Business Research Methods PPT - III

Business Research Methods
Unit - III
Dr. Ravindra
Associate Professor
Department of Commerce
Indira Gandhi University, Meerpur,
Rewari
6/22/2021 Dr. Ravindra, IGU, Meerpur 1

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Introduction
An appropriate statistical tests among variety of statistical
tests that are available for analyzing a given set of data is
selected on the basis of;
• Scale of measurement of the data.
• Dependence/independence of the measurements.
• Number of populations being studied.
• Specific requirements such as sample size and shape of
population distribution.
Generally, the distinction between parametric and non-
parametric statistical tests is based on (i) the scale of
measurement of the data and (ii) the assumptions regarding

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sampling distribution of sample statistic.
Certain statistical tests such as z, t, and F are used to
estimates and test of hypotheses about population
parameters. The use of these statistical tests;
• Requires specific scale of measurement such as interval
scale or ration scale to collect data.
• Involves hypothesis testing about specified parameter
values.
• Requires assumption of normality and whether standard
deviation of sampling population distribution is known or
not.

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In the absence of these requirements, statistical tests such
as z, t and F would not provide accurate conclusions about
population parameters. Thus, few other statistical tests
that do not require these conditions to be met are used to
test a hypothesis regarding population characteristics.
These statistical test are referred to as non-parametric
tests.
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The tests which can be used validly when the assumptions
needed for parametric testing cannot be met are known as
non-parametric test.

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Non-parametric tests (i) do not depend on the specific nature
of the population distribution from which samples are to be
drawn and (ii) use scale of measurement such as categorical or
ranks to collect data.
A non-parametric statistical test or method, also called
distribution free test, satisfies at least one of the following
criteria;
• The statistical test does not take into consideration population
parameters such as µ, ó and p.
• The statistical test is applied only on categorical data that are
non-numerical and frequency distribution of categories for

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one or more variables.
• The statistical test does not depend on the shape of population
distribution, in particular, normality.
Advantages and Limitation of Non-parametric Tests
Advantages
1. Non-parametric statistical tests can be applied to analyze
categorical (nominal scale) data, rank (ordinal scale) data,
interval and ration scale data.
2. Non-parametric statistical tests are easy to apply and take less
computation time when sample size is small.
3. Non-parametric statistical test are useful when the scale of

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measurement is weaker than required for parametric
statistical tests. Hence, these methods provide broad-based
conclusions.
4. Non-parametric statistical tests provide an approximate
solution to a population.
5. Non-parametrical statistical tests provide solution to
problems that do not depend on the shape of population
distribution, in particular, normality.
Limitations
1. Non-parametric statistical tests require more
computational time when sample size gets larger.

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2. Table values for non-parametric statistics are not as
readily available as of parametric test.
3. Non-parametric tests are usually not used widely.
Non-parametric Tests
In non-parametric tests category includes; Chi-Square test,
Sign test, Wilcoxen sign rank test, Mann-Whitney U-test,
Kruskal-Walis H-test, Median test, Run test, etc.
1. Chi-Square test: This test is already discussed in 1st
Semester in detail.
2. Sign test: The sign test is of two type; (i) One sample
sign test and, (ii) Two sample sign test.

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(i) One sample sign test: Based on the single random sample
coming from some continuous population, we wish to test
Ho: Population median = Mo against H1: Population
median > Mo, or H1: Population median < Mo, or H1:
Population median = Mo. Here, is Mo the hypothetical value
of the population mean. To calculate the value of sign test
following procedure will follow.
Procedure
1. In the sample, each observation greater than Mo is replaced
by ‘+’ sign and.
2. In the sample, each observation less then Mo is replaced by
a ‘ –’ sign.

