polygon 2009
DESCRIPTION
Polygon is a tribute to the scholarship and dedication of the faculty at Miami Dade College in interdisciplinary areas.TRANSCRIPT
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Mission of Miami Dade College The mission of the College is to provide accessible, affordable, high-quality education
that keeps the learner’s needs at the center of the decision-making process.
Miami Dade College District Board of Trustees
Helen Aguirre Ferré, Chair Peter W. Roulhac, Vice Chair
Armando J. Bucelo Jr. Marielena A. Villamil Mirta "Mikki" Canton
Benjamin León III Robert H. Fernandez
Eduardo J. Padrón, College President
Editorial Note
Barricade the Doors First: A Reflection on Zombie Fiction and its Critique of Post Modern Society.
Victor Calderín
Synthesis of Optimum Experimental Plans Ensuring Computation of Integral Parameters Using a Regression Equation.
Dr. Manuel Carames
Teaching Strategies Involving CAT’s and the Statistical Validation of the Results
Dr. Luis Martin & Dr. Manuel Carames
Incorporating Environmental Immersions to Learning Community linked courses in Mathematics and Geography
Dr. Jaime Bestard & Dr. Arturo
Rodriguez Using Beta-binomial Distribution in Analyzing Some Multiple-Choice Questions of the Final Exam of a Math Course, and its Application in Predicting the Performance of Future Students
Dr. Mohammad Shakil
The Magic Math David Tseng, (Graphed
by Nancy Liu)
Barricade the Doors First: A Reflection on Zombie Fiction and its Critique of Post Modern Society
Prof. Victor Calderin
English and Communincations Miami-Dade College
Hialeah Campus 1780 West 49th St., Hialeah 33012
E-mail: [email protected]
Abstract
The trope of the zombie is a staple of the contemporary horror genre. This paper will reflect on how the symbol of the zombie is a post modern refraction of the traditional resurrection story. In addition to addressing the concept of a resurrection, the zombie mythos also allows addresses the state of the family unit in the post modern world. Ultimately, the horror of zombie fiction is not the walking dead but what they say about us, the living.
Themes: Popular Culture
Key Words:Zombie Fiction, Resurrection, Post Modernism, George A. Romero
Barricade the Doors First: A Reflection on Zombie Fiction and its Critique of Post Modern Society
George A. Romero forever changed how we view the zombie in 1968 with the release of
Night of the Living Dead, adding a new symbolic depth to the undead. With his film, Romero ties
the metaphor of the zombie directly to the disconnection we feel in postmodern society. The
zombie is the embodiment of society sans soul, sans intellect, only driven by the urge to
consume. But these disparaging characteristics fly in the face of the metaphor that is at the core
of the zombie mythos: the resurrection. When we think about the concept of resurrection, two
key concepts are at play. First, the idea of a resurrection directly usurps the permanence of death,
contradicting most empirical data. But not only does the resurrection defeat death, it merges
spirit to transfigured flesh, creating the perfect condition for an afterlife. And it is this afterlife
that we find the second main factor; eternal reunion with family. Not only does the resurrection
offer beautified life after death, but it offers us the company of those we love. And so this
concept had endured for thousands of years, from the Elysian Mysteries to Christian Dogma,
until the advent of George A. Romero’s Night of the Living Dead.
Romero’s work defines our concept of the zombie; they are mindless automatons who
crave one thing: living flesh. And while they look human and have some semblance of
intelligence, as seen throughout the film when the spectator sees them pick rudimentary tools
like rocks and sticks, zombies do not have the basic requirements that would allow us to label
them as human. The classical definition of resurrection creates an image of perfect flesh merged
with spirit, but the postmodern condition transforms this idea in the zombie. The zombie is not
perfected flesh but rotten, dilapidated form that mimics the body. This is why in the opening of
the film Barbra and Johnny cannot immediately identify the zombie approaching, a critique on
our society and its zombie-esque movement. This is also why the scenes where the zombies are
aimlessly walking around the countryside are so disturbing. The horror isn’t just the gore factor
but also the social psychological danger of all of us becoming soulless machines. This lack of
any sense of spirit, of animus, is what really defines the zombie as a metaphor is direct
opposition to the resurrection. If the classical view allowed us to return in spirit and perfect flesh,
the postmodern view, which is based on lack of any connections, be they literal or symbolic,
creates a revised metaphor, a dark resurrection, where the flesh does return, but it is not perfect
since the spirit is completely missing. But this new symbol goes beyond this, as the individual is
affected by the zombie apocalypse, so is society as a whole.
In the very opening of Night of the Living Dead, we see how society has begun to already
fragment. One of the most fundamental units of society is the family, which will be test in
extreme ways throughout the film. Early in the film when the siblings, Barbra and Johnny, park
in the cemetery, they are bickering over the distance required for this ritual of family observance,
one which their mother insists upon. Their exchange touches on the theme of family discord.
Johnny does not see the purpose of leaving the flower of the relative’s grave, even joking about
the caretaker making a profit by reselling the ornament. Barbra defends her mother’s request,
later on praying to show respect. While the relative is never really mentioned, it is important to
note that it is a male relative. As Johnny begins to tease his sister, he is attacked by a male
zombie, and the sibling urges his sister to run for safely while he unsuccessfully tries to fend off
the ghoul. Barbra and Johnny make up the first familiar subunit of the film. The other units are
met in the house where the majority of the film takes place.
At the secluded farmhouse, we discover the other two family subunits. The first is the
Coopers, composed of the husband, wife, and infirm daughter, and the other is the young couple
from the town, Tom and Judy. Both groups were hiding from the zombies in the basement of the
house when Ben, the protagonist and lone figure of the film, and Barbra, who is catatonic after
her brother’s death, were barricading the home. Both groups show the family at different points
in time. The young couple represents the initial phase of the union. Tom and Judy are effectively
inseparable, which is a condition that will ironically lead to their inevitable demise. The Coopers
signify the family in the latter phase where they have a child. While Tom and Judy express love
for each other, Mr. and Mrs. Cooper do not, as seen when Mr. Cooper tells his wife that even
though she is not happy with him, they must stay together and survive for their daughter’s sake.
Despite all affection and intentions, none of these groups survive.
The idea of the resurrection promises a reunion with lost loved one, but in The Night of
the Living Dead, this reunion is a frustrated one. The family is shattered and left in pieces.
Barbra has already lost her brother, and at the end of the film, when he reappears as a zombie, it
is Johnny who drags his sister away into the zombie horde. The Coopers meet their end in the
basement. When their daughter turns, being previously bitten, she murders her mother, who
discovers her eating the flesh of her father, who dies from the gun wound inflicted by Ben in
their final scuffle. Tom and Judy meet their end during the botched attempt to fuel the truck,
where out of loyalty the boyfriend stays in the combusting truck trying to save his girlfriend as it
explodes. All the family units fall apart or are destroyed by the zombie threat, which is
completely fitting, for as the resurrection promises reunion, zombification promises separation.
In addition to this dichotomy, as the zombie nightmare destroys the family, the individual is the
only one able to survive, and this is true in the case of Ben.
Ben is able to endure the night because he is able to think clearly. Upon discovering the
house, he secures it, searches it for possible threats, and devises an escape plan and attempts to
executes it, which should be noted fails not because of him but because of Tom. Also, Ben is the
only one to survive the night, but in an ironic twist he is shot by the search party securing the
area. The one human to survive the zombies is senselessly killed by other humans, leaving us
with a question: Who are the real monsters?
The symbol of the zombie is a troubling one. It is a metaphor that not only denies us the
spirit and intellect that make us human, but it also destroys the family unit that is essential to
humanity. Besides these practical applications, the myth of the zombie completely subverts the
concept of a resurrection. As the resurrection story is one that offers hope against the inevitable,
the zombie narrative is one that destroys hope, leaving us alone, uncertain, and lost.
Works Cited
The Night of the Living Dead. Dir. George A. Romero. Perf. Duane Jones. 1968.
Optimal Experimental Plans
SYNTHESIS OF OPTIMUM EXPERIMENTAL PLANS ENSURING COMPUTATION OF INTEGRAL PARAMETERS USING A REGRESSION EQUATION.
(SOME EXAMPLES)
Dr. Manuel Carames Assistant Professors
Mathematics Department Miami‐Dade College, North Campus
11380 NW 27 Avenue, Miami, Florida 33167, USA Emails: [email protected]
ABSTRACT
This is an attempt to share with my colleagues some experiences related to Optimal Experimental Plans and particularly to ones that were introduced by me and have the general name of I plans (I because they are plans to estimate integral indexes of the response function).
Theme: Optimal Experimental Plans Key words: Experiment, estimation, statistic, system, model
Introduction: In different scenarios, engineers, researchers, etc need to estimate not the values of the response functions in certain points, they need to estimate some integral indexes of this function. To this idea is dedicated this paper
Body:
What is the researched object?
Pr
Xr Y
r Object
Zr
1
Optimal Experimental Plans
The set of parameters which define the state of the object is divided in the following groups:
ector of input variables) Input variables (v ( )mxxxx Lr
21,=
max
. In this group we have the
controllable parameters of the object. The values of these parameters are inside given intervals and they are given by the schedule of the technological process or technical
constrains and are of the form min ii xxix ≤≤ ; i 2,1 m,,L= ;
Output variables (vector of output variables) ( )ryyyy Lr
,, 21= . In this group we
the output
In group
have
the variables that contain information about quantity and quality characteristics ofproduct.
zr we in vaclude the riables that we can control but we cannot manipulate.
( )szLr
, zzz , 21= .
In group pr we in time
clude variables that we cannot control neither manipulate. In this group we
have noises and we do not know the points where they are applied, we neither know the
characteristics nor the power of then. ( )lpppp Lr
21= .
It is clear from the way that we defined the universe of signals that only the once that belong to xr could be manipulated.
In case of active experiment all the variables that belong to pr will be represented by equivalent addition noise e applied to the output.
Variables from zr , whose characteristics during the experiment are known will be considered as
variables that belong to group xr .
Let us assume that we have one output signal, then the response function can be represented as
( ) exy += θηrr
, where
e is the noise with the following properties:
( ) 0=eE the mean of the noise is zero. E is the mean operator ;
( ) jieeE ji ≠∀= 0.ie je
this means in different point of the factor space where the output is
measured and are not correlated ;
ji xx ,
( ) ieD i ∀= 2σ this means that dispersion of the noise in all points of factor space is the same. ixr
2
Optimal Experimental Plans
The last two conditions can be written other way around { } WeeE 2. σ=′ where
W is the unit matrix with dimension equals to nn× ;
is the amount of experiments (amount of measures of the output; n
),,,( 21 neeee L=′ is the vector of the values of the errors in the points where the measures were
taken;
E is the mean operator;
( ) ( )∑==
k
iii xfx
1,
rrr θθη ;
iθ ‐ coefficients of the model;
( )xfir
‐ given functions of input variables.
The schema of the object can be given now by the following sketch:
e
xr y
What do we want to do?
Sometimes we want to find not the discrete value of the response function and instead we want to find some integral indexes of the function, it means
( ) ( )[ ] ( ) ( ) ( )∫ ∫ ∫ ⎥⎦⎤
⎢⎣⎡∑ +=+==
Ω Ω Ω =xdxexfxdxexxdxy j
k
iiijjj
rrrrrrrrr ωθωθηωμ1
,
The estimation of the indexes we will find:
( ) ( ) ( )∫ ∑==Ω =
k
iiijj xdxxfE
1
ˆˆ rrr ωθμμ ;
where jμ is the estimated index;
Object +
3
Optimal Experimental Plans
y is the output parameter (output variable);
iθ̂ is the estimation of the coefficient iθ of the model;
( )xjrω given not random weight function that in general depends of the input factors;
Ω is the given region of factor space where we calculate the integral idexes;
MMj ;,1_____
= is the amount of calculated integral indexes.
