
9 10
r= 10
CV= 10
01
6
02
16
The CV here are (10)falls between the critical value.cannot be regarded as
significant.so we accept H0.
There is no significant difference between attitude of boys and girls.
RUN TEST IN LARGE SAMPLE
if the members of samples are divided only in 2 categories and the sample
size is atleast 1 through ie,n1 or n2 is large the sampling distribution of run
'r' follows a no with mean and SD are given by
2n1+n2
Mean(Mr)
+1
n1+n2
SD=√2 n1n2(2n1n2-n1-n2)
( n1+n2)²(n1+n2-1)
Z=r-Mr
SD
The null hypothesis is using the test statistic as above equation.The
normal probability area shown in runs table can be used to find the critical
value of z. if the the cv is greater than critical value we reject H0.
Q: In tossing a coin 50 times following and the sequence of outcome.
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HHTTTTHTTHHHTTHTTTTTHHTHTTTHHHHH
1
2
3 4
5
6 7
8
9 10 11 12
13
TTTTTTHTTHHHHHHTTT
14 15 16 17
18
Apply the run test whether the coin is unbiased in other words can it be
concluded that the occurrence of it and random in nature.
H0: H and T are random in nature.
r = 18
N1=22
N2=28
2𝑛1𝑛2
Mr= 𝑛1+𝑛2+1
2(22 × 28)
= 22 + 28 +1
= 25.64
SD = √2 n1n2 (2 n1n2-n1-n2)
(n1+n2)²(n1+n1-1)
=3.447
Z=r-Mr
SD =
18−25.64
3.447
= -2.216
|x| 2.21
2.21 >1.96 Tv= 1.96
Cv>Tv
so we reject H0. , H and T are not random in nature.
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MODULE - 4
FACTORIAL DESIGN
A. BASICS OF FACTORIAL DESIGNS
⚫ A factorial design is an experimental study in which two or more
categorical variables are simultaneously manipulated or observed in
order to study their joint influence (interaction effect) and separate
influences (main effects) on a separate dependent variable.
⚫ Factorial experiment is organized when the effect of more than one
factor is investigated on the dependent variable simultaneously.The
factorial experiment is represented by m', ('=P) where "p' and 'm'
are number of factors and number of levels, respectively.
⚫ The factorial experiments are represented by mp where ‘p’ be the
number of factors and ‘m’ be number of levels.
Example :
1. In a 2² factorial experiment has two factors, each at two levels.
Thus there are four treatment conditions.
2. In a 32 factorial experiment, it has two factors each at three level,
so there are 9treatment conditions.
⚫ In example 1, if A and B are two factors each having two levels, then
the factorial experiment will have four treatments, namely A1B2,
A1B2, A2B1 and A2B2. “In a factorial design, each level of one
independent variable is combined with each level of the other to
produce all possible combinations. Each combination, then, become
a condition in the experiment.
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ADVANTAGES OF FACTORIAL DESIGN
⚫ It increases the scope of the experiment and its inductive value and
it does so mainly by giving information, not only on the main factors
but on their interaction.
⚫ The various levels of one factor constitute replication of other factors
and increase the amount of information obtained on all factors.
⚫ Factorial designs allow additional factors to be examined at no
additional cost.
⚫ Factorial designs are more ethical than single factor experiments
⚫ In the factorial design the effect of a factor in estimated at several
levels of other factors and the conclusion hold over a wide range of
conditions.
B. FACTORIAL EXPERIMENTS& USES IN PSYCHOLOGICAL
STUDIES
⚫ A factorial design is necessary when interactions may be present to
avoid misleading conclusions. Factorial designs allow the effects of a
factor to be estimated at several levels of the other factors, yielding
conclusions that are valid over a range of experimental conditions.
⚫ Factorial design involves having more than one independent
variable, or factor, in a study. Factorial designs allow researchers to
look at how multiple factors affect a dependent variable, both
independently and together. Factorial design studies are named for
the number of levels of the factors.
C. 2² FACTORIAL DESIGN
⚫ In 2²factorial design, we have two factors, each at two level (0,1) so
that there are 2 x 2 = 4treatment combinations. For the analysis of
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factorial design, we use the yates’ notions. Let A and B denote the
names of the two factors under study and let a, b denote the levels of
the factors.
