Non Parametric Statistics

65
PART III
NON-PARAMETRIC
STATISTICS

66
CHAPTER I CHI SQUARE TEST
The Chi Square Distribution
 The test of difference between the observed frequencies and the expected
frequencies
 Written as X2 and read as the chi square distribution
 X is the Greek letter “chi” pronounced “ki”
 Has only one parameter called degrees of freedom
3 Unique Functions of Chi Square
 Test of Goodness of fit
 Test of Homogeneity
 Test of Independence
Test of Goodness of fit
 This is to test the difference between the observed frequencies and the expected
frequencies
FORMULA
X2=Σ (O−E)2
E
Where:
X2= the Chi Square test
O = the observed frequencies
E = the expected frequencies
Example:
A certain machine is supposed to mix almonds, hazelnuts, cashews and pecans in
the ratio of 5:2:1:3. A can containing 600 of these mixed nuts was found to have 150
almonds, 270 hazelnuts, 70 cashews and 110 pecans. At the 0.05 level of significance test
the hypothesis that the machine is mixing the nuts at the ratio of 6:2:4:3.
SOLVING BY THE STEPWISE METHOD
I. Problem: is the machine mixing the nuts at the ratio of 6:2:4:3?
II. Hypotheses:
H0: The machine is mixing the nuts at the ratio of 6:2:4:3
H1: The machine is not mixing the nuts at the ratio of 6:2:4:3
III. Level of Significance
α : 0.05
df = h – 1
= 4 – 1
= 3
X2
at 0.05 = 7.815 tabular value

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IV. Statistics
Chi Square test, test if goodness of fit
Nuts Ratio Observed Expected
Almonds 6 150 240
Hazelnuts 2 270 80
Cashew 4 70 160
Pecans 3 110 120
TOTAL 15 600 600
600 ÷ 15 = 40
For Expected:
40 x 6 = 240
40 x 2 = 80
40 x 4 = 160
40 x 3 = 120
Using the formula of Chi Square
X2=Σ (O−E)2
E
=
(150−240)2
240
+ (270−80)2
80
+ (70−160 )2
160
+ (110 −120 )2
120
= 33.75 + 451.25 + 50.63 + 0.83
= 536.46
V. Decision Rule: If the chi square computed value is greater than the chi square tabular
value, reject H0.
VI. Conclusion: The chi square computed value of 536.36 is greater than the chi square
tabular value of 7.815 at 0.05 level of significance with 3 degrees of freedom, so the
research hypothesis is accepted which means that the machine is not mixing the nuts in
the ratio of 6:2:4:3. It implies that the machine is not in good order because it does not
mix the nuts as expected.
Test of Homogeneity
 Concerned with 2 or more samples and is used to determine if two or more
variables are homogenous.
FORMULA
X2=
푁(푎푑−푏푐)2
푘푙푚푛

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Where:
X2= the chi square test
N = the Grand Total
klmn = the product of the rows and columns
Example 1:
To illustrate this, we can evaluate the attitude of a sample of Lakas and Laban
parties on the issue of peace and order in Mindanao. To carry out such study, a separate
random sample of members of each party is drawn from the nationwide population of
Lakas and Laban and each indivisual in both samples responds to the scale. Scores are
then classified into “Favorable” or “Unfavorable” categories.Use 0.05 level of
significance The following frequencies are obtained :
Favorable Unfavorable TOTAL
Lakas 75 75 150
A b k
Laban 95 65 160
C d l
TOTAL 170 140 310
M n N
SOLVING BY STEPWISE METHOD
I. Problem:Is there a significant difference between the attitude of the two political
parties on the issue of peace and order in Mindanao?
II. Hypotheses:
H0 : There is no significant difference between the attitudes of the two political
parties on the issue of peace and order in Mindanao
H1: There is significant difference between the attitude of the two political parties
on the issue of peace and order in Mindanao.
α = 0.05
df = (c – 1)(r – 1)
df = (2 – 1)(2 – 1)
df = 1

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IV. Statistics
CHI SQUARE TEST OF HOMOGENEITY
Favorable Unfavorable TOTAL
Lakas 75 75 150
A b k
Laban 95 65 160
C d l
TOTAL 170 140 310
M n N
FORMULA
X2=
푁(푎푑−푏푐)2
푘푙푚푛
X2=
310((75)(65))−((75)(95))2
(150)(160)(170)(140)
=
310(4875−7125)2
571200000
=
310(5062500)
571200000
= 2.75
V. Decision Rule: If the chi square computed value is greater than the tabular value,
reject Ho.
VI. Conclusion: Since the chi square computed value of 2.75 is less than the chi square
tabular value of 3.481 at 0.05 level of significance with 1 degree of freedom the research
hypothesis is accepted. This means that there is no significant difference between the
attitudes of the two political parties on the issue of peace and order in Mindanao. It
implies that the Lakas group has unfavorable attitude while the Laban group has the
favorable on the said issue.
Test of Independence
(One Sample, Two Criterion Variable)
 The samples used in this test is consists of randomly selected members drawn
from the same population.