We will look for the estimators of the coefficients of the model inside linear class of estimators, what means:
yTr=θ̂ ;
Where is the vector of the values of response function in different points ( nyyyy L,, 21=′ ) ixr of
the factor space;
n is the amount of experiments and
T is a matrix with dimension nk ×
We will look for estimators with the following characteristics
‐ ( ) realE μμrr
=ˆ what means that their expected value will be the same as the real value of the
integral indexes;
‐ ( ) ( ) 0ˆˆlim =⎥⎦
⎤⎢⎣
⎡≥−
′−
∞→ξμμμμ realnreal
np
rrrr what means that the estimation converge
from the probabilistic point of view to the real values of the estimated parameters. Index
nify that estimation
nsig nμ̂
r was obtained after nmeasures and ξ any in front given
positive number and
is
‐ ( ) ( )0ˆˆ μμrr
DD ≤ where ( )μ̂rD covariance matrix and ( )0μ̂r
D is the covariance matrix of
any unbiased estimations of 0μr
.
These estimators are known as the best linear estimators and the estimations that we obtain are the best linear estimation of the integral index of the response function.
4
Optimal Experimental Plans
It is known that in order for us to improve the quality of the properties of the statistic estimations we need, somehow, to select points in the factor space to do our measures, it means, WE NEED TO PLAN THE EXPERIMENT!!
Today we can find in the literature different type of experimental plans.
When we plan an active experiment, the necessary statistic material for the estimation of the parameters is collected following a define research program. Research program is the experimental plan and it satisfies certain criterion of optimality.
About Experimental Plans
They are divided in exact and continuous plans.
Exact plans are optimal for a given number of observations N
The task of finding an optimal exact plan is done by finding where, in the plan region, measurements should be taken to satisfy the given criterion of optimality.
If the obtained plan is concentrated in Nn ≤ points them we can define the observation frequency in
point l as Nrl
l =ξ , where is the number of observations done in point . From what was
said we have where
lr lx
∑ ==
n
ll
11ξ lξ proportion of observations that were done on point considering
the total amount of done observations as the unit.
lx
The main characteristic for exact plans is that Nr ll ξ= where is a positive integer. lr
Continuous plans are not related to a specific number of observations. These plans are given by a positive probability metric , that is , . A continuous norm plan ε is the following set of magnitudes :
( )∫(x) dξξ =x 1 x
⎭⎬⎫
⎩⎨⎧
=nnxxxξξξ
ε...21
...21
where are points of the spectra of the plan ix
and where is the planning region; XXix ∈
is the frequency of observations in corresponding points of the plan. iξ
We can find a correspondence between Norm Plans, Norm Information Matrix of the Plan and Norm Covariance Matrix.
5
Optimal Experimental Plans
Norm Information Matrix of the Plan is given by: ( ) ( ) ( ) ( )∫ ′=x
xdxfxfL ξε where
is the base in which the response function is decomposed. ( ) ( ) )( ) ( )( xkfx ,,L
In the case that the metric is contained in a finite number of points we have :
And the Norm Covariance Matrix is:
Statistic properties of the estimation of integral index of response functions
Let us assume that ( )θηrr
,x is a function linear related to parameters, this is: ( ) ( )xfx rrr θθη ′=,
where ( ) ( )( ) ( )( xfxxf )fxf kr
Lrr
,2=′ r,1 .
In the points were done independently measures with nxxx L,, 21
dispersions equal to .
fxfxf 2,1=′
yy ,2,1 L
n2,22
12 L
( ) ( ) ( )∑=
′=n
iixfixfiL
1ξε
( ) ( )εε 1−= LD
ny ,
,, σσσ
We will analyze only linear estimation for what means that we are looking for such estimations that
could be represented as where ( )nyyyy L,, 21=′ and T is a matrix with dim
n
θensions
k × .
Ty=θ̂
It is known from literature the following theorem:
The best linear estimation for the unknown parameters θ are where matrix is equal to
and is not a degenerated matrix; . Covariance matrix of
estimation is equal to
yB 1ˆ −=θ
ii2−= σ
B
( ) ( )∑ ′==
n
iiii xfxfvB
1
rr v
( ) 1−= BˆD θ .
One corollary of the above theorem is that the estimation of any linear combination θCt = will be
θ̂ˆ Ct = . Covariance matrix of estimations t̂ is equal to ( ) ( )CCDtD ′= θ̂ˆ
6
Optimal Experimental Plans
Let us demonstrate that our estimations of μr (integral indexes of response function) are linear
combination of the coefficient of the model:
( ) ∑=x,θ ( )=
k
jxjfj
1
rrr θη
Using matrix notation we can represent the last expression as:
Where
It is known too that the best linear estimation for θ̂ minimizes the sum of the weight squares of the difference between the real value and the one that is calculated by the model (Lease Square Method, LSM)
( ) ( )[ ]∑ ′−==
n
iiii xfyvS
1
2θθ r
[ ] ( ) ( )∫Ω
== xdxxllE rrrr θηωμ , ( ) ( )∫Ω
=∑=
xdxjfx
j jxlrrr
1θω
( ) ( ) ( ) ( )∫Ω
+∫= xdfkx +Ω
xkxldxfxlr
Lrrr ω r rωθ θ11
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
∫Ω
∫Ω
∫Ω
∫Ω
=
mdxmfldxfl
dxmfdxf
θ
θ
ωω
ωω
μ M
L
MLM
L
r 1
1
111
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=
lμ
μμ Mr 1
7
Optimal Experimental Plans
If we put together what was said in the statistic properties of the estimation and what LSM says we can conclude that covariance matrix for our integral indexes is the following:
( ) φθφμ ′= )ˆ(ˆ DD where
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
∫∫
∫∫
=
ΩΩ
ΩΩ
xdxfxxdxfx
xdxfxxdxfx
mm
k
rrrL
rrrMMM
rrrL
rrr
11
111
ωω
ωω
φ
Why is actual the problem?
In different situation scientists, engineers and researchers could find one or more of the following real problems:
- estimation of the mean value of the response function and not the function itself;
- estimation of the volume of raw material in a mine;
- estimation of the amplitude of harmonics in a complex signal;
- estimation of some ordinates of transformed function;
- estimation of the power of a signal for given frequencies.
What is the objective of the experiment?
Estimate ; [ ]lE μ
Where [ ] ( ) ( )dxxxllE ∫=
Ω
θωμ ,ηrrr
And is the response function ( ) ( )∑
=
+=+k
i
exifiex
1
,rrr θ θη
which is known except for the numerical values of the parameters.
Is the given region of factor space; Ω
number of integral indexes that need to be calculated and l 1= MM ;,____
8
Optimal Experimental Plans
given non‐random weight function of input factors. ( )xlrω
What do we need to find?
We need to find: that belongs to the class of continuous and norm plans such that produces the minimum value of variance of the estimator .
*ε Eμr
Definition of Experimental Optimal Plan DIPlan is an optimal plan if and only if it minimizes the determinant of the Covariance Matrix, that is,
*ε DI
or
Where is the plan Information Matrix
is the plan which minimizes the volume of the dispersion ellipsoid.
Some examples:
The response function mean value estimation. Application # 1
We will assume that we have a weight function then:
, is the mean value of the response function;
Meaning: in different technological objects the inputs can change
we need to estimate the mean value of the output index.
Estimation of the amplitude of given harmonic of the response function. Application # 2
The given information is the same as before with the exception of weight functions. In this problem they are the same as the Fourier series,
11 ≡ω
maxmin ixixix ≤≤
( ) ( ) ( ) ( )
rrr
( ) ( ) ( )[ ]MxMxxxxxx cos,sin,,2cos,2sin,cos,sin Lr
=ω
( )∫
Ω
=⎥⎦⎤
⎢⎣⎡ dxxE θημ
rr,
1
( ) ⎟⎞
⎜⎛ μ̂
⎠=
∈*detˆdetmin
εεμ
⎝ε
rrDD
E( ) ( )*
ˆdet*ˆdetmax εμεμr r
LL =Eε∈
L
DI
9
Optimal Experimental Plans
( )∫
−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡ L
L
dxxx
L
k
L
E θηπ
μ ,cos11
( )∫
−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡ L
L
dxxx
L
k
L
E θηπ
μ ,sin12
Meaning: a parametric signal which describes complex periodic movement is given and we need to select some of its components
Estimation of some ordinates of a given transform function. Application # 3
Given: parametric signal in the region . ( )θηrr
,x XA transformation to region is performed. MO
This means
where
and weight functions that areused to go from region to region .
Significance: if then we are applying Fourier Transformation to the input signal and we want to estimate the power of this signal for given frequencies.
Analytic synthesis of a plan
Given:
We need to find a plan which belongs to class norm and continuous plans and it will give the minimum dispersion for the estimation of mean value of the response function in the region
( ) ( )βωωηθη ( )rrrr ,~,MOxX ∈→∈x ( ) ( )∫=
xdxxxh θηωβω
r rrrη~ ,,,
( xh r, MO)ω X
( ) xiexh ωω =,
DI
( ) xx 10, θθθ +=r
[ ]1,1η ∈ Xx = −
⎭⎬⎫
⎩⎨⎧
−−
=∈11211
101*lll
Eε
EX
( ) minˆ →SD10
∈Eε
Optimal Experimental Plans
That is,
Let us find the Information Matrix of the Plan.
Vector in this case has the following form
Them
This result tells us that estimation dispersion of the mean of the response function does not depend of the frequencies of the plan, meaning that we can distribute the resources the way we want.
We can use 50% in the extremes (any person will think that this is the correct distribution and different type of optimal experimental plans recommended so) or you can put all your resources in the center of the plan.
The physical meaning of this result is that for accurate estimation of the mean value of a straight line you can distribute the resources uniformly at the ends of the plan region or you can put the resources all on the plan center to accurately estimate the free term.
Some conclusions
‐ It is really convenient to have optimal experimental plans that specifically deal with integral indexes of the response function.
- Some results were totally unexpected.
- There are many applications for these plans.
- There is a numerical method to synthesize these type of plans based on nonlinear optimization and use the Rosembrok’s algorithm on rotation of coordinates.
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−
−=
120
01
11
11100
01121
11
111 l
lll
φ ( )0,2=φ
( ) ( ) ( )∑=
=′=1i
ixfixfilL ε3
( ) 40
2
12
10
010,2 =⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟
⎠
⎞
⎜⎜
⎝
⎛l( ) ( ) ( )[ ] =′−=′= φεφφθφ 1ˆˆ LDSD
11
Optimal Experimental Plans
12
References
• Doctoral Thesis “Synthesis Optimal Experimental Plans for Calculation Integral Indexes of the Response Function” by Manuel Caramés, Moscow 1983 (in Russian);
• Letski E., Caramés M. Experimental Plans for Accelerated Life Testing. News for Machinery Industry. !979, No 2. (in Russian)
• Experiment Theory by Nalimov V.V. Nauka 1971 (in Russian);
• Experimental Optimal Theory by Fedorov V.V. Nauka 1971 (in Russian);
• Experimental Design for Researching Technological Processes. Hartman K., Letski E., Shefer V. Mir 1977 (in Russian);
• Krug G. K. Doctoral Thesis “Methods of Statistic Analysis for Objects that need to be Controlled” Moscow 1968.
