Module in Inferential Statistics
by
Fernando P. Tataro
Introduction
Module has been around for decades as an aid to the delivery of teaching. With
the proliferation of the Covid 19, module has become a necessity. In fact, many
authors have already written this type of manuscript. As the Covid 19 spreads out
worldwide, it causes many institutions/sectors (private or state) to be locked down. Its
spread prompted many countries in almost all parts of the world to issue ECQ, GCQ,
or Modified GCQ or whatever they may call it. The top 5 countries hardly hit by this
Pandemic include the United States (USA), Brazil, Russia, India and the United
Kingdom (UK). The number of infected (total cases) in the USA has already reached
more than 2.6 million and is still growing. In the Philippines, though not that serious
compared to the above-mentioned countries, for fear that the Medical Sector may not
be able to handle the great number of infected persons and to prevent the rapid
increase of infection, the government, as advised by the Department of Health (DOH),
has issued a lockdown as well.
One of the sectors greatly impacted by this Pandemic Covid 19 is the
Educational System. The issuance of lockdown due to the virus, hindered educators
and students to meet face-to-face because they have to stay at home. But all
concerned individuals must not succumb to the adversities brought about by it. Thanks
to the new technology at hand. The advent of this technology has somehow enabled
us to cope up with the current challenges, especially that caused by Covid 19
pandemic.
The influx of computers, TVs, cellphones, etc., has grown in gigantic numbers,
and so with the application programs / softwares (You-Tubes, Zoom, Google
Classroom, Google Forms, etc.), running in various platforms. The role of Modules
however will remain essential to the Educational System.
General Objectives: After finishing Modules 1 to 3, you are expected to
1.
2.
3.
4.
5.
6.
7.
Distinguish between parametric and nonparametric tests
Categorize the type of statistics you will be using
Calculate and compare results to tabular values to make a conclusion
Arrive at the conclusion using p-value instead of the computed value
Organize results in tables or graphs
Interpret tables or graphs and describe its characteristics
Achieve mastery in following the Stepwise Method in Solving Statistical
Problems
8. Apply to gained knowledge to real life situations
Module 1
SKEWNESS AND KURTOSIS
Skewness and kurtosis are measures that give information whether the
distribution is normal or abnormal. When skewness is positive, the distribution
is said to be positively skewed or skewed right, meaning, the right tail of the
distribution is longer than the left. When the skewness is negative, the
distribution is negatively skewed or skewed left, which means that the left tail
of the distribution is longer than the right. And If skewness is zero, the
distribution is perfectly symmetrical. The curve representing the distribution is
bell shape.
When the skewness is zero (perfectly symmetrical) and kurtosis is 0.265
(mesokurtic), the distribution is said to be normal.
These two measures are very important in the succeeding lessons on
Inferential Statistics. These enable us to determine if the distribution is normal
or non-normal.
Objectives:
When you have completed the lessons, you will be able to:
•
•
•
•
Calculate quantities such as the mean, median and standard deviation for
use in the skewness.
Use skewness to determine whether the distribution of data is positively
skewed or negatively skewed (abnormal) or neither (normal).
Calculate the quartiles, quartile deviation, percentiles and kurtosis to
distinguish whether the distribution of data is mesokurtic, leptokurtic and
platykurtic then tell the abnormality or normality of the distribution based on
these criteria.
Identify the different types of graphs of skewness and kurtosis.
LESSON 1 - The SKEWNESS
The formula for skewness is
𝑆𝑘 =
3(𝑋̅ − Md)
SD
Where:
𝑋̅= Mean
Md=Median
SD= Standard Deviation
Example 1. Consider whether this frequency distribution is normal or abnormal.
Scores
f
45-50
3
40-44
6
35-39
9
30-34
11
25-29
12
20-24
5
15-19
4
n=50
Expanded Class Frequency Table
Scores
𝑓
45-50
40-44
35-39
30-34
25-29
20-24
15-19
3
6
9
11
12
5
4
50
Class
Mark
x
47.5
42.5
37.5
32.5
27.5
22.5
17.5
a) Calculation of the Mean:
2
𝑓𝑥
𝑓𝑥
142.5
255
337.5
357.5
330
112.5
70
1605
6768.75
10837.5
12656.25
11618.75
9075
2531.25
1225
54712.5
Class
Boundary
Lb
Ub
44.5
50.5
39.5
44.5
34.5
39.5
29.5
34.5
24.5
29.5
19.5
24.5
14.5
19.5
Cum.
Frequency
CF
50
47
41
32
21
9
4
∑𝑓𝑥
𝑋̅ = 𝑛
=
1605
50
𝑋̅=32.1
b) Calculation of the Median, Md:
Using the formula:
𝑛
−𝐶𝐹
Md=Lme+2𝐹𝑚𝑒 𝑖
𝑛
To determine values of Lme, CF, and Fme, first calculate 2.
𝑛
2
50
=
2
= 25
Then, identify the value of Lme under the Cumulative Frequency (CF) Column.
Starting from the bottom, search for the first value that is equal to or greater the
𝑛
calculated 2 = 25.
This value is 32. The value 32 enables us to locate the Median Class. Thus, the
Median Class is 29.5 – 34.5. So, Lme=29.5, CF=21, and Fme=11. The class interval i is
5, the difference between two subsequent values, say 45 and 40. So 45-40 =5.
i =5
Substituting values in the formula for Median above:
Thus
𝑛
−𝐶𝐹
𝑀𝑑 = 𝐿𝑚𝑒 + 2𝐹𝑚𝑒 𝑖
𝑀𝑑 = 29.5 +
25−21
11
(5)
𝑀𝑑 = 31.318
c) Calculation of the Standard Deviation:
From the formula:
2
2 (∑ 𝑓𝑥)
∑ 𝑓𝑥 −
𝑛
𝑆𝐷 = √
𝑛−1
16052
50
54712.5−
𝑆𝐷 = √
50−1
𝑆𝐷 = 8.071
d) Calculation of the SK
𝑆𝐾 =
𝑆𝐾 =
3(𝑋̅ −𝑀𝑑)
𝑆𝐷
3(32.1−31.318)
8.071
𝑆𝐾 = 0.291
The SK value of 0.291 indicates that the distribution is abnormal. Being positive, the
distribution is said to be positively skewed or skewed right.
DIFFERENT GRAPHS OF SKEWNESS
Skewness is the shape of the distribution. The distribution is negatively
skewed when the thinner tail is deviated to the left side. It is positively skewed when
the thinner tail is deviated to the right, but it is normal when it is a bell shape. (Broto,
2006)
SK=+
SK=0
SK=-
Positively Skewed
Normal
Negatively Skewed
Normal
LESSON 2 - THE KURTOSIS
Kurtosis is a statistical measure used to describe the degree to which scores
cluster in the tails or the peak of a frequency distribution. The peak is the tallest part
of the distribution, and the tails are the ends of the distribution. There are three types
of kurtosis: mesokurtic, leptokurtic, and platykurtic.
The distribution is said to be normal when the value of kurtosis is equal to 0.265
mesokurtic. When its value is less than 0.265 leptokurtic, and when its value is greater
than 0.265 platykurtic. Both leptokurtic and platykurtic types of kurtosis indicate that
the distribution is abnormal.
The formula for kurtosis:
𝑄𝑑
𝐾𝑢 = (𝑃
90 −𝑃10 )
Where:
Ku=Kurtosis
Q=Quartile Deviation
𝑃90 = 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 90
𝑃10 = 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 10
Example:
Consider the frequency distribution of example 1 for the problem in Skewness.
Determine whether it is normal or abnormal by solving the value of Kurtosis.
a) Solve for Qd.
𝑄3 − 𝑄1
2
But Qd has the formula above. And Q1 and Q3 must be calculated first.
𝑄𝑑 =
For Q1:
𝑄1 = 𝐿𝑄1 +
1) Determine
𝑛
4
=
50
4
𝑛
−𝐶𝐹
4
𝐹𝑄1
𝑖
= 12.5.
2) Identify the first quartile class:
Look for the first value under Cumulative Frequency Column from the above
table greater than or equal to 12.5, this value is 21.
The value 21 is (21>=12.5).
This value is located in the third row from the bottom of the table. The 1st
Quartile Class (24.5 – 29.5) belongs to this row. LQ1 therefore is 24.5; CF=9,
the cumulative frequency before; the FQ1 =12; and i=5.
𝑄1 = 24.5 +
12.5 − 9
(5)
12
Q1 = 25.958
For Q3:
𝑄3 = 𝐿𝑄3 +
1) Determine
3𝑛
4
=
3(50)
4
3𝑛
−𝐶𝐹
4
𝐹𝑄3
𝑖
= 37.5.
2) Identify the third quartile class:
Look for the first value under Cumulative Frequency Column from the above
table greater than or equal to 37.5, this value is 41. The value 41 is (41 >= 37.5).
This value is located in the fifth row from the bottom of the table. The Quartile
Class (34.5 – 39.5) belongs to this row. LQ3 therefore is 34.5; CF=32, the
cumulative frequency before; the FQ3 =9; and i=5.
𝑄3 = 34.5 +
37.5 − 32
(5)
9
Q3 =37.556
𝑄𝑑 =
37.556−25.958
2
𝑄𝑑 = 5.799
For P10:
𝑃10 = 𝐿𝑃10 +
1) Determine
10𝑛
100
=
10(50)
100
10𝑛
−𝐶𝐹
100
𝐹𝑃10
𝑖
= 5.
2) Identify the P10 class:
Look for the first value under Cumulative Frequency Column from the above
table greater the or equal to 5, this value is 9. The value 9 is (9 >=5).
This value is in the 2nd row from the bottom of the table. The P10 Class (19.524.5) belongs to this row. LP10 is 19.5, CF=4, and FP10 = 5.
Substituting values,
𝑃10 = 19.5 +
𝑃10 = 20.5
5−4
5
(5)
90𝑛
𝑃90 = 𝐿𝑃90 + 100
𝐹
P90:
−𝐶𝐹
𝑃90
90𝑛
1) Determine 100 =
90(50)
100
𝑖
= 45.
2) Identify the P90 class:
Look for the first value under Cumulative Frequency Column from the above
table greater the or equal to 45, this value is 47. The value 47 is (47 >=45).
This value is in the 2nd row from the top of the table. The P90 Class (39.5-44.5)
belongs to this row. LP10 is 39.5, CF=41, and FP90 = 6.
Substituting values,
𝑃90 = 39.5 +
45−41
6
(5)
𝑃90 = 42.833
Since the item values in the formula for Ku were all determined, thus
𝐾𝑢 = (𝑃
𝑄𝑑
90 −𝑃10 )
5.799
𝐾𝑢 = (42.833−20.5)
𝐾𝑢 = 0.260
The value of Ku equal to 0.260 is less than 0.265. This means that, the
distribution is Leptokurtic or the data cluster to the peak. The distribution is abnormal.
Note: The arrangement of the data in the Frequency Distribution Table is in
descending order. The cumulative frequencies obtained started from the frequency
below the table towards the top by adding them together one after the other. Had the
data were arranged in ascending order, cumulative frequencies shall be obtained in
reverse order (i.e., by starting from the top).
Exercise:
Consider the following Frequency Distribution. Determine if the distribution is normal
or abnormal by solving for skewness and kurtosis.
Scores
f
16-20
2
21-25
5
26-30
11
31-35
13
36-40
10
41-45
6
46-50
3
n=50
LESSON 3 - INTRODUCTION TO INFERENTIAL STATISTICS
Inferential statistics deals with the analysis and interpretation of data based
on a sample or representative of the population. This statistic consists of differential
statistical tools/tests used in the analysis of interval, ratio, nominal and ordinal data.
These tests are used in making inferences about populations using data drawn from
the populations. The extent to which the use of this statistics can be done with
accuracy depends on the goodness of samples. The sampling techniques /
procedures are also of great importance with regard to the use of these different
statistical tests.
Objectives:
When you have completed the Lessons, you will be able to:
•
•
•
•
•
•
•
•
•
Become familiar with the kinds of statistical tests.
Illustrate the difference between a Null Hypothesis and an Alternative or
Research Hypothesis.
Differentiate a Type I error from a Type II Error.
Explain the meaning and the use of the Level of Significance.
Identify the conditions that will require the use of either a One-tailed Test or
a Two-tailed Test.
Explain the concept of the critical or rejection region in a Normal Curve.
Identify which formula to apply when the sample size is Large and when the
size is Small; alternatively, when to use a Z test or a t test.
State the conclusion to a Test of Hypothesis of rejecting or accepting the
Null Hypothesis.
Become familiar with solving the problems by Stepwise Method.
KINDS OF STATISTICAL TESTS
Statistical tests can be grouped into two. The parametric and the
nonparametric tests.
The parametric tests. To use the parametric tests, there are some
conditions that should be met. The data must be normally distributed and the level of
measurement must be either interval or ratio.
The data are said to be normal when the value of skewness equals zero
and the value of kurtosis is 0.265.
The interval data provide numbers that reflect difference among items.
With interval scales, the measurement units are equal. Examples are scores of
intelligence tests, and time reckoned from the calendar. They have no true zero value.
The ratio scale is the highest type of scale. The basic difference between
the interval and the ratio scales is that, the interval scale has no true zero value while
the ratio scale has an absolute zero value. Common ratio scales are measures of
length, width, weight, capacity and loudness and others.
The nonparametric tests. The nonparametric tests do not require
normality of the distribution. Skewness and Kurtosis are the measures that tell whether
the data is normal or abnormal. If the value of skewness is either positive or negative,
distribution is said to be abnormal. When the Kurtosis is greater than or less than
0.265, which means, it is not equal to that value, the distribution is abnormal as well.
Under these tests, the levels of measurement are the nominal and ordinal
data. Nominal data are data such as: male or female, yes or no responses, political
affiliations like, LP, LDP, Lakas, and religious groupings, Christians and Non-Christian
and other organizations.
Ordinal data are data such as: Strongly Agree, Agree, No Opinion,
Disagree and Strongly Disagree and also other data which employ rankings.
LESSON 4 - TESTS OF HYPOTHESES (Garcia, 2003)
The Null Hypothesis
A hypothesis is a statement about our expectation with regard to some
characteristic of a population. It is something that we assume about a population. With
the passing of time, however, population characteristics as measured by the mean,
can change. This is where we test a hypothesis about the population.
The hypothesis is a statement that, at the end of the solution to the problem,
is either rejected or accepted. This statement is called the Null Hypothesis and is
denoted by Ho. On the other hand, an alternative claim is the Alternative Hypothesis
and uses the symbol Ha. The options with regard to these two hypotheses are:
•
•
If Ho is rejected, then Ha is accepted
If Ho is accepted, then Ha is rejected
Put another way, when we reject Ho, we affirm the statement that is the
alternative to Ho (that is, we accept Ha). On the other hand, if we accept Ho, we affirm
the statement expressed by the Null Hypothesis (therefore, we reject Ha).
The Null Hypothesis is a statement of our expectation with regard to the
population and is expressed in the negative form. The following examples will illustrate
this point.
•
Expectation: The IQ level of first year college students in a particular
university is still the same.
Ho: The average IQ level of first year college students has not changed.