⚫ The following are the treatment combinations:
a0b0 or 1: A, B at the first level
a1b0 or a: A at the 2nd level and B at the 1st level
a0b1 or b: A at the 1st level and B at the 2nd level
a1b1 or ab: A and B both at 2nd level
⚫ The factor treatment combinations can be compared by laying out
the experiment as a two –way ANOVA with r – replications, each
replicate containing 4 units.
1. YATES METHOD FOR 2² EXPERIMENT
1.
TREATMENT
2.
TOTALYIELD
3.
4.
5.
1
[1]
[1] + [a]
[1] + [a] + [b] + [ab]
G
a
[a]
[b] + [ab] [ab] - [b] + [a] – [1]
[A]
b
[b]
[a] – [1]
[ab] + [b] - [a] – [1]
[B]
ab
[ab]
[ab] – [b] [ab] – [b] – [a] + [1]
[AB]
STEP 1: In the first column, write the treatment combinations 1, a, b, ab.
STEP 2: Against each treatment combinations, write the corresponding
totalyields from all the replications.
STEP 3: The entries in the third column is split into levels. The first half is
obtained by writing down in order, the pairwise sums of the
valuesin column 2 and the second half is obtained by writing in
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the sameorder the pairwise differences of the values in the
second column.
STEP 4: To complete the 4th column, the whole procedure in step 3 is
repeated on column 3.
D. 2³ FACTORIAL DESIGN
⚫ In 2³factorial experiment, we consider 3 factors A, B, and C, each at
two levels say (a0, a1),(b0, b1) and (c0, c1) respectively so that there
are 2³ = 8 treatment combinations. The eight treatment
combinations are, 1, a, b, ab, c, ac, bc, abc (or) a0b0c0, a1b0c0, a0b1c0,
a1b1c0, a0b0c1, a1b0c1, a0b1c1, a1b1c1.
⚫ A 2³factorial experiment can be performed as a one – way ANOVA
with 8 treatments or two –way ANOVA with r replications.
In a 2³factorial experiment, we split up the treatment s.s with 7 df into 7
components corresponding to the 3 main effects A, B, and C; three 2 factor
interaction AB, AC, and BC and one 3 factor interaction ABC each carrying
1 df.
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ANOVA TABLE
Q: An Experiment was planned to study the effect of potash and super phosphate on the yield of potatoes.All the combination of 2level of potash
and super- phosphate were studied in Randomized block design with 4
replications for each.Analyse the data and give year conclusions.
Blocks
Yields
I
(1)
23
K
25
P
22
kp
38
II
p
40
(1)
26
k
36
kp
38
III
(1)
29
k
20
kp
30
p
20
IV
kp
34
K
31
P
24
(1)
28
Total S.S = Σi Σj Yi²j-CF
Block S.S =Σi
Yi2
Treatment S.S =Σj
4
− CF
ΣjYi2
{𝑘}2
S.S due to ‘k’ = 4𝑟
4
− CF
[𝑝2 ]
S.S due ‘p’ = 4𝑟
S.S due ‘k’ =
[𝑘𝑝2 ]
4𝑟
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BLOCK
TREATMENTS
(1)
K
TOTAL
P
kp
I
23
25
22
38
108
II
26
36
40
38
140
III
29
20
20
30
99
IV
28
31
24
34
117
TOTAL
106
112
106
140
464
r=4
N = 16
𝑇 2 4642
C.F = =
𝑁
16
= 13456
Total S.S =23²+25²+.…+342 =660
Block S.S =
1082 + 1402 + 992 + 1172
4
= 232.