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3. The observations which are equal to Mo are ignored and
sample size is reduced.
4. Count the total of ‘+’ and ‘-’ sign.
5. Calculate less frequent sign and denoted it by ‘S’.
6. Calculate the critical value ‘K’ using the formula;
(n – 1)
K = - (0.98) √n
2
7. Compare the value of S with the value of K, if S > K the null

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hypothesis is accepted otherwise rejected.
Illustration 1: The production manager of a large
undertaking randomly paid 10 visits to the worksite in
month. The number of workers reported late for duty was
found to be 2, 4, 5, 1, 6, 3, 2, 1, 7 and 8 respectively.
Using sign test verify the claim of the production
supervisor that on an average, not more than 3 workers
report late for duty. Using 5% level of significance.
Solution: Let Ho: µ =3 against H1 =µ > 3 at 5% level of
significance.

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Determination of Signs
X Signs
(X = 3)
2 -
4 + Total No. of plus sign = 5
5 + Total No. of minus sign = 4
1 - No. of Zero = 1
6 + Total no. of sign = n = 9
3 0
2 - Less frequent sign (s) = 4
1 -
7 +
8 +

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From the above table we get total no. of sign = 9
Less frequent sign (S) = 4
The calculation of value of K
n – 1 14 - 1
K = - 0.98 √n = - 0.98√14
2 2
= 6.5 – 3.666 = 2.83
Since S > K, the null hypothesis is accepted.

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(ii) Sign Test for Paired Data
The paired sample sign test is based upon the sign of
difference in paired observations say (x, y) where x is the
value of an observation from population A and y is the
value of an observation from population B. This test
assumes that these pairs (x, y) of values are independent
and that the measurement scale within each pair is at least
ordinal.
For comparing two populations, the sign test is stated in
terms of probability that values in the pair (x, y) from
population A are greater than the values from population
B. The sign of the difference in values, plus (+) or minus

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( -), is also recorded for each pair. The probability that a
value, x from population A will be greater than a value, y
from population B is denoted by p(x > y). Every pair of
values in a sample is written as (x, y), where x > y is
denoted plus (+) sign and x < y is denoted by minus ( - )
sign. Possibility that x = y is ignored and is denoted by 0
(zero). The other procedure is same as in case of single
sample.
Illustration 2: Verify whether there is a difference
between the number of days until collection of an account
receivable before and after a new collection policy. Use
the ɑ = 0.05 significance level.

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Solution: Let us take the null hypothesis that the difference
between number of days until collection of an account
receivable before and after a new collection policy is
insignificant.
Before 30 28 34 35 40 42 33 38 34 45 28 27 25 41 36
After 32 29 33 32 37 43 40 41 37 44 27 33 30 38 36

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Determine the Sing
Total plus sign = 6
Total minus sign = 8
No. of zero sign = 1
Total no. of sign = 14
Less frequent sign (+) = 6
Before 30 28 34 35 40 42 33 38 34 45 28 27 25 41 36
After 32 29 33 32 37 43 40 41 37 44 27 33 30 38 36
Sign - - + + + - - - - + + - - + 0

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Let us take the null hypothesis that the difference between
number of days until collection of an account receivable
before and after a new collection policy is significant.
n – 1 14 - 1
K = - 0.98 √n = - 0.98 √14 = 7 – 3.79
2 2
K = 3.21
Since S > K, the null hypothesis Ho is accepted at 5%
level of significance.

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3. Wilcoxon Signed Rank Sum Test: Sign test does not
give any weightage to the magnitude of the observation.
So, the results of Sign test are much weaker. The
Wilcoxon signed rank sum test is used in place of one-
sample t-test and paired t-test. This test can also be used
for ordered categorical data where a numerical scale is
inappropriate but it is possible to rank the observation.
(i) Wilcoxon Signed Rank Sum Test (Single Sample): In
Wilcoxon signed rank test (single sample), we test the
same hypothesis, i.e., Ho: Population median = Mo,
against H1: Population median > Mo, or H1: Population