Ha: The average IQ level of first year college students has changed.
•
Expectation: Despite a number of economic problems, the average
annual income of Filipino families is still the same.
Ho: The average annual income of Filipino families has not changed.
Ha. The average annual income of Filipino families is lower than that of the
previous years.
•
Expectation: The weight (in grams) of a can of biscuits is consistent with
the manufacturer’s claim in the label.
Ho: The average weight of a can of biscuits has not changed.
Ha: The average weight of a can of biscuits is less than what is claimed by
the manufacturer.
•
Expectation: The average monthly allowance of the morning and
afternoon students in a high school are the same.
Ho: There is no significant difference in the average monthly allowance of
the morning and afternoon students.
Ha: There is a significant difference in the average monthly allowance of
the morning and afternoon students.
Types of Test
There are a number of tests involved when a hypothesis is being tested. In
this chapter, we shall discuss two types. These are:
1. Test of Means
2. Test of Difference of Means
The first type aims to find out if a population characteristic, as indicated by the Mean,
has changed. The second seeks to determine if the same characteristic between two
populations is significantly different.
Types of Error
The Null Hypothesis as stated, can either be correct (that is, the statement
is true) or wrong (that is, the statement is false). Two types of errors can be committed
when a conclusion is made after a hypothesis is tested. These are:
•
•
Type I error (or alpha error): The Null Hypothesis is rejected when, in
fact, it is true.
Type II error (or beta error): The Null Hypothesis is accepted when, in
fact, it is false.
Since we can never be completely certain about the claim made in the Null
Hypothesis and the corresponding conclusion, therefore, there is always the risk of
making either a Type I or a Type II error. Between these two types, the more serious
error is Type II. We certainly would avoid making a Type II error, if any error should be
made at all. The chance of making an error of this type can be minimized by using
what is referred to as the Level of Significance.
The Level of Significance
One reason for taking a sample from a population is to compute for the
mean of the sample and to use this mean as an estimate of the population mean.
There are many samples that one can get from a given population. For some of these
samples: the computed mean will be close to the population mean; for some, it may
not be very close; and for quite a few exceptional samples, the computed Mean may
be quite far from the population mean, making the sample mean a poor (or even
wrong) estimate of the population mean. Despite the effort to choose a representative
sample, such as situation, although unlikely, is still possible.
In Statistics, the dividing line between what is acceptable and not
acceptable is the Level of Significance. It is a level of probability that separates the
sample results that are acceptable and those that are not. More specifically, it is the
level of probability of making a Type I error. The customary Levels of Significance are
5 % and 1 %.
A level of Significance of 5% means, that there are 5 chances in 100 of
rejecting the Null Hypothesis when in fact it is true. Correspondingly, we are 95%
confident that we are making the right decision (in terms of our conclusion). Reducing
the Level of Significance to 1 % makes us 99% confident that we are doing the right
or correct decision. This is especially true in the medical profession, where “life-anddeath” situation is at stake. The probability of making a Type I error could be minimized
if we lower the Level of Significance to 1% instead of 5%.
A basic principle in Statistics states that the mean of each of the so many
samples that can be taken from a population, when organized into a Frequency
Distribution, tends to approximate a Normal Curve. This is called the Central Limit
Theorem which states that: when all possible samples of a given size are taken from
any population, the distribution of the means of Samples will approximate a Normal
Distribution. This approximation improves as the Sample Size becomes larger. The
Means of most samples will be clustered around the population mean in the Normal
Curve. For a few exceptional samples, that, by chance, yield sample means that are
far from the population mean, their means are close to the tail ends of the Normal
Curve. The region where such values are situated is called the Critical Region.
Critical or
rejection
region
Critical or
rejection
region
Level of Significance
As shown above, the two critical regions on both ends of the Normal Curve
are also called the rejection regions. The area in the Middle is the acceptance region.
The Level of Significance is the dividing line between the acceptance region and the
rejection region.
Important Note: Another way of knowing when to reject or accept the Null Hypothesis
is to solve for the p-value. When the p-value is less than the Level of Significance,
then the Null Hypothesis is rejected.
Use of Symbols
We need to make a distinction between the symbols used for a population
and those for a sample. These symbols are used in the formulas applied in testing a
Hypothesis. Let us summarize them below:
Sample
Population
Size
n
N
Mean
𝑋̅
𝜇 (mu)
Standard Deviation
s
𝜎 (𝑠𝑖𝑔𝑚𝑎)
(small letter s)
One-Tailed and Two-Tailed Tests
In testing a hypothesis, we have to decide whether to use a One-tailed Test
or a Two-tailed Test. Whether it is one-tailed or two-tailed will depend on the nature of
the problem which, in turn, will determine how the Null Hypothesis (Ho) and Alternative
Hypothesis (Ha) are stated.
Before we cite examples to determine which test to use, let us point out that
in a two-tailed test, the critical regions are on both sides of the Normal Curve, that is:
Critical or
rejection
region
Critical or
rejection
region
On the other hand, in one-tailed test, the critical region is only on one side
of the Normal Curve, either on the left or right side of it, that is:
Critical or
rejection
region
And
Critical or
rejection
region
Let us now cite examples to illustrate when to use each type of test. Without solving
the problem, the following example will illustrate the use of a Two-tailed Test:
•
A machine is used to produce a precision part of an automobile. The diameter of
one particular item should be 5 centimeters with a slight tolerable allowance on
both sides, that is, the diameter can be slightly wider or narrower but for as long as
the average diameter of the parts being produced stays at 5 cm, the machine is
functioning well. Since the matching has already been used for several years, the
manufacturer feels that it is time to check the machine. If it is no longer functioning
well, the average diameter could be wider or narrower than 5 cm. One way of
checking is to test is a hypothesis by taking a random sample of the parts being
manufactured.
This problem will require a Two-tailed Test with critical regions on both sides of the
Normal Curve because if the machine is not functioning properly it will produce
parts whose diameters will be wider and narrower. It is not a case of the
malfunctioning machine producing parts with consistently narrower diameters only
(or with consistently wider diameters only).
This situation determines now how the Null and Alternative Hypotheses are
stated, which are:
Null Hypothesis: The average diameter of the part being manufactured has not
changed (that is, Ho: 𝜇 = 5 𝑐𝑚).
Alternative Hypothesis: The average diameter has changed (that is, Ha: 𝜇 ≠ 5 𝑐𝑚).
Note: The manner in which the Alternative Hypothesis is stated as an inequality,
confirms the Two-tailed Test that is to be used.
The following example will illustrate the use of a One-tailed test (where the
critical region is on the right side only of the Normal Curve), that is, specifically a Righttailed test:
•
A barangay chairman feels that, on the whole households in his barangay, are
better-off financially now compared to several years ago. He feels this because he
has observed that mothers seem to be visiting supermarkets more often and that
he has been receiving less complaints. Furthermore, he knows that a quite a
number of husbands in his barangay have gone abroad to work as OFWs
(overseas foreign workers) where incomes are much higher than when you are just
in the Philippines. He knows from a survey conducted several years back that the
average monthly income in his barangay was Php 20,000. He decides to find out
if the average monthly income may have possibly increased by taking a sample of
households. He tests his hypothesis.
The Null and Alternative hypotheses for his problem are stated as follows:
Null Hypothesis: The average monthly family income in the barangay has not changed.
(that is, Ho: 𝜇 = 𝑃ℎ𝑝 20,000).
Alternative Hypothesis: The average monthly family income has increased (that is, Ha:
𝜇 > 𝑃ℎ𝑝 20,000).
Note: The inequality statement of the Alternative Hypothesis confirms a Right-tailed
test. Clearly, the indicators to the barangay chairman suggest that incomes may have
possibly increased. There is no suggestion at all that incomes may have decreased.
The following example will illustrate a one-tailed (specifically a Left-tailed)
test.
•
A university requires an IQ test for all incoming freshmen students and accepts
only those who qualify, with an IQ above the minimum IQ requirement. It is known
that the average IQ of the university’s freshmen students is 95. However, for
economic reasons (on the part of the university and on the part of the families of
students), the IQ test for incoming first-year students has been suspended for the
past three years. Lately, some teachers have been complaining that students in
their freshmen classes have been getting somewhat lower grades and this appears
to be a persistent claim to find out if the IQ level of their first-year students has
decreased. They take a sample of freshmen students and test a hypothesis.
Let us state the Null and Alternative Hypotheses:
Null Hypothesis: The average IQ of freshmen students has not changed (that is, Ho:
𝜇 = 95).
Alternative Hypothesis: The average IQ of freshmen of students has decreased (that
is, Ha: 𝜇 < 95).
Note: The evidence based on claims suggests that the IQ has decreased. It cannot be
stated as the inequality (𝜇 ≠ 95). Definitely, neither (𝜇 > 95). Thus, this Alternative
Hypothesis suggests a Left-tailed test.
Solving Problems by Stepwise Method: (Broto, 2006)
Here are the steps:
I.
II.
III.
Problem:
Hypothesis:
hypothesis.
Level of Significance:
Formulate the Statement of the Problem.
State the null hypothesis and/or the research
Specify the level of significance, denoted alpha 𝛼.
Usually taken as 0.05 or 0.01 depending on the
problem or study.
IV.
V.
VI.
Statistics:
Identify the Statistics Tools to use.
Decision Rule:
In general, the Decision Rule goes this way: If the
computed value of the statistics is greater than or beyond the tabular
value/critical value, reject the null hypothesis.
Conclusion:
State the conclusion based on the analysis
and interpretation arrived.
MODULE 2
THE PARAMETRIC TEST
Parametric tests require normal distribution and the levels of measurement
are expressed in interval or ratio data.
Types of parametric tests are the t-test, z-test, F-test, analysis of variance
for the test of difference and the Pearson Product Moment Coefficient of Correlation
for the tests of relationship / association, and the tests for prediction and forecasting
are the Simple Linear Regression Analysis and the Multiple Regression Analysis.
Objectives:
When you have completed the chapter, you will be able to:
•
•
•
Use lessons in module 1 to determine the types of distribution of data.
Distinguish parametric from nonparametric tests and their relationship to normality
of distribution.
Apply Stepwise Method in solving problems using parametric tests.
LESSON 5 - THE t-TEST
The t-test is used to compare two means, the means of two independent
samples or groups, and the means of correlated samples before and after the
treatment. Ideally the t-test is used when there are less than 30 samples, but some
researchers use the t-test even if there are more data 30 samples. The formula for the
t-test is
𝑡=
̅̅̅̅
̅̅̅̅
𝑋1 −𝑋
2
𝑆𝑆1 +𝑆𝑆2
1
1
)( + )
√( 𝑛 +𝑛
1
2 −2 𝑛1 𝑛2
Where:
t
= the t-test
̅̅̅1 = the mean of group 1
𝑋
̅̅̅2 = the mean of group 2
𝑋
𝑆𝑆1= sum of squares of group 1
𝑆𝑆2 = sum of squares of group 2
𝑛1 = number of observations in
group 1
𝑛2 = number of observations in
group 2
Example 1. The following are the scores of 10 male and 10 female BSBM students in
mathematics. Test the null hypothesis that there is no significant difference between
the performance of male and female BSMB students in the said test. Use the t-test at
0.05 level of significance.
Male
15
19
17
15
6
12
14
10
11
16
Male
𝑥12
225
361
289
225
36
144
196
100
121
256
1953
𝑥1
15
19
17
15
6
12
14
10
11
16
135
Female
13
9
12
5
9
3
8
4
6
13
Female
𝑥2
𝑥22
13
169
9
81
12
144
5
25
9
81
3
9
8
64
4
16
6
36
13
169
82
794
From the table above,
̅̅̅1 = ∑ 𝑥1 = 13.5
∑ 𝑥1 = 135, ∑ 𝑥1 2 = 1953, n1 = 10, 𝑋
𝑛1
̅̅̅2 =
∑ 𝑥2 = 82, ∑ 𝑥2 2 = 794, n2 = 10, 𝑋
2
𝑆𝑆1 = ∑ 𝑥1 −
2
(∑ 𝑥1 )2
𝑆𝑆2 = ∑ 𝑥2 −
𝑛1
(∑ 𝑥2 )2
𝑛2
= 1953 −
(135)2
= 794 −
10
822
10
∑ 𝑥2
𝑛2
= 130.5
= 121.6
= 8.2
̅̅̅1 − 𝑋
̅̅̅2
𝑋
𝑡=
√(
𝑡=
𝑆𝑆1 + 𝑆𝑆2
1
1
𝑛1 + 𝑛2 − 2)(𝑛1 + 𝑛2 )
13.5 − 8.2
√( 130.5 + 121.6)( 1 + 1 )
10 + 10 − 2 10 10
𝑡=
5.3
√252.1 ( 1 + 1 )
18 10 10
𝑡=
5.3
√14.0056(0.2)
𝑡=
5.3
√2.80112
𝑡=
5.3
1.6737
𝑡 =3.167
Adopting the Stepwise Method briefly delineated above:
I.
II.
III.
Problem: Is there a significant difference between the performance of the
male and female students in mathematics?
Hypothesis:
Ho:
There is no significant difference between the performance of
̅̅̅2 ).
male and female BSBM students in mathematics( ̅̅̅̅
𝑋1 = 𝑋
H1: There is a significant difference between the performance of male and
̅̅̅̅
̅̅̅
female BSBM students in mathematics ( 𝑋
1 ≠ 𝑋2 ).
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = 𝑛1 + 𝑛2 − 2 =10+10-2=18
From the table, 𝑡0.05 = 2.101. This value was obtained by pairing the level
of significance, 𝛼 and the degree of freedom, df, (0.05, 18).
IV.
V.
VI.
Statistics: t-test for Two Independent Samples
Decision Rule: If the t-computed value is greater than the t-tabular /critical
value, reject the null hypothesis Ho.
Conclusion: Since the t-computed value of 3.167 is greater than the t-tabular
value of 2.101, at 0.05 level of significance with 18 degrees of freedom, the
null hypothesis is rejected in favor of the research hypothesis. This means
that, there is a significant difference in the performance of the male and
female BSMB students in mathematics. It implies that the male performed
better than the female students, considering that the mean or average score
of 13.5 obtained by the male students is greater than the mean score of the
female students of 8.2.
Exercise:
Two groups of experimental rabbits were injected with tranquilizer at 1.0 mg and
1.5 mg dose respectively. The time given in seconds that took them to fall asleep
is hereby given. Use the t-test for two independent samples at 0.01 to test the null
hypothesis that the difference in dosage has no effect on the length of time it took
them to fall asleep.
1.0 mg
5.3
3.4
7.2
6.7
5.6
3.1
4.8
7.8
13.1
8.2
6.4
12.3
1.5 mg
12.1
7.8
15.4
14.2
13.6
9.7
10.4
17.2
20.3
19.7
LESSON 6 - THE t-TEST FOR CORRELATED SAMPLES
The t-test for correlated samples is used when comparing the means before
and after the treatment such as pretest and posttest. The formula is,
𝑡=
̅
𝐷
2
2 − (∑ 𝐷)
∑
𝐷
√
𝑛
𝑛(𝑛 − 1)
Where:
̅
𝐷 = the mean difference between the
pretest and the posttest.
