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MTE 3105: STATISTIK
SEMESTER
JAN-JUN 2012
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COURSE INTRODUCTION
SINOPSIS
HASIL PEMBELAJARAN
KANDUNGAN KURSUS PENILAIAN
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SINOPSIS
Mengimbas kembali konsep yangberkaitan dengan kebarangkalian
Menerokai statistik inferens
ujian-t, ujian khi kuasa dua, analisis varians(ANOVA) dalam pengujian hipotesis dan
regresi linear dalam menganalisis
perhubungan linear dalam dua
pembolehubah.
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SINOPSIS
Penggunaan teori pensampelan dan
penganggaran untuk menganggar min
populasi.
Kepentingan menggunakan kaedah
statistik yang sesuai dalam penyelesaian
masalah harian adalah dititikberatkan.
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Hasil Pembelajaran
Menerangkan aspek teori dan empirikal yangmendasari kebarangkalian
Mengaplikasikan teori persampelan dan
anggaran dalam menganggarkan min populasi Mengaplikasikan statistik inferens seperti ujian
khi kuasa dua, ANOVA dan regresi linear dalamujian hipotesis.
Menganalisis dan menyelesaikan masalahharian menggunakan kaedah statistik.
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THE TOPICS
Probability (3)
Sampling and estimation theory (9)
Hypothesis testing (12) Inferential statistics
The chi-square test (9)
The analysis of variance [ANOVA] (6) Linear regression (6)
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ASSESSMENT
50% COURSEWORK
50% EXAMINATION
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PROBABILITY
1. Introduction to
probability
Theoretical
Probability
Empiricalprobability
2. Compound eventsIndependent events
Mutually exclusiveevents
3. Addition Rule &
Multiplication Rule
4. Probability tree
diagrams
5. Conditional
Probability
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Experiment & outcomes
Event
any collection of outcomes froman experiment eg tossing a coin
Sample space
Set of all possible outcomes
IMPORTANT DEFINITIONS
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Estimating probability
Conduct (or observe) an experiment alarge number of times, and count thenumber of times event A actuallyoccurs, then an estimate of P(A) is
EXPTAL PROBABILITY
P(A) = Number of times A occurredTotal number of trials
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NOTATION: SOME REVISION
P :denotes a probability
A, B, : denote specific events
P(A) : denotes the probabilityof event A occurring
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The Classical approach(equally likely outcomes)
If a trial has S different possible outcomes,each with an equal chance of occurring, thenthe probability that event A occurs is
P(A)= number of ways A can occurtotal number of outcomesn(A)
n(S) =
THEORETICAL PROBABILITY
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EXPERIMENTAL VS THEORETICAL
Experimental probability uses therelative frequency approach and it isan approximation or estimation
Theoretical probability uses theclassical approach it is the actualexpected probability if the experimentis repeated over a large number oftimes.
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LAW OF LARGE NUMBERS
As a procedure is repeated again and
again, the relative frequency probability
(experimental probability) of an event
tends to approach the actual (theoretical)
probability.
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PROBABILITY LIMITS
The probability of an impossibleevent is 0.
The probability of a certain eventis 1.
0 P(A) 1
Impossibleto occur
Certainto occur
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PROBABILITY VALUES
Certain
Likely
50-50 Chance
Unlikely
Impossible
1
0.5
0
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The complement of event A,denoted by A, consists of all
outcomes in which event A doesnot occur.
COMPLEMENTARY EVENTS
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THE COMPLEMENT OF EVENT A
Total Area = 1
P (A)
P (A) = 1 - P (A)
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Any event combining two ormore simple events
COMPOUND EVENTS
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When finding the probability thatevent A occurs or event B occurs,
find the total number of ways A
can occur and the number of
ways B can occur, but ensure
that no outcome is counted morethan once.
PROB OF COMPOUND EVENTS
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P(A or B) = P(A) + P(B) - P(A and B)
where P(A and B) denotes the
probability that A and B bothoccur at the same time.
THE ADDITION RULE
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Kebarangkalian peristiwasaling eksklusif
P(A B) =P(A) +P(B) - P(A B)
Jika A dan B saling eksklusif,
makaP(A B) = P(A) + P (B)
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MUTUALLY EXCLUSIVE EVENTS
Events A and B are mutually
exclusiveif they cannot occur
simultaneously.
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Sets Presentations
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Sets Presentations
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NON-MUTUALLY EXCLUSIVE
Total Area = 1
P(A) P(B)
P(A and B)
Overlapping Events
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APPLYING THE ADDITION RULE
P(A or B)Addition Rule
Are
A and B
mutually
exclusive
?