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median < Mo or H1: Population median = Mo. The test is
described as below;
(i) We first obtain the difference of each observation from
hypothetical Ho. If some differences are zero, we discard
them and reduce the sample size accordingly.
(ii) Now we assign the ranks 1, 2, . . . .n to the absolute
differences (ignoring the sign) from smallest to largest. Do
not lose the track of the signs of differences.
(iii) If you have tie in ranks, assign the average value of those
ranks that would be assigned in case of no tie.
(iv) Calculate W+ = Sum of the ranks corresponding to positive
differences and W- = Sum of the ranks corresponding to

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negative differences. [As a check the total, w+ + w- =
should be equal to n(n +1)/2].
It is expected under the null hypothesis that the
distribution of the differences is almost symmetric around
zero and the distribution of positives and negatives to be
distributed at random among the ranks. We use this
concept to make the decision.
(v) We reject null hypothesis at ɑ × 100% level of
significance if;

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W- ≤ Wa, for a right tailed test
W+ ≤ Wa for a left tailed test.
W- ≤ Wa/2 for a two tailed test.
Note: In case of large samples, the distribution of;
T – n(n +1)/4
√n (n + 1) (2n + 1)/24
(vi) When the sample size is less then 25, then we assigned
symbol ‘T’ to the least sum of the two i.e. W+ or W-.
(vii) See the value of T from the table.

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Illustration 3: The production manager of a large
undertaking randomly paid 10 visits to the worksite in
month. The number of workers reported late for duty was
found to be 2, 4, 5, 1, 6, 3, 2, 1, 7 and 8 respectively.
Using sign test verify the claim of the production
supervisor that on an average, not more than 3 workers
report late for duty. Using 5% level of significance.
Solution: Let Ho: µ =3 against H1 =µ > 3 at 5% level of
significance.

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X d = X-3 I d I Ranks
(± Ignored)
Signed Rank
W+ W-
2 -1 1 2 - 2
4 1 1 2 2 -
5 2 2 5 5 -
1 -2 2 5 - 5
6 3 3 7 7 -
3 0 0 - - -
2 -1 1 2 - 2
1 -2 2 5 - 5
7 4 4 8 8 -
8 5 5 9 9 -
Total N = 9 ∑ W+= 31 ∑ W- = 14

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From the previous slid, it must be seen that total number
of ranks or n = 9 (i.e. < 25).
Since n < 25, the test statistics is given by:
T = smallest of the two sums of the signed ranks = 14
looking at Wilcoxon T table at 5% level for one tailed test
at n = 9 we get the value of T is 8.
Since the calculated value of T (14) is greater than the
critical value (8), the null hypothesis is accepted.

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(ii) Wilcoxon Signed rank Sum Test (Paired Samples) or
Wilcoxon Matched Pairs Test: This test is analogous to
paired t-test and stronger than the sign test for paired
observation. We wish to test Ho: M1 = M2 against H1: M1
= M2. Here, are the notation M1 and M2 for population
medians.
For each pair, obtain the difference by subtracting the 2nd
score from the 1st . Use the obtained differences and
follow the procedure of Wilcoxon signed rank sum test for
one sample using hypothetical median as zero. The mehod
is illustrated using an example.

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Illustration 4: Verify whether there is a difference
between the number of days until collection of an account
receivable before and after a new collection policy. Use
the ɑ = 0.05 significance level.
Solution: Let us take the null hypothesis that the
difference between number of days until collection of an
account receivable before and after a new collection
policy is insignificant.
Before 30 28 34 35 40 42 33 38 34 45 28 27 25 41 36
After 32 29 35 32 37 43 40 41 37 44 27 33 30 38 36

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X Y d = X - Y I d I Rank
(± ignoring)
Signed Ranks
W+ W-
30 32 -2 2 6 - 6
28 29 -1 1 3 - 3
34 33 1 1 3 3 -
35 32 3 3 9 9 -
40 37 3 3 9 9 -
42 43 -1 1 3 - 3
33 40 -7 7 14 - 14
38 41 -3 3 9 - 9
34 37 -3 3 9 - 9
45 44 1 1 3 3 -
28 27 1 1 3 3 -
27 33 -6 6 13 - 13
25 30 -5 5 12 - 12
41 38 3 3 9 9 -
36 36 - - - - -
Total n = 14 ∑ W+ = 36 ∑ W- = 69

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From the above table, it must seen that the total number of
ranks n = 14 (n < 25).
T = Smaller of two sums of the signed ranked = 36
looking at Wilcoxon T table at 5% level for a two tailed
test at n = 14, we get the critical value of T = 21.
Since, the calculated value of T is greater than the critical
value, null hypothesis is accepted.