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 A test used to look into whether measures taken on two criterion variables were
either independent or associated with one in a given population
FORMULA
X2=Σ (O−E)2
E
Where:
X2 = Chi Square test
O = Observed Frequency
E = Expected Frequency
Σ = Summation
Example:
One hundred individuals, male and female were given an test in psychomotor
skills and their scores were classified into high and low. Using the X2 test of
independence at 0.05 level of significance, the table is shown as follows:
SCORE
GENDER HIGH
O E
LOW
O E
TOTAL
Male 35 19 54
Female 30 16 46
Total 65 35 100
SOLVING STEPWISE METHOD:
I. Problem: Is there a significant relationship between gender and scores in psychomotor
skill?
II. Hypotheses:
Ho = There is no significant relationship between gender and scores in
psychomotor skill.
H1 = There is a significant relationship between gender and scores in psychomotor
skill.
III. Level of Significance:
α = 0.05
df = (c – 1)(r – 1)
= (2 – 1) (2 – 1)
= 1
X2 at 0.05 = 3.841 tabular value

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IV. Statistics
X2 for Independence
GENDER HIGH
O E
LOW
O E
TOTAL
Male 35 (35.1) 19 (18.9) 54
Female 30 (29.44) 16 (16.1) 46
Total 65 35 100
For expected values we multiply the column total to the row total and divide the product
by the grand total.
65 x 54
100
= 35.1
35 x 54
100
= 18.9
65 x 46
100
= 29.44
35 x 46
100
= 16.1
Using Chi Square Formula
X2=Σ (O−E)2
E
X2=
(35−35.1)2
35.1
+ (30−29 .44 )2
29 .44
+ (19−18.9)2
18.9
+ (16−16.1)2
16.1
X2=0.0002849 + 0.01065 + 0.00053 + 0.00062
X2= 0.0115
V. Decision Rule: If the x2 computed value is greater than the x2 tabular value, reject Ho.
VI. Conclusion: The x2 computed value of 0.0115 is less than the x2 tabular value of
3.841 at 0.05 level of significance with one degree of freedom. This leads to the
confirmation of the research hypotheses which means that there is no significant
relationship between gender and scores in psychomotor skill.

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Example 2:
Two lots of 50 experimental guinea pigs were used in testing the effectiveness of
the new serum to cure the illness. Both were inoculated with the new organism but only
one lot was previously given the preventive serum. Is the serum effective? Use 0.01
level of significance.
SOLVING BY THE STEPWISE METHOD
SERUM NO SERUM TOTAL
RECOVERED 15 12 27
DIED 7 16 23
TOTAL 22 28 50
I. Problem: Is the serum effective?
II. Hypotheses:
Ho: The serum is not effective
H1: The serum is effective
α = 0.05
df = (c – 1) (r – 1)
= (2 – 1)(2 – 1)
= 1
X2 at 0.01 = 6.635
IV. Statistics
x2 test of difference. When the df is equal to one and any expected frequency is small,
less than 10 the test of difference is being used.
FORMULA:
X2 = Σ (|O− E|−0.5)2
E
X2=
푁(|푎푑−푏푐|−N/2)2
푘푙푚푛
SERUM
O E
NO SERUM
O E
TOTAL
RECOVERED 15 a 11.88 12 b 15.12 27
DIED 7 c 10.12 16 d 12.88 23
TOTAL 22 28 50

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Computations:
X2 = Σ (|O− E|−0.5)2
E
X2 =
(|15−11.88| −0.5)2
11 .88
+ (|7−10.12|− 0.5)2
10 .12
+ (|12− 15.12| −0.5)2
15 .12
+ (|16 −12.88|−0.5)2
12.88
X2 = 0.58 + 0.68 + 0.45 + 0.53
X2 = 2.24
X2=
푁(|푎푑−푏푐|−N/2)2
푘푙푚푛
X2=
50(|240−84|−50/2)2
(22)(28)(27)(23)
X2= 50(17161)
382536
X2=2.24
V. Decision Rule: If the x2 computed value is greater than the x2 tabular value, reject the
null hypothesis
VI. Conclusion: The serum is therefore found effective. It can be said that in the lot with
serum 15 had recovered and 7 died. Likewise, in the lot without serum 16 died and only
12 had recovered.

74
CHAPTER II Wilcoxon Rank-Sum Test or Wilcoxon Two-Sample Test
 Used to compare if there is a significant difference between two independent
groups.
 Counterpart of t-test under parametric test
 It is non-parametric test
 used when the sample sized is very small
 distribution is normal
 the means of two independent groups are compared
 It is appropriate test of difference between two groups if the distribution is
abnormal
 Appropriate test for a very small sample size
How do we use?
 Rank the observation from lowest value to the highest value of both groups
 After ranking, assign the rank to respective observation
 Add the ranks of group 1, W₁
 Add the ranks of group 2, W₂
 Determine the number of observation in group 1 and group 2 that is n₁ and n₂
respectively.
 Use the formula
푈1 = 푊1 −
푛₁(푛1 + 1)
2
푈2 = 푊2 −
푛₂(푛2 + 1)
2
Where:
푈₁ =Wilcoxon Rank-Sum Test
푊₁ = sum of ranks of group 1
푛₁ = sample size of group 1
푈₂ = Wilcoxon Rank-Sum Test
푊₂ = sum of ranks of group 2
푛₂ = sample size of group 2
Example 1:
One of 18 selected patients who had advanced stage of leukemia, 10 were treated
with a new serum and 8 were not. The survival time, in years, was reckoned from the
time experiment was conducted. Using the Wilcoxon rank-sum test at α=.05 to test
whether the serum is effective, consider the following data.
Treatment 2.9 3.1 5.3 4.2 4.5 3.9 2.0 3.7 4.1 4.0
No
treatment
1.9 .5 .9 2.2 3.1 2.0 1.7 2.5

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SOLVING BY STEPWISE METHOD
I. Problem: is the new serum effective in treating leukemia?
II. Hypotheses
Ho: the new serum is not effective
Ha: the new serum is effective
III. Level of significance
α=.05
df= n₁=10 n₂=8
U.05=17
IV. Statistics:
Wilcoxon rank-sum test
Arrange the data of both groups from the lowest to highest value and rank them.
With treatment rank No treatment rank
2.9 9 1.9 4
3.1 10.5 0.5 1
5.3 18 0.9 2
4.2 16 2.2 7
4.5 17 3.1 10.5
3.9 13 2.0 5.5
2.0 5.5 1.7 3
3.7 12 2.5 8
4.1 15 푊2 = 41
4.0 14
푊1 = 130
Ranking the data from the lowest to the highest value
No. Observation from both
groups
Rank
1 .5 1
2 .9 2
3 1.7 3
4 1.9 4
5 2.0 5.5
6 2.0 5.5
7 2.2 7
8 2.5 8
9 2.9 9
10 3.1 10.5
11 3.1 10.5
12 3.7 12
13 3.9 13
14 4.0 14