Treatment S.S=
− 𝐶𝐹
10822 + 14022 +9922 + 11722
4
− 𝐶𝐹 = 198
Error S.S = Total S.S – Treatment S.S – Block S.S
=660-235.5-198=229.5
YATES TABLE
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1
2
3
4
(1)
106
218 (106
+112)
464 (218 +246) G
k
112
246 (106
+140)
40 (6 +34) [k]
p
106
6 (112 – 106)
28 (246 – 218) [p]
kp
140
34 (140 –
106)
28 (34 – 6) [kp]
402
s.s due to k =4 × 4 × 100
s.s due to p = 282
4×4
= 49
We have to test,
(1) There is no significant difference in the effect of bocks
(2) There is no significant difference in the main effect
(3) There is no significant difference in the interaction effect
Anova Table
SOURCE
SUM OF
SQUARE
S
Df
MEAN
SUM OF
SQUARE
S
FRATIO
TABLED
F
Total
660
-
-
-
Block
232.5
16 – 1 = 15
77.5
3.04
F(3,9) =
3.86
Treatment
198
4–1=3
66.0
2.59
F(3,9) =
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3.86
k
100
4–1=3
100
3.92
F(1,9) =
5.12
p
49
1
49
1.92
F(1,9) =
5.12
kp
49
1
49
1.92
F(1,9) =
5.12
Error
229.5
3(4 – 1) =
9
25.5
-
Inference:
(1) There is no significance difference in the effect of blocks (F < Fa, accept
H0)
(2) There is no significance difference in the main effect of k and p (F < Fa)
(3) There is no significance difference in the effect of kp (F < Fa).
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MODULE 5
PREPARATION OF QUESTIONNAIRE
Questionnaire, in general , maybe referred to as a device or tool consisting
of some systematically planned questions in the shape of an enquiry form
which the subjects of the study fill in by themselves for providing their
written responses to the questions included in the form.
The success in getting desired information or data from the respondents of
a given research study with the use of questionnaire as a tool depends to a
large extent on the quality of the questionnaire. The task of constructing or
developing a questionnaire by researchers may be found to involve a series
of well thought out sequential steps presented here.
Steps involved in the construction of a questionnaire
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Taking decision about the content of the questions
Deciding about the structure or format of the questions
Paying attention to the language and wordings of the question
Paying attention to the length of the questionnaire
Ordering or sequencing of the questions
Writing the instructions for responding to questions
Seeking opinion of the experts and colleagues
Preliminary try out of the questionnaire
Establishing reliability and validity of the questionnaire
Giving questionnaire its final form
1. Taking decision about the content of the questions
This type of information or data a researcher needs to extract from the
subjects of his study in the form of their responses to the items of the
questionnaire must thus be a genuine deciding factor for the content of
the questions included in the questionnaire. The researcher, therefore,
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while developing a questionnaire for his research study, should always
include such questions in his questionnaire that can help him to extract
relevant data from the subjects of the study for answering his research
questions.
2. Deciding about the structure or format of the questions
The choice of a particular structure or format of the questionnaire is
based upon
i.
The type of information needed from the respondents for meeting
the research objectives in a particular situation at a particular
time.
ii.
The way of analysing the collected data: Generally, two types of
formats - closed and open ended questions.
Close ended questions: close ended questions are meant for yielding
quantitative data on the different levels of measurements (nominal,
ordinal, interval and ratio). They are dichotomous or forced choice
questions, may also include multiple choices, rank ordering, rating scales or
ratio data questions.
Open ended questions: Open ended questions provide qualitative data
as they are responded through a lot of words expressed by the respondents
for answering them. It uses lengthy space for detailed comments.
3. Paying attention to the language and wordings of the
questions
The wordings or phrasing of the questions must be in a proper way. As
far as possible, the questions and statements used in a questionnaire
needs to employ the necessary simplicity and clarity in its wordings.
Each question or statement must be clear in its purpose.
Language and wordings of the questionnaire - Do's and Don'ts
•
•
Use simple language instead of difficult one.
Use complete sentences instead of fragments.
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•
•
•
•
•
•
•
•
•
•
Avoid misinterpretation of terms or language.
Avoid use of double negatives eg: Do you not approve the idea that a
college girl should not engage herself in domestic affairs?
Avoid use of double barreled questions- It leads to ambiguity.
Avoid suggestive or leading questions.
Avoid embarrassing questions – Avoid private or personal questions
which they do not want to make public.
The questions must be arranged in a logical manner.
Objective answering - Questions should be capable of objective
answering. Avoid answers of opinion and keep to questions of fact.
Should be attractive.
Questions requiring calculations should be avoided.