1
Teaching Strategies Involving CAT’s and the Statistical Validation of the Results
Dr. Luis Martin & Dr. Manuel Carames
Assistant Professors
Mathematics Department
Miami-Dade College, North Campus
11380 NW 27 Avenue, Miami, Florida 33167, USA
Emails: [email protected]
ABSTRACT
During the 2008 fall term, the authors conducted an experiment to study the effect of the
application of CAT’s on the rate of success of our students. Two MAC 1114 day classes
were chosen, with similar characteristics in their composition. A set of CAT’s were given
to one of the groups (experimental group) and no CAT’s whatsoever were given to the
other group (control group). At the end of the semester the information regarding the
behavior of both groups was gathered and compared by using a hypotheses testing
procedure, which was the well-known and documented inference about the difference
between two sample proportions. The results of this experiment as well as our
recommendations are described in this paper.
Theme: Educational Research
Key words: Statistics, Hypothesis Testing, Classroom Assessment Techniques.
2
1. Introduction
How the students perform at Miami-Dade College is the main concern for both
administrators and faculty members in this institution. As instructors, it is very important
for us that the students learn the material and consequently pass our classes with the
highest possible grade.
Classroom Assessment Techniques (CAT’s) have being extensively reported in the
literature (1), and also extensively applied by the instructors in their classes. On the
other hand, statistical procedures are available to assess and validate the results
obtained from the application of these CAT’s. We consider that the hypothesis test for
two proportions constitutes a very powerful tool. In this paper we report on the
application of this technique to validate the results obtained from the piloting of two
MAC 1114 classes during the 2008 fall semester.
2. Body
The CAT’s used in this experiment were the Minute Paper, the Muddiest Point, and the
RSQC2 (Recall, summarize, question, connect, and comment) (1).
The format of the actual surveys given to the students is given below.
Minute Paper
Course: _____________________ Date: _________________
This survey is anonymous
At this point I want to evaluate your learning for a reason other than to assign a
grade. I want to assess how much and how well you are learning so I can help you learn
better.
The analysis of the results of this assessment will permit me to learn more about
how you are learning in order to improve it.
1) What was the most important thing you learned today?
2) What questions remain uppermost in your mind as we conclude this class
session?
3
Muddiest Point
Course: ________________ Date: _________________
This survey is anonymous
At this point I want to evaluate your learning for a reason other than to assign a
grade. I want to assess how much and how well you are learning so I can help you learn
better.
The analysis of the results of this assessment will permit me to learn more about
how you are learning in order to improve it.
What was the muddiest point of my lecture today?
RSQC2
Course: _____________________ Date: _________________
This survey is anonymous
At this point I want to evaluate your learning for a reason other than to assign a
grade. I want to assess how much and how well you are learning so I can help you learn
better.
The analysis of the results of this assessment will permit me to learn more about
how you are learning in order to improve it.
1) Make a list – in words or simple phrases – of what you recall as the most
important or meaningful points from the previous class.
2) Use a sentence to summarize the essence of the previous class.
4
Class Selection and Processing the Results of each CAT
A total of five CAT’s were applied to the MAC 1114 experimental group. The control
group did not receive any CAT. Both are morning classes with similar characteristics
and enrollment. The results of each of these five CAT’s were summarized, tabulated,
and discussed with the students the day after its application. This contributed to improve
the communication with the students. We consider of high importance that the results of
the CAT’s should be ready for the next meeting with the students. Also, every single
aspect that was pointed out by the students in their surveys must be discussed and
clarified. If they see that we don’t put too much attention to this, they will lose interest.
Actual student survey papers are available from the authors upon request.
Statistical Validation of Results
The final passing rate was greater in the experimental group, 72% Vs 46% in the control
group. Now, the research question is whether the fact that the passing rate in the
experimental group was greater that the passing rate of the control group is due to
chance or sampling, or the application of CAT’s does help the students to better their
performance. To answer this question we decided to run a hypothesis test procedure.
Description of the Statistical Tool
The hypothesis testing procedure utilized in this experiment was the well-known and
documented inference about two proportions, described in (2), (3), (4), and (5). A
significance level of α = 0.05 was considered, which is the most frequently used in
studies like this.
The notation and formulas that are used in this work are described next.
Notation
21,nn Sample sizes (enrolment of each class)
21, xx Number of successes (# of students who passed in each class)
1
11
ˆn
xp Sample proportion of successes in sample 1
5
2
22
ˆn
xp Sample proportion of successes in sample 2
11ˆ1ˆ pq Sample proportion of failures in sample 1
22ˆ1ˆ pq Sample proportion of failures in sample 2
21
21
nn
xxp Pooled sample proportion of successes
pq 1 Pooled sample proportion of failures
p = population proportion (p1 = p2: The two population proportions are assumed to be
equal in the Null Hypothesis)
21
2121 )()ˆˆ(
n
qp
n
qp
ppppz = Test Statistic for Two Proportions Hypothesis Test
α = Significance Level of the Test
P-Value = Probability of getting a value of the test statistic that is at least as extreme as
the one representing the sample data, assuming that Ho is true.
Sample Requirements
Samples should be big enough: at least 10 successes and at least 10 failures (5). Other
authors require that the four outcomes 11 pn , 11qn , 22 pn , and 22qn are greater than 5 (2).
Samples must be random and independent samples (4) & (5).
6
All these requirements are fulfilled in our samples. This means that we expect that the
sampling distribution of the differences between the proportions of the two samples is
approximately normal.
Sample Data
The data gathered from the two classes at the end of the semester is summarized in the
following table.
Ref 475113 (TR) Ref 475112 (MWF) (Control Group)
n1 = 29 n2 = 28
x1 = 21 x2 = 13
72413.0ˆ1p 46429.0ˆ
2p
Hypotheses
The application of CAT’s should enhance the learning capabilities of the students and
therefore, it should contribute to better the performance throughout the semester and as
a consequence, it should contribute to increase the passing rate. The formal
hypotheses will be expressed as follows:
210 : ppH (Null Hypothesis)
211 : ppH (Alternate Hypothesis) (Original Claim)
The general approach, as in all hypotheses testing, is to assume that 0H is true, and
then see if the sample evidence goes against this assumption.
7
Calculations
The different formulas were fed with the sample data, and the calculations were
performed as shown below.
59649.057
34
2829
1321
21
21
nn
xxp Pooled Sample Proportion of Successes
Pooled Sample Proportion of Successes
00.2
28
40351.059649.0
29
40351.059649.0
46429.072413.0z Test Statistic
P-Value = 2 (Area to the right of test statistic)
= 2 (1 – 0.9772) = 0.0456 (This can be interpreted as the probability of getting a value of
the test statistic that is at least as extreme as the one representing the sample data, assuming that 0H is
true).
Critical Value = 1.645 (Corresponding to a right-tailed test with a significance level of 0.05, when
using the z-distribution).
Decision Rule
If we use the traditional method of testing hypotheses, the decision rule would be
expressed as follows: Reject 0H if the test statistic is greater than or equal to the critical
value. In our case, since 2.00 > 1.645, we conclude that we will reject 0H .
If we use the P-Value method, the decision rule is the following: Reject 0H if the P-Value
is less than or equal to the significance level of the test. In our case, since 0.046 < 0.05,
we conclude one more time in that we will reject 0H .
8
3) Interpretation of the Results on 0H , Conclusions, and Future Work
We report that at the significance level of 0.05, there is sufficient sample evidence to
support the claim that the student passing rate in MAC 1114 is increased when CAT’s
are applied, and this result is statistically significant, according to (2).
On the other hand, there is a maximum probability of 0.05 of making a Type I Error,
which is the one that is committed when a true null hypothesis is rejected.
Based on this experiment and the subsequent statistical analysis of the results, we have
much better guidance when recommending a treatment based on the application of
CAT’s.
We are also considering extending this experiment to math preparatory courses.
REFERENCES
(1) T.A. Angelo & K.P. Cross, “Classroom Assessment Techniques: A Handbook for
College Teachers”, 2nd Edition, 1993, Jossey-Bass, A Wiley Imprint.
(2) A. Naiman, et.al., “Understanding Statistics”,
(3) L.J. Kitchens, “Exploring Statistics”,
(4) M. Triola, “Elementary Statistics”, 10th Edition, 2006, Pearson, Addison Wesley.
(5) De Veaux, et.al., “Statistics: Data and Models”,
“Incorporating Environmental Immersions to Learning Community linked courses in
Mathematics and Geography”
Dr. Jaime Bestard, Mathematics, MDC- Hialeah Campus
Dr. Arturo Rodriguez, Earth Sciences, MDC-North Campus.
Abstract:
The learning community between GEO2000 “Physical Geography” and MAC1105 “College
Algebra” during the fall semester 2008-01 at MDC Hialeah Campus linked two disciplines and
two departments from different campuses and the cooperation as a facilitator of the Earth Ethics
Institute in an instructional practice that is intended to reinforce the competencies of the two
courses by the application of the concepts in a real case scenario of principles of measuring,
analysis of variable interaction and the observation of the environment. The students performed
practical approach to scenarios explained in the classroom, including a service learning activity
in favor of the community and the motivation of the students produced a positive effect in the
academic results that in similar courses taught separately was not experienced and not expected
given the typical performance for the instructors’ courses taught independently. The differences
between the mean of responses and the percent of positive responses in the student feedback
survey as well as the differences in the grade distribution review are significant at the 0.05 level,
what supports the effectiveness of the practice of curricular link of mathematics courses to other
disciplines with quantitative reasoning objectives.
Theme: Educational Research
Key Words: Learning Community, Environmental Immersions, STEM Education
1) INTRODUCTION
There is certain trend to advise students to take the mathematics requirements very independent
from the science or related courses, due to the belief that students with poor expectations in
quantitative reasoning may perform low in a semester with another science or related course to
mathematics.
The belief that the inclusive practice of the principles of a mathematics course in a same
semester curricular arrange with another discipline which uses quantitative reasoning to
argument may produce good result, brought the idea to form the learning community of
MAC1105 “College Algebra” with GEO2000”Physical Geography” in the MDC -Hialeah
Campus fall semester 2008-01.
The instructors usually teach the courses independently and the match occurred when two
AMATYC Institute projects were developed by them.
Final Report, March 2008
Foundations for Success
National Mathematics Advisory Panel
• ……”Limitations in the ability to keep many things in mind
(working-memory) can hinder mathematics performance.
- Practice can offset this through automatic recall, which
results in less information to keep in mind and frees
attention for new aspects of material at hand.
- Learning is most effective when practice is combined with
instruction on related concepts.
- Conceptual understanding promotes transfer of learning to
new problems and better long-term retention”…..
http://www.ed.gov/MathPanel
During two years, the Math advisory Panel worked in the nation: Review of 16,000 research
studies and related documents. Public testimony gathered from 110 individuals.
Review of written commentary from 160 organizations and individuals, 12 public meetings held
around the country. Analysis of survey results from 743 Algebra teachers
There is enough scientific evidence cumulated that lead to the previous point, just consider the
following research that occurred in our country during the last three years: Foundations for
Success National Mathematics Advisory Panel.
2) Methods
This experimental study is intended to:
To increase student engagement in the instructional process by facilitating the association
of the learning outcomes in different disciplines reinforcing the common topics in
competencies.
To increase motivation of students and the acquisition of quantitative reasoning skills,
when the solution of an example in class has potential environmental, technical and social
extended impact in a real life application.
To improve the student success rate as well as the student pass rates in critical courses, by
the effect of the multidisciplinary approach to the solution of problems and cooperative
discussions.
Improve retention and enrollment in critical courses by effective multidisciplinary and
college – wide coordinate interaction.
