75
Lesson 2

Comparison of
Means
Comparison of Means
1. T-test Dependent- comparing means of one sample measured twice ex
degree of friendliness of students before and after viewing a film on
friendliness
2. T-test Independent-comparing two different groups ex. Louisians and
Lormanians
3. Analysis of Variance-ANOVA- used to compare 3 groups ex. Atenean, UPians
and Lassalians
Follow-up tools:
Tukeys and Scheffe Methods- determine where significant differences lie
T-TEST DEPENDENT
T-Test Dependent
The dependent t-test (also called the paired t-test or paired-samples ttest) compares the means of two related groups to determine whether there is
a statistically significant difference between these means.
It is comparing means of one sample measured twice example ex. degree
of friendliness of students before and after viewing a film on friendship
Steps:
a. Find the mean for each point in time
̅ =
̅
=
b. Find the standard deviation for the difference between time 1 and
time 2 using the formula
SD = √
( ̅
̅ )
SEME 112 – Advanced Statistics
Module III
76
c. Find the standard error of the mean difference using the formula
̅ =
√
d. Translate the sample mean difference into units of standard error of
the mean difference using the formula
t=
̅
̅
̅
e. Find the degrees of freedom using the formula
df= n-1 where n is the number of cases
f. Compare the computed t value with the appropriate t ctitical in the
T table (Table A) considering degrees of freedom and and alpha value
( ) which is .05. Use also two-tailed when the research question is
non-directional and one-tailed if the research question is directional.
Illustrative Example:
Below are results of test before and after the remedial class in statistics. Is
there a significant difference in the performance of students before and after
remediation?
Before
After
58
66
63
68
66
72
70
76
63
78
51
56
44
69
58
55
50
55
Solution:
Construct the research question, null and alternative hypothesis.
RQ: Is there a significant difference in the performance of students before and
after the remediation?
Ho: There is no significant difference in the performance of students before
and after the remediation.
H1: There is a significant difference in the performance of students before and
after the remediation.
It is two-tailed.
The level of significance is =.05
The statistical tool is t test dependent.
SEME 112 – Advanced Statistics
Module III
77
Respondent
Before (x1)
58
63
66
70
63
51
44
58
50
After (x2)
66
68
72
76
78
56
69
55
55
∑x1= 523
∑x2 =595
Difference
(D) =(x1-x2)
D squared (D2)
(x1-x2)2
-8
-5
-6
-6
-15
-5
-25
3
-5
2
64
25
36
36
225
25
625
9
25
∑(x1-x2) =∑ D
=1070
2
Steps:
a. Find the mean for each point in time (Before and After)
̅ =
̅
̅ =
=
̅
=
58.11
̅
=
̅
=
66.11
b. Find the standard deviation for the difference between time 1 and
time 2 using the formula
SD = √
( ̅
SD = √
(
̅ )
)
SD = 7.41
c. Find the standard error of the mean difference using the formula
̅ =
̅ =
√
√
̅ =
SEME 112 – Advanced Statistics
Module III
78
d. Translate the sample mean difference into units of standard error of
the mean difference using the formula
t=
̅
̅
̅
t=
t=
Take the absolute value of t, then
t=
e. Find the degrees of freedom using the formula
df= n-1 where n is the number of cases
df=9-1=8
g. Compare the computed t value with the appropriate t ctitical in the
T table (Table A) considering degrees of freedom and alpha value ( )
which is .05 . Use also two-tailed since the research question is nondirectional.
Note: If computed t (t stat/t value) is greater than the t critical, then reject
the null hypothesis
Variables Means
Before
After
58.11
66.11
T value/ T critical
t stat
3.05
2.306
Df
Decision
Remarks
8
Reject
Ho
Significant
Conclusion: The students performed significantly better after the remediation,
hence, the remediation is effective.
T-TEST INDEPENDENT
The independent t-test, also called the two sample t-test, independentsamples t-test or student's t-test, is an inferential statistical test that
determines whether there is a statistically significant difference between the
means in two unrelated groups.
It is comparing two different groups ex. Louisians and Lormanians
SEME 112 – Advanced Statistics
Module III
79
Steps:
a. Find the mean for each sample
̅ =
̅ =
b. Find the variance for each sample
S 12 =
S2 2 =
- ̅
- ̅
c. Find the standard error of difference between means
= √(
d. t stat =
̅
)(
)
̅
e. degrees of freedom df= N1 + N2 -2
f. Determine t critical
Using the t table , = .05 and degrees of freedom, get the t
critical.
Use also two-tailed when the research question is non-directional
and one-tailed if the research question is directional.
g. Compare, decide, interpret
Illustrative Example:
Ex. The data below is the result of the quiz in Chemistry of two groups of
students subjected to two varied teaching techniques, the CAI and the
traditional method. Test if there exists a significant difference in their
performances.
CAI
13
Trad 7
15
9
11
6
14
7
10
5
12
6
10
5
10
4
11
5
14
6
Solution:
Construct the research question, null and alternative hypothesis.
RQ: Is there a significant difference in the quiz scores of students under the CAI
and traditional method?
Ho: There is no significant difference in the quiz scores of students under the
CAI and traditional method.
SEME 112 – Advanced Statistics
Module III
80
H1: There is a significant difference in the quiz scores of students under the
CAI and traditional method.
It is two-tailed.
The level of significance is =.05
The statistical tool is t test independent.
CAI (x1)
13
15
11
14
10
12
10
10
11
14
Trad (x2)
7
9
6
7
5
6
5
4
5
6
120
60
(x1)2
(x2)2
169
225
121
196
100
144
100
100
121
196
1472
49
81
36
49
25
36
25
16
25
36
378
a. Find the mean for each sample
̅ =
̅ =
̅ =
̅ =
̅ =
̅ = 6
b. Find the variance for each sample
S 12 =
- ̅
S2 2 =
- ̅
S 12 =
-
S 22 =
-
S12 = 147.2 -144
S22 =37.8 -36
S12 = 3.2
S22 = 1.8
SEME 112 – Advanced Statistics
Module III
81
c. Find the standard error of difference between means
= √(
)(
(
=√(
)
(
)
)
)(
(
)
)
= .75
d. t stat =
t stat =
t stat = 8
̅
̅
-
e. Determine degrees of freedom df= N1 + N2 -2
df= 10 + 10 -2
df = 18
f. Determine t critical
Using the t table , = .05 and degrees of freedom = 18.
Use also two-tailed since the research question is non-directional.
t critical is 2.101
Variables Means
CAI
Trad
T value/ T critical
t stat
12
6
8
2.101
df
18
Decision
Remarks
Reject
Ho
Significant
Conclusion: Since, there is a significant difference in the performance of
students under CAI and Traditional teaching method, CAI method is
significantly more effective than traditional method.
ANALYSIS OF VARIANCE
Analysis of Variance-ANOVAThe one-way analysis of variance (ANOVA) is used to determine whether
there are any statistically significant differences between the means of two or
more independent (unrelated) groups (although you tend to only see it used
when there are a minimum of three, rather than two groups). For example, you
could use a one-way ANOVA to understand whether exam performance differed
SEME 112 – Advanced Statistics
Module III
82
based on test anxiety levels amongst students, dividing students into three
independent groups (e.g., low, medium and high-stressed students). It is used
to compare 3 or more groups ex. Atenean, UPians and Lassalians
a. Find the mean for each sample
̅ 1=
̅2 =
̅3 =
b. 1. ∑xtotal = ∑x1+∑x2 +∑x3
2. ∑x2total = ∑x1 2+∑x2 2 +∑x32
3. Ntotal = N1 + N2 + N3
4. ̅ total =
c. SStotal = ∑x2total - Ntotal ̅ 2total
d. SSwithin = ∑x2total -∑(Ngroup ̅ 2group)
e. SSbetween =∑(Ngroup ̅ 2group) - Ntotal ̅ 2total
f. dfbetween = k-1
g. dfwithin = Ntotal –k
h. MSwithin =
i. MSbetween =
j. F =
Illustrative Example:
Ex. The data below show the number of children of families in 3 different
religions. Test if religious affiliation has an effect on family size.
Protestant
2
5
4
3
5
Catholic
6
7
8
6
4
Jewish
3
2
4
4
3
Solution:
Construct the research question, null and alternative hypothesis.
SEME 112 – Advanced Statistics
Module III
83
RQ: Is there a significant difference in the number of children of families
coming from the protestant, catholic and jewish ?
Ho: There is no significant difference in the number of children of families
coming from the protestant, catholic and jewish.
=
H1: There is a significant difference in the number of children of families
coming from the protestant, catholic and jewish .
It is two-tailed.
The level of significance is =.05
The statistical tool is Analysis of Variance.
Assume the number of children of protestant as x 1, catholic x2 and jewish x3
Protestant (x1)
2
5
4
3
5
∑x1=19
(x1)2
4
25
16
9
25
2
∑(x1) =79
Catholic (x2)
6
7
8
6
4
∑(x2)=31
(x2)2
36
49
64
36
16
2
∑(x2) =201
a. Find the mean for each sample
̅ 1=
̅2 =
̅ 1=
̅2 =
̅ 1=
̅ 2 = 6.2
Jewish (x3)
3
2
4
4
3
∑(x3)=16
(x3)3
9
4
16
16
9
3
∑(x3) =54
̅3 =
̅3 =
̅ 3 = 3.2
b. 1. ∑xtotal = ∑x1+∑x2 +∑x3
∑xtotal = 19 + 31 + 16
∑xtotal = 66
2. ∑x2total = ∑x1 2+∑x2 2 +∑x32
∑x2total = 79 + 201 + 54
∑x2total = 334
3. Ntotal = N1 + N2 + N3
Ntotal = 5 + 5 + 5
Ntotal = 15
SEME 112 – Advanced Statistics
Module III
84
4. ̅ total =
̅ total =
̅ total = 4.4
c.
SStotal = ∑x2total - Ntotal ̅ 2total
2
SStotal = 334 – 15 (4.4)
SStotal = 43.6
d.SSwithin = ∑x2total -∑(Ngroup ̅ 2group)
SSwithin = 334 – [5(3.8)2 + 5(6.2)2 + 5(3.2)2]
SSwithin = 18.4
e.SSbetween =∑(Ngroup ̅ 2group) - Ntotal ̅ 2total
SSbetween = [5(3.8)2 + 5(6.2)2 + 5(3.2)2]- 15 (4.4)2
SSbetween = 25.2
f. dfbetween = k-1 where k is the number of groups
dfbetween = 3-1
dfbetween = 2
g. dfwithin = Ntotal –k
dfwithin = 15-3
dfwithin = 12
h. MSwithin =
MSwithin =
MSwithin = 1.53
SEME 112 – Advanced Statistics
Module III
85
i. MSbetween =
MSbetween =
MSbetween = 12.6
j. F =
F=
F = 8.24
SS
25.2
18.4
Between
Within
Df
2
12
MS
12.6
1.53
Fvalue
8.24
Fcrit
3.88
Decision
Reject
Ho
(To find the Fcritical, use the anova table or F table (Table B) using
degrees of freedom and alpha which is .05.)
Since Fv>Fc, Reject Ho or the null hypothesis. There is a significant
difference among the number of children of families from different religious
sect. Therefore, religious affiliation has an effect on family size.