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4. Mann Whitney U-test: This test is an analogous of t-test
for two independent samples. Here, we test Ho: M1 = M2
against H1: M1 > M2 or H1: M1 < M2 or H1: M1 = M2
on the basis of two independent samples drawn from
continuous population.
To perform this test, we first of all rank the data jointly,
taking them as belonging to a single sample in either an
increasing or decreasing order of magnitude. We usually
adopt low to high ranking process which means we assign
rank 1 to an item with lowest value, rank 2 to next higher
item and so on. In case there are ties, then we would

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assign each of the tied observation the mean of the ranks
which they jointly occupy. For example, if sixth, seventh
and eighth values are identical, we would assign each the
rank (6 + 7 + 8)/3 = 7. After this we find the sum of the
ranks assigned to the values of the first sample (call it R1)
and also the sum of the ranks assigned to the values of the
second sample (and call it R2). Then we work out the test
statistics, i.e. U, which is a measurement of the difference
between the ranked observations of the two samples as
under:

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n1(n1 + 1)
U = n1 . n2 + - ∑R1
2
Where n1 and n2 are the sample sizes and R1 is the sum of
ranks assigned to the values of the first sample. (In
practice, whichever rank sum can be conveniently
obtained can be taken as R1, since it is immaterial which
sample is called the first sample.)
In applying U-test we take the null hypothesis that the two
samples come from identical populations. If this
hypothesis is true, it seems reasonable to suppose that the

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that the ranks assigned to the values of the two samples should
be more or less the same. Under the alternative hypothesis, the
means of the two population are not equal and if this is so,
then most of the smaller ranks will go to the values of one
sample while most of the higher ranks will go to those of the
other sample.
If the null hypothesis that the n1 + n2 observation came from
identical populations is true, the said ‘U’ statistic has a
sampling distribution with
n1 . n2
Mean = µu =
2

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and Standard Deviation (or standard error)
= όu = √n1n2 (n1 + n2 + 1)
12
If n1 and n2 are sufficiently large (i.e., both greater than
8), the sampling distribution of U can be approximated
closely with normal distribution and the limits of the
acceptance region can be determined in the usual way at a
given level of significance. But if either n1 or n2 is so
small that the normal curve approximation to the sampling
distribution of U cannot be used. In such a case, we use

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the exact distribution based on the values.
Illustration 5: The values in one sample are 53, 38, 69,
57, 46, 39, 73, 48, 73, 74, 60 and 78. In another sample
they are 44, 40, 61, 52, 32, 44, 70, 41, 67, 72, 53 and 72.
Test at the 10% level, the hypothesis that they come from
population with the same mean. Apply U-test.
Solution: First of all we assign ranks to all observations,
adopting low to high ranking process on the presumption
that all given items belong to a single sample. By doing so
we get the following:

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Table
Size of sample items in
ascending order
Rank Name of
sample
Size of sample items in
ascending order
Rank Name of
sample
32 1 B 57 13 A
38 2 A 60 14 A
39 3 A 61 15 B
40 4 B 67 16 B
41 5 B 69 17 A
44 6.5 B 70 18 B
44 6.5 B 72 19.5 B
46 8 A 72 19.5 B
48 9 A 73 21.5 A
52 10 B 73 21.5 A
53 11.5 B 74 23 A
53 11.5 A 78 24 A