The following factors produce interest in the experimental study:
Underprepared students in need of intensive instructional techniques, produce a demand
for motivational instructional engines.
In order to investigate avenues to better serve the students, faculty at MDC research best
practices in teaching learning strategies.
According to the MDC Mathematics Discipline Annual Report for the academic year
2007-08(Fig 1) the performance of several math courses is affected to levels of the pass
rate under 60 %.
While MAC1105 “College Algebra” is not currently the worst case scenario; it is not
showing a consistent positive trend.
Fig 1. Pass Rate(%) MDC Mathematics Discipline 2007-08 Academic Annual Report
There is certain trend to advice students not to take their mathematics requirements with
science or other math related courses, due to the belief that students with poor expectations in
quantitative reasoning may perform low in a semester with another science or related course
to mathematics.
The idea to link a math course with a science course using quantitative reasoning became
the foundation of our learning community MAC1105 “College Algebra” with
GEO2000”Physical Geography” MDC -Hialeah Campus fall 2008-01.
From this perspective, other authors studied the meta-reflective interactions in the problem
space, the cognitive demands, as follows:
Meta-reflective interactions in the problem space. Cognitive Demands and Second-Language
Learners: A Framework for Analyzing Mathematics Instructional Contexts, Campbell, et all.
Mathematical Thinking & Learning, 2007, Vol. 9 Issue 1, p3-30, 28p, 1 chart, 3 diagrams p9
The authors of this study participated in several previous projects related to Math across the
Curriculum, sponsored by the AMATYC Summer and Winter Institutes
The preparation of the exercises to extend the interaction of the quantitative reasoning
competencies to the multidisciplinary learning community was outlined and constructed
in such an institute.
The framework to produce the Learning Community was possible with the support of the
college-wide interaction, the collaborative work of the instructors and the support of the
Earth Ethics Institute.
Analysis of the competencies:
MAC1105
Competency 5: The student will demonstrate knowledge of functions from a
numerical, graphical, verbal and analytic perspective.
GEO2000
Competency 5: The student will be able to analyze the regional concept by:
a. demonstrating knowledge the area analysis tradition.
b. identifying the region’s locations, spatial extent and boundaries.
c. evaluating the factors that differentiate the regions as functional or formal.
Note: GEO 2000 Competencies 2, 3, 4 show compatibility with MAC1105 Competency
5, as well.
3) Data Analysis
3.1)The study consisted in the learning community for the MAC1105 ref # 481380 and
GEO2000 ref # 500955 as an experimental group of common students and control groups
of independent instruction of the courses in MAC1105 ref# 481373 and GEO2000 ref#
343421
The Action Plan had a delay due to the complexity of the interaction to make the learning
community with the targeted population, which made up, conveniently last Fall 2008.
The population under study was a learning community to serve dual enrollment students
from MATER Academy. The learning community includes the presence of both
instructors in both disciplines. The study group consisted in the learning community for
the MAC1105 ref # 481380 and GEO2000 ref # 500955 as an experimental group of
common students and control groups of independent instruction of the courses in
MAC1105 ref# 481373 and GEO2000 ref# 343421.
The experimental group was formed as described in the following figures 3 and 4.
It is remarkable the consistency of the targeted population with the MDC-Hialeah
Campus typical students.
Fig 3. Gender structure in targeted population
Fig 4. Ethnicity structure of the target population
3.2)Assessment of the students’ opinions
The intentionality of the activity was declared to the students who prepared their
experiment log.
The conditions of the instructional time of the activity was a key factor
The students were assessed by a pre and post survey, showing the logical change of the
population across the semester in Fig 5
0
5
10
15
20
25
30
35
Sophomore Junior Senior
Male
Female
0
10
20
30
40
50
60
70
Series1
Fig 5. Gender structure of the pre- post survey
A Pre-Post survey of the participating students’ opinions was applied in compliance with
the AMATYC, MAC3
project.
The survey includes twenty one questions related to the students’ perception of skills and
or learning outcomes. Those questions were responded in the Pre and the Post
applications, as well as ten questions related to the understanding and gains or
improvements during the course.
Also, the evaluators recorded gender, age group and ethnicity, keeping the privacy but
recording apart of their identification number that lead to matching the pre and post
surveys.
3.3)Summary of the AMATYC Survey
According to ANOVA (two factors) at 5% significance level there is no significant
difference among the questions.
There is a high significance among the responses to each question.
It implies that the questions were not weighed, while the answers obviously show the
students opinion.
3.4)Teaching strategies
During the fall replication, integrative instructional methods were applied to the students
in the combined subjects.
0
10
20
30
40
50
60
70
80
Male Female
Pre
Post
The environmental immersion consisted in a data collection of the velocity of the wind,
air humidity, temperature and position, and the observational study of the sea grass
Humidity Velocity Wind (mph)
Water Temp. (F)
Shore sand Temp. (F)
Air Temp (F)
Position in shore (ft)
Amount of life collected( units)
87 12 71 84 74 0 15
87 14 72 83 75 150 22
86 14 72 83 77 300 28
85 17 77 85 80 450 35
88 18 78 88 82 600 30
The definition of Quantitative Reasoning might not be interpreted related to a course like
GEO2000, but the reality of the situations that students find when they entered in the
environmental immersion is definitely a real application of the following definition
The application of mathematical concepts and skills to solve real-world problems. In
order to perform effectively as professionals and citizens, students must become
competent in reading and using quantitative evidence and applying basic quantitative
skills to the solution of real- life problems( 3).
3.5)Students Feedback Analysis
A comparative analysis (t – test )of the average response and the positive responses was
conducted between the experimental and the control groups
The results are significant at the level of 0.05 with p-values of 0.028 and 0.014,
respectively for the sample size of 27 students under analysis, which support that the
opinion of the students favor the application of the immersion.
A comparative analysis (t – test )of the average response and the positive responses was
conducted between the experimental and the control groups
The results are significant at the level of 0.05 with p-values of 0.028 and 0.014,
respectively for the sample size of 27 students under analysis, which support that the
opinion of the students favor the application of the immersion.
3.5) Analysis of the grades
Grade Distribution ReviewCourse Reference # Pass Rate % Success
Quotient %
Retention
Rate %
MAC1105(LC) 481380 75 78 96
GEO2000
(LC)
500955 100 100 100
MAC1105
(control)
481373 61 80 76
GEO2000
(control)
343421 92 92 100
Fig 6. Descriptive comparative Grade Distribution Review for the experimental and for
the control groups
Fig. 7 Grade Distribution
MAC1105 W/o LC GEO2000 W/o LC
MAC1105 W/ LC GEO2000 W/ LC
Ref# 481373 Ref # 343421
Ref# 481380
A 1
9
0
24
B 3
2
4
1 C 16
1
14
0
D 2
0
5
0 F 3
1
0
0
W 8
0
1
0 W/I 0
0
1
0
The differences are significant even between categories of grades, at the 0.05 level
The display of the general grade distribution offers the effect of the immersion over the learning
in the subject compared versus the case in which the subject is taught separately and without
immersions.
4) Concluding Remarks:
The study shows the influence of the instructional effective interaction over the
quantitative reasoning skills, and this effect can be measured from the both sides of the
learning process as follows:
IMPACT ON STUDENTS
Decrease math anxiety
Positive learning attitude
Recognition of math in daily life
Lifelong appreciation of math
Positive experience in math courses
Perception that math is relevant
Development of quantitative reasoning
IMPACT ON INSTRUCTORS
Collaboration- cooperation
Enhance of departmental campus vision
Improve teaching strategies.
Generalization of conclusions
RECOMMENDATIONS:
To introduce the Method of Test Item analysis to consolidate the results of the opinion
surveys
To replicate the study involving a sample in the IAC to extend the results with double
interactions of MAC1105 and MAT1033 together with GEO 2000 and PSC1515
To continue incorporating the Earth Ethics Institute environmental immersions to courses
with potential risk in motivation of students
REFERENCES:
1. Bestard, Rodriguez: “Course syllabi for MAC1105, GEO2000 MDC-Hialeah Campus, Fall
2008.
2. Bestard, J: “The T-link Project” AMATYC Summer Institute, Leavenworth, WA, Aug,
2006.
3. Diefenderfer, et all: “Interdisciplinary Quantitative Reasoning” Hollins University, VA,
2000.
4. Rodriguez, Bestard: “The Global Warming awareness and the environmental math
instruction at the MDC-Hialeah Campus. AMATYC Winter Institute, Miami Beach, Fl
January 2007)
Using Beta-binomial Distribution in Analyzing Some Multiple-Choice
Questions of the Final Exam of a Math Course, and its Application in
Predicting the Performance of Future Students*
Dr. Mohammad Shakil
Department of Mathematics
Miami-Dade College
Hialeah Campus
1780 West 49
th
St., Hialeah 33012
E-mail: [email protected]
Abstract
Creating valid and reliable classroom tests are very important to an instructor for assessing student
performance, achievement and success in the class. The same principle applies to the State Exit and
Classroom Exams conducted by the instructors, state and other agencies. One powerful technique
available to the instructors for the guidance and improvement of instruction is the test item analysis. This
paper discusses the use of the beta-binomial distribution in analyzing some multiple-choice questions of
the final exam of a math course, and its application in predicting the performance of future students. It is
hoped that the finding of this paper will be useful for practitioners in various fields.
Key words: Binomial distribution; Beta distribution; Beta-binomial distribution; Goodness-of-fit;
Predictive beta-binomial probabilities; Test item analysis.
2000 Mathematics Subject Classification: 97C30, 97C40, 97C80, 97C90, 97D40
*Part of this paper was presented on Conference Day, MDC, Kendall Campus, March 05,
2009.
1
1. Introduction: Creating valid and reliable classroom tests are very important to an instructor for
assessing student performance, achievement and success in the class. One powerful technique available to
the instructors for the guidance and improvement of instruction is the test item analysis. If the probability
of success parameter, p, of a Binomial distribution has a beta distribution with shape parameters α > 0 and
β > 0, the resulting distribution is known as a beta binomial distribution. For a binomial distribution, p is
assumed to be fixed for successive trials. For the beta-binomial distribution, the value of p changes for
each trial. Many researchers have contributed to the theory of beta binomial distribution and its
applications in various fields, among them Pearson (1925), Skellam (1948), Lord (1965), Greene (1970),
Massy et. al. (1970), Griffiths (1973), Williams (1975), Huynh (1979), Wilcox (1979), Smith (1983), Lee
and Sabavala (1987), Hughes and Madden (1993), and Shuckers (2003), are notable. Since creating valid
and reliable classroom tests are very important to an instructor for assessing student performance,
achievement and success in the class, this paper discusses the use of the beta-binomial distribution in
analyzing some multiple-choice questions of the final exam of a math course, and its application in
predicting the performance of future students. It is hoped that the finding of this paper will be useful for
practitioners in various fields. The organization of this paper is as follows. Section 2 discusses some well
known distributions, namely, binomial and beta. The beta-binomial distribution is discussed in Section 3.
In Section 4, the beta-binomial distribution is used to analyze multiple-choice questions in a Math Final
Exam, with application in predicting the performance of future students. Using beta-binomial distribution,
a diagnosis of some failure questions in the said exam is provided in Section 5. Some concluding remarks
are presented in Section 6.
2. An Overview of Binomial and Beta Distributions: This section discusses some well known
distributions, namely, binomial and beta.