A Multiple Comparison of Means
Tukey’s HSD (honest significant difference) and Scheffee Methods are
used only after a significant F ratio has been obtained to determine where
significant differences lie
Tukey’s – used when there are the same number of respondents in each group
Scheffee – used when the number of respondents are equal or not
HSD = q √
Where:
is the number of subjects per group
q table value (Table C)
Note: Get the difference between means. If the difference is greater than the
HSD, then the difference is significant.
Illustrative Example:
Using the result of the Analysis of Variance in the example above, since
there is a significant difference among the number of children of families from
SEME 112 – Advanced Statistics
Module III
86
different religious sect, it is a requirement to determine where the differences
are by using HSD.
Step 1. Get the q value in the Q table for HSD using df within, alpha which is
.05 and k as the number of treatment.
Df within= 12
k=3
q= 3.77
2.Compute for the HSD
HSD = q √
Where q= 3.77 (Table C)
MSwithin = 1.53
Ngroup = 5
HSD = 3.77 √
HSD = 2.08
̅ 3 =3.2 (Jewish)
̅ 3 =3.2
̅ 1 = 3.8
̅ 2 = 6.2
----------------------------
̅1 =
3.8(Protestant)
.6
-------------------
̅ 2 = 6.2 (Catholic)
3.0
2.4
--------
Differences
̅ 1 and ̅ 2 = 2.4
̅ 1 and ̅ 3 =.6
̅ 2 and ̅ 3 = 3
3. Compare differences with HSD
To interpret: If the difference is greater than HSD, then the difference is
significant.
Between Protestant and Catholic: ̅ 1 and ̅ 2 = 2.4 > HSD, then significant
Between Protestant and Jewish: ̅ 1 and ̅ 3
=.6 < HSD, then not
significant
Between Catholic and Jewish:
̅ 2 and ̅ 3 = 3 > HSD, then significant
Conclusion: There is a significant difference in the number of children
between families of Protestants and Catholics as well as Jewish and
Catholic but no significant difference between Protestant and Jewish.
SEME 112 – Advanced Statistics
Module III
88
Lesson 3

Linear Correlation
LINEAR CORRELATION- is a tool specifically used to measure degree of
relationship or association between two variables
Research studies deal with relationship between two or more variables.
Ex educational attainment and teaching competence.
Kinds of Relationship
1. Positive (Direct)
A direct or positive relationship between two variables means that
an increase in value of one variable corresponds to an increase in the
value of the other variable. If one variable increases the other also
increases and when one variable decreases the other also decreases.
For example, height of the person and the length of the shadow. The
taller the person is the longer is the length of the shadow. Other
variables which highly correlate are aptitude and academic
achievement, educational attainment and socioeconomic status. (The
graph is shown below.) The graph is inclined upward to the right.
2. Negative (Inverse)
An inverse or negative relationship between two variables means
that an increase in the value of one variable corresponds to a
decrease in the value of the other. As one variable increases, the
other variable decreases and vice versa. For example as supply
increases, price decreases and conversely. The graph is inclined
upward to the left.
3. No Correlation or Zero Relationship
A zero relationship between two variables means that if one
variable has a high value, the other variable may either have a high
or a low value. If there is no relationship between the two variables
such that the value of one variable changes and the other variable
remains constant, it is called no or zero correlation. This means that
the two variables have no association. For example height has
nothing to do with teaching competence. It does not necessarily
SEME 112 – Advanced Statistics
Module III
89
mean that if the teacher is tall, then he/she is highly competent in
teaching. The points are scattered and do not cluster to a specific
line.
https://www.emathzone.com/tutorials/basic-statistics/positive-and-negativecorrelation.html
The graph of the relationship of two variables is called scatter diagram.
( Refer to the graph above) The x axis includes the independent variable and
the y axis is the dependent variable. The graph will only utilize quadrant 1 in
the rectangular coordinate since the values of x and y are both positive. (
There are no negative lengths, distance or teaching competence.)
Coefficient of correlation –a numerical measure of the linear relationship
between two variables. It is denoted as ―r‖ (It can be positive or negative.)
Scale in interpreting coefficient of correlation:
1
Perfect relationship
0.91- 0.99
Very high correlation
0.71- 0.90
High correlation
0.51- 0.70
Moderate Correlation
0.31- 0.50
Low correlation
0.01- 0.30
Negligible Correlation
0
No relationship
Method 1. Pearson Product Moment Correlation ( rxy)
By assuming linear relationship between two quantities x and y, the
famous British statistician, Carl Pearson derived a formula for finding the
correlation between x and y expressed as a number. The formula is named in
his honor, the Pearson Product Moment Correlation
SEME 112 – Advanced Statistics
Module III
90
(
rxy = r =
√
)
(
(
) (
)
)
–(
)
Where: x = observed data for the independent variable
y = observed data for the dependent variable
n = sample size
r = degree of relationship between x and y
t significance of r
This tool is used to determine if r is significant.
 t= r √
In testing the significance of r-value, compare the computed t value
with that of the tabular value at .05 level of significance with df = n-2.
Illustrative Example:
Given the Algebra and Statistics grades of 10 students, determine if
there exists a significant relationship between the grades in the two subjects.
Algebra
Statistics
75
82
80
78
93
86
65
72
87
91
71
80
98
95
68
72
84
89
77
74
Step 1. Construct the research question, null and alternative hypothesis.
RQ: Is there a significant relationship between the Algebra and Statistics
grades of students?
Ho: There is no significant relationship between the Algebra and
Statistics grades of students.
H1: There is a significant relationship between the Algebra and Statistics
grades of students.
Step 2. Complete the column for x2, y2 and xy
Algebra (x)
75
80
93
65
87
71
98
x2
5625
6400
8649
4225
7569
5041
9604
SEME 112 – Advanced Statistics
Statistics (y)
82
78
86
72
91
80
95
y2
6724
6084
7396
5184
8281
6400
9025
Xy
6150
6240
7998
4680
7917
5680
9310
Module III
91
68
84
77
4624
7056
5929
72
89
74
5184
7921
5476
4896
7476
5698
∑x=798
∑x2 = 64722
∑y =819
∑y2 =67675
∑xy = 66045
Step 3. Use the formula to solve for the coefficient of correlation
(
rxy = r =
√
)
(
(
)
√
(
)
–(
(
rxy = r =
) (
)
) (
(
)
)
) (
)
(
) –(
)
rxy = r =
rxy = r = 0.87
There is a high positive relationship between Algebra and Statistics grades of
students.
Testing the significance of r
Although the value or r obtained is high (0.87 or 87%), this is not yet an
assurance that it is statistically ssignificant. Hence, it needs to be tested using
t-significance of r
 t= r √
r= .87
n=10
 t= r √
SEME 112 – Advanced Statistics
Module III
92
 t= 0.87 √
 t= 4.99
Step 4. Obtain the t critical value using the T table (Table A)
With df= n-2 and =.05, the t critical = 2.306
Step 5. Compare the t stat or the computed t with the t critical
Since tstat > tcrit, the decision is to reject the null hypothesis
(Ho).
Step 6. Formulate the conclusion.
Algebra grade is highly and positively correlated with the
Statistics grade. This implies that the students who have high grades in
Algebra have also high grades in Statistics and conversely. This is
indicative further that if a student performs well in Algebra, he/she will
likely to perform well also in Statistics. The two subjects are areas of
Math. If a student is good in one Math subject, there is a high probability
that he will also be good in all the other Math subjects. Meanwhile, for
the students who do not perform well in basic mathematics, they will
not also perform well in all the other higher Math subjects.
Method 2. Spearman rank correlation coefficient or Spearman rho
Spearman rank correlation coefficient or Spearman rho may be used to
correlate the scores. It is named after Charles Spearman, is a nonparametric
measure of rank correlation.
The spearman rank correlation coefficient is based on the ranked values
for each variable rather than the raw data. It is often used to evaluate
relationships involving ordinal variables. Used it when two variables are
ranked.
The formula is
rs = 1-
–
Where: Σ D2 =the sum of the squared difference between ranks
N
= the total number of cases
SEME 112 – Advanced Statistics
Module III
93
Steps:
1. Construct the research question, null and alternative hypothesis.
2. Rank the scores in each group from highest to lowest.
3. Solve for D column. D is the difference in ranks.
4. Square D and get the sum.
5. Solve for r and interpret.
6. Test for the significance of r
7. Formulate conclusion
Illustrative Example:
The following are the scores of students in a Math and Physics tests. Using
Spearman Rank Order Correlation, compute the coefficient of correlation (r s),
determination (r2) and alienation √(
) . Interpret the results.
Math
Physics
60 54 53 49 49 47
60 68 40 52 51 38
46 45 45 45
51 32 39 41
SOLUTION:
1. Construct the research question, null and alternative hypothesis.
RQ: Is there a significant relationship between the scores of students in
Mathematics and Physics?
Ho: The is no significant relationship between the scores of students in
Mathematics and Physics.
H1: There is significant relationship between the scores of students in
Mathematics and Physics.
2. Rank the scores in each group from highest to lowest.
3. Solve for D column. D is the difference in ranks.
4. Square D and get the sum.
Individual Math
Rankmath Physics
1
2
3
4
5
6
7
1
2
3
4.5
4.5
6
7
60
54
53
49
49
47
46
SEME 112 – Advanced Statistics
60
68
40
52
51
38
51
Rankphysics Difference Square of
in
ranks the
(D)
difference
in ranks
(D2)
2
1
1
1
1
1
7
4
16
3
1.5
2.25
4.5
0
0
9
3
9
4.5
2.5
6.25
Module III
94
8
9
10
45
45
45
9
9
9
32
39
41
10
8
6
1
1
3
1
1
9
∑D2 =46.5
5. Solve for r and interpret.
rs = 1rs = 1-
–
(
)
–
rs = 1- 0.281818…
rs = 0.72
Math and Physics are positively and highly correlated. Those with
high Math grades have also high Physics grades and vice versa.
6. Test for the significance of rs using
 t= rs √
 t= .72 √
(
)
 t= 2.93
Obtain the t critical value using the T table (Table A)
With df= n-2 =8 and =.05, the t critical = 2.306
Since tstat > tcrit, the decision is to reject the null hypothesis (Ho).
7. Conclusion:
Mathematics score is highly and positively correlated with Physics
score. This implies that the students who have high scores in
Mathematics have also high scores in Physics and conversely. This is
indicative further that if a student performs well in Mathematics, he/she
will likely to perform well also in Physics. If a student is good in Math,
there is a high chance that he will also be good in Physics. Meanwhile,
for the students who do not perform well in Mathematics, they will not
SEME 112 – Advanced Statistics
Module III
95
also perform well in Physics. Mathematics is the language of Physics.
One has to understand Mathematics well to understand Physics
competently.
The scatter diagram shows a highly positive relationship between math
and physics. Those who are good in Math are also good in Physics.
THINK
Answer the following exercises using the appropriate
statistical tool.
1. A research study was conducted to determine the correlation between
students grades in English and their grades in Mathematics. A random sample of
10 students of Business Administration of a certain university were taken and
the results of the sampling are tabulated below.