P(A or B) = P(A)+ P(B) - P(A and B)
P(A or B) = P(A) + P(B)Yes
No
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Find the probability of randomly
selecting a man or a boy.
Contingency Table
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
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The events of randomly selecting a
man or a boy are mutually exclusive.
Contingency Table
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
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Find the probability of randomly
selecting a man or someone whosurvived.
Contingency Table
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
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The events of randomly selecting a
man or someone who survived arenot mutually exclusive.
Contingency Table
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
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P(man or survivor) = 1692 + 706 - 332 = 1756
2223 2223 2223 2223
Contingency Table
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
= 0.929
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The General Addition Rule
P ( A B) = P(A) + P(B) P(A B)
Non-mutual exclusive
The Special Addition Rule
P ( A B) = P(A ) + P(B )
Mutual exclusive
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Kebarangkalian peristiwa takbersandar
Kebarangkalian bagi peristiwa A dan Bberlaku ialah
dengan syarat A dan B tak bersandar
P(A B) =P(A) x P(B)
I d d d
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Independent andConditional Events
The conditional probability of anevent is the probability that the
event occurs under the assumptionthat another event has occurred.
Kebarangkalian bhw peristiwa B
berlaku diberi bahawa A telahberlaku ialah
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Conditional Probability
P( B l A) = The probability of B
given A= Kebarangkalian
peristiwa B berlaku diberi bhwperistiwa A telah berlaku
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Example
A fair die is thrown, if it is known that the
number obtained is an even number, what
is the probability that the number is 4?
d l P b b l /
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Conditional Probability/Kebarangkalian Bersyarat
)(
)()1(
)(
)()1(
AP
ABpABP
BP
BApBAP
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Let A = a 4 obtained
B = an even number is obtained
P ( A given B has occurred)
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Conditional Probability
so,)(
)()1(
)(
)()1(
APABpABP
BP
BApBAP
)()1()()1( APABPBPBAP
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Conditional Probability
If A and B are mutually exclusiveevents, then
and = 0, then
= 0
)( ABp BAP
(
)()1()()1( APABPBPBAP
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APPLYING THE MULTIPLICATION RULE
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APPLYING THE MULTIPLICATION RULE
P(A or B)Multiplication Rule
Are
A and B
independent
?
P(A and B) = P(A) P(B A)
P(A and B) = P(A) P(B)Yes
No
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FORMAL MULTIPLICATION RULE
P(A and B) = P(A) P(B|A)
If A and B are independentevents, P(B|A) is really the
same as P(B)
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INDEPENDENT EVENTS
P (A and B) = P (A B)
= P (event A occurs in a first trial andevent B occurs in a second trial)
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COMPUTING PROBABILITY
A student answers two questions, of which
the first is a true (T) or false (F) item
followed by a multiple choice item with 5
choices (A, B, C, D, and E)
Find the probability that the students
answers both questions correctly
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THE MULTIPLICATION RULE
TATB
TC
TD
TEFA
FB
FC
FD
FE
AB
C
D
E
A
B
C
D
E
T
F
P(T) = P(C) = P(T and C) =12
1
5
1
10
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PROBABILITY OF AT LEAST ONE
At least one is equivalent toone or more.
The complementof getting atleast one item of a particular type
is that you get no items of thattype.
PROBABILITY OF AT LEAST ONE
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PROBABILITY OF AT LEAST ONE
The complementof getting at leastone item of a particular type is thatyou get no items of that type
If P(A) = P(getting at least one),
then P(A) = 1 - P(A) where
P(A) is P (getting none)
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PROBABILITY OF AT LEAST ONE
Find the probablility of a couplehave at least 1 girl among 3children.
If P(A) = P(getting at least 1 girl), then
P(A) = 1 - P(A)
where P(A) is P(getting no girls)
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PROBABILITY OF AT LEAST ONE
If P(A) = P(getting at least 1 girl), then
P(A) = 1 - P(A)
where P(A) is P(getting no girls)
P(A) = (0.5)(0.5)(0.5) = 0.125
P(A) = 1 - 0.125 = 0.875
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Q1BA10
1(a) An experiment consists of threerepetitions of a Bernoulli trial with theprobability of success equal to 0.25 and
the probability of failure equal to 0.75.(i) Draw a tree diagram to illustrate
the situation.
(ii) What is the probability that nofailure is obtained in the three trials.
(iii) What is the probability that exactly
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(iii) What is the probability that exactlyone failure is obtained in the threetrials?
(b) There are 78 tennis players in a
competition. Table 1 gives detailinformation about them
56 school-level
players
29 male 7 left-handed
27 female 3 left-handed
22 state-level
players
12 male 6 left-handed
10 female 2 left-handed
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