Cross checks- If possible, one or more cross-checks should be
incorporated into the questionnaire to determine whether the
respondent is answering atleast the important questions correctly.
4. Paying attention to the length of the questionnaire
Number of questions should be small (15-20). The researcher should
keep the length of the questionnaire as short as possible by including
only what is absolutely necessary for gathering relevant information for
the research study. Lengthy questions make unfavorable reactions in
respondents. So the researcher should bring reasonable limits in the
questionnaire.
5. Ordering or sequencing of the questions
In this step, the questionnaire developer tries to provide a proper
sequence and order to the questions by taking care of the following
thingsa)
The questionnaire should begin with a few non-threatening and easy
to answer items. If the initial items are too difficult or threatening,
there is little chance that the respondents will complete the
questionnaire.
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b)
c)
d)
The most important items should appear in the first half of the
questionnaire so that the partially completed questionnaire may still
bring important information
There should be logical and psychological sequencing among the
items of the questionnaire. Each question should follow comfortably
from the previous questions.
It is better to put demographic questions like age, marital status,
occupation etc towards the end. This is partly because they are
uninteresting and partly, being sensitive, the subject maybe
resented.
6. Writing the instructions for responding to questions
Researcher should provide directions to respondents. The directions
may include:
a) Stating the objectives or purpose of the study in clear terms.
b) Requesting and motivating respondents to respond to the
questionnaire.
c) In the end, thanking the respondents for responding to the
questionnaire.
d) Assuring the respondents to maintain confidentiality of their
responses.
7. Seeking opinion of the experts and colleagues
At this stage, the rough draft of the questionnaire prepared by the
researcher should be necessarily submitted to the group of experts and
professionals by making personal and individual contacts with them for
seeking their opinion regarding the language, format or otherwise
appropriateness of the items included in the questionnaire.
8. Preliminary try out for pilot study of the questionnaire
The draft of the prepared questionnaire after being screened by the team
of experts and modified by the researcher in the light of their
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recommendations must be subjected at this stage to administration on a
small sample of the respondents of the study for the purpose of
examining the appropriateness of the questionnaire as a tool and the
purpose of diagnosing its inherent weaknesses in connection with the
wordings of the question, response rate etc. In the light of the outcomes
of preliminary try out, necessary modifications should be introduced in
the final questionnaire.
9. Establishing reliability and validity of the questionnaire
In this step, reliability and validity of the questionnaire may be
established by following the methods utilised for establishing reliability
and validity in other psychological test and data gathering instruments.
10. Giving the questionnaire its final form
In this final stage of the questionnaire development, the questionnaire is
in its final form and can be given to the respondents.
SCORES AND SCALES OF MEASUREMENT
SCORES OF MEASUREMENT
The rules for assigning values within the scale depend upon the type of
scoring mechanisms used. Dichotomous and polytamous are the two most
common scoring mechanisms.
Dichotomous scoring: It refers to the assignment of one of two possible
values based on a person’s performance or response to a test question. A
simple example is a use of correct and incorrect to score an item response.
Polytomous scoring: It refers to the assignment of 3 or more possible
values for a given test question or item. Example: incorrect, partially
correct and fully correct.
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SCALES/ LEVELS OF MEASUREMENT
Scaling is progressively arranged series of items according to value or
magnitude into which an item can be placed according to its qualification.
Data can be classified on basis of 4 scales.
1. Nominal scale
• Lowest level of measurement
• Numbers are used to name, identify or classify persons, objects,
groups etc
• Eg: classification based on gender, caste, political party etc
• In a nominal measurement, members of any two groups are never
equivalent but all the members of any one group are always
equivalent.
2. Ordinal scale
• Second level of measurement
• In ordinal measurement numbers denote the rank order of the objects
or the individuals
• Numbers are arranged from highest to lowest or from lowest to
highest
• Eg: grouping in terms of academic achievements, social status etc.
• In ordinal scale besides the relationship of equivalence, relationship
of 'greater than' or 'less than' exist because all members of a
particular subclass are equivalent to each other and at the same time
greater than or less than the member of the subclasses.
• Properties of magnitude is present but equal interval and absolute
zero are absent.