English grade
93 89 84 91 90 83 75 81 84 77
Math grade
91 86 80 88 89 87 78 78 85 76
SEME 112 – Advanced Statistics
Module III
98
Lesson 4

Linear Regression
LINEAR REGRESSION
Linear regression is a basic and commonly used type of predictive
analysis. The overall idea of regression is to examine two things: (1) does a set
of predictor variables do a good job in predicting an outcome (dependent)
variable? (2) Which variables in particular are significant predictors of the
outcome variable, and in what way do they–indicated by the magnitude and
sign of the beta estimates–impact the outcome variable?
̂=a +bx
where: ̂ = predicted value
a = y intercept
b= slope of the line (regression coefficient)
To find the y-intercept (a)
a = ̅ – b ̅ where ̅ = mean of x values
̅ = mean of y values
To find the slope (b)
b=
–(
(
)(
)–(
)
)
Example 1. Using the data below, what would be the estimated grade of a
student in Mathematics if his grade in English is 90? What regression equation
can be used?
English Grade (x)
Math Grade (y)
Xy
93
91
8463
8649
89
86
7654
7921
84
80
6720
7056
91
88
8008
8281
90
89
8010
8100
83
87
7221
6889
75
78
5850
5625
81
78
6318
6561
84
85
7140
7056
77
76
5852
5929
847
838
71236
72067
English Grade (x) 9389
Math Grade (y) 91 86
84
80
∑x=847
n=10
∑xy=71236
SEME 112 – Advanced Statistics
91
88
90
89
83
87
75
78
81
78
84
85
77
76
Module III
99
∑x2 = 72067 ∑y=838
b= 0.79
̅= 84.7
̅=83.8
a= 16.89
Given the formula for the slope
b=
b=
–(
(
(
)(
)–( )
)–(
(
)–(
)
)(
)
)
b= 0.79
To solve for a, use the formula
a= ̅–b ̅
a= 83.8- 0.79(84.7)
a = 16.89
Using a and b, and the regression equation ̂ = a+bx, then the regression
equation is
̂ = 16.89 + 0.79x
Therefore, ̂ = 16.89 + 0.79x
but x=90
y= 16.89 + 0.79 (90)
y= 87.99 or 88 predicted grade in Mathematics
The graph is presented below:
SEME 112 – Advanced Statistics
Module III
100
Standard Error of Estimate
The standard error of estimate is the measure of variation of an
observation made around the computed regression line. It is used to check the
accuracy of predictions made with the regression line.
The smaller the value of a standard error of estimate, the closer are the
dots to the regression line and the better is the estimate based on the equation
of the line.
If the standard error is zero, then there is no variation corresponding to
the computed line and the correlation will be perfect.
Syx =√
(
)
Where: y = each of the value in the given
ŷ = predicted y using regression equation
English
(x)
93
89
84
91
90
83
75
81
84
77
Grade
Math
(y)
91
86
80
88
89
87
78
78
85
76
Grade
̂
90
87
83
89
88
82
76
81
83
78
(y- ̂)2
y- ̂
1
-1
-3
-1
1
5
2
-3
2
-2
2
1
1
9
1
1
25
4
9
4
4
∑(y- ̂) = 59
Using the formula for the standard error of estimate
Syx =√
(
)
Syx =√
Syx = 2.72 or 3
SEME 112 – Advanced Statistics
Module III
MODULE 4
NON-PARAMETRIC STATISTICS
Lesson 1
Chi-Square
Lesson 2
Mann-Whitney U test
Lesson 3
Wilcoxon Matched Pair Test
Lesson 4
Kruskall Wallis
Lesson 5
Friedman’s ANOVA
120

This module includes nonparametric statistical tools which are used in
research. These are chi-square, mann-whitney test, wilcoxon signed rank test,
kruskall wallis and friedman’s analysis of variance. These are the statistical
tools which can be considered as the counterpart of the parametric statistical
tools which are used in research. The use of technology and software in
Statistics will already be integrated in each of the lessons.
OBJECTIVES
After studying the module, you should be able to:
1. Determine the uses of each non parametric statistical tool.
2. Utilize the appropriate statistical tool in solving a problem.
3. Determine the existence of a relationship between/among two or
more variables
4. Compare two or more group scores on a variable
5. Utilize the appropriate table to obtain the critical value
6. Based on computed statistic and the specified critical value, make
a decision on Ho.
7. Formulate conclusion and interpret the results of a problem after
making a decision.
8. Utilize Megastat in treating data using non-parametric statistics.

DIRECTIONS/ MODULE ORGANIZER
There are 5 lessons in the module. Read each lesson carefully, follow the
steps in the given illustrative examples, then answer the exercises/activities to
find out how much you have learned from it. Work on these exercises carefully
and submit your output to your teacher.
In case you encounter difficulty, discuss this with your teacher during
the face-to-face meeting.
Good luck and enjoy solving!!!
SEME 112 – Advanced Statistics
Module IV
121
Lesson 1

Chi-Square
THE CHI-SQUARE TEST (X2)
It is particularly useful in tests involving cases where persons, events or
objects are grouped in two or more nominal categories such as yes or no,
approve-undecided-disapprove or class A, B, C, D. It is used to compare the
observed and expected.It is also a tool used to determine relationship and
differences.
Formula:
(
)
x2 =
Where: O= observed frequency
E= expected frequency
Df= (r-1) (c-1)
where r= number of rows and c= number of
columns
Types of Chi-Square Tests
1. Goodness of-fit
This is used to test the null hypothesis that a random variable follows or
does not differ significantly from an expected or hypothesized distribution. It
may therefore be used to test the representativeness of a sample in which
certain population values are known or estimated.
Ex. A group of 130 high school students were asked to choose from the
following electives: Trigonometry, Oral English, Typing, Basic Computer,
Drafting. Their choices are given in the next table. Is there a significant
difference in the students’ choice of electives.
DATA ON STUDENTS’ CHOICE OF ELECTIVES
Trigonometry Oral
Cosmetology Basic
English
Computer
Observed 10
20
35
40
Drafting
Total
25
130
Research Question: Is there a significant difference in the students’
choice of electives.
Step 1. State the hypotheses.
Ho: There is no significant difference in the students’ choice of
elective. =
SEME 112 – Advanced Statistics
Module IV
122
H1: There is a significant difference in the students’ choice of
elective.
Step 2. Set the significance level.
Set the significance level at
= .05.
Step 3. Compute for the test statistic (x2)
a. Determine the expected value.
In the said problem, the expected is there is an equal distribution
of students in the different electives to say that the electives are
equally preferred.
Since the total number of students is 130 and there are 5
electives, divide 130 by 5. Hence, the expected for each elective is 26.
Trigonometry Oral
English
Observed 10
20
Expected 26
26
Cosmetology Basic
Computer
35
40
26
26
Drafting
Total
25
26
130
130
b. Compute for the x2 using the formula
x2 =
(
Hence, x2 =
)
(
)
+
(
)
+
(
)
+
(
)
(
+
)
x2 = 21.92
observed
10
20
35
40
25
130
expected
26.000
26.000
26.000
26.000
26.000
130.000
O–E
-16.000
-6.000
9.000
14.000
-1.000
0.000
(O - E)² / E
9.846
1.385
3.115
7.538
0.038
21.923
chi21.92 square
Step 4. Find the critical value
To find the critical value, solve for the degree of freedom with the
formula
df= (number of columns-1) or number of electives -1
df= 5-1
df = 4
SEME 112 – Advanced Statistics
Module IV
123
Using the chi-square table (Table D), get the intersection of row 4 and
column .05 or 5%. The critical value is 9.49.
Step 5. Make a decision.
The chi- square value is 21.92 and the critical value is 9.49, so the null
hypothesis is rejected.
Step 6. Formulate the conclusion.
There is a significant difference in the students’ choice of elective.
Hence, the electives are not equally preferred. They have higher preference of
basic computer and cosmetology while less preference of trigonometry and
drafting.
THE USE OF MEGASTAT IN SOLVING PROBLEMS ON CHI-SQUARE FOR
GOODNESS OF FIT
For non-parametric statistics, the Data Analysis Toolpak can no longer be
used since its functions do not include non-parametric statistics. The Megastat,
on the other hand has all the functions for non-parametric statistics.
For chi-square goodness of fit, just follow the following steps:
1. Open Microsoft excel spreadsheet and encode the data.
2. Decide for the expected values based on the given problem.
3. Click on add-ins, then megastat.
4. Choose the function chi-square goodness of fit.
5. Click on the observed values for the input.
6. Click on the expected values for the input.
7. Then click ok. The output will be reflected in another sheet.
Refer to the images below for the steps:
SEME 112 – Advanced Statistics
Module IV
124
SEME 112 – Advanced Statistics
Module IV
125
The computed value is 21.92 which is the same as the value obtained in
the long method or manual computation. In Megastat, it is the p value which is
reflected and not the critical value. But the conclusion is the same that there
is a significant difference in the students’ choice of elective. Hence, the
electives are not equally preferred. They have higher preference of basic
computer and cosmetology while less preference of trigonometry and drafting.
2. Test of Independence
The chi-square statistic is also used to determine whether two variables
of classification are related or independent of each other.
Ex. test if sex is related to students’ course preference or whether
students’course preference is independent of sex.
DATA ON STUDENTS’ CHOICE OF COURSE AND THEIR SEX
Sex
Engineering
Non-Engineering
Male
150
200
Female
150
350
Total
300
550
SEME 112 – Advanced Statistics
Total
350
500
850
Module IV
126
Research Question: Is sex independent of the students’ course preference?
Step 1: State the hypotheses.
Ho: The students’ course preference is independent of sex.
H1: The students’ course preference is dependent of sex.
Step 2: Set the significance level.
Let the significance level be 5%.
Step 3. Compute the test statistic (x2)
This kind of test is chi-square test of independence because we are
testing the dependence of the students’ course preference to their sex.
c. Compute for the expected value using the formula
E=
(
)(
Sex
Male
Female
Total
)
Engineering
150/E1
150/E2
300
Non-Engineering
200/E3
350/E4
550
Total
350
500
850
Try the 1st E1. The row total is 350 and column total is 300 and the
grand total is 850.
E1=
(
)(
)
E1 = 123.53
Compute for E2 to E4 and verify if your answer is correct in the
completed table below.
d. Compute for the difference of observed and expected with
the formula
O-E
In the first box the observed is 150 and the expected is 123.53.
Hence the difference of observed and expected is
O – E = 150-123.53 = 26.47
SEME 112 – Advanced Statistics
Module IV
127
e. Solve for the square of the difference of the observed and the
expected
(O - E)²
Since O – E = 26.47
(O - E)² = 700.6609
f. Compute for the quotient of the difference of the observed
and the expected and the expected using the formula
(
)
(
)
=
= 5.67
Compute for the rest and verify if your answer is correct in the completed
table below.