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15 4.1 15
16 4.2 16
17 4.5 17
18 5.3 18
푈₁ = 푊₁ −
푛₁(푛₁ + 1)
2
= 130 −
10(10 + 1)
2
= 130 −
110
2
= 130 − 55
푈₁ = 75
푈₂ = 푊₂ −
푛₂(푛₂ + 1)
2
= 41 −
8(8 + 1)
2
= 41 −
72
2
= 41 − 36
푈2 = 5
V. Decision rule: select the smaller value from U₁ and U₂. If U computed value is less
than or equal to the tabular value, disconfirm the Ho.
VI. Conclusion: since the U₂ computed value is less than U tabular at .5 level of
significance with the degree of freedom n₁=10 and n₂=8, disconfirm null hypothesis.

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CHAPTER III The Kruskal-Wallis Test or H-test
 a nonparametric test which does not require normal distribution.
 This test is used to compare 3 or more independent groups.
 H-test is an alternative for the F-test (ANOVA) in parametric test.
FORMULA:
퐻 =
12
푛(푛 + 1)
Σ
푅푖 2
푛푖
− 3(푛 + 1)
Where:
H= Kruskal Wallis test
n= the number of observation
12= constant
3= constant
Example: Random samples of 3 brands of cigarettes were tested for tar content. The
following figures show the milligrams of tar found in the 15 cigarettes tested.
BRAND
I V Y
13 17 10
15 20 9
18 14 12
17 12 14
16 21 11
Use the kruskal-wallis test, at the .05 level of significance, to test whether there is
a significant difference in tar content among the 3 brands of cigarettes.
Brand I R1 Brand V R2 Brand Y R3
13 6 17 11.5 10 2
15 9 20 14 9 1
18 13 14 7.5 12 4.5
17 11.5 12 4.5 14 7.5
16 10 21 15 11 3
n1=5 Σ R1=49.5 n2=5 Σ R2=52.5 n3=5 Σ R3=18
Arrange the tar content jointly from the lowest to the highest, and then rank them.

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Number Observation Rank
1 9 1
2 10 2
3 11 3
4 12 4.5
5 12 4.5
6 13 6
7 14 7.5
8 14 7.5
9 15 9
10 16 10
11 17 11.5
12 17 11.5
13 18 13
14 20 14
15 21 15
STEPWISE METHOD
I. Problem: Are there significant difference in the average tar content of the 3
brands of cigarettes?
II. Hypotheses:
Ho: There is no significant difference in the average tar content of the 3 brands
of cigarettes.
Ha: There is significant difference in the average tar content of the 3 brands of
cigarettes.
III. Level of significance:
α= .05 x2=5.991
df= h-1
=3-1
=2
IV. Statistics: H-test
Computation:
H = 12
푛(푛+1)
Σ 푅푖2
푛푖
− 3(푛 + 1)
H =
12
15(15 + 1)
(49.5)2
(
5
+
(52.5)2
5
+
(18)2
5
) − 3(15 + 1)
H =
12
15(16)
(
2450.25
5
+
2756.25
5
+
324
5
) − 3(16)
H =
12
240
(490.05 + 551.25 + 64.8) − 48

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H =
12
240
(1106.1) − 48
H = 0.05(1106.1) − 48
H = 7.305
V. Decision Rule: If the H-computed value is greater than the x2 tabular value,
reject Ho.
VI. Conclusion: Since the H-computed value of 7.305 is greater than the x2
tabular value of 5.991 at 0.05 level of significance with 2 degree of freedom,
the researcher hypothesis is accepted. These mean that there is a significant
difference in the average tar content of the 3 brands of cigarettes. It can also
be conclude that the 3 brands are not equally same tar content.

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CHAPTER IV The Spearman Rank Order Coefficient of Correlation 풓풔
 Also called Spearman’s rho
 Named after the statistician Charles Spearman and often denoted by the Greek
letter þ (rho) or as 푟푠.
 It is a non-parametric measure of statistical dependence between two variables.
 It assesses how well the relationship between two variables can be described
using a monotonic function. Monotonic function, as the value of one variable
increases, so does the value of the other variable increased.
 Spearman's Rank Correlation Coefficient is a measure of the association between
the rankings of two samples and can used to test for independence of the samples.
FORMULA
푟푠=1
6 Σ 퐷2
푛(푛2 − 1 )
Where:
푟푠 = The Spearman Rank Order Coefficient of Correlation
Σ 퐷2= sum of the squares of the difference between the rank x and rank y
n = sample size
6 = constant
Example:
The table associates the IQ of each adolescent in a sample with the number of
hours they listen to rap music per month. Determine the strength of the correlation
between IQ and rap music using Spearman’s rank correlation at 0.05 level of
significance.
NUMBER OF HOURS LISTENED TO RAP MUSIC
EVERY MONTH
IQ
3 100
2 50
5 89
45 105
25 90
18 58
29 118
13 45
37 76

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How to calculate Spearman
1. Draw your data table. This will organize the information you need to calculate
Spearman's Rank Correlation Coefficient. You will need:
 6 Columns, with headers as shown below.
 As many rows as you have pairs of data.
2. F
i
l
l
i
n the first two columns with your pairs of data.
Data 1 Data 2 Rank 1 Rank 2 D D2
3 100
2 50
5 89
45 105
25 90
18 50
29 118
13 45
37 76
3. In your third column, rank the data in your first column from 1 to n (the
number of data you have). The lowest number was rank as last and the highest
number rank first.
3 100 8
2 50 9
5 89 7
45 105 1
25 90 4
18 50 5
29 118 3
13 45 6