2.1 Binomial Distribution: The binomial distribution is used when there are exactly two mutually
exclusive outcomes of a trial. These outcomes are often called successes and failures. The binomial
probability distribution is the probability of obtaining x successes in n trials. It has the following
probability mass function:
,10,...,,1,0,1,;
pnxpp
x
n
npxb
xn
x
(1)
where p is the probability of a success on a single trial,
x
n
is the combinatorial function of n things
taken x at a time, and the mean and standard deviation are
pn
and
ppn 1
respectively.
For example, the following Figure 1 depicts the binomial probabilities of x successes in n = 30 trials,
when p = 0.25 is the probability of a success on a single trial.
2
Figure 1: The PDF of a Binomial Distribution: n = 30 trials, and p = 0.25
2.2 Beta Distribution: The Beta distribution is a continuous distribution on the interval [0, 1],
with shape parameters α > 0 and β > 0. Letting p have a Beta distribution, its probability density
function is given by:
,0,0,10,
),(
)1(
),|(
11
p
B
pp
pf (2)
where
)(
)()(
),(
B denotes the complete beta function. Taking values on the interval
[0,1], the distribution is unimodal if α > and β > 1. If both α and β are 1, then the beta
distribution is equivalent to the continuous uniform distribution on that interval. If only one of
these parameters are less than 1, then the distribution Is J-shaped or reverse J-shaped. If both are
less than 1, the distribution is U-shaped. The effects of various values of α and β on the shapes of
the Beta distribution are given in the following Figure 2. The mean and the variance for a Beta
random variable are given by
, and
1
2
, respectively.
3
p
Figure 2: The PDF of Beta Distribution for Various Values of α and β (Note that A =
α and B = β in the Figure) (Source: http://www.itl.nist.gov/).
3. The Beta-Binomial Distribution: This section discusses the beta-binomial distribution.
For the sake of completeness, the beta-binomial distribution is derived. If the probability of success
parameter, p, of a Binomial distribution has a beta distribution with shape parameters α > 0 and β
> 0, the resulting distribution is known as a beta binomial distribution. For a binomial
distribution, p is assumed to be fixed for successive trials. For the beta-binomial distribution, the
value of p changes for each trial. Suppose a continuous random variable Y has a distribution with
parameter θ and pdf
yg
. Let
h
be the prior pdf of
. Then the distribution associated
with the marginal pdf of Y, that is,
dyghyk
1
,
is called the predictive distribution because it provides the best description of the probability on
Y. Accordingly, by Bayes’ theorem, the conditional (that is, the posterior) pdf
yk
of
,
given Y = y, is given by:
yk
hyg
yk
1
.
Note that the above formula can easily be generalized to more than one random variable. For a
4
nice discussion, please visit Hogg, et al. (2005). In what follows, using Bayes’ Rule, the
derivation of beta-binomial distribution is given. For details, see, for example, Schuckers (2003),
Lee (2004), and Hogg, et al. (2005, 2006), among others.
3.1 Derivation of Beta-Binomial Distribution: Suppose that there are m individual Test items
and each of those individual test items is tested n times. Let
iii
pnBinpnX ,~,
, where X
i
is the
number of successes, and
nipp
x
n
xXP
ii
xn
i
x
i
i
i
,...,3,2,1,)1()(
. (3)
Supposing the prior pdf of each of the parameter p
i
in equation (3) to be the beta pdf (2), the joint
pdf is given by:
,
)1(
),|(),|(),,|,(
11
1
B
pp
x
n
pfnpxfnpxf
ii
xn
i
x
i
m
i
i
, (4)
where
T
m
pppp ,,,
21
, and
T
m
xxxx ,,,
21
. It is evident from equation (4) that, in
drawing inference from the beta-binomial probability model, the selection of the parameters α
and β is crucial, since they define the overall probability of success. Thus integrating the
equation (4), a joint Beta-binomial distribution or product Beta-binomial distribution is obtained
as follows:
pdpfnpxfpdnpxfnxf ),|(),|(),,|,(),,|(
.,...2,1,0,
),(
),(
1
nx
B
xnxB
x
n
i
ii
m
i
i
(5)
The equation (5), denoted as
nBetabinnX
i
,,~,,
, is called the predictive distribution
because it provides the best description of the probabilities on
m
XXX ,,,
21
. Taking m = 1 in
equations (4) and (5), we easily get the following:
(i) The Predictive Beta-Binomial Distribution
.,...,2,1,0,
),(
),(
)(
1
nx
B
xnxB
x
n
xk
5
(ii) The Posterior of the Binomial Distribution with Beta Priors:
nxp
xnxB
pp
xpk
xnx
,...,2,1,0,10,
),(
)1(
11
,
which is a beta pdf with parameters
x
, and
xn
. Clearly prior is conjugate since both
posterior and prior belong to the same class of distributions (that is, beta).
3.2 Mean and Variance: The equation (5), denoted as,
nBetabinnX
i
,,~,,
, is called
the predictive distribution because it provides the best description of the probabilities on
m
XXX ,,,
21
. Its mean and variance are given by
n
n
XE
i
, and
Cn
nn
XVar
i
1
1
2
respectively, where
, and
1
n
C
. The beta-binomial distribution is also known as an extravariation model,
because it allows for greater variability among the x
i
's, than the binomial distribution. The
additional term, C, allows for additional variability beyond the
1
that is found under the
binomial model. Note that the variance of a binomial random variable is ppn 1 .
4. Beta-Binomial Distribution Analysis of the Mathematics Exam Questions: Using the beta-
binomial distribution, this section analyzes the multiple-choice questions of the final exam of a
math course taught by me during the Fall 2007-1 term. For analysis, the data obtained from the
ParSCORETM Item Analysis Report of the exam under question has been considered. The
ParSCORETM item analysis consists of three types of reports, that is, a summary of test
statistics, a test frequency table, and item statistics. The test statistics summary and frequency
table describe the distribution of test scores. The item analysis statistics evaluate class-wide
performance on each test item. Some useful item analysis statistics are following. For the sake of
completeness, the details of these are provided in the Appendix A.
Item Difficulty
Item Discrimination
Distractor Analysis
Reliability
The test item statistics of the considered math final exam – version A are summarized in the
following Tables 1 and 2. It consisted of 30 items. A group of 7 students took this version A of
the test. Another group of 7 students took the version B of the test. It appears from these
statistical analyses that a large value of KR-20 = 0.90 for version B indicates its high reliability
in comparison to version A, which is also substantiated by large positive values of Mean DI =
0.450 > 0.30 and Mean Pt. Bisr. = 04223, small value of standard error of measurement (that is,
SEM = 1.82), and an ideal value of mean (that is, µ = 19.57 > 18, the passing score) for version
B. For details on these, see Shakil (2008).
6
Table 1: Test Item Statistics of the Math Final Exam – Version A
Table 2: A Comparison of Exam Test Item Statistics – Versions A & B
4.1 Goodness-of-Fit of Binomial and Predictive Beta-Binomial Distributions: Maple 11 has
been used for computing the data moments, estimating the parameter (by employing the method
of moments), and chi-square test for goodness-of-fit. The data moments are computed as:
5714267.0
1
and
421756.0
2
. The observed, expected binomial and expected predictive
beta-binomial frequencies of the performance of the questions (that is, successful questions) in
7
the considered math exam (version A) data have been provided in the following Table 3, along
with a plot of the corresponding histogram given in the Figure 3.
Table 3: Observed and Expected Binomial and Predictive Beta-Binomial Frequencies
x Obs Bin BBD
0 1 7.97E-02 3.212274
1 5 0.743555 3.026111
2 2 2.974236 3.037186
3 4 6.609452 3.138957
4 6 8.812655 3.330805
5 2 7.050165 3.661291
6 5 3.133425 4.301067
7 5 0.596846 6.29231
Figure 3: Frequency Distributions of Successful Math Exam Questions
8
The estimation of the parameters and chi-square goodness-of-fit test are provided in Tables 4 and
5 respectively.
Table 4: Parameter Estimates of the Binomial and Predictive Beta-Binomial Models for the
Success of the Math Exam Question (Version A) Data
Model
Parameter Binomial Predictive
Beta-Binomial
p
0.57143
0.8981194603
0.6735948342
Table 5: Comparison Criteria (Chi-Square Test for Goodness-of-Fit)
Model
Binomial Predictive
Beta-Binomial
Test Statistic 74.458 6.673302289
Critical Value 14.06714058 14.06714058
p-value 0 0.4636692440
From the chi-square goodness-of-fit test, we observed that the Predictive Beta-Binomial Model
fits the Successes of the considered Math Exam Questions Data (Version A) reasonably well.
The Predictive Beta-Binomial Model produces the highest p-value and therefore fitted better than
Binomial distributions. Also, from the Histograms for the Observed, Expected Binomial and
Expected Predictive Beta-Binomial Frequencies of Successful Math Exam Questions Data
(Version A) plotted, for the parameters estimated in Table 4, as given in the Figures 4 – 6 below,
we observed that the Predictive Beta-Binomial Model fits the Successes of the considered Math
Exam Questions Data (Version A) reasonably well.
9
Figure 4: Binomial Probabilities of k Successes per Question
Figure 5: Fitting the Beta Posterior Distribution PDF to the Considered Math Exam Questions–
Ver. A
Figure 6: Comparison of Prior and Posterior Beta Distributions
10
4.2 Data Analysis: This section discusses various data analysis of the considered Math Exam Question
Success Data, which are presented in the following Tables 6, 7 and 8. The computations are done by
using Maple 11 and R Software.
Table 6: Summary Statistics of Different Posterior Beta-Binomial
Distributions for the Considered Math Exam Question Success Data (Version A)
No.
of
Questions
No. of Trials
(Students)
Per
Question
No. of
Successes
Per
Question
Point
Estimate
of Prior
Beta
Mean
Point
Estimate
of
Posterior
Beta-
Binomial
Mean
Posterior
Beta-
Binomial
Median
Posterior
Beta-Binomial
Variance
90 %
C. I. Estimate
of Posterior
Beta-
Binomial
Mean
m
n
k
beta
~
binombeta
~
30 7 0 0.5714267 0.1047771 0.07506755 0.009799589202 (0.004539054,
0.306892967)
1 0.5714267 0.2214399 0.19921830 0.01801184815 (0.04340072,
0.47599671)
2 0.5714267 0.3381027 0.32500740 0.02338026884
(0.1100912,
0.6112220)
3 0.5714267 0.4547654 0.45109090 0.02590485125
(0.1958725,
0.7263161)
4 0.5714267 0.5714282 0.57722770 0.02558559538
(0.2980277,
0.8248512)
5 0.5714267 0.6880910 0.70327590 0.02242250125
(0.4171535,
0.9067242)
6 0.5714267 0.8047538 0.82891440 0.01641556884
(0.558081,
0.968254)
7 0.5714267 0.9214166 0.95171830 0.007564798163
(0.7406081,
0.9986905)
11
Table 7: Predictive Beta-Binomial Probabilities for Future (Simulated) Sample of n = 20
Students who take the Same Math Exam (Version A)
No. of
Questions
No. of Future
Sample of Trials
(Students) Per
Question
No. of Successes in the
Previous Sample of n =
7
Most Likely No.
of Successes in
the Future
Sample of n = 20
Predictive Beta-
Binomial
Probabilities
m n
k
~
k
30 20 0 0 0.3146720
1 0.2119047
2 0.1483370
3 0.1048905
1 1 0.1164135
2 0.1299005
3 0.1283373
4 0.1178286
5 0.1026085
2 5 0.1023370551
6 0.1027100402
3 8 0.0938969431
9 0.0950385713
10 0.0918930055
4 11 0.0940656041
12 0.0960800571
13 0.0936068326
5 14 0.1029871
15 0.1068208
16 0.1045330
6 16 0.1144824
17 0.1319770
18 0.1430934
19 0.1402707
20 0.1085313
7 18 0.1301244
19 0.2119595
20 0.4232004
12
Table 8: Predictive Beta-Binomial Probability of At Least 18 Successes out of Future
(Simulated) Sample of n = 20 Students who take the Same Math Exam (Version A)
No. of
Questions
No. of Future Sample
of Trials (Students)
Per Question
No. of Successes in
the Previous Sample
of n = 7
Predictive Beta-Binomial
Probability of at least 18
successes out of Future
Sample of n = 20
m n
k
18k
30 20 0 0.00001583399
1 0.0002678766
2 0.002198643
3 0.01199997
4 0.0487983
5 0.1553201
6 0.3918954
7 0.7652843
5. Diagnosis of Failure Questions of the Said Math Exam and Some Recommendation:
Using the Predictive Beta-Binomial Probabilities, this section discusses the diagnosis of some failure
questions of the considered Math Exam - Version A, that is, having Success Rate < 60 %. Some
recommendations are also given based on this analysis.