Compute for the chi square value using the formula
x2 =
(
)
x2 = 5.67 + 3.97 + 3.09 + 2.17 = 14.90
Male
Observed
Expected
O-E
(O - E)² / E
Female Observed
Expected
O-E
(O - E)² / E
Total
Observed
Engineering
150
123.53
26.47
5.67
150
176.47
-26.47
3.97
300
NonEngineering
200
226.47
-26.47
3.09
350
323.53
26.47
2.17
550
Total
350
350.00
0.00
8.77
500
500.00
0.00
6.14
850
14.90 chi-square
1 Df
Step 4. Find the critical value
To find the critical value, solve for the degree of freedom with the
formula
df= (number of rows-1) (number of columns-1) except total row and
column
df= (2-1) (2-1)
df =1
SEME 112 – Advanced Statistics
Module IV
128
Using the chi-square table (Table D), get the intersection of row 1 and
column .05 or 5%. The critical value is 3.84
Step 5. Make a decision
The chi- square value is 14.90 and the critical value is 3.84, so the null
hypothesis is rejected.
Step 6. Formulate the conclusion.
The students’ choice of course is dependent on their sex. Males have
greater probabilities of choosing the engineering course while females would
highly prefer the non-engineering courses.
THE USE OF MEGASTAT IN SOLVING PROBLEMS ON CHI-SQUARE FOR TEST OF
INDEPENDENCE
Practically the steps are the same except that the statistical tool is chisquare contingency table.
Refer to the images below for the steps to be followed:
1. Open Microsoft excel spreadsheet and encode the data except the
total row and column.
2. Click on add-ins, then megastat.
3. Choose the function chi-square contingency table.
4. Click on the table values (or highlight) for the input. Then click the
(
)
other values such as expected, O-E, (O-E)2 and
.
5. Then click ok. The output will be reflected in another sheet.
SEME 112 – Advanced Statistics
Module IV
129
SEME 112 – Advanced Statistics
Module IV
130
Notice that the chi-square value obtained is the same as the computed
chi-square value in the long method or manual computation. The p value is
.0001 which is less than .05, hence significant. The yellow color would indicate
right away that the null hypothesis is rejected. Therefore, the same conclusion
holds. The students’ choice of course is dependent on their sex. Males have
greater probabilities of choosing the engineering course while females would
highly prefer the non-engineering courses.
3. Test of Homogeneity
The chi-square test is frequently used to determine if two or more
population are homogeneous. By this, it means that the data distributions are
similar with respect to a particular criterion variable.
SEME 112 – Advanced Statistics
Module IV
131
Ex. Given the data below, can we infer that male and female adults differ in
attitude toward birth control.
In favor
3
10
13
Male
Female
Total
Against
9
4
13
Total
12
14
26
Research Question: Is there a significant difference in the attitude of males and
females toward birth control?
Step 1: State the hypotheses.
Ho: There is no significant difference in the attitude of males and
females toward birth control. =
H1: There is a significant difference in the attitude of males and
females toward birth control.
Step 2: Set the significance level.
Let the significance level be 5%.
Step 3. Compute the test statistic (x2)
This kind of test is chi-square test of homogeniety because we are
testing if males and females have the same attitude toward birth control.
a. Compute for the expected value using the formula
E=
(
)(
In favor
3/E1
10/E2
13
Male
Female
Total
)
Against
9/E3
4/E4
13
Total
12
14
26
Try the 1st E1. The row total is 12 and column total is 13 and the
grand total is 26.
E1=
(
)(
)
E1 = 6
Compute for E2 to E4 and verify if your answer is correct in the
completed table below.
SEME 112 – Advanced Statistics
Module IV
132
b. Compute for the difference of observed and expected with the formula
O-E
In the first box the observed is 3 and the expected is 6. Hence the
difference of observed and expected is
O – E = 3-6 = -3
c. Solve for the square of the difference of the observed and the expected
(O - E)²
Since O – E = -3
(O - E)² = 9
d. Compute for the quotient of the difference of the observed and the
expected and the expected using the formula
(
)
(
)
=
= 1.5
Compute for the rest and verify if your answer is correct in the completed
table below.
Compute for the chi square value using the formula
x2 =
(
)
x2 = 1.5 + 1.29 + 1.5 + 1.29
x2 = 5.58
Male
Observed
Expected
O-E
(O - E)² / E
Female Observed
Expected
O-E
(O - E)² / E
Total
Observed
Expected
O-E
(O - E)² / E
In favor
3
6.00
-3.00
1.50
10
7.00
3.00
1.29
13
13.00
0.00
2.79
Against
9
6.00
3.00
1.50
4
7.00
-3.00
1.29
13
13.00
0.00
2.79
Total
12
12.00
0.00
3.00
14
14.00
0.00
2.58
26
26.00
0.00
5.58
chi5.58 square
SEME 112 – Advanced Statistics
Module IV
133
Step 4. Find the critical value
To find the critical value, solve for the degree of freedom with the
formula
df= (number of rows-1) (number of columns-1) except total row and
column
df= (2-1) (2-1)
df =1
Using the chi-square table (Table D), get the intersection of row 1 and
column .05 or 5%. The critical value is 3.84
Step 5. Make a decision
The chi- square value is 5.58 and the critical value is 3.84, so the null
hypothesis is rejected.
Step 6. Formulate the conclusion.
There is a significant difference in the attitude of males and females
toward birth control. Looking at the data, the females tend to favor birth
control while the males are against birth control. The females experience more
difficulties in giving birth and taking care of children than the males, hence,
this finding.
THE USE OF MEGASTAT IN SOLVING PROBLEMS ON CHI-SQUARE FOR TEST OF
HOMOGENIETY
Practically the steps are the same with that of the test of independence
including the statistical tool to be used which is chi-square contingency table.
1. Open Microsoft excel spreadsheet and encode the data except the total
row and column.
2. Click on add-ins, then megastat.
3. Choose the function chi-square contingency table.
4. Click on the table values (or highlight) for the input. Then click the
(
)
other values such as expected, O-E, (O-E)2 and
.
5. Then click ok. The output will be reflected in another sheet.
The output is reflected below:
SEME 112 – Advanced Statistics
Module IV
136
Lesson 2

Mann-Whitney U Test
Mann-Whitney U Test
This statistical tool is used to compare two small independent samples. (
Counterpart of t-test independent)
The Mann-Whitney U test is used to compare differences between two
independent groups when the dependent variable is either ordinal or
continuous, but not normally distributed. The Mann-Whitney U test is often
considered the nonparametric alternative to the independent t-test although
this is not always the case.
Steps in solving:
1. Rank all the scores from highest to lowest.
2. Get the sum of ranks of each group.
3. Solve U for each group using the formula
= n1 n 2 +
(
)
-
= n1 n 2 +
(
)
-
4. Get the smaller U and compare with critical value in Table E
5. Reject Ho if U< c.v.
Sample Problem:
1. Given the level of anxiety of the respondent groups, test if Ph.D. and
MA/MS graduates differ in their level of anxiety.
Ph.D 31
MS/MA 30
22
26
46
24
54
13
28
18
21
21
33
29
41
30
48
25
29
26
Solution:
Formulate first the null and alternative hypothesis basing from
the research question
RQ: Is there a significant difference in the level of anxiety of
Ph. D. and MA/MS graduates?
Ho: There is no significant difference in the level of anxiety of
Ph. D.
and MA/MS graduates. =
SEME 112 – Advanced Statistics
Module IV
137
H1: There is a significant difference in the level of anxiety of
Ph. D. and
MA/MS graduates.
Compute for the test statistic using the different steps presented
above.
Steps:
1. Rank all the scores from highest to lowest.
2. Get the sum of ranks of each group.
Ph.D
31
22
46
54
28
21
33
41
48
29
Rank
6
16
3
1
11
17.5
5
4
2
9.5
∑ = 75
MS/MA
30
26
24
13
18
21
29
30
25
26
Rank
7.5
12.5
15
20
19
17.5
9.5
7.5
14
12.5
∑
= 135
3. Solve U for each group using the formula
= n1 n 2 +
(
)
-
= n1 n 2 +
(
)
-
U1 = (10) (10) +
(
)(
)
-75
(
)(
)
-135
U1 = 80
U2 = (10) (10) +
U2 = 20
4. Get the smaller U and compare with critical value
Smaller U is 20
SEME 112 – Advanced Statistics
Module IV
138
Critical value at alpha .05 is 23 using the mann-whitney table (Table E)
5. Reject Ho if U< c.v.
Decision: Reject Ho. The Ph.D and MS/MA graduates differ in their level of
anxiety. It can be noted that the Ph.D graduates have significantly higher level
of anxiety. The fact that they have higher educational attainment, they are
given more designations, hence, higher responsibilities as compared to the
MS/MA graduates. This brings them a higher anxiety level.
THE USE OF MEGASTAT IN SOLVING PROBLEMS ON MANN-WHITNEY U TEST
1. Open Microsoft excel spreadsheet and encode the data
2. Click on add-ins, then megastat.
3. Choose the function non-parametric test and select Wilcoxon MannWhitney U test.
4. Click on the values (or highlight) for the input. Then click ok. The output
will be reflected in another sheet.
Refer to the images below for the steps:
SEME 112 – Advanced Statistics
Module IV
142
Lesson 3

Wilcoxon MatchedPair Signed-Ranks Test
Wilcoxon’s Matched-Pairs Signed-Ranks Test
It is use to compare before and after a treatment is applied. (
Counterpart of t-test dependent).
The Wilcoxon signed-rank test is the nonparametric test equivalent to
the dependent t-test. It is used to compare two sets of scores that come from
the same participants. This can occur when we wish to investigate any change
in scores from one time point to another, or when individuals are subjected to
more than one condition.
For example, you could use a Wilcoxon signed-rank test to understand
whether there was a difference in smokers' daily cigarette consumption before
and after a 6 week hypnotherapy programme (i.e., your dependent variable
would be "daily cigarette consumption", and your two related groups would be
the cigarette consumption values "before" and "after" the hypnotherapy
programme).
https://statistics.laerd.com/spss-tutorials/wilcoxon-signed-rank-testusing-spss-statistics.php
Steps:
1. Obtain difference between each pair of scores.
2. Rank the absolute values of these differences. ( ―Absolute‖ means to
disregard signs in our ranking)
3. Add the ranks according to the sign of the differences.
4. The smaller of these sum is taken as the Wilcoxon’s T statistic ( or the sum
of the fewer sign
5. Reject Ho if Tv > c.v. Otherwise accept Ho.
Ex. Given the data below, determine if there a significant difference in the
degree neighborliness of students before and after the viewing the film on
neighborliness.
After
Before
16
4
12
18
22
10
16
14
14
12
10
14
20
10
18
12
10
4
22
12
Solution:Formulate first the null and alternative hypothesis basing from
the research question
SEME 112 – Advanced Statistics
Module IV
143
RQ: Is there a significant difference in the degree of
neighborliness of students before and after viewing the film on
neighborliness ?
Ho: There is no significant difference in the degree of
neighborliness of students before and after viewing the film on
neighborliness. =
H1: There is a significant difference in the degree of
neighborliness of students before and after viewing the film on
neighborliness.
Compute for the test statistic using the different steps presented above.