D = the difference between the pretest and the posttest
n = the sample size
Example 1. An experimental study was conducted on the effect of programmed
materials in Mathematics on the performance of 20 selected college students. Before
the program was implemented, pretest was administered and after 3 months the same
instrument was used to get the posttest result. The following is the result of the
experiment:
PRETEST
X1
20
35
15
16
18
17
23
22
19
25
28
22
12
15
21
28
25
16
34
32
POSTTEST
X2
24
38
20
25
27
24
35
27
23
28
32
28
25
26
32
39
36
28
41
38
D
-4
-3
-5
-9
-9
-7
-12
-5
-4
-3
-4
-6
-13
-11
-11
-11
-11
-12
-7
-6
∑ 𝐷 = -153
̅ = −7.65
𝐷
Substituting values in the formula below,
D2
16
9
25
81
81
49
144
25
16
9
16
36
169
121
121
121
121
144
49
36
∑ 𝐷2 =1389
𝑛 = 20
𝑡=
𝑡=
̅
𝐷
(∑ 𝐷)2
2
√∑ 𝐷 − 𝑛
𝑛(𝑛 − 1)
7.65
(−153)2
√1389 −
20
20(19)
t =-10.087
10.087
Solving by the Stepwise Method:
I.
II.
III.
IV.
V.
VI.
Problem: Is there a significant difference in the performance of 20 selected
college students in Mathematics before and after the program (Pretest and
Posttest) was implemented?
Hypothesis:
Ho: There is no significant difference between the performance of
the
20 selected students in the Pretest and Posttest.
H1: There is a significant difference between the performance of
the 20 selected students in the Pretest and Posttest.
Level of Significance:
𝛼 = 0.05
df = n-1
df = 19
t0.05=1.729
Statistics: t-test for correlated samples.
Decision Rule: If the t-computed value is greater than or beyond the critical
value, reject Ho.
Conclusion: Inasmuch as the t-computed value of 10.087 is greater the tcritical value of 1.729 at 0.05 level of significance with 19 degrees of
freedom, the null hypothesis is therefore rejected in favor of the research
hypothesis. This means that the posttest performance of the students
surpassed their pretest performance. It implies that the use of the
programmed materials in Mathematics is effective.
Pretest
16
18
16
24
Posttest
20
20
24
28
20
25
22
18
15
15
20
30
23
24
19
15
Exercise:
Ten subjects were given an attitude test on a controversial issue. Then,
they were shown a film favorable to the ten subjects and the same attitude
test was administered. Make a directional test at 𝛼 = 0.05.
THE Z-TEST
The z-test is another test under parametric statistics requiring normality of
the distribution. It utilizes the two population parameters 𝜇 𝑎𝑛𝑑 𝜎. It is used to compare
two means: the sample mean, and perceived population mean. It is also used to
compare the two-sample means reckoned from the same population. It is used when
the samples are equal to or greater 30. The z-test can be applied in two ways: the
One-Sample Mean Test and the Two-Sample Mean Test.
The tabular value of z-test at 0.01 and 0.05 level of significance are shown
below.
Test
One-tailed
Two-tailed
Level of Significance
0.01
0.05
± 2.33
± 1.645
± 2.575
± 1.96
LESSON 7 - THE ONE-SAMPLE MEAN TEST
The One-Sample Mean Test is used when the sample mean is being
compared to the perceived population mean. Nevertheless, in the absence of the
population standard deviation, the sample standard deviation can be used for this
value. The formula is
𝑧=
(𝑥̅ − 𝜇)√𝑛
𝜎
Where:
𝑥̅ = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛
𝜇 = ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
of the population mean
𝜎 = population standard deviation
n = sample size
Example 1. XYZ company claims that the average lifetime of a certain tire is at least
26,500 km. To check the claim, a taxi company puts 40 of these tires on its taxis and
gets a mean lifetime of 24,430 km. With a standard deviation of 1240 km, is the claim
true? Use z-test at 0.05.
Solving by Stepwise Method:
I.
II.
III.
Problem: Is the claim true that the average lifetime of a certain tire is at least
26,500 km?
Hypothesis:
Ho: The average lifetime of a certain tire is 26,500 km.
Ha: The average lifetime of a certain tire is not 26,500 km.
Level of Significance:
𝛼 = 0.05
𝑧 = ±1.645
IV.
Statistics: z-test for one-tailed test
Computation:
𝑧=
=
V.
VI.
(𝑥̅ − 𝜇)√𝑛
𝜎
(24430−26500)√40
1240
= −10.56
Decision Rule: If the z-computed value is greater than the z-tabular value,
then reject the null hypothesis.
Conclusion: Since the z-computed value of 10.56 is greater than the ztabular value of 1.96, at 0.05 level of significance, the research hypothesis
is accepted. This means that the average lifetime of a certain tire is no
longer 26,500 km. It implies that the average lifetime of these tires is
already less than this value.
LESSON 8 - THE TWO-SAMPLE MEAN TEST
The two-sample mean test is used when comparing two separate samples
drawn at random, from a normal population. To test whether the difference between
̅̅̅1 and 𝑋
̅̅̅2 is significant or can be attributed to chance, the formula
the two values 𝑋
used is:
𝑧=
̅̅̅̅
̅̅̅̅
𝑋1 −𝑋
2
2
2
𝑛1
𝑛2
𝑠
𝑠
√ 1+ 2
Where:
̅̅̅1 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 1
𝑋
̅̅̅2 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 2
𝑋
𝑠12 = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 1
𝑠22 = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 2
𝑛1 = 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 1
𝑛2 = 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 2
Example 1. An admission test was administered to incoming freshmen in the
Colleges of Marketing Management and Information and Computer Technology with
100 students. Each was randomly selected. The mean scores of the given samples
̅̅̅2 = 75 and the variances of the test scores were 38 and 32,
were ̅̅̅
𝑋1 = 85 𝑎𝑛𝑑 𝑋
respectively. Is there a significant difference between the two groups? Use 0.01 level
of significance.
Solve by Stepwise Method:
I.
II.
Problem: Is there a significant difference between the two groups?
Hypotheses:
Ho: There is no significant difference between the two groups.
Ha: There is a significant difference between the two groups.
III.
Level of Significance:
𝛼 = 0.01
𝑧 = ±2.575
IV.
Statistics:
z-test for two-tailed test
𝑧=
̅̅̅̅
̅̅̅̅
𝑋1 −𝑋
2
2
2
𝑛1
𝑛2
𝑠
𝑠
√ 1+ 2
𝑧=
85−75
38 32
+
100 100
√
𝑧 =11.95
V.
VI.
Decision Rule: If the z-computed value is greater than the z-tabular
value, then reject the null hypothesis.
Conclusion:
Inasmuch as the z-computed value of 11.95 is greater
than the z-tabular value of 2.575 at 0.01 level of significance, so that, the
null hypothesis is rejected. This means that the two groups have different
performances. It implies that the incoming freshmen in the College of
Marketing Management has performed better than the incoming freshmen
in the College of Information and Computer Technology in the entrance
examination administered to them.
THE F-TEST
The F-test otherwise known as the Analysis of Variance (ANOVA) is used in
comparing the means of two or more independent groups. One-way ANOVA is used
when there is only one variable involved. The Two-way ANOVA is used when two
variables are involved: the column and row variables. The researcher is interested to
know if there are significant differences between and among column and row
variables. This is also used to determine if there is an interaction effect between the
variables being analyzed.
Like the t-test, the F-test is also a parametric test, which requires that the samples
are normally distributed and that the data are expressed as interval or ratio. This test
is more efficient than the other tests of difference.
LESSON 9 - THE F-TEST (ONE-WAY-ANOVA)
Example 1. A certain grocery store is selling 4 brands of soap. The owner is
interested if there is a significant difference in the average sales for one week. The
following data were recorded.
Brand
A
B
C
D
8
10
2
4
5
12
3
6
4
9
5
7
7
8
6
9
8
7
4
3
5
10
3
4
4
11
3
6
Perform the analysis of variance and test the hypothesis at 0.05 level of
significance that the average sales of the four brands of soap are equal.
Solving by Stepwise Method:
I.
II.
Problem: Is there a significant difference in the average sales of the four
brands of soap?
Hypotheses:
Ho:
There is no significant difference in the average sales of
the four brands of soap.
Ha:
There is a significant difference in the average sales of the
four brands of soap.
III.
Level of Significance:
𝛼 = .05
𝑑𝑓 = 3 𝑎𝑛𝑑 24
IV.
Statistics
One-Way-Analysis of Variance (F-test) computation:
A
B
C
2
2
𝑋1
𝑋
𝑋
𝑋1
𝑋2
𝑋32
2
3
8
64
10
100
2
4
5
25
12
144
3
9
4
16
9
81
5
25
7
49
8
64
6
36
8
64
7
49
4
16
5
25
10
100
3
9
D
𝑋4
4
6
7
9
3
4
𝑋42
16
36
49
81
9
16
4
41
16
259
11
67
̅̅̅1=41 = 5.857
𝑋
7
121
659
3
26
̅̅̅2=67 = 9.571
𝑋
7
9
108
̅̅̅3=26 = 3.714
𝑋
7
6
39
36
243
̅̅̅4=43 = 5.571
𝑋
7
Steps in Solving for F-test (One-Way ANOVA):
Calculate the following:
1. Correction Factor, 𝐶𝐹 =
CF =
(41+67+26+39)2
=
28
(∑ 𝑋)2
(173)2
28
𝑁
= 1068.893
2. Total Sum of Squares, 𝑇𝑆𝑆 = ∑ 𝑥12 +∑ 𝑥22 +∑ 𝑥32 +∑ 𝑥42 − 𝐶𝐹
TSS =259 + 659 + 108 + 243 − 1068.893 = 200.107
(∑ 𝑋1 )2
3. Between Sum of Squares, BSS=
BSS =
412
7
672
+
7
+
262
7
392
+
7
𝑛1
+
(∑ 𝑥2 )2
𝑛2
+
(∑ 𝑋3 )2
𝑛3
+
(∑ 𝑋4 )2
𝑛4
− 𝐶𝐹
− 1068.893=126.393
4. Within Sum of Squares, 𝑊𝑆𝑆 = 𝑇𝑆𝑆 − 𝐵𝑆𝑆
WSS = 200.107 − 126.393 = 73.714
Analysis of Variance Table
Sources of
Variation
Between
Groups
Within
Group
Total
V.
VI.
Degrees of
Freedom
Sum of
Squares
k-1=4-1=3
126.393
27-3=24
73.714
N-1=28-1=27
200.107
Mean Squares
126.393
= 42.131
3
73.714
= 3.071
24
F-Value
Computed
Tabular
42.131
3.01
= 13.719
3.071
Decision Rule: If the F-computed value is greater than the F-tabular value,
reject the null hypothesis.
Conclusion: Since the F-computed value of 13.719 is greater than the Ftabular value of 3.01 at 0.05 with 3 and 24 degrees of freedom, so that the
null hypothesis is rejected. This means that there is a significant difference
in the average sales of the four brands of soap. It implies that soap B has
the greatest average sales.
Note: When it was found out that Ho was rejected or that there was a significant
difference in the computed means among groups, use Scheff𝑒́ s Test to identify
which groups have significant differences.
SCHEFF𝑬́S TEST
𝐹′ =
̅̅̅1 − 𝑋
̅̅̅2 )2
(𝑋
𝑆𝑊 2 (𝑛1 + 𝑛2 )
𝑛1 𝑛2
Where:
F’ = Scheffe’s Test
𝑋1 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑔𝑟𝑜𝑢𝑝 1
𝑋2 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑔𝑟𝑜𝑢𝑝 2
𝑛1 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒𝑠 𝑖𝑛 𝑔𝑟𝑜𝑢𝑝 1
𝑛2 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒𝑠 𝑖𝑛 𝑔𝑟𝑜𝑢𝑝 2
𝑆𝑊 2 = 𝑤𝑖𝑡ℎ𝑖𝑛 𝑚𝑒𝑎𝑛 𝑠𝑞𝑢𝑎𝑟𝑒𝑠
Brand A vs B:
F’ =
(5.857−9.571)2
3.071(7+7)
7(7)
Brand A vs C:
F’ =
(5.857−3.714)2
3.071(7+7)
7(7)
F’ = -4.233
F’ = 5.234
Brand A vs D:
Brand B vs C:
F’ =
(5.857−5.571)2
3.071(7+7)
7(7)
F’ =
(9.571−3.714)2
3.071(7+7)
7(7)
F’ = 0.000
F’ = 39.097
Brand B vs D:
Brand C vs D:
F’ =
(9.571−5.571)2
3.071(7+7)
7(7)
F’ = 18.235
F’ =
(3.714−5.5712
3.071(7+7)
7(7)
F’ = 3.930
Comparison of the Average Sales of the Four Brands of Soap
Between Brand
F’
A vs B
A vs C
A vs D
B vs C
B vs D
C vs D
4.233
5.234
0.000
39.097
18.235
3.930
(F0.05)
(k-1)
(3.071) (3)
9.213
9.213
9.213
9.213
9.213
9.213
Interpretation
Not Significant
Not Significant
Not Significant
Significant
Significant
Not Significant
The above table shows the F’-computed values for all the four brands of soap under
comparison.
Since the F’-computed values of brands A vs B, A vs C, A vs D, and C vs D are all
less than the F-tabular value of 9.213 at 0.05 level of significance and 3 degrees of
freedom, thus it can be said that the means for these brands of soap have no
significant differences. On the other hand, inasmuch as the F’-values of 39.097 and
18.235 for B vs C and B vs D respectively are both greater than the F-tabular values,
hence the means for brands B and C, and B vs D have significant differences. It
implies that the brand B of soap is more saleable than the brands C and D.
Exercise. The following data represent the operating time in hours of the 3 types of
scientific pocket calculators before a recharge is required. Perform the Analysis of
Variance (F-test) at 0.05 level of significance.
A
4.8
5.6
4.7
6.2
4.4
6.9
Brand
B
6.7
6.9
7.2
5.8
5.4
6.3
4.8
C
7.6
7.2
7.6
6.8
6.3
LESSON 10 - THE F-TEST (TWO-WAY-ANOVA WITH INTERACTION EFFECT)
In statistics, the two-way analysis of variance (ANOVA) is an extension of the oneway ANOVA that examines the influence of two different categorical independent
variables on one continuous dependent variable.
The Two-Way ANOVA uses the following formulas to ultimately calculate the values
of F for: Between Columns, Rows and Interaction.