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4. In your fourth column do the same as in step 3, but instead rank the second
column.
3 100 8 3
2 48 9 8
5 89 7 5
45 105 1 2
25 90 4 4
18 50 5 7
29 118 3 1
13 45 6 9
37 76 2 6
5. In the "d" column, calculate the difference between the two numbers in each
pair of ranks. That is, if one is ranked 8 and the other is 3 the difference would
be 5. (The sign does not matter, since the next step is to square this number.)
3 100 8 3 5
2 48 9 8 1
5 89 7 5 2
45 105 1 2 -1
25 90 4 4 0
18 50 5 7 -2
29 118 3 1 2
13 45 6 9 -3
37 76 2 6 -4
6. Square each of the numbers in the "d" column and write these values in the
"d2" column
3 100 8 3 5 25
2 48 9 8 1 1
5 89 7 5 2 4
45 105 1 2 -1 1
25 90 4 4 0 0
18 50 5 7 -2 4
29 118 3 1 2 4
13 45 6 9 -3 9
37 76 2 6 -4 16

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7. Add up all the data in the "d2" column. This value is Σd2.
ΣD2=25+1+4+1+0+4+4+9+16
ΣD2 =64
SUBSTITUTE TO THE FORMULA
풓풔=ퟏ − ퟔ Σ 푫ퟐ
풏(풏ퟐ −ퟏ )
풓풔=ퟏ − ퟔ(ퟔퟒ)
ퟗ(ퟗퟐ−ퟏ )
풓풔=ퟏ − ퟑퟖퟒ
ퟗ(ퟖퟏ−ퟏ )
ퟗ(ퟖퟎ )
ퟕퟐퟎ
풓풔=ퟏ − ퟎ. ퟓퟑퟑퟑ
풓풔=ퟎ. ퟒퟔퟔퟕ
Interpret your result. It can vary between -1 and 1.
 Close to -1 - Negative correlation.
 Close to 0 - No linear correlation.
 Close to 1 - Positive correlation.
Stepwise Method
I. Problem: Is there a significant relationship between the number of hours spent in
studying in English and the corresponding grades in the midterm examination
II. Hypotheses:
H0: There is no significant relationship between the number of hours spent in
studying English and the corresponding grades in midterm examination.
H1: There is a significant between the number of hours spent in studying
English and the corresponding grades in midterm examination.
휶 = .05
df = n-1
=9-1
df=8
풓풔 풄풐풎풑풖풕풆풅= ퟎ. ퟒퟔퟔퟕ

84
IV. Statistics:
풓풔 Spearman Rank Order Coefficient Correlation
3 100 8 3 5 25
2 48 9 8 1 1
5 89 7 5 2 4
45 105 1 2 -1 1
25 90 4 4 0 0
18 50 5 7 -2 4
29 118 3 1 2 4
13 45 6 9 -3 9
37 76 2 6 -4 16
ΣD2=25+1+4+1+0+4+4+9+16
ΣD2 =64
SUBSTITUTE TO THE FORMULA
풓풔=ퟏ − ퟔ Σ 푫ퟐ
풏(풏ퟐ −ퟏ )
풓풔=ퟏ − ퟔ(ퟔퟒ)
ퟗ(ퟗퟐ −ퟏ )
ퟗ(ퟖퟏ−ퟏ )
ퟗ(ퟖퟎ )
ퟕퟐퟎ
풓풔=ퟏ − ퟎ. ퟓퟑퟑퟑ
풓풔=ퟎ. ퟒퟔퟔퟕ 풓풔−풕풂풃풖풍풂풓 = .643
V. Decision Rule: If the 푟푠 computed value is greater than 푟푠 tabular value, reject H0.
VI. Conclusion: Since the 푟푠 computed value of .47 is less than the 푟푠 tabular value of
.643 at .05 level of significance with 8-degrees of freedom, alternative hypothesis is
accepted. There is no significant relationship between the number of hours spent in
studying English and the corresponding grades in midterm examination.

85
Example:
A musical (solo vocal) talent contest where 15 competitors are evaluated by two
judges, A and B. Usually judges award numerical scores for each contestant after his/her
performance. The following are numerical scores of 15 competitors evaluated by two
judges. Use 푟푠at 0.05 level of significance to test the null hypothesis if the two judges
differ most in their opinions about the competitors.
Judge A Judge B
9 10
8 6
8 9
10 8
9 7
6 8
4 6
7 6
5 4
3 5
2 5
1 7
7 10
3 6
9 8
Solving Stepwise Method
I. Problem:
Is there a significant relationship between the opinions of the two judges
in 15 solo vocal competitors?
II. Hypothesis
H0: The two judges differ most in their opinions in 15 solo vocal
competitors.
H1: The two judges do not differ most in their opinions in 15 solo vocal
competitors.
휶 = 0.05
Df= n-1
=15-1
=14

86
IV. Statistics
푟푠 Spearman Rank Order Coefficient Correlation
Computation
Z Judge B Rank A Rank B D D2
9 10 2 1 1 1
8 6 5 9 -4 16
8 9 5 3 2 4
10 8 1 4 -3 9
9 7 2 7 -5 25
6 8 9 4 5 25
4 6 11 9 2 4
7 6 7 9 -2 4
5 4 10 15 -5 25
3 5 12 13 -1 1
2 5 14 13 1 1
1 7 15 7 8 64
7 10 7 1 6 36
3 6 12 9 3 9
9 8 2 4 -2 4
Σ 퐷2= 1+16+4+9+25+25+4+4+25+1+1+64+36+9+4
Σ 푫ퟐ = 228
Substitute to the formula
풓풔=ퟏ −
ퟔ Σ 푫ퟐ
풏(풏ퟐ − ퟏ )
풓풔=ퟏ −
ퟔ(ퟐퟐퟖ)
ퟏퟓ(ퟏퟓퟐ − ퟏ )
풓풔=ퟏ −
ퟏퟑퟔퟖ
ퟏퟓ(ퟐퟐퟒ)
풓풔=ퟏ −
ퟏퟑퟔퟖ
ퟑퟑퟔퟎ
풓풔=ퟏ − ퟎ. ퟒퟏ
풓풔= . ퟓퟗ

87
V. Decision Rule: If the 풓풔 computed value is greater than 풓풔 tabular value, reject
H0.
VI. Conclusion: Since the 풓풔 computed value of .59 is greater than the 풓풔 tabula of .456
at 0.05 level of significance with 14 degrees of freedom, the research hypothesis is
accepted. There is a significant relationship between the opinions of the two judges in 15
solo vocal competitors.