Steps:
1. Obtain difference between each pair of scores.
2. Rank the absolute values of these differences. ( ―Absolute‖ means to
disregard signs in our ranking)
3. Add the ranks according to the sign of the differences.
After
16
12
22
16
14
10
20
18
10
22
Before
4
18
10
14
12
14
10
12
4
12
Differences
12
-6
12
2
2
-4
10
6
6
10
Rank of Difference
9.5
5
9.5
1.5
1.5
3
7.5
5
5
7.5
R (+)
9.5
R (-)
5
9.5
1.5
1.5
3
7.5
5
5
7.5
∑R (+)=47
____
ΣR (-)=8
4. The smaller of the sum of ranks is taken as the Wilcoxon’s T statistic ( or
the sum of the fewer sign
5. Reject Ho if Tv > c.v. Otherwise accept Ho.
Tv= 8
Using the table (Table F), at ά= .05 two tailed t.v. = 8
Tv > c.v.
Decision: Reject Ho. Hence there is a significant difference in the
degree of neighborliness of students before and after viewing the film on
neighborliness. This implies further that there is a significant improvement of
students on their attitudes and perceptions toward neighborliness after viewing
the film. The film is effective in changing the attitude of students on
neighborliness.
SEME 112 – Advanced Statistics
Module IV
148
Lesson 4

Kruskall Wallis One- Way
Analysis of Variance by Ranks
KRUSKAL-WALLIS ONE-WAY ANOVA BY RANKS
The Kruskal-Wallis H test (sometimes also called the "one-way ANOVA on
ranks") is a rank-based nonparametric test that can be used to determine if
there are statistically significant differences between two or more groups of an
independent variable on a continuous or ordinal dependent variable. It is
considered the nonparametric alternative to the one-way ANOVA, and an
extension of the Mann-Whitney U test to allow the comparison of more than
two independent groups.
For example, you could use a Kruskal-Wallis H test to understand
whether exam performance, measured on a continuous scale from 0-100,
differed based on test anxiety levels (i.e., your dependent variable would be
"exam performance" and your independent variable would be "test anxiety
level", which has three independent groups: students with "low", "medium" and
"high" test anxiety levels).
This is used to compare three different groups ( independent)
(Counterpart of ANOVA- one way analysis of variance)
Steps:
1. The scores of the three sets are combined. Rank from lowest to greatest.
2. Get the sum of ranks per group.
3. Using the formula, compute for H.
H =
(
)
-3(N+1)
Where N= the number in all samples combined
Rj= sum of ranks and Nj the numbers in j samples
4.Get the critical value using the chi-square table (Table D) with df=k-1, where
k is the number of groups.
5.Make a decision and interpret. If H> critical value, reject Ho otherwise accept
Ho
SEME 112 – Advanced Statistics
Module IV
149
Example: Three groups of students took an exam in Mathematics and their
scores are indicated below. Determine if there exists a significant differences
in their scores.
X
12
16
14
2
12
Y
13
18
14
13
8
7
6
4
Z
13
14
7
8
4
3
2
5
9
Solution:Formulate first the null and alternative hypothesis basing from
the research question
RQ: Is there a significant difference in the scores of the three
groups of students in the Mathematics exam?
Ho: There is no significant difference in the scores of the three
groups of students in the Mathematics exam.
= =
H1: There is a significant difference in the scores of the three
groups of students in the Mathematics exam.
Compute for the test statistic using the different steps presented above.
Steps:
1. The scores of the three sets are combined. Rank.
2. Get the sum of ranks per group.
X
12
16
14
2
12
Rx
13.5
21
19
1.5
13.5
______ _____
nx=5 ΣRx=68.5
Y
13
18
14
13
8
7
6
4
Ry
16
22
19
16
10.5
8.5
7
4.5
_____ _____
ny=8 ΣRy= 103.5
SEME 112 – Advanced Statistics
Z
Rz
13
16
14
19
7
8.5
8
10.5
4
4.5
3
3
2
1.5
5
6
9
12
____ _____
nz=9 ΣRz=81
Module IV
150
3. Using the formula, compute for H.
H =
H =
(
-3(N+1) where N=22,
)
(
(
)
)
+
+
]-3(22+1)
H = 71.30- 69
Hence, H = 2.3
If the number of cases per group is from one to five, special tables are used in
the interpretation of H. When each group contains five or more cases, H is
interpreted as chi-square with the number of groups minus 1 as the degree of
freedom.
Therefore, df=2.
c.v. = 5.99
The
Decision: Fail to reject Ho. Hence, there is no significant difference in the
scores of students in the Mathematics exam.
THE USE OF MEGASTAT IN SOLVING PROBLEMS ON KRUSKALL WALLIS ONE
WAY ANOVA BY RANKS
1. Open Microsoft excel spreadsheet and encode the data
2. Click on add-ins, then megastat.
3. Choose the function non-parametric test and select Kruskall Wallis.
4. Click on the values (or highlight) for the input. Then click ok. The output
will be reflected in another sheet.
Refer to the images below for the steps:
SEME 112 – Advanced Statistics
Module IV
156
Lesson 5

Friedman’s
ANOVA
FRIEDMAN’S ANOVA BY RANKS
This is used to compare 3 or more related samples. For example, one
group would rate three different books in algebra or three groups would rate
would rate the same intervention.
The Friedman
test is
a non-parametric statistical
test developed
by Milton Friedman. Similar to the parametric repeated measures ANOVA, it is
used to detect differences in treatments across multiple test attempts. The
procedure involves ranking each row (or block) together, then considering the
values of ranks by columns. Applicable to complete block designs, it is thus a
special case of the Durbin test.
Classic examples of use are:


n wine judges each rate k different wines. Are any of the k wines ranked
consistently higher or lower than the others?
n welders each use k welding torches, and the ensuing welds were rated on
quality. Do any of the k torches produce consistently better or worse welds?
The Friedman test is used for one-way repeated measures analysis of variance
by ranks. In its use of ranks it is similar to the Kruskal–Wallis one-way analysis
of variance by ranks.
Friedman’s test is a non-parametric test for finding differences in
treatments across multiple attempts. Nonparametric means the test doesn’t
assume your data comes from a particular distribution (like the normal
distribution). Basically, it’s used in place of the ANOVA test when you don’t
know
the
distribution
of
your
data.
https://www.statisticshowto.com/friedmans-test/
Steps:
1. Rank per score horizontally. ( According to the raters perception)
2. Add rank per treatment.
3. Determine k, df, c.v and ά.
4. Substitute the values in the formula and solve.
The formula is
X2 = ( ) ∑Rj 2 – 3N (k+1)
Where k = number of groups/treatments
SEME 112 – Advanced Statistics
Module IV
157
N= number of cases
Rj=sum of ranks of scores in a treatment
df= k-1
Ex. Three modules A, B, C were rated independently by 10 raters. The ratings
are shown below. Do the raters rate the modules differently?
Rater
Module A
Module B
Module C
A
16
20
17
B
15
14
19
C
10
19
15
D
15
25
21
E
12
20
18
F
17
20
21
G
20
18
17
H
14
23
21
I
13
21
19
J
16
24
21
Solution: Formulate first the null and alternative hypothesis basing from
the research question
RQ: Is there a significant difference in the ratings given by the
10 raters to the three modules?
Ho: There is no significant difference in the ratings given by
the 10 raters to the three modules. = =
H1: There is a significant difference in the ratings given by the
10 raters to the three modules.
Compute for the test statistic using the different steps presented above.
Steps:
1. Rank per score horizontally. ( According to the raters perception)
2. Add rank per treatment.
Rater
A
B
C
D
E
F
G
H
I
J
A
16
15
10
15
12
17
20
14
13
16
rA
B
3
20
2
14
3
19
3
25
3
20
3
20
1
18
3
23
3
21
3
24
____
rA=27
SEME 112 – Advanced Statistics
rB
C
1
17
3
19
1
15
1
21
1
18
2
21
2
17
1
21
1
19
1
21
____
rB=14
rC
2
1
2
2
2
1
3
2
2
2
____
rC= 19
Module IV
158
3.Determine k, df, c.v and .
k= 3
df= k-1 or 2 critical value at =.05 is 6.20 (Use the table for chi
square to find the critical value if k is greater than 5 or n is >13, otherwise
use the table below) and n=10
Friedman’s ANOVA by Ranks Critical Value Table
Three tables according by “k”.
If your k is over 5, or your n is over 13, use the chi square critical value table to get the critical value.
k=3
N
α <.10
α ≤.05
α <.01
3
6.00
6.00
—
4
6.00
6.50
8.00
5
5.20
6.40
8.40
6
5.33
7.00
9.00
7
5.43
7.14
8.86
8
5.25
6.25
9.00
9
5.56
6.22
8.67
10
5.00
6.20
9.60
11
4.91
6.54
8.91
12
5.17
6.17
8.67
13
4.77
6.00
9.39
∞
4.61
5.99
9.21
N
α <.10
α ≤.05
α <.01
2
6.00
6.00
—
3
6.60
7.40
8.60
4
6.30
7.80
9.60
5
6.36
7.80
9.96
6
6.40
7.60
10.00
7
6.26
7.80
10.37
8
6.30
7.50
10.35
∞
6.25
7.82
11.34
N
α <.10
α ≤.05
α <.01
3
7.47
8.53
10.13
4
7.60
8.80
11.00
5
7.68
8.96
11.52
∞
7.78
9.49
13.28
k=4
k=4
SEME 112 – Advanced Statistics
Module IV
159
Reference:
Friedman’s Two-way Analysis of Variance by Ranks — Analysis of k-Within-Group Data with a
Quantitative Response Variable. Retrieved 7-17-2016 from: http://psych.unl.edu/psycrs/handcomp/hcfried.PDF
CITE THIS AS:
Stephanie Glen. "Friedman’s Test / Two Way Analysis of Variance by Ranks" From StatisticsHowTo.com: Elementary Statistics for
the rest of us! https://www.statisticshowto.com/friedmans-test/
4. Substitute the needed values in the formula and solve.
The formula is
x2 = ( ) ∑Rj 2 – 3N (k+1)
Where k = number of groups/treatments
N= number of cases
Rj=sum of ranks of scores in a treatment
df= k-1
x2 =
( )(
)
(
+
)– 3(10) (3+1)
x2 = 8.6
Since the x2 value is 8.6 and and the critical value is 5.99 (Table G), then
x2 > c.v.
Decision: Reject Ho
Hence, the raters rated the modules differently.
THE USE OF MEGASTAT IN SOLVING PROBLEMS ON KRUSKALL WALLIS ONE
WAY ANOVA BY RANKS
1. Open Microsoft excel spreadsheet and encode the data
2. Click on add-ins, then megastat.
3. Choose the function non-parametric test and select Friedmans test.
4. Click on the values (or highlight) for the input. Then click ok. The output
will be reflected in another sheet.