(𝐺𝑇)2
•
Correction Factor, 𝐶𝐹 =
•
Total Sum of Squares, 𝑆𝑆𝑇 = ∑ 𝑥 2 − 𝐶𝐹
•
Within Sum of Squares, 𝑆𝑆𝑊 = ∑ 𝑥 2 − ∑
•
Column Sum of Squares, 𝑆𝑆𝑐 = ∑
𝑁
2
(∑ 𝑋𝑤 )
𝑛𝑤
,w = within number
2
(∑ 𝑋𝑐 )
𝑁𝑐
− 𝐶𝐹, c=column number
2
(∑ 𝑋𝑟 )
•
Row Sum of Squares, 𝑆𝑆𝑟 = ∑
•
Interaction Sum of Squares, SScr = SST-SSW-SSc-SSr
𝑁𝑟
− 𝐶𝐹, r=row number
Degrees of Freedom:
•
•
•
•
•
•
Total Degree of Freedom, dft = N-1
Within Degree of Freedom, dfw = k(ni-1)
Column Degree of Freedom, dfc= c-1
Row Degree of Freedom, dfr= r-1
Interaction Degree of Freedom, dfcr = (c-1)(r-1)
𝑆𝑆
Mean of Square, MS = 𝑑𝑓
•
F-test, F =
𝑀𝑆𝑎𝑛𝑦
𝑀𝑆𝑤
GT= x grand total, ∑ 𝑥 2 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑥, ∑ 𝑋𝑖 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑜𝑓 𝑥 𝑤𝑖𝑡ℎ𝑖𝑛
∑ 𝑋𝑐 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑜𝑓 𝑥 𝑖𝑛 𝑎 𝑐𝑜𝑙𝑢𝑚𝑛 𝑐, ∑ 𝑋𝑟 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑜𝑓 𝑥 𝑖𝑛 𝑎 𝑟𝑜𝑤
Example 1. Thirty language students were randomly assigned to one of three
instructors and to one of the two methods of teaching. Achievement was measured
on a test administered at the end of the term. Use a Two-Way ANOVA with
interaction effect at 0.05 level of significance to test the following null hypotheses:
1. There is no significant difference in the performance of the three groups of
students under three different instructors.
2. There is no significant difference in the performance of the two groups of
students under two different methods of teaching.
3. Interaction effect is not present.
TWO-WAY ANOVA with Significant Interaction
Teacher Factor (Column)
Method of
Teaching 1 (row)
Method of
Teaching 2 (row)
A
38
42
44
37
38
41
40
38
37
39
B
52
53
47
49
46
47
45
43
44
39
C
39
43
44
38
39
52
45
43
46
42
Solving by Stepwise Method:
I.
II.
III.
IV.
Problem:
1. Is there a significant difference in the performance of the students under
three different teachers?
2. Is there a significant difference in the performance of the students under
two different methods of teaching?
3. Is there an interaction effect between teachers and methods of teaching
factors?
Hypotheses: (Ho)
1. There is no significant difference in the performance of the students under
three different teachers.
2. There is no significant difference in the performance of the students under
two different teachers.
3. There is no significant effect between teachers and methods of teaching.
Level of Significance:
𝛼 = 0.05
Total Degree of Freedom, dft = N-1 = 30-1 = 29
Within Degree of Freedom, dfw = k(ni-1) = 6(5-1) = 24
Column Degree of Freedom, dfc= c-1 = 3-1 = 2
Row Degree of Freedom, dfr= r-1 = 2-1 = 1
Interaction Degree of Freedom, dfcr = (c-1)(r-1) = 2*1=2
Statistics: F-test Two-Way ANOVA
With interaction effect.
Teacher Factor (Column)
Method of
Teaching 1
(row)
Total
Method of
Teaching 2
(row)
Total
Total
A
38
42
44
37
38
199
41
40
38
37
39
195
394
B
52
53
47
49
46
247
47
45
43
44
39
218
465
C
39
43
44
38
39
203
52
45
43
46
42
228
431
Total
649
641
`1290
Extend the steps in One-Way ANOVA
Steps: Calculate
1. CF=
(1290)2
30
= 55470
2. SST = 382+422+442+…+462+422- CF
= 56080 - 55470 = 610
3. SSW = 56080 - (1992+2472+2032+1952+2182+2282) /5 = 56080 - 55870.4 = 209.6
4. SSc = (3942+4652+4312)/10 - CF = 55722.2 - 55470 = 252.2
5. SSr = (6492+6412)/15 - CF = 55472.13 – 55470 = 2.13
6. SScr = SST-SSW-SSc-SSr = 610 – 209.6 – 252.2 – 2.13 = 146.07
ANOVA TABLE (TWO-WAY) for Teachers and Methods of Teaching Factors
Sources of
Variation
Between
Columns
(Teacher)
Rows
(Methods)
Interaction
Within
Total
V.
VI.
F-value
Tabular
SS
df
MS
252.20
2
126.10
14.44
3.40
Significant
2.13
1
2.13
0.24
4.26
Not Significant
146.70
209.60
610.00
2
24
29
73.35
8.73
8.40
3.40
Significant
Computed
Interpretation
Decision Rule: If the F-computed value is greater than the F-tabular value,
then reject the null hypothesis.
Conclusion: The above table shows ANOVA table for Teachers and Methods
of Teaching Factors.
It can be deduced from this table, that since the F-computed value of 14.44
for the row (teacher) is greater than the F-tabular value of 3.40 at 0.05 level of
significance with 2 and 24 degrees of freedom, hence the null hypothesis is
rejected. This means that there is a significant difference in the performance
of students under three different teachers. Teacher factor affects the
performance of the students. This implies that students under teacher B have
performed better than those under teacher A and teacher C. While students
under teacher A have the poorest performance.
On the other hand, since the F-computed value of 0.24 for row is less than the
F-tabular value of 4.26 with 1 and 24 degrees of freedom, the null hypothesis
is accepted. This indicates that as far as methods of teaching is concerned,
the performance of the students is unaffected. Methods of teaching doesn’t
matter.
The F-computed value of 8.40 against the F-tabular value of 3.40 indicates
that there is an interaction effect between teachers and their methods of
teaching. Students under teacher B have better performance under methods
of teaching 1 while students under teacher C have better performance under
methods of teaching 2.
LESSON 11 - THE PEARSON PRODUCT MOMENT COEFFICIENT OF
CORRELATION, r
The Pearson Product Moment Coefficient r is an index of relationship between
two variables. The independent variable can be represented by x while the
dependent variable can also be represented by y. The value of r ranges from -1 to
+1. If the of value of r is +1 or -1, there is a perfect correlation between x and y.
Which means that x influences y or y depends upon the value of x. However, if r
equals zero, then x and y are independent of each other.
Consider the graph below.
High
r=+
x
Low
High
If the trend of the line graph is going upward, the value of r is positive. This
indicates that as x increases, the value of y also increases. Likewise, if x decreases,
the value of y also decreases, the x and y are positively correlated.
High
r=-
x
Low
High
If the trend of the line graph is going downward, the value of r is negative. It
indicates that as x increases, the corresponding value of y decreases, x and y are
negatively correlated.
High
Low
High
If the trend of the line graph cannot be established either upward or downward
(i.e., when the plotted points are so scattered), then r=0. This indicates that there is
no correlation between the x and y variables.
The formula for the Pearson Product Moment Coefficient of Correlation, r is:
𝑟=
𝑛 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
√(𝑛 ∑ 𝑥 2 −(∑ 𝑥)2 )(𝑛 ∑ 𝑦 2 −(∑ 𝑦)2 )
Where:
r = the Pearson Product Moment Coefficient of
Correlation
n = sample size
∑ 𝑥𝑦 = the sum of the product x and y
∑ 𝑥 ∑ 𝑦 = the product of the sum of x and the sum of y
∑ 𝑥 2 = sum of squares of x
∑ 𝑦 2 = sum of squares of y
Example 1. Below are the midterm (x) and final (y) grades.
x 75 70 65 90 85 80 70 65 90 88
y 80 77 65 94 88 85 88 76 72 91
Solving by Stepwise Method
I.
II.
III.
Problem: Is there a significant relationship between the midterm and the
final examinations of 10 students in Statistics.
Hypotheses:
Ho: There is no significant relationship between the midterm grades and
the final examination/grades of 10 students in Statistics.
Ha: There is a significant relationship between the midterm grades and the
final examination/grades of 10 students in Statistics.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = 𝑛 − 2
= 10 − 2
=8
𝑟0.05 = 0.632:
IV.
Statistics: Pearson Product Moment Coefficient of Correlation
Computation:
No
x
1
75
2
70
3
65
4
90
5
85
6
80
7
70
8
65
9
90
10
88
Total
778
𝑟=
y
80
77
65
94
88
85
88
76
72
91
816
x2
5625
4900
4225
8100
7225
6400
4900
4225
8100
7744
61444
y2
6400
5929
4225
8836
7744
7225
7744
5776
5184
8281
67344
xy
6000
5390
4225
8460
7480
6800
6160
4940
6480
8008
63943
10(63943) − (778)(816)
√[10(61444) − (778)2 ][10(67344) − (816)2 ]
𝑟 = 0.55
V.
VI.
Decision Rule: If the computed r value is greater than the r tabular value,
reject Ho.
Conclusion/Implication:
Since the r-computed value of 0.55 is less than the r-tabular value of 0.632
at 0.05 level of significance with 8 degrees of freedom, hence, the null
hypothesis is accepted. This means that there is no significant relationship
between the midterm grades of the students and the final grades. It implies
that the midterm grade has no say with the final grade.
Coefficient of Determination
The coefficient of determination (CD) is r2 times 100%. This statistics enables
us to explain the extent to which the independent variable x influences the
dependent variable y.
Example. What is the coefficient of determination when r = 0.55?
CD = (0.55)2x (100%)
CD = 30.25 %
This value of 30.25 % indicates that the final examination grade does not
depend on the midterm grade, implying that the final grade is not influenced by the
midterm grade.
LESSON 12 - THE SIMPLE LINEAR REGRESSION ANALYSIS
The simple linear regression analysis is used when there is a significant
correlation between x and y variables. The established linear equation is used to
predict the value of y given the value of x. The formulas for the simple linear
regression analysis are illustrated below.
𝑦̂ = 𝑎 + 𝑏𝑥
Where:
𝑦̂ = 𝑡ℎ𝑒 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑜𝑟 𝑡ℎ𝑒 𝑝𝑟𝑒𝑑𝑖𝑐𝑡ed value of y
x = the independent variable
a = the y-intercept
b = the slope of the line
𝑏=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑥 2 − (∑ 𝑥)2
𝑎 = 𝑦̅ − 𝑏𝑥̅
𝑥̅ = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑥, 𝑦̅ = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑦
Example. A study was made by SM Bicutan to determine the relationship between
weekly sales and advertising expenditures. The following data were recorded. Using
r at 0.05 level of significance, determine the prediction equation if x and y have
significant relationship.
Advertising Cost
Sales
(in thousand Php)
x
4.8
3.2
3.6
3.3
5.2
5.6
3.3
4.3
3.8
4.6
(in thousand Php)
y
42.5
38.6
40.2
38.5
45.4
48.8
40.0
38.4
42.4
40.7
Solve by Stepwise Method:
I.
Problem: Are weekly sales influenced by the advertising expenditures?
II.
III.
IV.
Hypotheses:
Ho: Weekly sales are not influenced by the advertising expenditures.
Ha: Weekly sales are influenced by the advertising expenditures.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = 𝑛 − 2 = 10 − 2 = 8
𝑟0.05 = 0.632
Statistics:
Pearson Product Moment Coefficient of Correlation
Computation:
No
1
2
3
4
5
6
7
8
9
10
Total
X
4.8
3.2
3.6
3.3
5.2
5.6
3.3
4.3
3.8
4.6
41.7
𝑟=
y
42.5
38.6
40.2
38.5
45.4
48.8
40.0
38.4
42.4
40.7
415.5
x2
23.04
10.24
12.96
10.89
27.04
31.36
10.89
18.49
14.44
21.16
180.51
y2
1806.25
1489.96
1616.04
1482.25
2061.16
2381.44
1600
1474.56
1797.76
1656.49
17365.91
xy
204
123.53
144.72
127.05
236.08
273.28
132
165.12
161.12
187.22
1754.11
10(1754.11) − (41.7)(415.5)
√[10(180.51) − (41.7)2 ][10(17365.91) − (415.5)2 ]
𝑟 = 0.827
V.
VI.
Decision Rule: If the computed value is greater than the tabular value, reject
null hypothesis.
Conclusion: Since the r-computed value of 0.827 is greater than r-tabular
value of 0.632 at 0.05 level of significance with 8 degrees of freedom, hence,
the null hypothesis is rejected in favor of the research hypothesis. This means
that the advertising cost is related to the sales. The sales are influenced by
the advertisement, implying that the higher the cost of advertisement, the
higher the sales.
And since the variable advertising cost, denoted x, is significantly related to
the sales, denoted y, Simple Linear Regression can be used.
Solving for:
1. 𝑏 =
𝑛 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑥 2 −(∑ 𝑥)2
=
10(1754.11)−(41.7)(415.5)
10(180.51)−(41.7)2
, 𝑏 = 3.243
2. 𝑎 = 𝑦̅ − 𝑏𝑥̅ , 𝑥̅ = 4.17, 𝑦̅ = 41.55, 𝑎 = 41.55 − 3.243(4.17) = 28.026
3. Thus, the Linear Regression Equation is
𝑦̂ = 28.026 + 3.243𝑥
4. Check: by solving values for some y’s: when x = 4.8, y = 43.59, and when
x = 3.2, y = 38.40. The values computed for y are closed to the values:
42.5 and 38.6, respectively.
This implies that the established Simple Linear Regression Equation,
𝑦̂ = 28.026 + 3.243𝑥, can be used to predict values of y, given values of x.
LESSON 13 - THE MULTIPLE REGRESSION ANALYSIS
The multiple regression analysis is used for predictions. The dependent
variable can be predicted given several independent variables. For instance, we can
make better predictions of the performance of newly hired workers if we consider not
only their education but also their years of experience, attitudes, and some other
variables that may influence performance.
Many mathematical formulas can be used to express relationship among two
or more variables, but the most commonly used in statistics are linear equations.
𝑦 = 𝑏0 + 𝑏1 𝑥1 + 𝑏2 𝑥2 + ⋯ + 𝑏𝑛 𝑥𝑛
Where:
y = the dependent variable to be predicted
x1, x2, x3, …, xn = the known independent variables that may
influence y.
b0, b1, b2, …, bn = the arbitrary constants whose values can
be determined from the observed data.
When there are two independent variables x1 and x2 and we want to fit the
equation in the equation model, we use the equation,
𝑦 = 𝑏0 + 𝑏1 𝑥1 + 𝑏2 𝑥2
We must solve the three normal equations:
∑ 𝑦 = 𝑛𝑏0 + ∑ 𝑥1 𝑏1+∑ 𝑥2 𝑏2
(1)
∑ 𝑥1 𝑦 = ∑ 𝑥1 𝑏0 + ∑ 𝑥12 𝑏1 +∑ 𝑥1 𝑥2 𝑏2
(2)
∑ 𝑥2 𝑦 = ∑ 𝑥2 𝑏0 +∑ 𝑥1 𝑥2 𝑏1 + ∑ 𝑥22 𝑏2
(3)
Example 1. The following are data on the ages and salaries of a random sample of 6
executives working at BBC corporation and their academic achievements while in
college.
Income
(in thousand Php)
y
82.3
75.6
85.4
78.8
73.2
69.3
Age
Academic
Achievement
x2
1.75
2.0
1.5
2.0
2.25
2.5
x1
39
33
42
39
30
29
a) Find the equation of the form 𝑦 = 𝑏0 + 𝑏1 𝑥1 + 𝑏2 𝑥2 based on the data shown
in the table.
b) Use the equation obtained in (a) to estimate the average income of a 36-year
old executive with 1.5 academic achievement.