(i) Number of Failure Questions (that is, Questions having Success Rate < 60 %) is 18n .
(ii) Number of Failure Questions with Point Biserial 46.0
pbis
r is 3k .
(iii) Suppose p denotes the probability of failure in the 18 failure items due to the
0
pbis
r
or low
positive value of
pbis
r
or poor construction of exam questions.
(iv) After analyzing the said Math Exam Questions (Version A) Item Analysis Data, it is found that
the number of failure questions out of the total number of exam questions 30m
,
having the
Point Biserial 46.0
pbis
r , is given by 3k . Thus, about 10 % (that is, 10.0p ) failure of
the total number of exam questions 30m
in Version A have Point Biserial 46.0
pbis
r .
Since
10 p
, letting
p
have a Beta distribution with shape parameters
0,0
, we
have
10.0
. (6)
Further, if we consider the first two failure questions in the above 18 Failure Exam
Questions and consider that one of these two failure
questions has Point Biserial 46.0
pbis
r
,
then the probability of the second exam
13
question failure, having Point Biserial
46.0
pbis
r
, is increased to about 90 % (that is,
90.0p ). Consequently, using the following formula for the posterior mean of the beta-
binomial distribution
nk
n
k
p
betabinom
,,3,2,1,
~
,
the posterior estimate of p with one failure after one trial is given by
90.0
1
1
. (7)
Solving the equations (6) and (7) by using Maple 11, the values of the parameters
and
are obtained as follows:
.1125000000.0,00125000000.0 and
(v) Now, updating the posterior probabilities
with 3 successes out of 3 trials, 4 successes out of 4
trials, and so on, in the remaining 15 failure items due to the
0
pbis
r
or low positive value of
pbis
r
or poor construction of exam questions, the posterior estimates of p are provided in the
following Table 9:
Table 9: Diagnosis of the Failure Questions in the said
Math Exam (Version A) using the Predictive Beta-Binomial Probabilities
Number of Trials
Posterior Beta-Binomial Estimate of
p
k
17,,3,
~
k
k
k
p
betabinom
3 0.9640000000
4 0.9727272727
5 0.9780487805
6 0.9816326531
7 0.9842105263
8 0.9861538462
9 0.9876712329
10 0.9888888889
11 0.9898876404
12 0.9907216495
13 0.9914285714
14 0.9920353982
15 0.9925619835
16 0.9930232558
17 0.9934306569
14
(vi) Recommendation: Using the updated posterior probabilities from the above Table F after the
3rd failure, then the 4th one, and so on, we have the following product:
8060295953.0
1125000000.000125000000.0
00125000000.0
17
3
17
3
kk
k
k
k
k
.
Thus, it is observed from the above analysis, the probability that all the remaining failure
questions having poor Point Biserial is about 80.60 %, which, I believe, is the needed value in
making our decision to revise the considered Math Exam Questions (Version A).
6. Concluding Remarks: This paper discusses the beta-binomial distribution, the use of the beta
binomial distribution to analyze questions of the state exit exams, and create valid and reliable classroom
tests. This paper discusses the use of the beta-binomial distribution to assess students` performance based
on questions in the state exit exams. It is hoped that the present study would be helpful in recognizing the
most critical pieces of the state exit test items data, and evaluating whether or not the test item needs
revision by taking different sample data for the considered math exam and applying the said technique to
analyze these data. The methods discussed in this project can be used to describe the relevance of test
item analysis to classroom tests. These procedures can also be used or modified to measure, describe and
improve tests or surveys such as college mathematics placement exams (that is, CPT), mathematics study
skills, attitude survey, test anxiety, information literacy, other general education learning outcomes, etc. It
is hoped that the finding of this paper will be useful for practitioners in various fields.
References
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Karian, Z. A., and Tanis, E. A. (1999). Probability and Statistics-Explorations with MAPLE. 3rd Edition,
Prentice- Hall, USA.
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Lee, J. C., and Sabavala, D. J. (1987). Bayesian Estimation and Prediction for the Beta-Binomial Model.
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523-529.
Shakil, M. (2008). Assessing Student Performance Using Test Item Analysis and its Relevance to the
State Exit Final Exams of MAT0024 Classes – An Action Research Project, Polygon, Vol. II,
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Appendix A
Review of Some Useful Item Analysis Statistics: An item analysis involves many statistics that can
provide useful information for determining the validity and improving the quality and accuracy of
multiple-choice or true/false items. These statistics are used to measure the ability levels of examinees
from their responses to each item. The ParSCORE
TM
item analysis generated by Miami Dade College –
Hialeah Campus Reading Lab when a Multiple-Choice Exam is machine scored consists of three types of
reports, that is, a summary of test statistics, a test frequency table, and item statistics. The test statistics
summary and frequency table describe the distribution of test scores. The item analysis statistics evaluate
class-wide performance on each test item. The ParSCORE
TM
report on item analysis statistics gives an
overall view of the test results and evaluates each test item, which are also useful in comparing the item
analysis for different test forms. In what follows, descriptions of some useful, common item analysis
statistics, that is, item difficulty, item discrimination, distractor analysis, and reliability, are presented
16
below. For the sake of completeness, definitions of some test statistics as reported in the ParSCORE
TM
analysis are also provided.
(I) Item Difficulty: Item difficulty is a measure of the difficulty of an item. For items (that is, multiple-
choice questions) with one correct alternative worth a single point, the item difficulty (also known as the
item difficulty index, or the difficulty level index, or the difficulty factor, or the item facility index, or the
item easiness index, or the
p
-value) is defined as the proportion of respondents (examinees) selecting the
answer to the item correctly, and is given by
n
c
p
where
p
the difficulty factor,
c
the number of respondents selecting the correct answer to an item,
and
n
total number of respondents. Item difficulty is relevant for determining whether students have
learned the concept being tested. It also plays an important role in the ability of an item to discriminate
between students who know the tested material and those who do not. Note that
(i) 10 p .
(ii) A higher value of
p
indicate low difficulty level index, that is, the item is easy. A lower
value of
p
indicate high difficulty level index, that is, the item is difficult. In general, an
ideal test should have an overall item difficulty of around 0.5; however it is acceptable
for individual items to have higher or lower facility (ranging from 0.2 to 0.8). In a
criterion-referenced test (CRT), with emphasis on mastery-testing of the topics covered,
the optimal value of
p
for many items is expected to be 0.90 or above. On the other
hand, in a norm-referenced test (NRT), with emphasis on discriminating between
different levels of achievement, it is given by 50.0p .
(iii) To maximize item discrimination, ideal (or moderate or desirable) item difficulty level,
denoted as
M
p
, is defined as a point midway between the probability of success, denoted
as
S
p
, of answering the multiple - choice item correctly (that is, 1.00 divided by the
number of choices) and a perfect score (that is, 1.00) for the item, and is given by
2
1
S
SM
p
pp
.
(iv) Thus, using the above formula in (iv), ideal (or moderate or desirable) item difficulty
levels for multiple-choice items can be easily calculated, which are provided in the
following table.
17
Number of Alternatives Probability of Success
(
S
p
)
Ideal Item Difficulty Level
(
M
p
)
2 0.50 0.75
3 0.33 0.67
4 0.25 0.63
5 0.20 0.60
(Ia) Mean Item Difficulty (or Mean Item Easiness): Mean item difficulty is the average of difficulty
easiness of all test items. It is an overall measure of the test difficulty and ideally ranges between 60 %
and 80 % (that is, 80.060.0 p ) for classroom achievement tests. Lower numbers indicate a difficult
test while higher numbers indicate an easy test.
(II) Item Discrimination: The item discrimination (or the item discrimination index) is a basic measure
of the validity of an item. It is defined as the discriminating power or the degree of an item's ability to
discriminate (or differentiate) between high achievers (that is, those who scored high on the total test) and
low achievers (that is, those who scored low), which are determined on the same criterion, that is, (1)
internal criterion, for example, test itself; and (2) external criterion, for example, intelligence test or other
achievement test. Further, the computation of the item discrimination index assumes that the distribution
of test scores is normal and that there is a normal distribution underlying the right or wrong dichotomy of
a student’s performance on an item. There are several ways to compute the item discrimination, but, as
shown on the ParSCORE
TM
item analysis report and also as reported in the literature, the following
formulas are most commonly used indicators of item’s discrimination effectiveness.
(a) Item Discrimination Index (or Item Discriminating Power, or D -Statistics), D : Let the students’
test scores be rank-ordered from lowest to highest. Let
groupupperinstudentsofNumberTotal
correctlyitemtheansweringgroupupperinstudentsofNo
p
U
%30%25
%30%25.
,
and
grouplowerinstudentsofNumberTotal
correctlyitemtheansweringgrouplowerinstudentsofNo
p
L
%30%25
%30%25.
18
The ParSCORE
TM
item analysis report considers the upper
%27
and the lower
%27
as the analysis
groups. The item discrimination index, D , is given by
LU
ppD
.
Note that
(i)
11 D
.
(ii) Items with positive values of
D
are known as positively discriminating items, and those
with negative values of D are known as negatively discriminating items.
(iii) If
0D
, that is,
LU
pp
, there is no discrimination between the upper and lower
groups.
(iv) If
00.1D
, that is,
000.1
LU
pandp
, there is a perfect discrimination between
the two groups.
(v) If 00.1D , that is, 00.10
LU
pandp , it means that all members of the lower
group answered the item correctly and all members of the upper group answered the item
incorrectly. This indicates the invalidity of the item, that is, the item has been miskeyed
and needs to be rewritten or eliminated.
(vi) A guideline for the value of an item discrimination index is provided in the following
table.
(vii)
Item Discrimination Index,
D
Quality of an Item
50.0D
Very Good Item; Definitely Retain
49.040.0 D
Good Item; Very Usable
39.030.0 D
Fair Quality; Usable Item
29.020.0 D
Potentially Poor Item; Consider Revising
20.0D
Potentially Very Poor;
Possibly Revise Substantially, or Discard
(b) Mean Item Discrimination Index,
D
:
This is the average discrimination index for all test items combined. A large positive value (above 0.30)
indicates good discrimination between the upper and lower scoring students. Tests that do not
discriminate well are generally not very reliable and should be reviewed.
19
(c) Point-Biserial Correlation (or Item-Total Correlation or Item Discrimination) Coefficient,
pbis
r :
The point-biserial correlation coefficient is another item discrimination index of assessing the usefulness
(or validity) of an item as a measure of individual differences in knowledge, skill, ability, attitude, or
personality characteristic. It is defined as the correlation between the student performance on an item
(correct or incorrect) and overall test-score, and is given by either of the following two equations (which
are mathematically equivalent).