Refer to the images below for the steps:
SEME 112 – Advanced Statistics
Module IV
164
TABLE A
t Distribution: Critical Values of t
Significance level
Degrees of
freedom
Two-tailed test:
One-tailed test:
10%
5%
5%
2.5%
2%
1%
1%
0.5%
0.2%
0.1%
0.1%
0.05%
1
2
3
4
5
6.314
2.920
2.353
2.132
2.015
12.706
4.303
3.182
2.776
2.571
31.821
6.965
4.541
3.747
3.365
63.657
9.925
5.841
4.604
4.032
318.309
22.327
10.215
7.173
5.893
636.619
31.599
12.924
8.610
6.869
6
7
8
9
10
1.943
1.894
1.860
1.833
1.812
2.447
2.365
2.306
2.262
2.228
3.143
2.998
2.896
2.821
2.764
3.707
3.499
3.355
3.250
3.169
5.208
4.785
4.501
4.297
4.144
5.959
5.408
5.041
4.781
4.587
11
12
13
14
15
1.796
1.782
1.771
1.761
1.753
2.201
2.179
2.160
2.145
2.131
2.718
2.681
2.650
2.624
2.602
3.106
3.055
3.012
2.977
2.947
4.025
3.930
3.852
3.787
3.733
4.437
4.318
4.221
4.140
4.073
16
17
18
19
20
1.746
1.740
1.734
1.729
1.725
2.120
2.110
2.101
2.093
2.086
2.583
2.567
2.552
2.539
2.528
2.921
2.898
2.878
2.861
2.845
3.686
3.646
3.610
3.579
3.552
4.015
3.965
3.922
3.883
3.850
21
22
23
24
25
1.721
1.717
1.714
1.711
1.708
2.080
2.074
2.069
2.064
2.060
2.518
2.508
2.500
2.492
2.485
2.831
2.819
2.807
2.797
2.787
3.527
3.505
3.485
3.467
3.450
3.819
3.792
3.768
3.745
3.725
26
27
28
29
30
1.706
1.703
1.701
1.699
1.697
2.056
2.052
2.048
2.045
2.042
2.479
2.473
2.467
2.462
2.457
2.779
2.771
2.763
2.756
2.750
3.435
3.421
3.408
3.396
3.385
3.707
3.690
3.674
3.659
3.646
32
34
36
38
40
1.694
1.691
1.688
1.686
1.684
2.037
2.032
2.028
2.024
2.021
2.449
2.441
2.434
2.429
2.423
2.738
2.728
2.719
2.712
2.704
3.365
3.348
3.333
3.319
3.307
3.622
3.601
3.582
3.566
3.551
42
44
46
48
50
1.682
1.680
1.679
1.677
1.676
2.018
2.015
2.013
2.011
2.009
2.418
2.414
2.410
2.407
2.403
2.698
2.692
2.687
2.682
2.678
3.296
3.286
3.277
3.269
3.261
3.538
3.526
3.515
3.505
3.496
60
70
80
90
100
1.671
1.667
1.664
1.662
1.660
2.000
1.994
1.990
1.987
1.984
2.390
2.381
2.374
2.368
2.364
2.660
2.648
2.639
2.632
2.626
3.232
3.211
3.195
3.183
3.174
3.460
3.435
3.416
3.402
3.390
120
1.658
1.980
2.358
2.617
3.160
3.373
SEME 112 – Advanced Statistics
Module IV
165
150
200
300
400
1.655
1.653
1.650
1.649
1.976
1.972
1.968
1.966
2.351
2.345
2.339
2.336
2.609
2.601
2.592
2.588
3.145
3.131
3.118
3.111
3.357
3.340
3.323
3.315
500
600
1.648
1.647
1.965
1.964
2.334
2.333
2.586
2.584
3.107
3.104
3.310
3.307

1.645
1.960
2.326
2.576
3.090
3.291
TABLE B
F Distribution: Critical Values of F (5% significance level)
v1
1
2
3
4
5
6
7
8
9
10
12
14
16
18
20
v2
1 161.45 199.50 215.71 224.58 230.16 233.99 236.77 238.88 240.54 241.88 243.91 245.36 246.46 247.32
248.01
2 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 19.38 19.40 19.41 19.42 19.43 19.44 19.45
3 10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.85 8.81 8.79 8.74 8.71 8.69 8.67 8.66
4 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00 5.96 5.91 5.87 5.84 5.82 5.80
5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.68 4.64 4.60 4.58 4.56
6
7
8
9
10
5.99
5.59
5.32
5.12
4.96
5.14
4.74
4.46
4.26
4.10
4.76
4.35
4.07
3.86
3.71
4.53
4.12
3.84
3.63
3.48
4.39
3.97
3.69
3.48
3.33
4.28
3.87
3.58
3.37
3.22
4.21
3.79
3.50
3.29
3.14
4.15
3.73
3.44
3.23
3.07
4.10
3.68
3.39
3.18
3.02
4.06
3.64
3.35
3.14
2.98
4.00
3.57
3.28
3.07
2.91
3.96
3.53
3.24
3.03
2.86
3.92
3.49
3.20
2.99
2.83
3.90
3.47
3.17
2.96
2.80
3.87
3.44
3.15
2.94
2.77
11
12
13
14
15
4.84
4.75
4.67
4.60
4.54
3.98
3.89
3.81
3.74
3.68
3.59
3.49
3.41
3.34
3.29
3.36
3.26
3.18
3.11
3.06
3.20
3.11
3.03
2.96
2.90
3.09
3.00
2.92
2.85
2.79
3.01
2.91
2.83
2.76
2.71
2.95
2.85
2.77
2.70
2.64
2.90
2.80
2.71
2.65
2.59
2.85
2.75
2.67
2.60
2.54
2.79
2.69
2.60
2.53
2.48
2.74
2.64
2.55
2.48
2.42
2.70
2.60
2.51
2.44
2.38
2.67
2.57
2.48
2.41
2.35
2.65
2.54
2.46
2.39
2.33
16
17
18
19
20
4.49
4.45
4.41
4.38
4.35
3.63
3.59
3.55
3.52
3.49
3.24
3.20
3.16
3.13
3.10
3.01
2.96
2.93
2.90
2.87
2.85
2.81
2.77
2.74
2.71
2.74
2.70
2.66
2.63
2.60
2.66
2.61
2.58
2.54
2.51
2.59
2.55
2.51
2.48
2.45
2.54
2.49
2.46
2.42
2.39
2.49
2.45
2.41
2.38
2.35
2.42
2.38
2.34
2.31
2.28
2.37
2.33
2.29
2.26
2.22
2.33
2.29
2.25
2.21
2.18
2.30
2.26
2.22
2.18
2.15
2.28
2.23
2.19
2.16
2.12
21
22
23
24
25
4.32
4.30
4.28
4.26
4.24
3.47
3.44
3.42
3.40
3.39
3.07
3.05
3.03
3.01
2.99
2.84
2.82
2.80
2.78
2.76
2.68
2.66
2.64
2.62
2.60
2.57
2.55
2.53
2.51
2.49
2.49
2.46
2.44
2.42
2.40
2.42
2.40
2.37
2.36
2.34
2.37
2.34
2.32
2.30
2.28
2.32
2.30
2.27
2.25
2.24
2.25
2.23
2.20
2.18
2.16
2.20
2.17
2.15
2.13
2.11
2.16
2.13
2.11
2.09
2.07
2.12
2.10
2.08
2.05
2.04
2.10
2.07
2.05
2.03
2.01
26
27
28
29
30
4.22
4.21
4.20
4.18
4.17
3.37
3.35
3.34
3.33
3.32
2.98
2.96
2.95
2.93
2.92
2.74
2.73
2.71
2.70
2.69
2.59
2.57
2.56
2.55
2.53
2.47
2.46
2.45
2.43
2.42
2.39
2.37
2.36
2.35
2.33
2.32
2.31
2.29
2.28
2.27
2.27
2.25
2.24
2.22
2.21
2.22
2.20
2.19
2.18
2.16
2.15
2.13
2.12
2.10
2.09
2.09
2.08
2.06
2.05
2.04
2.05
2.04
2.02
2.01
1.99
2.02
2.00
1.99
1.97
1.96
1.99
1.97
1.96
1.94
1.93
35
40
50
60
70
4.12
4.08
4.03
4.00
3.98
3.27
3.23
3.18
3.15
3.13
2.87
2.84
2.79
2.76
2.74
2.64
2.61
2.56
2.53
2.50
2.49
2.45
2.40
2.37
2.35
2.37
2.34
2.29
2.25
2.23
2.29
2.25
2.20
2.17
2.14
2.22
2.18
2.13
2.10
2.07
2.16
2.12
2.07
2.04
2.02
2.11
2.08
2.03
1.99
1.97
2.04
2.00
1.95
1.92
1.89
1.99
1.95
1.89
1.86
1.84
1.94
1.90
1.85
1.82
1.79
1.91
1.87
1.81
1.78
1.75
1.88
1.84
1.78
1.75
1.72
80
3.96
3.11
2.72
2.49
2.33
2.21
2.13
2.06
2.00
1.95
1.88
1.82
1.77
1.73
1.70
SEME 112 – Advanced Statistics
Module IV
166
90
100
120
150
3.95
3.94
3.92
3.90
3.10
3.09
3.07
3.06
2.71
2.70
2.68
2.66
2.47
2.46
2.45
2.43
2.32
2.31
2.29
2.27
2.20
2.19
2.18
2.16
2.11
2.10
2.09
2.07
2.04
2.03
2.02
2.00
1.99
1.97
1.96
1.94
1.94
1.93
1.91
1.89
1.86
1.85
1.83
1.82
1.80
1.79
1.78
1.76
1.76
1.75
1.73
1.71
1.72
1.71
1.69
1.67
1.69
1.68
1.66
1.64
200
250
300
400
500
3.89
3.88
3.87
3.86
3.86
3.04
3.03
3.03
3.02
3.01
2.65
2.64
2.63
2.63
2.62
2.42
2.41
2.40
2.39
2.39
2.26
2.25
2.24
2.24
2.23
2.14
2.13
2.13
2.12
2.12
2.06
2.05
2.04
2.03
2.03
1.98
1.98
1.97
1.96
1.96
1.93
1.92
1.91
1.90
1.90
1.88
1.87
1.86
1.85
1.85
1.80
1.79
1.78
1.78
1.77
1.74
1.73
1.72
1.72
1.71
1.69
1.68
1.68
1.67
1.66
1.66
1.65
1.64
1.63
1.62
1.62
1.61
1.61
1.60
1.59
600
750
1000
3.86
3.85
3.85
3.01
3.01
3.00
2.62
2.62
2.61
2.39
2.38
2.38
2.23
2.23
2.22
2.11
2.11
2.11
2.02
2.02
2.02
1.95
1.95
1.95
1.90
1.89
1.89
1.85
1.84
1.84
1.77
1.77
1.76
1.71
1.70
1.70
1.66
1.66
1.65
1.62
1.62
1.61
1.59
1.58
1.58
SEME 112 – Advanced Statistics