88
CHAPTER V A Sign Test for Two Independent Samples (Median Test Two-
Sample Test)
 -another test under nonparametric statistics
 -also known as median test
Why do we use?
-to compare two independent sample
The sample observation above is (+) sign while below is (-). A x² is used to
determine whether the observed frequencies of + and – signs differ significantly.
Formula 푥² =
푁(푎푑 )−(푏푐)²
푘푙푚푛
Where:
푥²= chi-square test
a and c = observed (+) frequencies
d and b = observed (-) frequencies
k and l = row total
m and n = column total
N = grand total
Example 1
Considered the test score of 9 males and 12 females student on spelling test.
Females 12 26 25 10 10 10 22 20 19 17 17 15
males 6 22 19 7 8 12 16 8 19
Solving by stepwise method
I. Problem: is there is significance difference between the performance of two
groups?
II. Hypotheses
Ho: there is no significance difference between the performances of two
groups
Ha: there is significance difference between the performances of two groups
III. Level of significance
α=.05
df= (c-1)(v-1)
푥². 05 = 3.841

89
IV. Statistics: median test for two independent sample
Computation
These data may be tabulated in the form a 2’2 table as follow.
+ - Total
Female 7(a) 5(b) 12(k)
Male 3(c) 6(d) 9(l)
Total 10(m) 11(n) 21(N)
푥² =
푁(푎푑) − (푏푐)²
푘푙푚푛
=
21(7)(6) − (5)(3)²
(12)(9)(10)(11)
=
21(27)²
11880
=
21(729)
11880
=
15039
11880
x² = 1.288
V. Decision Rule: if the 푥² computed value is greater than the 푥² tabular value,
reject Ho.
VI. Conclusion: since 푥² computed value of 1.288 is less than the 푥² tabular value
of 3.841 at .05 level of significant with the 1 degree of freedom, accept the
null hypothesis.

90
CHAPTER VI A SIGN TEST FOR K INDEPENDENT SAMPLES(THE
MEDIAN TEST: MULTI- SAMPLE CASE)
A sign test for K Independent Samples (The Median Test: Multi- Sample Case)
This test is under nonparametric tests. This is a straightforward extension of the
two median tests for two independent samples.
푿ퟐ = Σ
(푶 − 푬)ퟐ
푬
Where:
푿ퟐ = 풄풉풊 − 풔풒풖풂풓풆 풕풆풔풕
푶 = 풐풃풔풆풓풗풆풅 풇풓풆풒풖풆풏풄풊풆풔
푬 = 풆풙풑풆풄풕풆풅 풇풓풆풒풖풆풏풄풊풆풔
Example # 1
A sampling of the number of fish display recorded at three different types of stores
ownership 1.chain store 2. Privately owned 3. Cooperative. The following are the number
of fish display by the three different stores ownership.
CHAIN STORE PRIVATELY
OWNED
COOPERATIVE
30 35 45
12 15 25
15 9 30
9 8 15
5 10 5
29 35 38
32 25 23
17 21 29
25 27 26
15 11 22
Use the median test at 0.05 level of significance to test the null hypothesis that there is no
significant difference among the fish display of the three different types of stores
ownership
Solving the stepwise Method:
I. Problem: Is there a significant difference among the fish display of the three different
types of stores ownership?
II. Hypothesis:
H0: there is no significant difference among the fish display of the three different
types of stores ownership.

91
H1: There is a significant difference among the fish display of the three different
types of stores ownership.
훼 = 0.05
푑푓 = (푐 − 1)(푟 − 1)
푑푓 = (2 − 1)(3 − 1)
푑푓 = (1)(2)
푑푓 = 2 푥2
푡푎푏푢푙푎푟 = 5.991
IV. Statistics:
Median test for k independent samples
Computations:
First, arrange the data from the highest to lowest value.
45
26
38
25
35
25
35
25
32
21
30
23
30
22
29
17
29
15
27
15
15
15
12
11
10
9
9
8
5
5
Second, calculate the median score for the total sample(ignoring group membership)
Note: If there is an odd number in data values, the median is the middle most value.
If there is an even data values, the median is the average of the two middle-most
value.
In this example, have 30 data values, so the median is the average of the values of the two
ퟐퟏ+ퟐퟑ
middle scores. 23 and 22 is the two middle scores. We will find the average =
ퟐ
=22.
Therefore, 22 is the median score.
Third, assign a positive (+) to the values above the median and negative (-) sign to vlues
at or below the median.
Back to the data of three stores.