(a)
q
p
s
XX
r
TC
pbis
,
where
pbis
r
the point-biserial correlation coefficient;
C
X
the mean total score for examinees who
have answered the item correctly;
T
X
the mean total score for all examines;
p
the difficulty value
of the item; pq 1 ; and
s
the standard deviation of total exam scores.
(b)
qp
s
mm
r
qp
pbis
,
where
pbis
r
the point-biserial correlation coefficient;
p
m
the mean total score for examinees who
have answered the item correctly;
q
m
the mean total score for examinees who have answered the item
incorrectly;
p
the difficulty value of the item;
pq 1
; and
s
the standard deviation of total
exam scores.
Note that
(i) The interpretation of the point-biserial correlation coefficient,
pbis
r
, is same as that of the
D
-statistic.
(ii) It assumes that the distribution of test scores is normal and that there is a normal
distribution underlying the right or wrong dichotomy of a student performance on an
item.
(iii) It is mathematically equivalent to the Pearson (product moment) correlation coefficient,
which can be shown by assigning two distinct numerical values to the dichotomous
variable (test item), that is, incorrect = 0 and correct = 1.
(iv)
11
pbis
r
.
(v)
0
pbis
r
means little correlation between the score on the item and the score on the test.
20
(vi) A high positive value of
pbis
r indicates that the examinees who answered the item
correctly also received higher scores on the test than those examinees who answered the
item incorrectly.
(viii) A negative value indicates that the examinees who answered the item correctly received
low scores on the test and those examinees who answered the item incorrectly did better
on the test. It is advisable that an item with 0
pbis
r or with large negative value of
pbis
r
should be eliminated or revised. Also, an item with low positive value of
pbis
r
should be
revised for improvement.
(ix) Generally, the value of
pbis
r
for an item may be put into two categories as provided in the
following table.
Point-Biserial Correlation Coefficient,
pbis
r
Quality
30.0
pbis
r
Acceptable Range
1
pbis
r
Ideal Value
(x) The statistical significance of the point-biserial correlation coefficient,
pbis
r
, may be
determined by applying the Student’s
t
test.
Remark: It should be noted that the use of point-biserial correlation coefficient,
pbis
r
, is more
advantageous than that of item discrimination index statistics, D , because every student taking the test is
taken into consideration in the computation of
pbis
r
, whereas only 54 % of test-takers passing each item
in both groups (that is, the upper 27 % + the lower 27 %) are used to compute
D
.
(d) Mean Item-Total Correlation Coefficient,
pbis
r
: It is defined as the average correlation of all the
test items with the total score. It is a measure of overall test discrimination. A large positive value
indicates good discrimination between students.
(III) Internal Consistency Reliability Coefficient (Kuder-Richardson 20,
20
KR
, Reliability
Estimate): The statistic that measures the test reliability of inter-item consistency, that is, how well the
test items are correlated with one another, is called the internal consistency reliability coefficient of the
test. For a test, having multiple-choice items that are scored correct or incorrect, and that is administered
only once, the Kuder-Richardson formula 20 (also known as KR-20) is used to measure the internal
consistency reliability of the test scores. The KR-20 is also reported in the ParSCORE
TM
item analysis. It
is given by the following formula:
21
1
2
1
2
20
ns
qpsn
KR
n
i
ii
where
20
KR
= the reliability index for the total test;
n
= the number of items in the test;
2
s
= the
variance of test scores;
i
p
= the difficulty value of the item; and
ii
pq 1
.
Note that
(i)
0.10.0
20
KR
.
(ii)
0
20
KR
indicates a weaker relationship between test items, that is, the overall test
score is less reliable. A large value of
20
KR
indicates high reliability.
(iii) Generally, the value of
20
KR
for an item may be put into the following categories as
provided in the table below.
20
KR
Quality
60.0
20
KR
Acceptable Range
75.0
20
KR
Desirable
85.080.0
20
KR
Better t
1
20
KR
Ideal Value
(iv) Remarks: The reliability of a test can be improved as follows:
a) By increasing the number of items in the test for which the following
Spearman-Brown prophecy formula is used.
rn
rn
r
est
11
where
est
r
= the estimated new reliability coefficient; r = the
original
20
KR
reliability coefficient;
n
= the number of times the
test is lengthened.
22
b) Or, using the items that have high discrimination values in the test.
c) Or, performing an item-total statistic analysis as described above.
(IV) Standard Error of Measurement (
m
SE ): It is another important component of test item analysis to
measure the internal consistency reliability of a test. It is given by the following formula:
20
1 KRsSE
m
,
0.10.0
20
KR
,
where
m
SE
= the standard error of measurement;
s
= the standard deviation of test scores; and
20
KR
=
the reliability coefficient for the total test.
Note that
(i)
0
m
SE
, when
1
20
KR
.
(ii)
1
m
SE
, when
0
20
KR
.
(iii) A small value of
m
SE
(e.g.,
3
) indicates high reliability; whereas a large value of
m
SE
indicates low reliability.
(iv) Remark: Higher reliability coefficient (i.e.,
1
20
KR
) and smaller standard deviation
for a test indicate smaller standard error of measurement. This is considered to be more
desirable situation for classroom tests.
(v) Test Item Distractor Analysis: It is an important and useful component of test item analysis. A test
item distractor is defined as the incorrect response options in a multiple-choice test item. According to the
research, there is a relationship between the quality of the distractors in a test item and the student
performance on the test item, which also affect the student performance on his/her total test score. The
performance of these incorrect item response options can be determined through the test item distractor
analysis frequency table which contains the frequency, or number of students, that selected each incorrect
option. The test item distractor analysis is also provided in the ParSCORE
TM
item analysis report. A
general guideline for the item distractor analysis is provided in the following table:
Item Response
Options
Item Difficulty
p
Item Discrimination Index
D
or
pbis
r
Correct Response
85.035.0 p
(Better)
30.0D
or
30.0
pbis
r
(Better)
Distractors
02.0p
(Better)
0D
or
0
pbis
r
(Better)
23
(v) Mean: The mean is a measure of central tendency and gives the average test score of a sample of
respondents (examinees), and is given by
n
x
x
n
i
i
1
,
where
scoretestindividualx
i
,
scoretestindividualx
i
,
srespondentofnon .
.
(vi) Median: If all scores are ranked from lowest to highest, the median is the middle score. Half of the
scores will be lower than the median. The median is also known as the 50th percentile or the 2nd quartile.
(vii) Range of Scores: It is defined as the difference of the highest and lowest test scores. The range is a
basic measure of variability.
(viii) Standard Deviation: For a sample of
n
examinees, the standard deviation, denoted by
s
, of test
scores is given by the following equation
1
1
2
n
xx
s
n
i
i
,
where scoretestindividualx
i
and
scoretestaveragex
. The standard deviation is a measure of
variability or the spread of the score distribution. It measures how far the scores deviate from the mean. If
the scores are grouped closely together, the test will have a small standard deviation. A test with a large
value of the standard deviation is considered better in discriminating the student performance levels.
(ix) Variance: For a sample of
n
examinees, the variance, denoted by
2
s
, of test scores is defined as the
square of the standard deviation, and is given by the following equation
1
1
2
2
n
xx
s
n
i
i
.
(x) Skewness: For a sample of
n
examinees, the skewness, denoted by
3
, of the distribution of the test
scores is given by the following equation
24
n
i
i
s
xx
nn
n
1
3
3
21
,
where
scoretestindividualx
i
,
scoretestaveragex
and
scorestestofdeviationdardss tan . It measures the lack of symmetry of the distribution. The
skewness is 0 for symmetric distribution and is negative or positive depending on whether the
distribution is negatively skewed (has a longer left tail) or positively skewed (has a longer right tail).
(xi) Kurtosis: For a sample of
n
examinees, the kurtosis, denoted by
4
, of the distribution of the test
scores is given by the following equation
32
13
321
1
2
1
4
4
nn
n
s
xx
nnn
nn
n
i
i
,
where
scoretestindividualx
i
,
scoretestaveragex
, and
scorestestofdeviationdardss tan
. It measures the tail-heaviness (the amount of probability in the
tails). For the normal distribution,
3
4
. Thus, depending on whether
33
4
or
, a distribution is
heavier tailed or lighter tailed.
1
The Magic Math
Written by David Tseng
Graphed by Nancy Liu
Abstract: The content of this article provides quick and easy methods to make calculations
involving real numbers. These tricks involve the Chinese chopsticks method of multiplication,
the multiplication table of 9, the squares of 5, multiplication without calculator, and the
Pythagorean Theorem. These tricks should be very interesting for students taking mathematic
classes such as MAT0002 (basic arithmetics), MAT0020 (elementary algebra), MAT1033
(intermediate algebra), MAC1114 (trigonometry), and MAC1147 (pre-calculus), as they reveal
to be fast, reliable and time saving.
Introduction: In order to inspire students that math is not boring but instead can be
very interesting, this article demonstrates few examples to show how math can be
beautiful and intellectually challenging.
Chinese chopsticks method for the multiplication of numbers without a table of
multiplication
Once upon a time there was a Chinese farmer who never went to school; therefore he did not
learn how use his abacus. To sell his goods in the market, he used basic counting of chopsticks.
To help him know how much 23 x 12 is, we are going to use Chinese chopsticks to solve the
multiplication
1) Regular case
23 x 12 = ?
In this method, each number forms a group, and each group contains a number of chopsticks
equal to the number of the group they represent. Let’s call group one A
1
= 2 chopsticks, group
two A
2
= 3 chopsticks, group three A
3
= 1chopstick and finally group four A
4
= 2 chopsticks.
Each group is arranged in the following order: A
1
in the top of the page, A
2
in the bottom, A
3
to
the left, and A
4
to the right. Each group has to be disposed in a way that the chopsticks from
group A
1
and A
2
can be distinguished to be from two different groups; A
3
and A
4
should be
placed over A
1
and A
2
at a 90
o
angle, but also clearly distinguishable. Because of the way they
are placed, A
3
and A
4
must have points of intersection with A
1
and A
2
; the entire method relies
on these intersection points, and throughout this article, we will designate them with the +
symbol.
A
1
+A
3
is in the top left corner, A
1
+ A
4
in the top right corner, A
2
+ A
3
are in the bottom left
corner and A
2
+ A
4
in the bottom right corner. Each intersection has several points of contacts.
For example if A
1
and A
4
are put one on top of each other at a 90 degrees angle, they will have
2
four points of contact, as A
1
has 2 chopsticks and A
4
has also 2 chopsticks. In that same spirit, A
1
+A
3
have 2 points of contact, A
2
+ A
3
have 3 and A
2
+ A
4
have 6. However, each intersection is
not counted as a single entity. A
2
+ A
3
and A
1
+ A
4
, which are in the same diagonal, must be
counted together; their total amount of intersection points will thus be 7.
We finally have the result for our multiplication: the three single digit numbers 2, 6 and 7.
However we should arrange them accordingly to find out the final answer. The order is simple:
first we write the result from A
1
+A
3
which is 2
,
then the result for the diagonal A
2
+ A
3
and A
1
+
A
4
which is 7,
and then the result from A
2
+ A
4
which is 6. The result is thus 276! It can actually
be read from the indicated “bubbles” in the graph from left to right.
23 x 12 = 276
2) Digit carry-over case
What would happen if the numbers we have to multiply bring about intersections that have a
number of touching points with two digits? To solve this, we use the digit carry-over method.
Let’s multiply 24 x 31 = ?