Module IV
167
TABLE B (continued)
F Distribution: Critical Values of F (5% significance level)
v1
25
30
35
40
50
60
75
100
150
200
v2
1 249.26 250.10 250.69 251.14 251.77 252.20 252.62 253.04 253.46 253.68
2 19.46 19.46 19.47 19.47 19.48 19.48 19.48 19.49 19.49 19.49
3 8.63 8.62 8.60 8.59 8.58 8.57 8.56 8.55 8.54 8.54
4 5.77 5.75 5.73 5.72 5.70 5.69 5.68 5.66 5.65 5.65
5 4.52 4.50 4.48 4.46 4.44 4.43 4.42 4.41 4.39 4.39
6
7
8
9
10
3.83
3.40
3.11
2.89
2.73
3.81
3.38
3.08
2.86
2.70
3.79
3.36
3.06
2.84
2.68
3.77
3.34
3.04
2.83
2.66
3.75
3.32
3.02
2.80
2.64
3.74
3.30
3.01
2.79
2.62
3.73
3.29
2.99
2.77
2.60
3.71
3.27
2.97
2.76
2.59
3.70
3.26
2.96
2.74
2.57
3.69
3.25
2.95
2.73
2.56
11
12
13
14
15
2.60
2.50
2.41
2.34
2.28
2.57
2.47
2.38
2.31
2.25
2.55
2.44
2.36
2.28
2.22
2.53
2.43
2.34
2.27
2.20
2.51
2.40
2.31
2.24
2.18
2.49
2.38
2.30
2.22
2.16
2.47
2.37
2.28
2.21
2.14
2.46
2.35
2.26
2.19
2.12
2.44
2.33
2.24
2.17
2.10
2.43
2.32
2.23
2.16
2.10
16
17
18
19
20
2.23
2.18
2.14
2.11
2.07
2.19
2.15
2.11
2.07
2.04
2.17
2.12
2.08
2.05
2.01
2.15
2.10
2.06
2.03
1.99
2.12
2.08
2.04
2.00
1.97
2.11
2.06
2.02
1.98
1.95
2.09
2.04
2.00
1.96
1.93
2.07
2.02
1.98
1.94
1.91
2.05
2.00
1.96
1.92
1.89
2.04
1.99
1.95
1.91
1.88
21
22
23
24
25
2.05
2.02
2.00
1.97
1.96
2.01
1.98
1.96
1.94
1.92
1.98
1.96
1.93
1.91
1.89
1.96
1.94
1.91
1.89
1.87
1.94
1.91
1.88
1.86
1.84
1.92
1.89
1.86
1.84
1.82
1.90
1.87
1.84
1.82
1.80
1.88
1.85
1.82
1.80
1.78
1.86
1.83
1.80
1.78
1.76
1.84
1.82
1.79
1.77
1.75
26
27
28
29
30
1.94
1.92
1.91
1.89
1.88
1.90
1.88
1.87
1.85
1.84
1.87
1.86
1.84
1.83
1.81
1.85
1.84
1.82
1.81
1.79
1.82
1.81
1.79
1.77
1.76
1.80
1.79
1.77
1.75
1.74
1.78
1.76
1.75
1.73
1.72
1.76
1.74
1.73
1.71
1.70
1.74
1.72
1.70
1.69
1.67
1.73
1.71
1.69
1.67
1.66
35
40
50
60
70
1.82
1.78
1.73
1.69
1.66
1.79
1.74
1.69
1.65
1.62
1.76
1.72
1.66
1.62
1.59
1.74
1.69
1.63
1.59
1.57
1.70
1.66
1.60
1.56
1.53
1.68
1.64
1.58
1.53
1.50
1.66
1.61
1.55
1.51
1.48
1.63
1.59
1.52
1.48
1.45
1.61
1.56
1.50
1.45
1.42
1.60
1.55
1.48
1.44
1.40
80
90
100
120
150
1.64
1.63
1.62
1.60
1.58
1.60
1.59
1.57
1.55
1.54
1.57
1.55
1.54
1.52
1.50
1.54
1.53
1.52
1.50
1.48
1.51
1.49
1.48
1.46
1.44
1.48
1.46
1.45
1.43
1.41
1.45
1.44
1.42
1.40
1.38
1.43
1.41
1.39
1.37
1.34
1.39
1.38
1.36
1.33
1.31
1.38
1.36
1.34
1.32
1.29
200
250
300
400
500
1.56
1.55
1.54
1.53
1.53
1.52
1.50
1.50
1.49
1.48
1.48
1.47
1.46
1.45
1.45
1.46
1.44
1.43
1.42
1.42
1.41
1.40
1.39
1.38
1.38
1.39
1.37
1.36
1.35
1.35
1.35
1.34
1.33
1.32
1.31
1.32
1.31
1.30
1.28
1.28
1.28
1.27
1.26
1.24
1.23
1.26
1.25
1.23
1.22
1.21
600
750
1000
1.52
1.52
1.52
1.48
1.47
1.47
1.44
1.44
1.43
1.41
1.41
1.41
1.37
1.37
1.36
1.34
1.34
1.33
1.31
1.30
1.30
1.27
1.26
1.26
1.23
1.22
1.22
1.20
1.20
1.19
SEME 112 – Advanced Statistics
Module IV
168
TABLE B (continued)
F Distribution: Critical Values of F (1% significance level)
v1 1
2
3
4
5
6
7
8
9
10
12
14
16
18
20
v2
1 4052.18 4999.50 5403.35 5624.58 5763.65 5858.99 5928.36 5981.07 6022.47 6055.85 6106.32 6142.67 6170.10 6191.53 6208.73
2 98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 99.39 99.40 99.42 99.43 99.44 99.44 99.45
3 34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 27.35 27.23 27.05 26.92 26.83 26.75 26.69
4 21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66 14.55 14.37 14.25 14.15 14.08 14.02
5 16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16 10.05 9.89 9.77 9.68 9.61 9.55
6
7
8
9
10
13.75 10.92
12.25 9.55
11.26 8.65
10.56 8.02
10.04 7.56
9.78
8.45
7.59
6.99
6.55
9.15
7.85
7.01
6.42
5.99
8.75
7.46
6.63
6.06
5.64
8.47
7.19
6.37
5.80
5.39
8.26
6.99
6.18
5.61
5.20
8.10
6.84
6.03
5.47
5.06
7.98
6.72
5.91
5.35
4.94
7.87
6.62
5.81
5.26
4.85
7.72
6.47
5.67
5.11
4.71
7.60
6.36
5.56
5.01
4.60
7.52
6.28
5.48
4.92
4.52
7.45
6.21
5.41
4.86
4.46
7.40
6.16
5.36
4.81
4.41
11
12
13
14
15
9.65
9.33
9.07
8.86
8.68
7.21
6.93
6.70
6.51
6.36
6.22
5.95
5.74
5.56
5.42
5.67
5.41
5.21
5.04
4.89
5.32
5.06
4.86
4.69
4.56
5.07
4.82
4.62
4.46
4.32
4.89
4.64
4.44
4.28
4.14
4.74
4.50
4.30
4.14
4.00
4.63
4.39
4.19
4.03
3.89
4.54
4.30
4.10
3.94
3.80
4.40
4.16
3.96
3.80
3.67
4.29
4.05
3.86
3.70
3.56
4.21
3.97
3.78
3.62
3.49
4.15
3.91
3.72
3.56
3.42
4.10
3.86
3.66
3.51
3.37
16
17
18
19
20
8.53
8.40
8.29
8.18
8.10
6.23
6.11
6.01
5.93
5.85
5.29
5.18
5.09
5.01
4.94
4.77
4.67
4.58
4.50
4.43
4.44
4.34
4.25
4.17
4.10
4.20
4.10
4.01
3.94
3.87
4.03
3.93
3.84
3.77
3.70
3.89
3.79
3.71
3.63
3.56
3.78
3.68
3.60
3.52
3.46
3.69
3.59
3.51
3.43
3.37
3.55
3.46
3.37
3.30
3.23
3.45
3.35
3.27
3.19
3.13
3.37
3.27
3.19
3.12
3.05
3.31
3.21
3.13
3.05
2.99
3.26
3.16
3.08
3.00
2.94
21
22
23
24
25
8.02
7.95
7.88
7.82
7.77
5.78
5.72
5.66
5.61
5.57
4.87
4.82
4.76
4.72
4.68
4.37
4.31
4.26
4.22
4.18
4.04
3.99
3.94
3.90
3.85
3.81
3.76
3.71
3.67
3.63
3.64
3.59
3.54
3.50
3.46
3.51
3.45
3.41
3.36
3.32
3.40
3.35
3.30
3.26
3.22
3.31
3.26
3.21
3.17
3.13
3.17
3.12
3.07
3.03
2.99
3.07
3.02
2.97
2.93
2.89
2.99
2.94
2.89
2.85
2.81
2.93
2.88
2.83
2.79
2.75
2.88
2.83
2.78
2.74
2.70
26
27
28
29
30
7.72
7.68
7.64
7.60
7.56
5.53
5.49
5.45
5.42
5.39
4.64
4.60
4.57
4.54
4.51
4.14
4.11
4.07
4.04
4.02
3.82
3.78
3.75
3.73
3.70
3.59
3.56
3.53
3.50
3.47
3.42
3.39
3.36
3.33
3.30
3.29
3.26
3.23
3.20
3.17
3.18
3.15
3.12
3.09
3.07
3.09
3.06
3.03
3.00
2.98
2.96
2.93
2.90
2.87
2.84
2.86
2.82
2.79
2.77
2.74
2.78
2.75
2.72
2.69
2.66
2.72
2.68
2.65
2.63
2.60
2.66
2.63
2.60
2.57
2.55
35
40
50
60
70
7.42
7.31
7.17
7.08
7.01
5.27
5.18
5.06
4.98
4.92
4.40
4.31
4.20
4.13
4.07
3.91
3.83
3.72
3.65
3.60
3.59
3.51
3.41
3.34
3.29
3.37
3.29
3.19
3.12
3.07
3.20
3.12
3.02
2.95
2.91
3.07
2.99
2.89
2.82
2.78
2.96
2.89
2.78
2.72
2.67
2.88
2.80
2.70
2.63
2.59
2.74
2.66
2.56
2.50
2.45
2.64
2.56
2.46
2.39
2.35
2.56
2.48
2.38
2.31
2.27
2.50
2.42
2.32
2.25
2.20
2.44
2.37
2.27
2.20
2.15
80
90
100
120
150
6.96
6.93
6.90
6.85
6.81
4.88
4.85
4.82
4.79
4.75
4.04
4.01
3.98
3.95
3.91
3.56
3.53
3.51
3.48
3.45
3.26
3.23
3.21
3.17
3.14
3.04
3.01
2.99
2.96
2.92
2.87
2.84
2.82
2.79
2.76
2.74
2.72
2.69
2.66
2.63
2.64
2.61
2.59
2.56
2.53
2.55
2.52
2.50
2.47
2.44
2.42
2.39
2.37
2.34
2.31
2.31
2.29
2.27
2.23
2.20
2.23
2.21
2.19
2.15
2.12
2.17
2.14
2.12
2.09
2.06
2.12
2.09
2.07
2.03
2.00
200
250
300
400
500
6.76
6.74
6.72
6.70
6.69
4.71
4.69
4.68
4.66
4.65
3.88
3.86
3.85
3.83
3.82
3.41
3.40
3.38
3.37
3.36
3.11
3.09
3.08
3.06
3.05
2.89
2.87
2.86
2.85
2.84
2.73
2.71
2.70
2.68
2.68
2.60
2.58
2.57
2.56
2.55
2.50
2.48
2.47