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From the above we find that the sum of the ranks assigned
to sample one items or RA = 2 + 3 +8 + 9 + 11.5 + 13 + 14
+ 17 + 21.5 + 21.5 + 23 + 24 = 167.5 and similarly we
find that the sum of ranks assigned to sample two items or
RB = 1 + 4 + 5 + 6.5 + 6.5 + 10 + 11.5 + 15 + 16 +18 +
19.5 + 19.5 = 132.5 and we have n1 = 12 and n2 = 12
n1(n1 + 1)
U = n1 . n2 + - ∑R1
2

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12 (12 + 1)
= (12) (12) + - 167.5
2
= 144 + 78 – 167.5 = 54.5
Since in the given problem n1 and n2 both are greater than
8, so the sampling distribution of U approximates closely
with normal curve. Keeping this in view, we work out the
mean and standard deviation taking the null hypothesis
that the two samples come from identical populations as

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under:
n1 . n2 (12) (12)
Mean = µu = = = 72
2 2
όu = √n1n2 (n1 + n2 + 1) √(12) (12) (12 + 12 + 1)
12 12
= 72.32

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As the alternative hypothesis is that the means of the two
populations are not equal, a two tailed test is appropriate.
Now we calculate the test statistics
U - µu 54.5 - 72
Zc = = = - 1.0104
όu 17.32
The calculated value of statistics is – 1.0104 is less than
the table value 1.96 at 5% level of significance. Therefore,
the null hypothesis is accepted.

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5. Kruskul Wallis H-test: This test is conducted in a way
similar to the U test described in previous slides. This test
is used to test the null hypothesis that ‘k’ independent
random samples come from identical universes against the
alternative hypothesis that the medians of these universes
are not equal. This test is analogous to the one-way
analysis of variance (ANOVA), but unlike the latter it
does not require the assumption that the samples come
from approximately normal populations or the universes
having the same standard deviation.

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In this test, like the U test, the data are ranked jointly from
low to high or high to low as if they constituted a single
sample. The test statistic is H for this test which is worked
out as under:
12 R21 R22 R23 R2k
H = + + + . . - 3(n + 1)
n(n + 1) n1 n2 n3 nk
Where, n = n1 + n2 + n3 + . . . .+ nk and R1 being the sum
of the ranks assigned to n1 observations in the ith sample.

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If the null hypothesis is true that there is no difference
between sample medians and each sample has at least five
items, then the sampling distribution of H can be
approximated with a chi-square distribution with (k – 1)
degree of freedom. As such we can reject the null
hypothesis at a given level of significance if H value
calculated, as stated above, exceeds the concerned table
value of chi-square. For small sample the critical values
can be taken from the Table given in the last slide.
Illustration 6: Use the Kruskul-Wallis test at 5% level of
significance to test the null hypothesis that a professional

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bowler performs equally well with the four bowling balls,
given the following results;
Bowling Results in Five Games
Solution: To apply the H test or the Kruskul-Wallis test to
this problem, we begin by ranking all the given figures
With Ball No. A 271 282 257 248 262
With Ball No. B 252 275 302 268 276
With Ball No. C 260 255 239 246 266
With Ball No. D 279 242 297 270 258

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Table
Bowling
Results
Rank Name of ball
associated
Bowling
Results
Rank Name of ball
associated
302 1 B 262 11 A
297 2 D 260 12 C
282 3 A 258 13 D
279 4 D 257 14 A
276 5 B 255 15 C
275 6 B 252 16 B
271 7 A 248 17 A
270 8 D 246 18 C
268 9 B 242 19 D
266 10 C 239 20 C

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Ball A = R1 = 7 + 3 + 14 + 17 + 11 = 52
Ball B = R2 = 16 + 6 + 1 + 9 + 5 = 37
Ball C = R3 = 12 + 15 + 20 + 18 + 10 = 75
Ball D = R4 = 4 + 19 + 2 + 8 + 13 = 46
Now we calculate H statistics as under:
12 R21 R22 R23 R24
H = + + + - 3(n + 1)
n(n + 1) n1 n2 n3 n4
n = 5 + 5 + 5 + 5 = 20

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12 522 372 752 462
H = + + + - 3(20+ 1)
20(20 + 1) 5 5 5 5
= (0.02857) (2362.8) – 63 = 67.51 – 63 = 4.51
Degree of freedom is K – 1 or 4 – 1 = 3
With 3 degree of freedom at 5% level of significance, the
value of chi-square is 7.815. Since, the calculated value
4.51 is less than the value of chi-square, therefore, the null
hypothesis is accepted.