3
This time we will use B
1
= 2 chopsticks, B
2
= 4 chopsticks, B
3
=3 chopsticks and B
4
= 1 chopstick.
We arrange them the same way as we did previously. However, this time we do not count the
intersection points in any order. We should start with the bottom right corner or the B
2
+ B
4
intersection, followed by the diagonal formed with B
1
+ B
4
and B
2
+ B
3
and then finish counting
with the top left corner or B
1
+ B
3
intersection. From this we obtain that B
2
+ B
4
has 4 points of
contact, and the diagonal B
1
+ B
4
and B
2
+ B
3
has 14 points of contact. Here we pause a moment
to explain the necessity of the digit carry-over case. Since each intersection should give a single
digit number of points of contact, from the number 14, we take the four as being the result for the
intersection and the remainder 1 is carried-over and added to the counting of the next
intersection, in this case, the upper left corner or B
1
+ B
3
intersection. Therefore we no longer
have 6 as a result of the intersection we normally would, but seven. We thus have now the result
for our multiplication: 7, 4 and 4 which is correct.
24 x 31 = 744
4
The Magic Squares of 5
Calculating squares of numbers is fairly easy as long as we are taking the square of the numbers
between one and ten. When the square involves greater numbers it becomes a little bit more
complicated. However, there is a way to quickly calculate the square of numbers whose last digit
is 5.
As we know, 5
2
= 25. Let’s rewrite the single digit 5 as the two digit number 05; its square root
is still going to be 25. To investigate the squares of higher numbers, let’s analyze 25
2
= 225. The
squares of these numbers can be found in a very easy way. We separate the number to be squares
into two components: B which is the last digit of the number and A the remainder. As an
example, the number 05 would be separated into A = 0 and B= 5, 25 would be separated into
A=2 and B=5. Note that since the numbers we are dealing with are those ending with 5, B will
always be equal to 5.
The calculation of squares is as followed: A x (A+1) = C and B
2
= D will produce CD.
Illustrating this in 05
2
A = 0 therefore A+1 = 1 making A x (A+1) = 0 x 1= 0. Then we have B
2
=
5
2
= 25. Finally we get C= 0 and D= 25 therefore CD = 025 or commonly written as 25. In the
same manner, if we want to square the number 25, we separate it into two components: A = 2
and B= 5. B
2
=25 and A x (A +1) = 2 x 3 = 6. Therefore C = 6 and D = 25, making CD = 625
which is the result of 25
2
.
5
The Interesting Number 9
This case does not make any calculation easier but it is interesting to discover the nice properties
about the multiplication table of the number 9.
The number 9 has a multiplication table that is totally symmetrical. The multiples of 9 from one
to five are respectively 9, 18, 27, 36 and 45. It appears that the multiples of 9 from six to ten are
54, 63, 72, 81, and 90 respectively, which are the symmetric opposite of the first five numbers
09,18,27,36 and 45.
Also if we look at the table on a column, from top to bottom, we observe that the multiples of 9
from 1 to 10 are all made of both numbers from 0 to 9 as their first digit and 9 to 0 as their
second digit, showing again a geometrical symmetry.
Interesting Application of Multiplication
In elementary school, students learn the tables of multiplication. Unfortunately, due to the
constant use of calculators they stop using them; and instead of multiplying, say 4x6 in their
minds, they plug it into the calculator to obatin the answer. Although using the calculator may at
times to prove to be a faster method to get to the mathetical answer of a problem, it deprives the
students from learning some interesting ideas that can be unveiled from the tables of
multiplication. One such instance is the table of two, and its concept of doubling. The doubling
effect is better be explained by the following example,
As you walk along a lake, you noticed that lake is inhabited by just one lotus flower. The
next day, you walked by the lake, and noticed that there are two lotus flowers. The
following day, you see four lotus flowers instead of two. If you know the lake will be
completely covered on the 30
th
day, how long would it take for half of the lake to be
6
covered by lotus flowers? Keep in mind that you do not know the size of the lake, nor the
size of the lotus flower.
You can start this problem by noticing that as each day goes by the amount of
lotus flowers is doubled. Then you designate the intital lotus flower as . Since
the amount of flowers is doubled everyday, you would multiply it by two every
time. This means that on day one you have . On the second day you have
2x( 2 . And so on and so forth. If the doubling concept is kept in mind,
then it can be said that by the 29
th
day, the lake should be half occupied, and so
the next (the 30
th
), after it is doubled, the lake would be completely covered.
Another Way of Multiplying Without Using a Calculator
Really early in school life, students adopt the use of calculators in a very addictive way. Many
times they cannot solve the simplest calculation without plugging it in a calculator. There are
many ways to perform calculations without using a calculator; however they can be tedious at
times. Here we will be demonstrating a very simple and easy way to calculate squares of
numbers that are not multiples of five.
The steps to follow are simple: If you have a number you want to square, let’s call it x
2
. Since
x
2
= x
2
– y
2
+ y
2
, and x
2
– y
2
= (x + y) (x – y), therefore we obtain (x + y) (x – y) + y
2
, where x is
the number to be squared, and y is the difference between that number and the next perfect
square. This is the formula we are going to use to compute our squares without the calculator!
7
The Pythagorean Theorem
All students who have gone through Algebra, Geometry and Trigonometry, know about the
Pythagorean Theorem, as it is one of the first mathematical theorems that is taught. It states that
for a right triangle, the addition of the squares of the two shorter sides of the triangle is equal to
the hypotenuse (the longest side) of that same triangle. It is expressed by the equation A
2
+ B
2
=
C
2
.
The most common and well known Pythagorean triangles are the 3-4-5, the 5-12-13, and the 7-
24-25 triangles. Looking at them, we can discover a pattern. For all three of them the shortest
sides are equal to 3,5 and 7 respectivcely. They are all arithmetic sequences with a difference of
2. Also, another pattern that we can notice is that to find the value of the second shorter side of
the triangle. By multiplying the shortest with a sequenced number and then adding it to the
result: (3x1) +1=4; (5x2) +2 =12; (7x3) +3 =24. Finally, the hypotenuse is obtain by adding 1 to
the previously found side: 4+1 =5; 12+1 =13; 24 +1=25.
We can, therefore summerize our patterns into a formula. Whenever the shortest side of the right
triangle is an odd number, the second shorter side is found by first finding the component y that
will be plugged in later into the formula: xy +y = B. x is the starting number in the arithmetic
sequence, y is a multiplier that increases by 1 for each calculation. This calculation produces the
number for the hypotenuse.
8
What happens if the shortest side of the Pythagorean triangle is an even number? Well, the
answer is simple. We follow steps similar to the prior ones, having two corrections to them.
Whenver the shortest side is an even number, the second shorter side is computed with the
formula (xz – 1) where z as a multiplier that starts at 1 and decreases by 1 for next calculation.
When z
1
= 1, x
1
=4; when z
2
=2, x
2
= 8. We then find the regularity of the right triangle starting
with even numbers as being 4. We can predict any number down the road for the second shorter
side to be the previous + 4. The hypotenuse is then found by adding 2 to the value of the second
shorter side of the triangle.
The Magic Pascal Triangle
The Pascal triangle was created by the French mathematician Pascal in the mid seventeen
century. Looking at it, it resembles a simple triangle with numbers inside. But in reality, it is a
beautifully intriguing multiplication table. Within the triangle, there are many interesting
characteristics we can discover.
a) Symmetry in the diagonal direction:
Each diagonal of the triangle is formed by a set of consecutive numbers that have
a symmetric image on the other side of the triangle
b) Symmetry in the horizontal direction:
9
Imagine the triangle is vertically separated in half. Each horizontal line from one
side has its symmetric immage on the other side of the triangle. Take, for
example, the fifth horizontal line of our graph, if you cut it in half right through
the middle, you observe the same pattern of numbers to be exactly symmetric: 1-
5-10 is the exact reflection of 10-5-1.
c) Multiples of two in the horizontal direction:
If we add all the numbers of each horizontal line of the triangle, their results are
equal to exponentials of 2 in the perfect order. For example, the sum of the first
row is 1 and 2
0
= 1; the sum of the second row is 2 and 2
1
= 2, and the sum of the
third row is 4 and 2
2
= 4; this pattern goes on indefinitely.
d) Addition in L shape:
The fourth characteristic of the Pascal triangle resides in the addition of numbers
forming L-shapes. If we add any series of numbers in a diagonal line, starting
with the most outer number in that line, their total is equal to the number forming
the L arm. To illustrate that fact, let’s take for example the second diagonal from
our graph; the series starts with the most outer number 1 and if we add 1+2+3 =6
which is the number forming the L with these three numbers.
*
* Note: 2^x means 2
x
Shortcut to Obtain Partial Fractions
For students, the operation to calculate partial fractions can be tedious at times. Here we will
present the Heaviside method in three shortcut cases where calculating partial fractions is not
that hard. These methods can be used for College Algebra, Calculus or Differential Equations.
Case A
Let’s find the partial fractions for F(s) =
� � � �
(
�
�
� �
)
(�
�
� � )
F(s) =
� � � �
(
�
�
� �
)
(�
�
� � )
=
� � � �
�
�
� �
+
� � � �
�
�
� �
10
Let
�
(
� � �
)
(� � � )
=? (
�
� � �
−
�
� � �
) Let’s find ? =
�
�
Replace x = s
2
1
(�
�
− 2)(�
�
+ 4)
=
1
6
(
1
�
�
−
1
�
�
+ 4
)
Multiply by (3s +1)
F(s) =
�
�
�
� � � �
�
�
� �
−
� � � �
�
�
� �
�
Case B
Let’s find the partial fractions for F(s) =
� � � (� � � )
� (
(
� � �
)
�
� � )
Let F (s) =
� � � (� � � )
� (
(
� � �
)
�
� � )
=
�
�
+
�
(
� � �
)
� �
(� � � )
�
� �
[This is required for solving inverse Laplace Transforms]
F (s) =
� �
(
� � �
)
�
� � � � �
(
� � �
)
� � � �
� (
(
� � �
)
�
� � )
1 – s(4 – 3S) = A( (s – 1)
2
+ 1) + B(s – 1)s + sC
*The cover-up technique requires the effective elimination of more coefficients by letting
S=0
1 = 2A A=
�
�
Let s = 1 1 – 1(4 – 3) = A + C
0 = A + C
C = – A = −
�
�
Let s = -1 1+1(4 + 3) = 5A + B (- 2) (-1) – C
8 = 5A + 2B – C
2B = 8 – 5A + C
= 8 −
�
�
−
�
�
= 5
B =
�
�
F(s) =
�
� �
+
�
�
� � �
(� � � )
�
−
�
�
�
((� � � )
�
� � )
11
Case C
Let’s find the partial fraction for F(s) =
�
�
� � � � �
(� � � )
�
Let F(s) =
�
�
� � � � �
(� � � )
�
=
�
(� � � )
+
�
(� � � )
�
+
�
(� � � )
�
We modify the numerators by matching them with (s-2)
s
2
+ 4s – 5 = ( (s – 2) + 2)
2
+ 4( (s – 2) +2) – 5
= (s – 2)
2
+ 4 + 4(s – 2) + 4(s – 2) + 8 – 5
= (s – 2)
2
+ 8(s – 2) +7
Therefore
F(s) =
(� � � )
�
� �
(
� � �
)
� �
(� � � )
�
F(s) =
�
� � �
+
�
(� � � )
�
+
�
(� � � )
�
REF: “Joy of Math” by Arthur Benjamin
Biography:
F(s) =
�
� �
+
�
� (� � � )
−
�
� (
(
� � �
)
�
� � )