2.45
2.44
2.41
2.39
2.38
2.37
2.36
2.27
2.26
2.24
2.23
2.22
2.17
2.15
2.14
2.13
2.12
2.09
2.07
2.06
2.05
2.04
2.03
2.01
1.99
1.98
1.97
1.97
1.95
1.94
1.92
1.92
600
750
1000
6.68
6.67
6.66
4.64
4.63
4.63
3.81
3.81
3.80
3.35
3.34
3.34
3.05
3.04
3.04
2.83
2.83
2.82
2.67
2.66
2.66
2.54
2.53
2.53
2.44
2.43
2.43
2.35
2.34
2.34
2.21
2.21
2.20
2.11
2.11
2.10
2.03
2.02
2.02
1.96
1.96
1.95
1.91
1.90
1.90
SEME 112 – Advanced Statistics
Module IV
169
TABLE B (continued)
F Distribution: Critical Values of F (1% significance level)
v1 25
30
35
40
50
60
75
100
150
200
v2
1 6239.83 6260.65 6275.57 6286.78 6302.52 6313.03 6323.56 6334.11 6344.68 6349.97
2 99.46 99.47 99.47 99.47 99.48 99.48 99.49 99.49 99.49 99.49
3 26.58 26.50 26.45 26.41 26.35 26.32 26.28 26.24 26.20 26.18
4 13.91 13.84 13.79 13.75 13.69 13.65 13.61 13.58 13.54 13.52
5 9.45 9.38 9.33 9.29 9.24 9.20 9.17 9.13 9.09 9.08
6
7
8
9
10
7.30
6.06
5.26
4.71
4.31
7.23
5.99
5.20
4.65
4.25
7.18
5.94
5.15
4.60
4.20
7.14
5.91
5.12
4.57
4.17
7.09
5.86
5.07
4.52
4.12
7.06
5.82
5.03
4.48
4.08
7.02
5.79
5.00
4.45
4.05
6.99
5.75
4.96
4.41
4.01
6.95
5.72
4.93
4.38
3.98
6.93
5.70
4.91
4.36
3.96
11
12
13
14
15
4.01
3.76
3.57
3.41
3.28
3.94
3.70
3.51
3.35
3.21
3.89
3.65
3.46
3.30
3.17
3.86
3.62
3.43
3.27
3.13
3.81
3.57
3.38
3.22
3.08
3.78
3.54
3.34
3.18
3.05
3.74
3.50
3.31
3.15
3.01
3.71
3.47
3.27
3.11
2.98
3.67
3.43
3.24
3.08
2.94
3.66
3.41
3.22
3.06
2.92
16
17
18
19
20
3.16
3.07
2.98
2.91
2.84
3.10
3.00
2.92
2.84
2.78
3.05
2.96
2.87
2.80
2.73
3.02
2.92
2.84
2.76
2.69
2.97
2.87
2.78
2.71
2.64
2.93
2.83
2.75
2.67
2.61
2.90
2.80
2.71
2.64
2.57
2.86
2.76
2.68
2.60
2.54
2.83
2.73
2.64
2.57
2.50
2.81
2.71
2.62
2.55
2.48
21
22
23
24
25
2.79
2.73
2.69
2.64
2.60
2.72
2.67
2.62
2.58
2.54
2.67
2.62
2.57
2.53
2.49
2.64
2.58
2.54
2.49
2.45
2.58
2.53
2.48
2.44
2.40
2.55
2.50
2.45
2.40
2.36
2.51
2.46
2.41
2.37
2.33
2.48
2.42
2.37
2.33
2.29
2.44
2.38
2.34
2.29
2.25
2.42
2.36
2.32
2.27
2.23
26
27
28
29
30
2.57
2.54
2.51
2.48
2.45
2.50
2.47
2.44
2.41
2.39
2.45
2.42
2.39
2.36
2.34
2.42
2.38
2.35
2.33
2.30
2.36
2.33
2.30
2.27
2.25
2.33
2.29
2.26
2.23
2.21
2.29
2.26
2.23
2.20
2.17
2.25
2.22
2.19
2.16
2.13
2.21
2.18
2.15
2.12
2.09
2.19
2.16
2.13
2.10
2.07
35
40
50
60
70
2.35
2.27
2.17
2.10
2.05
2.28
2.20
2.10
2.03
1.98
2.23
2.15
2.05
1.98
1.93
2.19
2.11
2.01
1.94
1.89
2.14
2.06
1.95
1.88
1.83
2.10
2.02
1.91
1.84
1.78
2.06
1.98
1.87
1.79
1.74
2.02
1.94
1.82
1.75
1.70
1.98
1.90
1.78
1.70
1.65
1.96
1.87
1.76
1.68
1.62
80
90
100
120
150
2.01
1.99
1.97
1.93
1.90
1.94
1.92
1.89
1.86
1.83
1.89
1.86
1.84
1.81
1.77
1.85
1.82
1.80
1.76
1.73
1.79
1.76
1.74
1.70
1.66
1.75
1.72
1.69
1.66
1.62
1.70
1.67
1.65
1.61
1.57
1.65
1.62
1.60
1.56
1.52
1.61
1.57
1.55
1.51
1.46
1.58
1.55
1.52
1.48
1.43
200
250
300
400
500
1.87
1.85
1.84
1.82
1.81
1.79
1.77
1.76
1.75
1.74
1.74
1.72
1.70
1.69
1.68
1.69
1.67
1.66
1.64
1.63
1.63
1.61
1.59
1.58
1.57
1.58
1.56
1.55
1.53
1.52
1.53
1.51
1.50
1.48
1.47
1.48
1.46
1.44
1.42
1.41
1.42
1.40
1.38
1.36
1.34
1.39
1.36
1.35
1.32
1.31
600
750
1000
1.80
1.80
1.79
1.73
1.72
1.72
1.67
1.66
1.66
1.63
1.62
1.61
1.56
1.55
1.54
1.51
1.50
1.50
1.46
1.45
1.44
1.40
1.39
1.38
1.34
1.33
1.32
1.30
1.29
1.28
SEME 112 – Advanced Statistics
Module IV
170
Table C
Q Table for HSD
Q critical values for alpha = .05
df within
k= Number of Treatments
df↓ k → 2
3
4
5
6
7
8
9
10
5
3.64 4.60 5.22 5.67 6.03 6.33 6.58 6.80 6.99
6
3.46 4.34 4.90 5.30 5.63 5.90 6.12 6.32 6.49
7
3.34 4.16 4.68 5.06 5.36 5.61 5.82 6.00 6.16
8
3.26 4.04 4.53 4.89 5.17 5.40 5.60 5.77 5.92
9
3.20 3.95 4.41 4.76 5.02 5.24 5.43 5.59 5.74
10
3.15 3.88 4.33 4.65 4.91 5.12 5.30 5.46 5.60
11
3.11 3.82 4.26 4.57 4.82 5.03 5.20 5.35 5.49
12
3.08 3.77 4.20 4.51 4.75 4.95 5.12 5.27 5.39
13
3.06 3.73 4.15 4.45 4.69 4.88 5.05 5.19 5.32
14
3.03 3.70 4.11 4.41 4.64 4.83 4.99 5.13 5.25
15
3.01 3.67 4.08 4.37 4.59 4.78 4.94 5.08 5.20
16
3.00 3.65 4.05 4.33 4.56 4.74 4.90 5.03 5.15
17
2.98 3.63 4.02 4.30 4.52 4.70 4.86 4.99 5.11
18
2.97 3.61 4.00 4.28 4.49 4.67 4.82 4.96 5.07
19
2.96 3.59 3.98 4.25 4.47 4.65 4.79 4.92 5.04
20
2.95 3.58 3.96 4.23 4.45 4.62 4.77 4.90 5.01
24
2.92 3.53 3.90 4.17 4.37 4.54 4.68 4.81 4.92
30
2.89 3.49 3.85 4.10 4.30 4.46 4.60 4.72 4.82
40
2.86 3.44 3.79 4.04 4.23 4.39 4.52 4.63 4.73
60
2.83 3.40 3.74 3.98 4.16 4.31 4.44 4.55 4.65
120
2.80 3.36 3.68 3.92 4.10 4.24 4.36 4.47 4.56
infinity 2.77 3.31 3.63 3.86 4.03 4.17 4.29 4.39 4.47
SEME 112 – Advanced Statistics
Module IV
171
Table D
Source:https://www.mun.ca/biology/scarr/4250_Chisquare_critical_values.html
SEME 112 – Advanced Statistics
Module IV
172
TABLE E
MANN-WHITNEY U TEST
google.com/search?q=mann+whitney+u+test+table&tbm=isch&source=iu&ict
x=1&fir=koXDn8PciMkbDM%252C6ntOp75dyqfRJM%252C_&vet=1&usg=AI4_kSesFtasNTIbRyYVldF2PIEg0H5QA&sa=X&ved=2ahUKEwjFtaiG_v3qAhXTEnAKH
ZlxCAEQ9QEwAXoECAUQJQ&biw=1366&bih=657#imgrc=bsHMgOTCQ5AEEM
SEME 112 – Advanced Statistics
Module IV
173
TABLE F
WILCOXON SIGNED-RANKS TABLE
https://www.google.com/url?sa=i&url=http%3A%2F%2Fwww.realstatistics.com%2Fstatistics-tables%2Fwilcoxon-signed-rankstable%2F&psig=AOvVaw01XGkUSfG9sMNJlfFeO2nk&ust=1596508200056000&s
ource=images&cd=vfe&ved=2ahUKEwjcsbfS_v3qAhUBEKYKHXQ4BZ8Qr4kDegU
IARC4AQ
SEME 112 – Advanced Statistics
Module IV
174
TABLE G
Friedman’s ANOVA by Ranks Critical Value Table
Three tables according by “k”.
If your k is over 5, or your n is over 13, use the chi square critical value table to get the critical value.
k=3
N
α <.10
α ≤.05
α <.01
3
6.00
6.00
—
4
6.00
6.50
8.00
5
5.20
6.40
8.40
6
5.33
7.00
9.00
7
5.43
7.14
8.86
8
5.25
6.25
9.00
9
5.56
6.22
8.67
10
5.00
6.20
9.60
11
4.91
6.54
8.91
12
5.17
6.17
8.67
13
4.77
6.00
9.39
∞
4.61
5.99
9.21
N
α <.10
α ≤.05
α <.01
2
6.00
6.00
—
3
6.60
7.40
8.60
4
6.30
7.80
9.60
5
6.36
7.80
9.96
6
6.40
7.60
10.00
7
6.26
7.80
10.37
8
6.30
7.50
10.35
∞
6.25
7.82
11.34
N
α <.10
α ≤.05
α <.01
3
7.47
8.53
10.13
4
7.60
8.80
11.00
5
7.68
8.96
11.52
∞
7.78
9.49
13.28
k=4
k=4
Reference:
Friedman’s Two-way Analysis of Variance by Ranks — Analysis of k-Within-Group Data with a
Quantitative Response Variable. Retrieved 7-17-2016 from: http://psych.unl.edu/psycrs/handcomp/hcfried.PDF
CITE THIS AS:
Stephanie Glen. "Friedman’s Test / Two Way Analysis of Variance by Ranks" From StatisticsHowTo.com: Elementary
Statistics for the rest of us! https://www.statisticshowto.com/friedmans-test/
SEME 112 – Advanced Statistics
Module IV