Computations:
No
𝑦
𝑥1
𝑥2
𝑥12
𝑥22
𝑥1 𝑦
𝑥2 𝑦
𝑥1 𝑥2
1
2
3
4
5
6
Total
82.3
75.6
85.4
78.8
73.2
69.3
464.6
39
33
42
39
30
29
212
1.75
2
1.5
2
2.25
2.5
12
1521
1089
1764
1521
900
841
7636
3.0625
4
2.25
4
5.0625
6.25
24.625
3209.7
2494.8
3586.8
3073.2
2196
2009.7
16570
144.03
151.2
128.1
157.6
164.7
173.25
918.88
68.25
66
63
78
67.5
72.5
415.25
From (1)
∑ 𝑦 = 𝑛𝑏0 + ∑ 𝑥1 𝑏1+∑ 𝑥2 𝑏2
464.6 = 6𝑏0 + 212𝑏1 +12𝑏2
(1)
From (2)
and from (3)
∑ 𝑥1 𝑦 = ∑ 𝑥1 𝑏0 + ∑ 𝑥12 𝑏1 +∑ 𝑥1 𝑥2 𝑏2
16570 = 212𝑏0 + 7636𝑏1 +415.25𝑏2
∑ 𝑥2 𝑦 = ∑ 𝑥2 𝑏0+∑ 𝑥1 𝑥2 𝑏1 + ∑ 𝑥22 𝑏2
918.88 = 12𝑏0+415.25𝑏1 +24.625𝑏2
(2)
(3)
Solving the 3 equations (1), (2) and (3), you get
𝑏0 = 83.68
𝑏1 = 0.423
𝑏2 = −10.593
Take note that 3 linear equations with 3 unknowns can be solved by Elimination,
Substitution, or by Extended Matrix (Cramer’s Rule) or you can make use of your
scientific calculators with built-in functions dedicated for solving such equations.
Knowing the values for 𝑏0 , 𝑏1 , 𝑎𝑛𝑑 𝑏2 , the regression equation therefore is
a) 𝑦 = 83.68 + 0.423𝑥1 − 10.593𝑥2 and
b) 𝑤ℎ𝑒𝑛 𝑥1 = 36 𝑎𝑛𝑑 𝑥2 = 1.5, 𝑦 = 83.019
Using the regression equation of 𝑦 = 83.68 + 0.423𝑥1 − 10.593𝑥2 as established
above, based on the given data, when the age of an executive is 36 years old
with academic achievement 𝑜𝑓 1.5, then his predicted average salary will be Php
83,019.
Exercise: The following data were obtained for 8 students in Management:
midterm grades, average grade in quizzes and their final grades. (Antonio S.
Broto, Statistics Made Simple, pp 240, exercise 18).
Midterm
Grade
3.0
1.5
1.25
1.25
1.75
2.75
2.5
2.0
Average Grade in
Quizzes
2.5
1.5
1.5
1.25
2.0
2.75
2.5
2.25
Final Grade
2.75
1.5
1.25
1.25
1.5
2.5
2.5
2.0
a) Use the method of least squares (multiple regression) to fit an equation of the
form 𝑦 = 𝑏0 + 𝑏1 𝑥1 + 𝑏2 𝑥2 .
b) Predict the final grade of student whose midterm grade is 1.75 and the
average grade in the quizzes is 1.5.
MODULE 3
THE NONPARAMETRIC TEST
Nonparametric tests do not require a normal distribution. When the value of
skewness is not zero (either positive or negative), and the Kurtosis is greater or less
than 0.265, the distribution is said to be abnormal. When the value of the Kurtosis is
less than 0.265, the distribution is said to be leptokurtic but it is platykurtic when the
value is greater than 0.265.
Nonparametric tests also utilize both nominal and ordinal data. Nominal data
are expressed in categories, whereas, the ordinal data are expressed in ranking.
The most commonly used tests under the nonparametric tests are the ChiSquare test; U-test; H-test; Spearman Rank Order Coefficient of Correlation, rs; Sign
Test (Median Test); Mc Nemar’s Test, Friedman Test, Fr; and Kendall’s Coefficient of
Concordance, W.
Objectives:
When you have completed the module, you will be able to:
•
•
•
•
•
Distinguish different types of Chi-Square tests and their uses.
Identify parametric and nonparametric tests which are counterparts.
Recognize data appropriate for nonparametric tests.
Convert data suited for parametric tests into data for nonparametric tests.
Apply the Stepwise Method in solving statistical problems into nonparametric tests.
LESSON 14 - THE CHI-SQUARE TEST
This test is a test of difference between the observed and expected frequencies.
The Chi-Square is considered a unique test because it has 3 functions which are as
follows:
•
•
•
The test of goodness-of-fit
The test of homogeneity
The test of independence
THE CHI-SQUARE TEST OF GOODNESS-OF-FIT
This is a test of difference between the observed frequencies and expected
frequencies. As an illustration, let’s take for example the throwing of a coin, the coin
has only two faces, the head and the toe. If a coin is thrown, it may either fall the
head or the toe be on top. Which means that, theoretically, there is a fifty-fifty chance
for each of the face being on top. Now if you throw a coin 100 times, the expected
frequencies for the head being on top should be 50. And this is true for the other
face, the head. But the actual observed frequencies may be different from the
expected frequencies. The formula for the Chi-Square Test is:
𝜒2 = ∑
(𝑂 − 𝐸)2
𝐸
Where:
𝜒 2 = 𝑡ℎ𝑒 𝐶ℎ𝑖 − 𝑆𝑞𝑢𝑎𝑟𝑒 𝑡𝑒𝑠𝑡
𝑂 = 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
𝐸 = 𝑡ℎ𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
Example. A coin is thrown 100 times. The observed frequency for the head being on
top when it falls is 48, while the toe being 52. Using Chi-Square, determine if there is
a significant difference between the observed and expected frequencies. Use 0.05
level of significance.
Solving by Stepwise Method:
I.
II.
Problem: Is there a significant difference between the observed and
expected frequencies?
Hypotheses:
Ho: There is no significant difference between the observed and expected
frequencies.
Ha: There is a significant difference between the observed and expected
frequencies.
III.
Level of Significance:
IV.
𝛼 = 0.05
𝑑𝑓 = ℎ − 1 = 2 − 1 = 1
2
𝜒0.05 = 3.841 (𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑢𝑙𝑎𝑟 𝑣𝑎𝑙𝑢𝑒)
Statistics: Chi-Square Test of Goodness- of-Fit
Computation:
Face
Ratio
Head
Toe
Total
1
1
2
(Actual Result)
Observed
48
52
100
𝜒2 = ∑
𝜒2 =
(Theory)
Expected
50
50
100
(𝑂 − 𝐸)2
𝐸
(48 − 50)2 (52 − 50)2
+
50
50
𝜒 2 = 0.16
V.
VI.
Decision Rule: If the 𝜒 2 − 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 is greater than the 𝜒 2 −
𝑡𝑎𝑏𝑢𝑙𝑎𝑟 𝑣𝑎𝑙𝑢𝑒, the null hypothesis is rejected.
Conclusion: Since the 𝜒 2 − 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 of 0.16 is less than the 𝜒 2 −
𝑡𝑎𝑏𝑢𝑙𝑎𝑟 𝑣𝑎𝑙𝑢𝑒 of 3.481 at 0.05 level of significance with 1 degree of
freedom, the null is accepted. This means that there is no significant
difference between the observed and the expected frequencies. This
implies that the theory of a fifty-fifty chance for each face of the coin being
on top holds true inasmuch as the value of 𝜒 2 does not warrant the theory
to be rejected.
Exercise:
A certain machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in
the ratio of 4:3:2:1. A can containing 500 of these mixed nuts was found to have 275
peanuts, 105 hazelnuts, 76 cashews and 44 pecans. At 0.05 level of significance,
test the hypothesis that the machine is mixing the nuts at this specified ratio.
THE CHI-SQUARE TEST OF HOMOGENEITY
This test involves two or more samples, with only one criterion variable. This
is used to determine if two or more populations are homogeneous. The distribution of
data is similar with respect to a particular criterion variable. The formula is:
𝜒2 =
𝑁(𝑎𝑑 − 𝑏𝑐)2
𝑘𝑙𝑚𝑛
Where:
𝜒 2 = 𝑐ℎ𝑖 − 𝑠𝑞𝑢𝑎𝑟𝑒 𝑡𝑒𝑠𝑡
𝑁 = 𝑔𝑟𝑎𝑛𝑑 𝑡𝑜𝑡𝑎𝑙
𝑎, 𝑏, 𝑐 𝑎𝑛𝑑 𝑑 = 𝑡ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 𝑎𝑠 𝑏𝑒𝑙𝑜𝑤
𝑘𝑙𝑚𝑛 = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑡𝑜𝑡𝑎𝑙
Example. Evaluate the attitude of a sample of LDP and Nationalista parties on the
issue of peace and order in Mindanao. To carry out the study, a random sampling of
members of each party is drawn from the nationwide population of LDP and
Nationalista and each individual in both samples responds to the scale. Scores are
then classified into “Favorable” or Unfavorable” categories. The following frequencies
were recorded:
LDP
Nationalista
Total
Favorable
63
a
52
c
115
m
Unfavorable
37
b
48
d
85
n
Total
100 k
100 l
200
N
Solving by Stepwise Method:
I.
II.
III.
Is there a significant difference between the attitudes of the two political
parties on the issue of peace and order in Mindanao?
Hypotheses:
Ho: There is no significant difference between the attitudes of the two
political parties on the issue of peace and order in Mindanao.
Ha: There is a significant difference between the attitudes of the two
political parties on the issue of peace and order in Mindanao.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = (𝑐 − 1)(𝑟 − 1)
= (2 − 1)(2 − 1)
𝑑𝑓 = 1
2
𝜒0.05
= 3.841
IV.
Statistics: Chi-Square of Homogeneity
Computation:
𝜒2 =
𝜒2 =
𝑁(𝑎𝑑 − 𝑏𝑐)2
𝑘𝑙𝑚𝑛
200((63)(48)−(37)(52))2
(100)(100)(115)(85)
𝜒 2 =2.476
V.
VI.
Decision Rule: If the 𝜒 2 − 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 value is greater than the 𝜒 2 − 𝑡𝑎𝑏𝑢𝑙𝑎𝑟
value, reject null hypothesis.
Conclusion: Since the 𝜒 2 − 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 value of 2.476 is less than the 𝜒 2 −
𝑡𝑎𝑏𝑢𝑙𝑎𝑟 value of 3.841 at 0.05 level of significance with 1 degree of
freedom, hence the null 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.
Exercise. Using Chi-Square at 0.05 level of significance, determine the attitude of the
TCU community on the issue of charter change. To carry out such study, 100
samples from each teaching and non-teaching personnel were taken to respond to
the issue categorized as “YES” or “NO”. The following frequencies were obtained:
Teaching
Non-Teaching
Total
YES
34 (a)
55 (c)
89 (m)
NO
66 (b)
45 (d)
111 (n)
Total
100 (k)
100 (l)
200 (N)
THE CHI-SQUARE TEST OF INDEPENDENCE (ONE SAMPLE, TWO
CRITERION VARIABLES)
The one-sample test of independence is different from the test of
homogeneity. This sample consists of members randomly drawn from the same
population. This test is used to find out whether measures taken on two criterion
variables are either independent or associated with one in a given population using
such variables as say level of education and performance, income and years of
experience, etc. The calculation of this test is similar to the test of goodness-of-test
and the test of homogeneity. The formula is:
𝜒2 = ∑
(𝑂 − 𝐸)2
𝐸
Where:
𝜒 2 = 𝑡ℎ𝑒 𝐶ℎ𝑖 − 𝑆𝑞𝑢𝑎𝑟𝑒 𝑡𝑒𝑠𝑡
𝑂 = 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
𝐸 = 𝑡ℎ𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
𝐸𝑖𝑗 =
(𝑅𝑜𝑤 𝑇𝑜𝑡𝑎𝑙)𝑖 𝑋(𝐶𝑜𝑙𝑢𝑚𝑛 𝑇𝑜𝑡𝑎𝑙)𝑗
𝐺𝑟𝑎𝑛𝑑 𝑇𝑜𝑡𝑎𝑙
i = row number
j= column number
Example. 100 individuals, male and female, were given an IQ test and their scores
were classified into high and low. Using the 𝜒 2 -test of independence at 0,05 level of
significance, the table is shown as follows:
IQ
Sex
High
Male
Female
Total
O
23
32
55
Low
E
O
33
12
45
Total
E
Solving by Stepwise Method:
I.
II.
III.
Problem: Is there a significant relationship between sex and IQ?
Hypotheses:
Ho: There is no significant relationship between sex and IQ.
Ha: There is a significant relationship between sex and IQ.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = (𝑐 − 1)(𝑟 − 1)
= (2 − 1)(2 − 1)
=1
2
𝜒0.05
= 3.841
56
44
100
IV.
Statistics: Chi-Square of Independence
Computation:
IQ
Sex
High
O
23
32
55
Male
Female
Total
Low
E
30.80
24.20
55
O
33
12
45
𝐸11 =
(56)(55)
= 30.80
100
𝐸12 =
(56)(45)
= 25.20
100
𝐸21 =
(44)(55)
= 24.20
100
𝐸22 =
(44)(45)
= 19.80
100
𝜒2 = ∑
𝜒2 =
(23 − 30.80)2
30.80
+
Total
E
25.20
19.80
45
56
44
100
(𝑂 − 𝐸)2
𝐸
(33 − 25.2)2
25.2
+
(32 − 24.20)2
24.20
+
(12 − 19.80)2
19.80
𝜒 2 = 9.976
V.
Decision Rule: If the 𝜒2 − computed value is greater than the 𝜒 2 – tabular
VI.
value, reject the null hypothesis.
Conclusion: Inasmuch as the 𝜒 2 − computed value of 9.976 is greater than
the 𝜒 2 − tabular value of 3.841 at 0.05 level of significance with 1 degree
of freedom, the null hypothesis is rejected. This means that there is a
significant relationship between sex and IQ. It implies that the female has
a better IQ than the male counterpart.
Students’ Activity:
Two lots of 30 experimental guinea pigs were used in testing the effectiveness
of a new serum in combating a certain disease. 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.
Recovered
Died
Total
Serum
12
2
14
No Serum
3
13
16
Total
15
15
30
Note: When df is 1 and any expected frequency is small, less than 10, the 𝜒 2 − 𝑡𝑒𝑠𝑡,
using the Yate’s correction for lack of continuity will be applied because the distribution of
Chi-Square is discrete. Whereas the values obtained by the use of the formula result in a
continuous probability model. The formula used is
𝜒2 = ∑
(|𝑂−𝐸|−0.5)2
𝐸
or
𝑁 2
𝑁 (|𝑎𝑑 − 𝑏𝑐| − 2 )
𝜒2 =
𝑘𝑙𝑚𝑛
LESSON 15 - THE WILCOXON RANK-SUM TEST OR WILCOXON TWO-SAMPLE
TEST
The Wilcoxon Rank-Sum Test is commonly used for the comparison of two
groups of nonparametric (interval or not normally distributed) data, such as those
which are not measured exactly but rather as falling within certain limits (e.g., how
many animals died during each hour of an acute study).