92
The data may arrange in 2x3 tables as follows:
STORES
OWNERSHIP
Above 22
(+)
At or
below 22
(-)
푻풐풕풂풍풉풐풓풊풛풐풏풕풂풍 풇풓풆풒 .
Observed
freq.
Expected Observed
freq.
Expected
CHAIN 4 5 6 5 10
PRIVATELY
OWNED
4 5 6 5 10
COOPERATIVE 7 5 3 5 10
풕풐풕풂풍 풗풆풓풕풊풄풂풍 풇풓풆풒 . 15 15 Grand Total =30
Solve for Expected.
퐸 =
푇표푡푎푙 푣푒푟푡푖푐푎푙 푓푟푒푞. 푥 푇표푡푎푙 ℎ표푟푖푧표푛푡푎푙 푓푟푒푞.
퐺푟푎푛푑 푡표푡푎푙
퐸 = 15 푥 10
30
푬 = ퟓ
CHAIN
STORE
Sign PRIVATELY
OWNED
Sign COOPERATIVE Sign
30 + 35 + 45 +
12 - 15 - 25 +
15 - 9 - 30 +
9 - 8 - 15 -
5 - 10 - 5 -
29 + 35 + 38 +
32 + 25 + 23 +
17 - 21 - 29 +
25 + 27 + 26 +
15 - 11 - 22 -

93
FIND Median test for k independent samples or 풙ퟐ
풙ퟐ =
Σ(푶 − 푬)ퟐ
푬
푥 2 = (ퟒ−ퟓ)ퟐ
5
+
(ퟒ−ퟓ)ퟐ
5
+ (ퟕ−ퟓ)ퟐ
5
+
(ퟔ−ퟓ)ퟐ
5
+ (ퟔ−ퟓ)ퟐ
5
+
(ퟑ −ퟓ)ퟐ
5
x2 = (−ퟏ)ퟐ
5
+
(−ퟏ)ퟐ
5
+
(ퟐ)ퟐ
5
+
(ퟏ)ퟐ
5
+
(ퟏ)ퟐ
5
+
(−ퟐ)ퟐ
5
x2 = 0.2 + 0.2 + 0.8 + 0.2 + 0.2 + 0.8
퐱ퟐ = ퟐ. ퟒ
V. Decision Rule: If x2 computed value is greater than x2 tabular value, reject H0.
VI. Conclusion:
The x2 computed value of 2.4 is less than x2 tabular value of 5.991 at x2 tabular
value at 0.05 level of significance with 2 degrees of freedom;hence the research
hypothesis is rejected which means that there is no significant difference among the fish
display of the three different types of stores ownership.
Example #2
Consider the following data of 3 countries who mostly got the top five title for Miss
univers.
USA VENEZUEL A PUERTO RICO
15 20 10
25 30 10
18 22 15
20 19 10
10 14 5
significant difference among the three countries mostly get the top-five title for Miss
Universe.
Solving for the Stepwise method
I. Problem: Is there a significant difference in the three countries that mostly get the top
five titles for miss universe?

94
II. Hypothesis:
H0: There is no significant difference among the three countries mostly get the top
five titles for Miss Universe.
H1: There is a significant difference among the three countries mostly get the top
five titles for Miss Universe.
훼 = 0.05
푑푓 = (푐 − 1)(푟 − 1)
푑푓 = (2 − 1)(3 − 1)
푑푓 = (1)(2)
푑푓 = 2 푥2
푡푎푏푢푙푎푟 = 5.991
IV. Statistics:
Median test for k independent samples
Computations:
First, arrange the data from the highest to lowest value.
30
25
22
20
20
19
18
15
15
14
10
10
10
10
5
In this example, have 15 data values, so the median is the middle values. So, 15 is
middle data values. Therefore, 15 is our median score.
Third, assign a positive (+) to the values above the median and negative (-) sign to
values at or below the median.

95
Go back to the data of three countries.
US
A
Sign VENEZUE
L A
Sig
n
PUERTO
RICO
Sig
n
15 - 20 + 10 -
25 + 30 + 10 -
18 + 22 + 15 -
20 + 19 + 10 -
10 - 14 - 5 -
The data may arrange in 2x3 tables as follows:
COUNTRIES Above
15
(+)
At or
below 15
(-)
푻풐풕풂풍풉풐풓풊풛풐풏풕풂풍 풇풓풆풒 .
Observed
freq.
Expected Observed
freq.
Expected
USA 3 2.33 2 2.67 5
VENEZUELA 4 2.33 1 2.67 5
PUERTO RICO 0 2.33 5 2.67 5
풕풐풕풂풍 풗풆풓풕풊풄풂풍 풇풓풆풒. 7 8 Grand Total =15

96
SOLVE FOR EXPECTED:
퐸 =
푇표푡푎푙 푣푒푟푡푖푐푎푙 푓푟푒푞. 푥 푇표푡푎푙 ℎ표푟푖푧표푛푡푎푙 푓푟푒푞.
퐺푟푎푛푑 푡표푡푎푙
퐸 = 7푋5
15
= 2.33, 퐸 = 8푋5
15
= 2.67
FIND Median test for k independent samples or 풙ퟐ
풙ퟐ =
Σ(푶 − 푬)ퟐ
푬
푥 2 = (ퟑ−ퟐ.ퟑퟑ)ퟐ
2 .33
+
(ퟒ−ퟐ.ퟑퟑ)ퟐ
2.33
+ (ퟎ−ퟐ.ퟑퟑ)ퟐ
2.33
+
(ퟐ−ퟐ.ퟔퟕ)ퟐ
2.67
+ (ퟏ−ퟐ.ퟔퟕ)ퟐ
2.67
+
(ퟓ − ퟐ.ퟔퟕ)ퟐ
2.67
x2 = (ퟎ.ퟔퟕ)ퟐ
2.33
+
(ퟏ.ퟔퟕ)ퟐ
2.33
+
(−ퟐ.ퟑퟑ)ퟐ
2.33
+
(−ퟎ.ퟔퟕ)ퟐ
2.67
+
(−ퟏ.ퟔퟕ)ퟐ
2.67
+
(ퟐ.ퟑퟑ)ퟐ
2.67
x2 = 0.19 + 1.20 + 2.33 + 0.17 + 1.04 + 2.03
퐱ퟐ = ퟔ. ퟗퟔ
V. Decision Rule: If x2 computed value is greater than x2 tabular value, reject H0.
VI. Conclusion:
The x2 computed value of 6.96 is greater than x2 tabular value of 5.991 at
x2 tabular value at 0.05 level of significance with 2 degrees of freedom;hence the
research hypothesis is rejected which means that there is no significant difference among
the three countries mostly get the top five titles for Miss Universe.