3. Interval scale
• Third level of measurement
• Includes all characteristics of nominal and ordinal scale
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• Numerically equal interval on scale indicates equal distance to the
properties of objects being measured that is the unit of measurement
is equal and constant and hence known as equal intervals scale
• Zero point doesn't tell real absence of any property being measured
which means that the zero point does not indicate absolute zero but in
fact our arbitrary.
4. Ratio scale
• It is the highest level of measurement with all the properties of other
three scales in addition with the absolute zero point
• In a ratio scale, ratio of any two numbers is independent of the unit of
measurement and can be meaningfully equated.
Types of scale
Magnitude
Equal interval
Absolute zero
Nominal scale
No
No
No
Ordinal scale
Yes
No
No
Interval scale
Yes
Yes
No
Ratio scale
Yes
Yes
Yes
RELIABILITY
Reliability refers to ability of an instrument to produce consistent or stable
result. Reliability is a degree to which measures are free from error so that
they give same results when repeated under constant conditions.
Anastasi - "Reliability of a test refers to consistency of scores obtained by
the same person on same test in different administrations and different
occasions".
The 3 methods for testing reliability of an instrument are:
1. Test-Retest Reliability
2. Split-Half Reliability
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3. Equivalent form Reliability
1. Test- Retest Reliability
Test-retest reliability measures the consistency of results when you repeat
the same test on the same sample at a different point in time. You use it
when you are measuring something that you expect to stay constant in your
sample.
Example: A test of colour blindness for trainee pilot applicants should have
high test-retest reliability, because colour blindness is a trait that does not
change over time.
Split half reliability
A measure of consistency where a test is split into two equal or nearly equal
halves (such as odd and even items) and the scores for each half of the test
is compared with one another. If the test is consistent it leads the
experimenter to believe that it is most likely measuring the same thing.
Alternate form Reliability
Alternate form reliability, also known as parallel-forms reliability,
equivalent-forms reliability and the comparable-forms reliability, occurs
when an individual is given two different versions of the same test at
different times. The correlation between the scores obtained on the two
forms represents the reliability coefficient of the test.
VALIDITY
Validity refers to the degree to which a test measures what it intends to
measure. Validity is concerned with generalizability, that is, when a test is a
valid one, it can be generalized to a general population.
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There are varieties of ways to assess the validity of the test of measurement.
These are:
1. Face validity
2. Content validity
3. Criterion-related validity
4. Construct validity
Face Validity
Face validity is the mere appearance that a measure has validity. We often
say a test has face validity if the items seem to be reasonably related to the
perceived purpose of the test.
For example, a researcher may create a questionnaire that aims to measure
depression levels in individuals. A colleague may then look over the
questions and deem the questionnaire to be valid purely on face value.
Content Validity
When a test is constructed so that its content of term measures what the
whole test claims to measure, the test is said to have content or curricular
validity. Thus, content validity is concerned with the relevance of the
contents of the items, individually and as a whole.
Content validity is the degree to which a test measures an intended content
area. For example, suppose a professor wants to test the overall knowledge
of his students in the subject of elementary statistics. His test would have
content validity if:
➢ The test covers every topic of elementary statistics that he taught in
the class.
➢ The test does not cover unrelated topics such as history, economics,
biology, etc.
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Criterion-related Validity
Criterion related validity is one which is obtained by comparing (or
correlating) the test scores with scores obtained on a criterion available at
present or to be available in the future.
There are two types of criterion validity:
• Concurrent Validity: The test is correlated with a criterion which
is available at the present time ie., no time gap. Example: Scores on a
newly constructed intelligence test may be correlated with scores
obtained on an already standardized test of intelligence. The resulting
coefficient of correlation will be an indicator of concurrent validity.
• Predictive validity: A test is correlated against the criterion to be
made available sometime in the future. In other words, test scores are
obtained and then a time gap of months or years is allowed to elapse,
after which the criterion scores are obtained.
Construct Validity
A construct is a non-observable trait, such as intelligence, which explains
our behavior. Construct validity is aimed at determining the extent to which
the constructed tool is able to measure a construct, ie., a hypothesized idea
to describe or explain the behavior under measurement. Anastasi has
defined it as “the extent to which the test may be said to measure a
theoretical construct or trait”.
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