𝑈1 = 𝑊1 −
𝑛1 (𝑛1 + 1)
2
𝑈2 = 𝑊2 −
𝑛2 (𝑛2 + 1)
2
Where:
𝑈1 = 𝑊𝑖𝑙𝑐𝑜𝑥𝑜𝑛 𝑅𝑎𝑛𝑘 − 𝑆𝑢𝑚 𝑇𝑒𝑠𝑡
𝑊1 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑎𝑛𝑘𝑠 𝑜𝑓 𝑔𝑟𝑜𝑢𝑝 1
𝑛1 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑔𝑟𝑜𝑢𝑝 1
𝑈2 = 𝑊𝑖𝑙𝑐𝑜𝑥𝑜𝑛 𝑅𝑎𝑛𝑘 − 𝑆𝑢𝑚 𝑇𝑒𝑠𝑡
𝑊2 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑎𝑛𝑘𝑠 𝑜𝑓 𝑔𝑟𝑜𝑢𝑝 2
𝑛2 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑔𝑟𝑜𝑢𝑝 2
𝑈 = min (𝑈1 , 𝑈2 )
If U is less than the tabular U, reject Ho.
Example. Of the eighteen selected patients who were seriously infected of covid 19,
ten were treated with a new serum and eight were not. The number of days the
patient recovered were then recorded. Using the Wilcoxon-rank-sum test at 0.05
level of significance, test whether the serum is effective, consider the following data:
With
Treatment
No Treatment
6.5
5.4
15.2 18
5.3
7.2
4.5
8.4
6.8
7.3
13.6
12.2
13.1
12.0
17.6
19.4
6.7
8.2
Solving by the Stepwise Method:
I. Problem: Is the new serum effective in treating leukemia?
II. Hypotheses:
Ho: The new serum is not effective in treating leukemia.
Ha: The new serum is effective in treating leukemia.
III. Level of Significance:
𝛼 = 0.05
𝑑𝑓 = 𝑛1 𝑎𝑛𝑑 𝑛2
= 10 𝑎𝑛𝑑 8
𝑈0.05 = 17
IV. Statistics: U-test. Wilcoxon rank-sum test.
Computation: Arrange the data jointly from the lowest to the highest value and
rank them. Then indicate the ranks in the table below.The sum of ranks with
treatment will be 𝑊1 , similarly the sum of ranks without treatment will be 𝑊2 .
With Treatment
6.5
5.4
5.3
7.2
4.5
8.4
6.8
7.3
6.7
8.2
Total
Rank
4
3
2
7
1
10
6
8
5
9
𝑊1 =55
𝑈1 = 55 −
No Treatment
15.2
18
13.6
12.2
13.1
12
17.6
19.4
Rank
15
17
14
12
13
11
16
18
𝑊2 =116
10(10 + 1)
=0
2
𝑈2 = 116 −
8(8 + 1)
= 80
2
𝑈 = min(0,80) = 0
V. If the computed U is less than the tabular U, reject Ho.
VI. Since the smaller of U1=0 and U2=80, which is 0, is less than the tabular
value of 17, at 0.05 level of significance with 10 and 8 degrees of freedom,
hence the null hypothesis is rejected in favor of the research hypothesis. This
means that the serum is effective in the treatment of Covid 19. It implies that
patients with treatment of serum recover more rapidly than those patients
without treatment of the serum because it takes them less number of days to
recover.
LESSON 15 - THE KRUSKAL-WALLIS TEST, ALSO CALLED THE KRUSKALWALLIS H-TEST.
This test is used to compare 3 or more independent groups. This is a
nonparametric test which does not require normal distribution. This is a counterpart
of the F-test (ONE- WAY ANOVA) in parametric tests. It uses the table from ChiSquare test for the tabular values. The formula is:
12
𝑅𝑖2
𝐻=
∑
− 3(𝑛 + 1)
𝑛(𝑛 + 1)
𝑛𝑖
Where:
H=Kruskal Wallis test
N= number of observations
Example. Consider the examination scores of samples of College students
taught in English using three different methods: Method 1 (Face-to-Face
classroom teaching), Method 2 (On-line teaching), and Method 3 (Modular
teaching). Use the H-test at 0.05 level of significance to test the null
hypothesis that their means are not equal. Consider the following data:
Method 1
96
86
91
93
89
87
Method 2
87
88
89
85
81
86
80
Method 3
88
76
74
64
79
Solving by Stepwise Method:
I.
II.
III.
Problem: Are there significant differences in the average scores of these
students using the different methods of teaching English?
Hypotheses:
Ho: There are no significant differences in the average scores of students
using 3 methods of teaching English.
Ha: There are significant differences in the average scores of students using 3
methods of teaching English.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = ℎ − 1 = 3 − 1 = 2
2
𝜒0.05
= 5.991
IV.
Statistics:
Kruskal Wallis, H-test
Computation:
Method 1
96
86
91
93
89
87
R1
18
8.5
16
17
14.5
10.5
n1=6
84.5
Method 2
87
88
89
85
81
86
80
n2=7
R2
10.5
12.5
14.5
7
6
8.5
5
64
Method 3
88
76
74
64
79
R3
12.5
3
2
1
4
n3=5
22.5
12
𝑅𝑖2
𝐻=
∑
− 3(𝑛 + 1)
𝑛(𝑛 + 1)
𝑛𝑖
12
84.52 642 22.52
𝐻=
(
+
+
) − 3(18 + 1)
18(18 + 1) 6
7
5
𝐻 = 8.840
V.
VI.
Decision Rule: if the computed H is greater than the tabular value, reject Ho.
Conclusion: Since the computed H of 8.840 is greater the tabular value of
5.991 at 0.05 level of significance with 2 degrees of freedom, the null
hypothesis is rejected. This means that there are significant differences in the
average scores of 18 students using three different Methods of teaching. It
implies that Method 1 (face-to-face teaching) is more effective than the other
two methods (on-line and modular teaching, respectively).
Students’ Exercise:
The data on war on rape cases under the Duterte administration committed from
January to December in 3 cities in Metro Manila are as follows:
Month
January
February
March
April
May
June
July
August
September
October
November
December
City
B
6
7
8
3
5
7
4
3
4
5
4
2
A
5
6
4
3
4
5
4
3
7
5
6
4
C
6
7
8
9
7
10
9
6
7
9
10
8
Perform Kruskal-Wallis H-test to determine if the hypothesis that the average rape
cases in the 3 cities are significantly the same. Use 𝛼 = 0.05 level of significance.
LESSON 16 - THE SPEARMAN RANK ORDER COEFFICIENT OF
CORRELATION rs
The Spearman Rank Order Coefficient of Correlation is denoted rs. This test of
correlation does not require the strict assumption of normality like the Pearson
Product Moment Coefficient of Correlation, denoted r. The formula is:
𝑟𝑠 = 1 − 6
∑ 𝐷2
𝑛(𝑛2 − 1)
Where:
𝑟𝑠 = Spearman Rank Order Coefficient
Correlation
∑ 𝐷 2 = sum of the squares of the difference
between rank x and rank y
𝑛 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒
Example. The following are the number of hours spent by 12 students
studying for the final examination and the scores they obtained in Calculus.
Calculate 𝑟𝑠 at 0.05 level of significance.
Number of
Hours Spent
x
Midterm
Scores y
6
5
10
11
2.75
3
2
2
12
18
15
20
8
9
14
1.75 1.5 1.5 1.25 2.25 2.25 1.25
12
2
Solving by Stepwise Method:
I.
II.
III.
IV.
Problem: Is there a significant relationship between the number of hours spent
for studying Calculus and the scores students obtained in the Final
Examination?
Hypotheses:
Ho: There is no significant relationship between the number of hours spent for
studying Calculus and the scores students obtained in the Final Examination.
Ha: There is a significant relationship between the number of hours spent for
studying Calculus and the scores students obtained in the Final Examination.
Level of Significance:
𝛼 = 0.05
df = 𝑛 − 1 = 12 − 1 = 11
𝑟𝑠0.05 = 0.532
Statistics: Spearman Rank Order Coefficient of Correlation, rs.
Number
Midterm
of
Scores
Hours
y
Spent x
6
5
10
11
12
18
15
20
8
9
14
12
2.75
3
2
2
1.75
1.5
1.5
1.25
2.25
2.25
1.25
2
Rx
Ry
D
D2
11
12
8
7
5.5
2
3
1
10
9
4
5.5
11
12
7
7
5
3.5
3.5
1.5
9.5
9.5
1.5
7
0
0
1
0
0.5
-1.5
-0.5
-0.5
0.5
-0.5
2.5
-1.5
0
0
1
0
0.25
2.25
0.25
0.25
0.25
0.25
6.25
2.25
∑ 𝐷2 = 13
𝑟𝑠 = 1 − 6
𝑟𝑠 = 1 − 6
V.
VI.
∑ 𝐷2
𝑛(𝑛2 − 1)
13
= 0.955
12(122 − 1)
Decision Rule: If the computed rs is greater than the tabular value, reject Ho.
Conclusion:
The computed value of rs=0.955 being greater than the tabular value of
0.532 at 0.05 level of significance with 11 degrees of freedom, leads to the
rejection of the null hypothesis. This indicates that there is a significant
relationship between the number of hours spent for studying and the scores
the students obtained in Calculus. It implies that the more students spend
their time in studying Calculus, the higher the scores they obtain.
Students’ Exercise:
The following is the ranking of two judges given to the work of 10 artists.
Using rs at 0.05 level of significance, test the hypothesis that the two judges differ
most in their opinions about these artists.
Judge A 6
Judge B 5
7
9
8
10
5
4
9
8
3
5
2
2
1
1
4
6
10
7
LESSON 17 - A SIGN TEST FOR TWO INDEPENDENT SAMPLES (MEDIAN
TEST TWO-SAMPLE CASE)
This test is known as the Median Test under nonparametric statistics. This
test is used to compare the median of two independent samples. It’s the counter part
of the t-test under parametric test, though what is being compared by t-test are the
means of two independent groups or samples. The data consist of two independent
samples of n1 and n2 observations.
The procedure is to get the median of the data jointly. The data above this
median are assigned a (+) sign, while those at or below this median a (-) sign. Then
the number of + and – signs for each sample is obtained. A Chi-Square test is used
to determine whether the observed frequencies of +and – signs differ significantly.
𝜒2 =
𝑁(𝑎𝑑 − 𝑏𝑐)2
𝑘𝑙𝑚𝑛
Where:
𝜒 2 = 𝑐ℎ𝑖 − 𝑠𝑞𝑢𝑎𝑟𝑒 𝑡𝑒𝑠𝑡
𝑁 = 𝑔𝑟𝑎𝑛𝑑 𝑡𝑜𝑡𝑎𝑙
𝑎 𝑎𝑛𝑑 𝑐 = 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 (+)𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
𝑏 𝑎𝑛𝑑 𝑑 = 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 (−)𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
𝑘 𝑎𝑛𝑑 𝑙 = 𝑡ℎ𝑒 𝑟𝑜𝑤 𝑡𝑜𝑡𝑎𝑙𝑠
𝑚 𝑎𝑛𝑑 𝑛 = 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑡𝑜𝑡𝑎𝑙𝑠
Example. Consider the IQ test scores of 12 female and 10 male students.
Female 75
Male
68
98
87
87
93
65
54
102
85
95
93
110
78
132
106
127
115
96
96
106
118
Solving by Stepwise Method:
I.
II.
Is there a significant difference between the IQ test scores of the male and
female students?
Hypotheses:
Ho: There is no significant difference between the IQ test scores of the male
and female students.
III.
IV.
Ha: There is a significant difference between the IQ test scores of the male
and female students.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = (𝑐 − 1)(𝑟 − 1)
𝑑𝑓 = (2 − 1)(2 − 1) = 1
2
𝜒0.05 = 3.841
Statistics: Median Test for Two Independent Samples
Computation:
•
•
•
Determine the median. Arrange the data from lowest to highest. The median is
the middle item if there are odd number data, or the average of the two middle
items if there are even number data. The median is 95.5.
Mark or assign a plus (+) sign to those data above the median, while those at or
below will be marked or assigned a minus (-) sign.
Count each plus and minus sign under the female and male columns
respectively.
The observed frequencies are illustrated and summarized as follows:
Female
75
98
87
65
102
95
110
132
127
96
106
118
Female
Male
Total
•
Sign
+
+
+
+
+
+
+
+
Above (+)
8 (a)
3 (c)
11 (m)
𝜒2 =
VI.
At or Below (-)
4 (b)
7 (d)
11 (n)
Sign
+
+
+
Total
12 (k)
10 (l)
22 (N)
Use the formula for Chi-Square.
𝜒2 =
V.
Male
68
87
93
54
85
93
78
106
115
96
𝑁(𝑎𝑑 − 𝑏𝑐)2
𝑘𝑙𝑚𝑛
22(8𝑥7 − 4𝑥3)2
12(10)(11)(11)
= 2.933
Decision Rule: If the computed 𝜒 2 is greater than the tabular value, reject null
hypothesis.
Conclusion: Since the computed value 𝜒 2 of 2.933 is less than the tabular value 𝜒 2
of 3.841 at 0.05 level of significance with 1 degree of freedom, hence the null
hypothesis is accepted. This means that there is no significant difference between
the IQ scores of the female and male students.
Exercise. Consider the test scores of 25 students in spelling. The students are composed of
15 females and 10 males. The following are their scores:
Female 13
Male
9
14
16
17
12
24
15
16
8
15
9
12
17
16
11
22
24
18
16
25
23
14
19
20
LESSON 18 - A SIGN TEST FOR TWO CORRELATED SAMPLES (FISHER SIGN TEST)
This test is under nonparametric statistics. This is the counterpart of the t-test for
correlated sample under the parametric test. The Fisher Sign Test compares two correlated
samples and is applicable to data composed of N paired observations. The difference
between the paired observations is obtained. This test is based on the assumption that half
the difference between the paired observations will be positive and the other half will be
negative. The formula is:
𝑍=
|𝐷| − 1
√𝑁
Where:
Z = the Fisher Sign Test
D = the difference between the number of + and – signs.
Example. The pretest and the posttest results before and after the implementation of
the program are presented below:
Pretest x
Posttest y
16
18
20
24
30
28
34
32
12
12
10
8
18
21
16
14
11
15
15
17
Solving by Stepwise Method:
I.
II.
III.
Problem: Is there a significant difference between the pretest and posttest
results of the 10 students?
Hypothesis:
Ho: There is no significant difference between the pretest and posttest results
of the 10 students.
Ha: There is a significant difference between the pretest and posttest results
of the 10 students.
Level of Significance:
𝛼 = 0.05
𝑍0.05 = ±1.96
IV.
Statistics: Z-test (The Fisher Sign Test)
Computation:
Pretest
x
16
20
30
34
12
10
18
16
11
15
Posttest
y
18
24
28
32
12
8
21
18
15
17
Sign of x-y
D
+
+
0
+
-
In this example, there are 3 + signs, 6 – signs, and 1 zero. Zero is omitted.