97
CHAPTER VII The Mc Nemar’s test for correlated proportions
 This belongs to non – paramentric statistics. Mc Nemar’s test is design to test if
there is a significant change between the before and after situations. The formula
is:
풙ퟐ =
(풃 − 풄)ퟐ
풃 + 풄
Where:
풙ퟐ = chi – square test
b = the first cell of the 2nd column in a 2x2 table
c = the first cell of the 2nd row in a 2x2 table
Example
Data on drinking water before and after the meal is served for a sample of 100 customers
in an authentic restaurant.
Drinking water
regularly
before the meal
serve
Water drink regularly after the meal is served TOTAL
Yes No
Yes a = 50 b = 5 55
No c = 18 d = 14 32
TOTAL 68 19 100
SOLVING USING STEPWISE METHOD
I. Problem: Is there a significant difference between drinking water before and after the
meal serve in the costumers?
II. Hypothesis:
Ho = there is no significant difference between drinking water before and after the meal is
serve.
Hi = there is a significant difference between drinking water before and after the meal is
serve.
α = .05
df = (c – 1)(r – 1)
= (2 – 1)(2 – 1)
= (1)(1)
= 1
X2.05 = 3.841

98
IV. Statistics: The Mc Nemar’s test for correlated proportion
Computation:
풙ퟐ =
(풃 − 풄)ퟐ
풃 + 풄
풙ퟐ =
(ퟓ − ퟏퟖ)ퟐ
ퟓ + ퟏퟖ
풙ퟐ =
(−ퟏퟑ)ퟐ
ퟐퟑ
풙ퟐ =
ퟏퟔퟗ
ퟐퟑ
풙ퟐ = ퟕ. ퟑퟓ
V. Decision Rule: If the 푥 2 computed is greater than the 푥 2 tabular, reject Ho.
VI. Conclusion: Since the 푥 2 computed value of 7.347 is greater than the 푥 2 tabular
value of 3.841 at 0.05 level of significance. Therefore, we reject the null hypothesis.
Where there is a significant difference between drinking water regularly before and after
the meal is being served.

99
CHAPTER VIII FRIEDMAN Fr TEST FOR RANDOMIZED BLOCK DESIGN
 is a nonparametric test
 used for comparing the distributions of measurements for k treatments laid
out in b blocks using randomized block design
 the procedure is similar to Kruskal-Wallis H-Test
 we will use the chi-square tabular
Formula:
Fr= 12 ΣTi2- 3b(k+1)
bk(k+1)
Where:
Fr=Friedman test
b=number of blocks
k=number of treatments
i=1, 2,…k
Example:
1. A researcher is studying the effects of taking medicine in children, 6 samples of young
children were used as subjects to assess their reaction to the taste of the medicine. Their
response was measured from sad (low score) to happy (high score). The minimum was 0
and 10 was the maximum. The following data are recorded:
Medicine
Child 1 2 3 4 5
1 2.7 3.1 7.2 6.4 8.1
2 4.5 4.6 3.0 9.7 6.4
3 1.8 4.9 3.8 4.2 5.7
4 8.0 7.4 9.4 6.4 6.5
5 7.2 6.9 4.8 5.9 8.2
6 3.2 2.8 2.6 4.3 4.9
Solution using Stepwise method:
I. Problem: Is there a significant difference in the reaction of 6 young children
on 5 different medicines.

100
II. Hypotheses:
Ho: There is no significant difference in the reaction of 6 young children on 5
different medicines.
Ha: There is a significant difference in the reaction of 6 young children on 5
different medicines.
α= .05
df=k-1
=5-1
=4
X2.05=9.488, it is found at the chi-square tabular value
Medicine
Child 1 2 3 4 5
1 2.7 (1) 3.1 (2) 7.2 (4) 6.4 (3) 8.1 (5)
2 4.5 (2) 4.6 (3) 3.0 (1) 9.7 (5) 6.4 (4)
3 1.8 (1) 4.9 (4) 3.8 (2) 4.2 (3) 5.7 (5)
4 8.0 (4) 7.4 (3) 9.4 (5) 6.4 (1) 6.5 (2)
5 7.2 (4) 6.9 (3) 4.8 (1) 5.9 (2) 8.2 (5)
6 3.2 (3) 2.8 (2) 2.6 (1) 4.3 (4) 4.9 (5)
Rank Sum T1=15 T2=17 T3=14 T4=18 T5=26
Rank the 5 reactions (treatment) of every child (block) from lowest to highest. From the
first child, the lowest is 2.7 so it is rank 1 ant the highest is 8.1 that is why it is rank 5.
We will continue it until the 6th child. Next, add the corresponding rank of the 1st
medicine that is 1+2+1+4+4+3=15. You will continue until the 5th medicine.
Then, apply it to the formula:
Fr= 12 ΣTi2- 3b(k+1)
bk(k+1)
= 12 [152+172+142+182+262] – 3(6) (6)
(6) (5) (5+1)
= 12 [225+289+196+324+676] - 108
180
= (0.07) (1710) – 108
= 119.7 – 108
Fr = 11.7

101
V. Decision Rule:
If the value of Fr is greater than the X2 tabular value, reject Ho.
VI. Conclusion:
Since the Fr value of 11.7 is greater than the X2 tabular value of 9.488 at .05 level
of significance with 4 degrees of freedom, the null hypothesis of no significant difference
in the reaction of 6 young children of the 5 different medicines was rejected.
Since the Fr value of 7.32 is lesser than the X2 tabular value of 7.815 at .05 level
of significance with 3 degrees of freedom, the null hypothesis of no significant difference
in the scores of the 5 students in 4 subjects was accepted.