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6. Run Test for Randomness: A run is a sequence or order
of identical occurrences of elements/observations
(symbols or numbers) preceded and followed by different
occurrences of elements or by no element at all. The run
test helps to determine whether the order or sequence of
observations in a sample is random. The run test examines
the number of ‘runs’ of each of two possible
characteristics that sample elements may have. For
example, in tossing a coin, the outcome of three trials in
succession would constitute a run. To quantify how many

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runs are acceptable before questioning the randomness of the
process, a probability distribution is used that leads to a
statistical test for randomness.
Suppose a coin is tossed 20 times and produce the following
sequence of head (H) and tails (T).
HHH TT HH TTT HHH TT HHH T
1 2 3 4 5 6 7 8
In this example, the first run of HHH is considered as run 1,
the second run of TT as run 2 and so on, so that there are r =
8 runs in all. However, in this case rather than prefect
separation between H and T, it appears to be a cluster. This

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situation is not likely to have arisen by chance.
Small sample Run Test: In order to test the randomness,
let n1 = number of elements of type one, and n2 = number
of elements of type two. Then sample size is n = n1 + n2.
There are n1 =12 heads and n2 =8 tails in above
illustration. Let type one elements be denoted by plus (+)
sign and T by minus (-) sign. These plus (+) or minus (-)
signs indicate the direction of change from an existing
pattern. Accordingly, a plus (+) would be considered a
change from an existing pattern of values in one direction

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and a minus (-) would be considered a change in the other
direction.
If sample size is small, so that either n1 or n2 is less than
20, then statistical test is carried out by comparing the
actual number of runs, R to its critical value for the given
values of n1 and n2. The null and alternative hypotheses
stated as;
Ho: Observations in the samples are randomly generated.
H1: Observations in the samples are not randomly
generated.
Can be tested that the occurrences of plus (+) sign and

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minus (-) signs are random by comparing R-value with its
critical value at a particular level of significance.
Decision Rules
• If R ≤ C1 or R ≥ C2, then reject null hypothesis, Ho.
• Otherwise accept Ho.
Where, C1 and C2 are critical values of R obtained from
standard table with probability P(R ≤ C1) + P(R ≥) =ɑ.
Large Sample Run Test
If the sample size is large so that either n1 or n2 is more then
20, then the sampling distribution of R-statistics (i.e. run) can

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be approximated by the normal distribution. The mean and
standard deviation of the number of runs for the normal
distribution are given by;
2n1 . n2
Mean = µR = + 1
n1 + n2
Standard Deviation = όR = √2n1n2(2n1n2 – n1 – n2)
(n1 + n2)2 (n1 + n2 – 1)
Thus, the standard normal test statistics is given by

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R - µR r - µR
Z = =
όR όR
The critical z – value is obtained in the usual manner at a
specified level of significance.
Illustration 7: The following is an arrangement of 25 men
(M) and 15 women (W) lined up to purchase ticket for a
premier picture show:
M WW MMM W MM W M WWW MMM W
MM WWW MMMMM WWW MMMMM W MM
Test for randomness at 5% level of significance.