Thus,
𝑍=
𝑍=
|𝐷| − 1
√𝑁
|3 − 6| − 1
√9
𝑍 = 0.667
V.
VI.
Decision Rule: If Z-computed value is greater than the Z-tabular value, reject
the null hypothesis.
Conclusion:
Inasmuch as the Z-computed value of 0.667 is less than the Z-tabular value of
1.96 at 0.05 level of significance, thus the null hypothesis is accepted. This
means that there is no significant difference between the pretest and posttest
results of the 10 students.
Exercise: Perform a Fisher Sign Test for the example in t-test for correlated samples
then compare the results.
LESSON 19 - A SIGN TEST FOR K INDEPENDENT SAMPLES (THE MEDIAN
TEST: MULTI-SAMPLE CASE)
This test is under the nonparametric tests. This is a forthright extension of the
median test for two independent samples. The Chi-Square test formula is used for
this test.
Example. A sampling of the acidity of rain for 10 randomly selected rainfalls was
recorded at three different locations in the province of Albay: Guinubatan, Legaspi
City, and Polangui. The pH readings for these 24 rainfalls are shown in the table.
(Note: pH readings range from 0 to 14; 0 is acid, 14 is alkaline. Pure water falling
through clean air has a pH reading of 5.7).
Guinobatan
4.6
4.2
4.1
3.6
2.5
3.9
4.4
3.8
Legaspi City
4.5
4.8
4.5
3.5
2.8
4.3
4.9
3.5
Polangui
4.9
5.4
5.3
3.8
5.6
5.2
5.1
3.4
Use the Median test at 0.05 level of significance to test the hypothesis that there is
no significant difference among the pH readings of the 3 different municipalities/cities
of Albay.
Solving by Stepwise Method:
I.
II.
III.
Problem: Is there a significant difference among the pH readings of the 3
different municipalities/cities of Albay.
Hypotheses:
Ho: There is no significant difference among the pH readings of the 3 different
municipalities/cities of Albay.
Ha: There is a significant difference among the pH readings of the 3 different
municipalities/cities of Albay.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = (𝑐 − 1)(𝑟 − 1)
𝑑𝑓 = (3 − 1)(2 − 1) = 2
2
𝜒0.05
= 5.991
IV.
Statistics: Sign Test for K Independent Samples (Median Test: Multi-Sample
Case)
Computation:
Guinobatan
4.6
4.2
4.1
3.6
2.5
3.9
4.4
3.8
Sign
+
+
-
Legaspi City
4.5
4.8
4.5
3.5
2.8
4.3
4.9
3.5
Sign
+
+
+
+
+
Polangui
4.9
5.4
5.3
3.8
5.6
5.2
5.1
3.4
Sign
+
+
+
+
+
+
-
The median is 4.35 (obtained using MS-excel by the median function).
Above Md
At or Less
Total
Guinobatan
O
E
2
4.33
6
3.67
8
8
Legaspi City
O
E
5
4.33
3
3.67
8
8
Polangui
O
E
6
4.33
2
3.67
8
8
Total
13
11
24
(𝑂 − 𝐸)2
𝜒 =∑
𝐸
2
𝜒 2 = 4.36 or
p-value = 0.113
V.
VI.
Decision Rule: If the computed value is greater than the tabular value, reject
the null hypothesis.
Since the 𝜒 2 − 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 value of 4.36 is less than the 𝜒 2 − 𝑡𝑎𝑏𝑢𝑙𝑎𝑟 value of
5.991 at 0.05 level of significance with 2 degrees of freedom, the null
hypothesis is accepted. This means that there is no significant difference
among the pH readings of 3 different municipalities/cities of Albay.
Note: Another way to arrive at this decision and conclusion is to use the pvalue. You can obtain this value using MS-Excel. It is compared to the level of
significance used, usually, 𝛼 𝑖𝑠 0.05 𝑜𝑟 0.01. 𝑝 − 𝑣𝑎𝑙𝑢𝑒 =CHISQ.TEST(ofr,efr).
Where:
ofr = observed frequency range
efr = expected frequency range
CHISQ.TEST = the name of the function in MS-Excel
When using the p-value, the decision rule is: If the p-value is less than
or equal to the level of significance, reject the null hypothesis.
Since the p-value of 0.113 is greater than 0.05 level of significance, the null
hypothesis is accepted. As you can see, you arrived at same decision and
interpretation/conclusion.
LESSON 20 - THE MC NEMAR’S TEST FOR CORRELATED PROPORTIONS
This test belongs to the nonparametric statistics which doesn’t require normal
distribution of data. A Chi-Square test for the situations when samples are
matched should not be independent. This is a before and after design to test
whether there is a significant change between the before and after situations.
The formula is:
𝜒2 =
(𝑏 − 𝑐)2
𝑏+𝑐
Where:
𝜒 2 = 𝐶ℎ𝑖 − 𝑆𝑞𝑢𝑎𝑟𝑒
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 seat belt use before and after involvement in car accidents for a
sample of 120 accident victims.
Wore seat belt
regularly
before the
accident
Wore seat belt regularly after the accident
Total
Yes
a = 74
c = 23
97
86
34
120
Yes
No
Total
No
b = 12
d = 11
23
Solving by Stepwise Method:
I.
II.
Problem: Is there a significant difference in the use of seat belt before and
after involvement in the car accident?
Hypotheses:
Ho: There is no significant difference in the use of seat belt before and after
involvement in the car accident.
III.
IV.
Ha: There is a significant difference in the use of seat belt before and after
involvement in the car accident.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = (𝑐 − 1)(𝑟 − 1) = (2 − 1)(2 − 1) = 1
2
𝜒0.05
= 3.841
Statistics: Mc Nemar’s test for correlated proportion
Computation:
𝜒2 =
V.
VI.
(𝑏−𝑐)2
𝑏+𝑐
=
(12−23)2
12+23
= 3.457
Decision Rule: If the computed 𝜒 2 is greater than the tabular 𝜒 2 , reject Ho.
Conclusion: Since the computed 𝜒 2 of 3.457 is less than the tabular 𝜒 2 of
3.841 at 0.05 level of significance and 1 degree of freedom, the null
hypothesis is accepted. This means that there is no significant difference in
the use of seat belt before and after involvement in the car accident. This
implies that, their involvement in the car accident did not change their
attitudes towards wearing seat belts.
LESSON 21 - THE FRIEDMAN Fr TEST FOR RANDOMIZED BLOCK DESIGN
The Friedman test is a nonparametric statistical test developed by Milton
Friedman. This test is similar to the parametric repeated measures ANOVA used for
comparing the distributions of measurements for k treatments laid out in b blocks
using randomized block design. The procedure for conducting the test is similar to
that used for the Kruskal-Wallis H-test. When either the number of k treatments or
the number of b blocks is larger the five, sampling distribution of Fr test can be
approximated by a Chi-Square distribution with (k-1) df. The formula is:
𝐹𝑟 =
12
∑ 𝑇𝑖2 − 3𝑏(𝑘 + 1)
𝑏𝑘(𝑘 + 1)
Where:
𝐹𝑟 = Friedman test
𝑏 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑙𝑜𝑐𝑘𝑠
𝑘 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠
𝑇𝑖 = 𝑟𝑎𝑛𝑘 𝑠𝑢𝑚 𝑓𝑜𝑟 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝑖
𝑖 = 1,2,3, … , 𝑘
Example. In a study of the probability of a vaccine for children, 6 sample
healthy children were used as subjects to assess their reaction to the taste of
four vaccines. The children’s response was measured on a 10-point visual
digital scale incorporating the use of faces, from sad (low score) to happy
(high score). The minimum score was 0 and the maximum was 10. The
following data were recorded:
Vaccine
Children
1
2
3
4
5
6
1
5.6
8.8
5.4
7.6
4.8
4.1
2
2.4
9.2
2.8
9.6
7.8
8.3
3
6.8
6.7
3.6
5.2
2.7
3.4
4
6.5
9.6
6.7
8.9
2.8
3.4
Solving by Stepwise Method:
I.
II.
III.
IV.
Problem: Is there a significant difference in the reaction of 6 children on the 4
different vaccines?
Hypotheses:
Ho: There is no significant difference in the reaction of 6 children on the 4
different vaccines.
Ha: There is a significant difference in the reaction of 6 children on the 4
different vaccines.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = 𝑘 − 1 = 4 − 1 = 3
2
𝜒0.05
= 7.815
Statistics: Friedman Fr Test
Computation:
Vaccines
Children
1
2
3
4
5
6
Rank
Sum
1
Reaction
(cm)
5.6
8.8
5.4
7.6
4.8
4.1
Rank
2
2
3
2
3
3
15
2
Reaction
(cm)
2.4
9.2
2.8
9.6
7.8
8.3
Rank
1
3
1
4
4
4
17
3
Reaction
(cm)
6.8
6.7
3.6
5.2
2.7
3.4
Rank
4
1
2
1
1
1.5
10.5
4
Reaction
Rank
(cm)
6.5
3
9.6
4
6.7
4
8.9
3
2.8
2
3.4
1.5
17.5
𝐹𝑟 =
𝐹𝑟 =
12
∑ 𝑇𝑖2 − 3𝑏(𝑘 + 1)
𝑏𝑘(𝑘 + 1)
12
(152 + 172 + 10.52 + 17.52 ) − 3(6)(4 + 1)
(6)(4)(4 + 1)
𝐹𝑟 = 3.05
V.
VI.
Decision Rule: If the computed Fr is greater than the tabular Fr, reject Ho.
Conclusion: Inasmuch as the 𝐹𝑟 − 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 value of 3.05 is less than the
𝐹𝑟 − 𝑡𝑎𝑏𝑢𝑙𝑎𝑟 value of 7.815 at 0.05 level of significance with 3 degrees of
freedom, the null hypothesis of no significant difference in the reaction of 6
children on the 4 different vaccines was accepted.
Exercise: In a study of an antibiotics for children, 5 sample healthy children
were used as subjects to assess their reaction to the taste of four antibiotics.
The children’s response was measured by an apps developed available in the
Google playstore. The minimum score was 0 and the maximum was 10. The
following data were recorded:
Antibiotics
Children
1
2
3
4
5
1
4.3
6.8
5.4
7.2
6.1
2
8.8
9.2
8.5
9.6
9.3
3
5.4
6.7
4.3
5.2
3.9
4
6.5
9.4
8.4
9.3
7.6
Use the Friedman Fr test to assess the children’s reaction on the 4 different
antibiotics at 0.01 level of significance.
LESSON 22 - THE KENDALL’S COEFFICIENT OF CONCORDANCE, 𝒲
This statistic is another nonparametric test used to determine if there is an
agreement or concordance among the raters or judges of N objects or individuals.
When 𝒲 is 1, it is interpreted as ‘High agreement’ and when 𝒲 is, ‘No agreement’.
The formula is:
12(∑ 𝐷2 )
𝒲= 2
𝑚 (𝑁)(𝑁 2 − 1)
Where:
𝒲 = the coefficient of concordance
𝐷 = the difference between the individual sum of ranks of
the raters or
judges and the average of the sum of ranks of the
object or individuals.
∑ 𝐷2 = the sum of squares of the difference
𝑚 = number of Judges or raters
𝑁 = objects or individuals being rated or ranked
Example. Below are the recorded ranks of the 4 judges to 10 contestants.
Contestants
1
2
3
4
5
6
7
8
9
10
Judges
A
1
4
3
5
2
8
7
10
9
6
B
2
3
4
5
1
7
8
10
9
6
C
3
2
5
4
1
6
7
9
10
8
D
2
4
3
5
1
7
6
10
9
8
Solving by Stepwise Method:
I.
II.
III.
Problem: Is there an agreement or concordance in the rankings of the 4
judges to the 10 contestants?
Hypotheses:
Ho: There is no agreement or concordance in the rankings of the 4 judges to
the 10 contestants.
Ha: There is an agreement or concordance in the rankings of the 4 judges to
the 10 contestants.
Level of Significance:
𝛼 = 0.05
𝑑𝑓 = 𝑚 = 4 𝑎𝑛𝑑 𝑁 = 10
𝒲0.05 = 0.44
IV.
Statistics: Kendall’s Coefficient of Concordance, 𝒲
Computation:
Contestants
A
1
4
3
5
2
8
7
10
9
6
1
2
3
4
5
6
7
8
9
10
Total
Judges
B
C
2
3
3
2
4
5
5
4
1
1
7
6
8
7
10
9
9
10
6
8
𝑅̅ =
𝒲=
V.
VI.
D
2
4
3
5
1
7
6
10
9
8
Sum of |𝑅̅-Sum of Ranks|
Ranks
D
9
13.1
13
9.1
15
9.1
19
3.1
5
17.1
28
5.9
28
5.9
39
16.9
37
14.9
28
5.9
221
D2
171.61
82.81
82.81
9.61
292.1
34.81
34.81
285.61
222.01
34.81
1250.99
221
= 22.1
10
12(1250.99)
= 0.948
42 (10)(102 − 1)
Decision Rule: If the computed 𝒲 is greater than the tabular 𝒲, reject Ho.
Conclusion: The computed 𝒲 of 0.948 being greater than the tabular 𝒲 value
of 0.44 at 0.05 level of significance with 4 and 10 degrees of freedom, the null
hypothesis is accepted. This means that there is a significant agreement or
concordance in the rankings of the 4 judges to the 10 contestants.
Exercise. Three judges rank a group of 5 contestants in a beauty pageant. The
following are the ranks of the judges:
Students
Judges
A
B
C
a
2
1
2
b
2
1
3
c
4
3
2
d
3
4
5
e
5
5
4
Module 1 - Additional Exercises
1.1 The performance ratings of 50 policemen in the capital town of NCR are shown
below.
Classes
Frequency
70-74
5
75-79
11
80-84
13
85-89
15
90-94
6
Determine the value of the following measures:
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Mean 𝑥̅
Median 𝑥𝑑
Standard deviation SD
1st Quartile 𝑄1
3rd Quartile 𝑄3
10th Percentile 𝑃10
90th Percentile 𝑃90
Skewness SK
Kurtosis Ku
Tell and justify if the distribution of data is normal or abnormal based on the value
of SK and Ku
1.2 Consider the performance rating of 65 college professors in TCU.
Performance Rating
Frequency
65-69
70-74
75-79
80-84
85-89
90-94
95-99
3
8
13
17
13
8
3
A. Determine the value of the following measures:
1. 𝑥̅
2. Md
3. Q1
4. Q3
5. P10
6. P90
B. Find the value of the skewness and interpret the result.
C. Find the value of the kurtosis and interpret the result.
1.3 In a PE class, 50 students were made to run in a 100-meter dash. The following
results are presented in a frequency distribution.
Seconds
Frequency
10-11
2
12-13
8
14-15
14
16-17
16
18-19
8
20-21
2
a. Find the value of the kurtosis. Then interpret the result.
b. Graph the data and tell whether it is leptokurtic, mesokurtic or platykurtic.
Module 2 – Additional Exercises
2.1 During the lockdown due to pandemic, a certain livelihood program was given to
15 jeepney drivers to enhance their income. The data were recorded before and after
the implementation as shown below.