102
CHAPTER IX KENDALL’S COEFFICIENT
 Test used to find if there is an agreement or concordance among ratters’ and
judges of N objects or individuals.
FORMULA
푊 = 12 Σ 퐷2
푚2(푁) (푁2−1)
Where:
W= the coefficient of concordance
D= the difference between the individual sum of ranks of the ratters’ or judges
and the average of the sum of rank of the object or individuals
Σ 퐷2= the sum of squares of the difference judges or ratters’
m= judges or ratters’
N= objects or individuals being rated and ranked.
Example:
The data’s on the ranking of 10 portfolios by 4 judges.
Individual
projects
Judge’s Ranks
J E A N
1 2 1 6 5
2 3 6 2 9
3 6 3 9 8
4 5 9 3 2
5 9 2 5 4
6 1 5 10 3
7 10 4 8 7
8 4 8 7 6
9 8 7 1 10
10 7 10 4 1

103
STEPWISE METHOD
I. Problem: Is there an agreement or concordance of 4 judges regarding the 10
portfolio?
II. Hypotheses:
Ho: There is no agreement or concordance of the 4 judges regarding to the 10
portfolio.
Ha: There is an agreement or concordance of the 4 judges regarding to the 10
portfolio.
III. Level of significance:
α= 0.05 W.05=.44
df= m=4; N=10
IV. Statistics:
W coefficient of concordance
Individual
projects
Judge’s Ranks Sum of
ranks
푅̅
푅̅
-sum of
rank
D
J E A N D2
1 2 1 6 5 14 8 64
2 3 6 2 9 20 2 4
3 6 3 9 8 26 -4 16
4 5 9 3 2 19 3 9
5 9 2 5 4 20 2 4
6 1 5 10 3 19 3 9
7 10 4 8 7 29 -7 49
8 4 8 7 6 25 -3 9
9 8 7 1 10 26 -4 16
10 7 10 4 1 22 0 0
ΣR= 220 퐷̅
= 220
10
ΣD2= 180
퐷̅ = 22
Solution: Add the ranks of the 4 judges J, E, A, and N of the 10 individual portfolios;
place them under column Sum of Ranks. Get the summation of the Sum of Ranks by
dividing it by 10, the number of portfolio. The averages are 22 and subtract it from the
individual sum of ranks of the ten portfolios and place them under column D. Square the
difference and place them under D2.
To compute W, we used the formula:
W = 12 Σ D2
m2(N)(N2−1)
W = 12 (180)
42(10)(102−1)

104
W = 2,160
(16)(10)(99)
W = 2,160
158,404
W = 0.0136
V. Decision rule: If the computed W is greater than the tabular value reject Ho.
VI. Conclusion: The computed W of 0.0136 is lesser than the tabular value of
.44 at .05 level of significance with m=4 and N=4 degrees of freedom, the null
hypothesis is accepted in favour of the research hypothesis. This means that
there is no agreement or concordance of the 4 judges regarding the 10
portfolio. It implies that portfolio number 1 is rank 1 while portfolio number 7
is the last rank.

105
ASSESSMENT
TEST I
1. The nicotine content of two brands of cigarettes measured in milligrams, are as
follows:
Brand x 4.1 0.7 3.1 2.5 4.0 6.2
Brand y 2.1 4.0 5.4 4.8 3.3 1.6 1.7 5.4
Test the hypothesis, at .05 level of significance, that the average nicotine contents
of the two brands are equal.(use Wilcoxon Two-Sample Test)
2. A clinical trial is run to assess the effectiveness of a new anti-retroviral therapy for
patients with HIV. Patients are randomized to receive a standard anti-retroviral therapy
(usual care) or the new anti-retroviral therapy and are monitored for 3 months. The
primary outcome is viral load which represents the number of HIV copies per milliliter of
blood. A total of 30 participants are randomized and the data are shown below. α=0.05.
.(use Wilcoxon Two-Sample Test)
Standard therapy New therapy
7500 400
8000 250
2000 800
550 1400
1250 8000
1000 7400
2250 1020
6800 6000
3400 920
6300 1420
9100 2700
970 4200
1040 5200
670 4100
400 Undetectable
Total :15 Total :15

106
3. The data below are the sample from Data Set 16 in Appendix B. Test at .05 to see if
the braking distances have the same mediums..(use Wilcoxon Two-Sample Test)
4 cylinders 6 cylinders
136 131
146 129
139 127
131 146
137 155
144 122
133 143
144 133
129 128
144 146
130 139
140 136
135
4.(Use the spearman rank order)
The table below shows the scores of the students in their test in mathematics.
Grades of 1A Grades of 1B
20 9
19 8
18 10
17 12
16 14
15 15
12 13
11 8
10
5. (Use the spearman rank order)
The table below shows the amount of chocolate bar in two stores.
Store 1 5 10 15 20 25 30 8 7 3
Store 2 5 6 7 20 35 40 2

107
6. (Use the spearman rank order)
The table below shows the weight of female and male
Female 45 55 54 51 61 59 46 58 50 60
male 47 63 56 52 49 57 53 62 48
7. Consider the salary data of three kinds of job or work. (Use Sign Test for K
Independent Samples)
NURSE ENGINEER SEAMAN
8000 25000 50,000
5000 20000 85,000
5500 18000 100,000
8500 22000 120,000
7000 20000 75,000
9000 95,000
significant difference among the salary of three kind of job.
2. Consider the number of LET examiners of three different schools. (Use Sign Test for
K Independent Samples)
University of Santo Tomas CvSU UP
200 200 300
150 300 500
300 250 600
285 375 523
158 190 355
185 380
significant difference among the NUMBER OF LET Examiners of three different
schools.

Non Parametric Statistics

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