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Solution: Let us take the null and alternative hypotheses as
follow;
Ho: Arrangement is random.
H1: Arrangement is not random.
Given R = 17 runs with n1 = 25 (number of M) and n2 = 15
(number of W), so that n = n1 + n2 = 40. Since sample size is
large, calculating mean and standard deviation as follows:
2n1 . n2 2 × 25 × 15
Mean = µR = + 1 = + 1 = 19.75
n1 + n2 25 + 15

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Standard Deviation = όR = √2n1n2(2n1n2 – n1 – n2)
(n1 + n2)2 (n1 + n2 – 1)
= όR = √2× 25 × 15(2 × 25 × 15 – 25 – 15)
(25 + 15)2 (25 + 15 – 1)
= √750 (710) = 2.92
1600 × 39
R - µR 17 – 19.75
Z = = = - 0.94
όR 2.92

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Since, the calculated value of z is – 0.94 is less than the
critical value of Z = 1.96 at 5% level of significance.
Therefore, the null hypothesis is accepted, the arrangement is
random.
8. Kolmogorov- Smirnov Test (K – S Test): This test is used
in situations where a comparison has to be made between an
observed sample distribution and theoretical distribution.
K-S One Sample Test: This test is used as a test of goodness
of fit and is ideal when the size of the sample is small. It
compares the cumulative distribution function for a variable
with a specified distribution. The null hypothesis assumes no

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difference between the observed and theoretical
distribution and the value of test statistic 'D' is
calculated as;
Formula: D =Maximum |Fo (X) − Fr (X)|
Where;
Fo(X) = Observed cumulative frequency distribution of a
random sample of n observations.
and Fo(X)=k/n = (No. of observations ≤ X)/(Total no. of
observations).
Fr(X) = The theoretical frequency distribution.

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The critical value of D is found from the K-S table values
for one sample test.
Acceptance Criteria: If calculated value is less than
critical value accept null hypothesis.
Rejection Criteria: If calculated value is greater than
table value reject null hypothesis.
Illustration 8: In a study done from various streams of a
college 60 students, with equal number of students drawn
from each stream, are we interviewed and their intention
to join the Drama Club of college was noted.

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It was expected that 12 students from each class would join
the Drama Club. Using the K-S test to find if there is any
difference among student classes with regard to their intention
of joining the Drama Club.
Solution: Ho: There is no difference among students of
different streams with respect to their intention of joining the
drama club.
We develop the cumulative frequencies for observed and
theoretical distributions.
Classes B.Sc. B.A. B. Com M.A. M. Com
No. of students in
each class
5 9 11 16 19

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Table
Stream No. of students Interesting
to Join
FO(X) FT(X) |FO(X)−FT(X)|
Observed
(o)
Theoretical(T)
B.Sc. 5 12 5/60 12/60 -7/60
B.A. 9 12 14/60 24/60 10/60
B. Com 11 12 25/60 36/60 11/60
M.A. 16 12 41/60 48/60 7/60
M. Com 19 12 60/60 60/60 00/60
Total N = 60 60

Non-parametric Test
Test statistic D is calculated as;
D = Maximum |F0(X) − FT(X)|
=11/60 = 0.183
The table value of D at 5% significance level is given by
D0.05 = 1.36/√n = 1.36/√60 = 0.175
Since the calculated value is greater than the critical value,
hence we reject the null hypothesis and conclude that
there is a difference among students of different streams
in their intention of joining the Club.

Non-parametric Test
K-S Two Sample Test: When instead of one, there are
two independent samples then K-S two sample test can be
used to test the agreement between two cumulative
distributions. The null hypothesis states that there is no
difference between the two distributions. The D-statistic is
calculated in the same manner as the K-S One Sample
Test.
Formula: D=Maximum |Fn1(X) − Fn2(X)|
Where, n1 = Observations from first sample.
n2 = Observations from second sample.

Non-parametric Test
It has been seen that when the cumulative distributions show
large maximum deviation |D| it is indicating towards a
difference between the two sample distributions.
The critical value of D for samples where n1= n2 and is ≤ 40,
the K-S table for two sample case is used.
When n1 and/or n2 > 40 then the K-S table for large samples
of two sample test should be used. The null hypothesis is
accepted if the calculated value is less than the table value and
vice-versa.
Thus use of any of these nonparametric tests helps a
researcher to test the significance of his results when the
characteristics of the target population are unknown or no
assumptions had been made about them.

Non-parametric Test (Appendix)

Non-parametric Test
Thank You!

Business Research Methods PPT - III

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