Income
Before the Implementation
After the Implementation
(Php)
(Php)
6,000
7,000
8,000
7,600
9,000
10,500
8,500
9,500
8,800
9,000
9,200
10,400
9,500
10,500
11,300
12,000
8,600
9,500
7,200
8,700
9,400
10,500
10,500
10,000
11,600
12,500
12,000
13,500
15,000
15,800
Use the t-test for correlated samples at .05 level of significance to test whether there
were significant changes in the income of the jeepney drivers after the implementation
of the program.
2.2 The data below represent the number of hours of pain relief provided by two brands
of headache syrups administered to 20 individuals. These individuals were randomly
divided into two groups and each group was rated with a different brand.
Brand X
Brand Y
6
6
8
7
9
5
4
3
3
4
7
6
8
5
6
3
5
6
8
6
Use the t-test at 0.05 level of significance to test the null hypothesis that there is no
significant difference between the average number of hours of pain relief provided by
the two brands of headache syrups.
2.3 The following data show the weight losses (in mg) of certain machine parts due to
friction using two different lubricants.
Lubricant 𝐴
Lubricant 𝐵
10
7
13
6
12
6
11
5
13
7
14
10
7
9
10
4
8
11
11
6
12
13
Test at 0.01 level of significance whether the differences between the two-sample
means are significant.
2.3 The table shows the number of errors made on 10 occasions by two compositors
on their technical report. Is there a significant difference in the number of errors made
in general by the two compositors? Use the t-test for independent samples at 0.05
level of significance.
Composition 1
Composition 2
11
14
12
11
10
13
12
9
11
11
14
13
13
11
11
12
14
11
9
9
2.4 Ten samples were given an attitude test on a controversial issue. They were shown
a favorable film regarding the issue and the same attitude test was administered
before and after. Make a directional test at 0.05 level of significance.
Pretest
Posttest
15
19
19
21
15
24
23
27
21
22
24
28
23
24
17
23
16
19
14
15
2.5 A school administrator in a laboratory school claimed that the reading
comprehension of grade 12 students in their school have an average of 82.3 with a
standard deviation of 7.4. To test if the claim is valid, a researcher conducted a reading
comprehension test for 60 randomly selected grade 12 students and found out that
the average is 82.6. Use the Z-test at 0.05 level of significance to determine the validity
of the claim.
2.6 A law student wants to confirm the claim of RTC Judge that convicted illegal
gamblers spend on the average of 12.6 months in jail. A random sample of 50 cases
from court files, showed that 𝑥̅ = 13.2 and the SD = 3.3 months. Use the Z-test to test
the null hypothesis the 𝜇 = 12.6 months against the alternative hypothesis of 𝜇 ≠ 12.6
months at 0.05 level of significance.
2.7 Is there a significant difference between the average weight of females born from
the two different countries. A random sampling yielded the following results:
𝑛1 = 100
𝑥1 = 64.2
̅̅̅
𝑆𝐷1 = 2.48
𝑛2 = 120
𝑥1 = 63.8
̅̅̅
𝑆𝐷2 = 2.53
Use the Z-test at 0.05 level of significance.
2.8 A study was made to check whether the male average income is higher than the
female average income. Use Z-test at 0.05 level of significance.
Sex
Number of
Workers
Male
Female
Mean
80
75
SD
Php 1,218
Php 980
Php 42.30
Php 38.40
2.9 The table below shows the number of minutes that patients had to wait for their
appointment with 5 doctors.
Doctor
A
B
C
D
E
21
19
22
31
29
10
12
16
13
19
19
18
17
16
21
10
12
29
30
16
20
31
25
27
26
Use the F-test at 0.05 level of significance to test if there are significant differences
among the means of the samples.
2.10 The following are the data on homicide cases committed from January to
December in 3 cities in the National Capital Region are as follows:
City
January
A
B
C
February
March
April
May
June
July
August
September
November
December
4
5
3
6
4
7
8
5
3
2
5
7
6
7
3
2
5
3
7
3
6
5
5
7
9
8
10
5
6
8
Perform the one-way analysis of variance to test the null hypothesis that the average
homicide cases in the 3 cities are equal at 𝛼 = 0.05 level of significance.
2.11 A research study was conducted on 4 groups of students. The following number
of correct responses were recorded out of 10 trials.
Apply the analysis of variance to find out if the groups differed significantly in
their performance. Use 𝛼 = 0.05.
Group
Trial
1
2
3
4
5
6
7
8
9
10
A
B
C
D
8
6
9
10
11
10
8
10
6
8
7
8
9
10
9
12
14
10
14
9
3
4
6
5
6
6
3
2
4
3
2
7
9
8
7
8
6
4
6
2
2.12 The following are the test scores of 20 students under two teachers and two kinds
of modules used in teaching English. Apply two-way-ANOVA, at 0.05 level of
significance.
TEACHERS
Textbooks
A
B
A
5
6
3
2
4
9
8
3
8
9
B
6
9
3
2
1
7
8
8
9
5
Total
Sub Total
Total
2.13 Two sets of attitudinal scales were administered to two groups of students from
private and public schools. Perform the two-way-ANOVA at 0.05 level of significance
if there is significant difference between the two groups of students coming from two
different schools and two groups of students given Set A and Set B on attitudinal
scale?
SCHOOLS
Attitudinal
Scale
Set A
Private
Public
22
18
9
10
8
8
16
17
6
8
5
19
23
19
21
17
19
11
24
25
19
18
17
11
Sub Total
Set B
Total
Total
2.14 The table below shows the percentage of the votes predicted by a poll survey A
for 10 candidates for the senate on different cities and the percentage of the votes
which they actually received B.
Poll Survey
A
Actually Received
B
42
47
32
38
26
35
39
51
19
25
48
50
37
38
25
34
47
53
31
27
Use the Pearson Product Moment Correlation Coefficient at 0.05 level of significance
to determine if there is a significant relationship between the poll survey and the actual
votes received.
2.15 The following are the scores of 10 students in the final examination in
mathematics of investment (MOI) and accounting.
MOI (x)
Accounting (y)
81
64
65
74
75
76
77
80
83
88
84
69
66
64
77
81
76
84
96
84
Find the value of r and interpret the result at 0.05 level of significance.
2.16 A study was made by the SM Bicutan Hypermarket to determine the
relationship between weekly sales and advertising expenditures. The following data
were recorded. Use 𝑟 at 0.05 level of significance.
Adverting Cost
(in thousand pesos)
5.0
2.5
3.0
2.5
5.5
6.0
2.0
3.5
2.5
3.0
Sales
(in thousand pesos)
37.5
40.2
38.4
35.1
45.6
56.7
23.5
42.5
43.8
46.3
2.17 The following midterm grade, average grade in quizzes and final grade of 8
college students in Civil Engineering were obtained by a certain professor in TCU.
Midterm Grade
𝑥1
3.0
1.75
1.5
1.25
2.0
2.75
2.5
1.75
Average Grade in
Quizzes
𝑥2
Final Grade
y
2.75
1.5
1.5
1.25
2.25
2.5
2.5
2.0
2.75
1.5
1.5
1.25
2.0
2.5
2.5
1.75
a. Use the method of least squares to fit an equation of the form
𝑦 = 𝑏0 + 𝑏1 𝑥1 + 𝑏2 𝑥2
b. Predict the final grade of student whose midterm grade is 1.5 and the average
grade in the quizzes is 1.75.
2.18 The table below displays the area of the lot, the number of bedrooms, and the
prices at which 10 one-family cottages sold at DMCI subdivision:
Area in Square
Meters
𝑋1
250
200
Number of
Bedrooms
𝑋2
4
3
Prices
y
(In Million Php)
1.7
1.5
160
180
240
320
270
300
250
220
2
2
4
5
3
4
3
3
0.9
1.2
1.8
2.6
1.8
2.2
1.6
1.4
Determine a linear equation which will enable us to predict the average sale price of
one - family cottage in terms of the area in square meters and the number of
bedrooms.
Module 3 – Additional Exercises
3.1 In 120 tosses of a coin, 64 heads and 56 tails are recorded. Is this a balanced
coin? Use 𝜒 2 − 𝑡𝑒𝑠𝑡 at 0.05 level of significance.
3.2 The grades in Mathematics in the Modern World for a particular semester are as
follows:
Grades
Observed
1.25
1.50
1.75
2.00
2.25
15
19
32
22
17
Use the 𝜒 2 − 𝑡𝑒𝑠𝑡 at 0.05 and test the hypothesis that the distribution of grades is
uniform.
3.3 A random sample of 270 voters classified according to the political affiliation were
asked if they were in favor of the ongoing peace negotiation in selected cities/towns in
Mindanao.
Political Affiliation
Favor
Not in Favor
Total
NP
LP
PDP-LABAN
38
53
54
52
37
36
90
90
90
Total
145
125
270
Use Χ 2 − 𝑡𝑒𝑠𝑡 at 0.05 level of significance to test the hypothesis that the sample
belongs to the same population.
3.4 A random sample of 60 adults are grouped according to sex and their opinion
regarding the ceasefire between the government and the CPP-NPA at Christmas.
Ceasefire
Sex
Yes
No
Total
Male
Female
19
21
11
9
30
30
Total
40
20
60
Use Χ 2 − 𝑡𝑒𝑠𝑡 at 0.05 level of significance to determine if there is no significant
difference between the opinion of male and female adults on the issue of ceasefire
between the government and the CPP-NPA.
3.5 A random sample of 400 adults are classified according to their age bracket and
drinking habits.
Drinking Habits
Nondrinkers
Moderate drinkers
Heavy Drinkers
20-29
30-39
28
33
53
40-above
75
74
63
19
38
17
Test the hypothesis that age bracket is dependent of drinking habit. Use 𝑋 2 − 𝑡𝑒𝑠𝑡 at
0.05 level of significance.
3.6 The following data were taken from 200 individuals in a study to determine the
dependence of lung cancer and smoking habits,
Nonsmokers
Moderate
Smokers
Heavy
Smokers
With Cancer
No Cancer
26
49
48
21
46
10
Total
75
69
56
Test the hypothesis that the lung cancer is independent of smoking habits. Use 0.05
level of significant.
3.7 From 18 students of Math class, 9 students are selected at random and given
additional instruction by the teacher. The rest of the students were given no additional
instruction. The results on the final examination were as follows:
Grades in the Final Examination
With Additional
Instruction
86
91
85
82
85
86
80
84
86
No Additional
Instruction
76
80
80
84
73
82
84
78
82
Use the Wilcoxon rank-sum test at 0.05 level of significance if the additional instruction
affects the average grade.
3.8 Electric cables are being manufactured by two companies. To determine if there
is a difference in the mean breaking strength of the cables, 8 pieces from each
company are selected at random and tested for breaking strength. The results are:
Company A
Company B
10.5
11.2
8.9
8.6
10.7
9.6
9.8
7.5
11.4
9.6
11.5
10.8
12.6
11.6
9.7
7.4
Use the Wilcoxon rank-sum test at 0.05 level of significance if there is a difference in
the mean breaking strength of the cables manufactured by two companies.
3.9 Random sample of 3 brands of cigarettes were tested for nicotine content. The
following figures show the milligrams of nicotine found in the 15 cigarettes tested.
Brand
A
B
C
16
15
14
17
13
18
17
19
20
21
12
11
10
9
11
Use the Kruskal-Wallis test, at the 0.05 level of significance, to test whether there is a
significant difference in nicotine content among the 3 brands of cigarettes.
3.10 The following data represent the operating time in hours for 4 Brands of midrange
cellphones before a recharge is required.
Brand
Samsung
Oppo
Vivo
Xiaomi
12.8
13.2
13.6
13.4
13.3
13
11.6
12.7
12.4
12.5
12.5
12.4
12.8
12.6
12.2
11.6
12.4
12.6
11.8
12.2
Use the Kruskal-Wallis test, at the 0.01 level of significance to test the hypothesis that
the operating times for all 4 brands of cellphones are equal.
3.11 Two judges of a city fiesta parade in NCR ranked 10 floats in the following order:
Judge A
Judge B
8
5
10
9
3
4
6
7
2
1
5
7
9
5
2
4
8
6
1
3
Use Spearman’s Rank correlation coefficient to test if there is a significant correlation
in the ranking of two judges.
3.12 Two groups of experimental dogs are given two brands of vitamins and the
following weight gains in grams were obtained.
Group 1
Group 2
250
67
206
75
106
138
280
180
212
193
106
48
112
78
142
82
168
67
185
88
Apply the sign test to determine if the two samples come from population with the
same median.
3.13 The following are the scores of 10 male and 10 female students in Psychological
laboratory examination.
Male
22
Female 46
83
94
51
62
67
75
74
56
79
42
46
84
20
92
37
49
69
96
Test the significant difference between the scores of the male and female students
using sign test.
3.14 The following are the weights of 15 women enrolled in a 10-week slimming
program. Their weights were taken before and after the program.
Before
After
120
130
140
125
140
115
135
146
141
130
126
114
115
120
136
120
125
115
126
145
100
120
126
134
129
124
112
106
112
121
Use the sign test at 0.05 level of significance to test if there is a significant difference
in the weights of women enrolled in the programs before and after.
3.15 The following are the scores obtained by the groups of 6 subjects each given with
3 different methods of teaching in Algebra.
Method 1
Method 2
Method 3
38
37
30
40
36
34
40
38
50
46
32
48
70
40
70
69
50
39
Apply a median test at 0.05 level of significance.
3.16 The following data were obtained before and after a televised debate on charter
change for a sample of 60 registered voters.
After the Debate
Before the
Debate
Yes
No
Total
Yes
20
15
35
No
12
13
25
Total
32
28
60
Apply the Mc Nemar’s test at 0.05 level of significance.
3.17 The following data were recorded for 6 subjects exposed to 4 different treatments.
Treatments
Subjects
T1
T2
T3
T4
1
2
3
4
5
6
10
9
8
10
5
9
10
5
7
9
6
9
4
6
9
8
3
8
6
3
10
7
4
7
Apply the Friedman rank test at 0.05 level of significance.
3.18 The following data were obtained for 5 subjects repeatedly measured under 3
different conditions.
Condition
Subject
1
2
3
1
2
3
4
5
11
29
22
3
13
14
34
30
5
11
13
41
18
17
16
Use the Friedman rank test at 0.05 level of significance.
3.19 Three judges ranked 5 contestants in a beauty contest. The following were
obtained:
Students
Judge
A
B
C
D
E
X
2
3
1
4
5
Y
2
1
4
3
5
Z
1
3
2
5
4
Compute the coefficient of concordance 𝑊 at 0.05 level of significance to determine
if there is an agreement among the ranks of the three judges on the 5 students
.
Works Cited
(n.d.). Retrieved from https://www.sciencedirect.com>topics
Bayanito, M. R. (2015). The Contemporary World. Taguid City: National Book Store.
Bibliography
(n.d.). Retrieved from https://www.sciencedirect.com>topics
Broto, A. S. (2006). Statistics Made Simple. Mandaluyong City: National Book Store.
Garcia, G. A. (2003). Fundamental Concepts and Methods in STATISTICS (Part 1). Manila, Philippines:
University of Santo Tomas Publishing House.