BIODATA SINGKATBIODATA SINGKAT NAMA : Prof. DR. SAMSUBAR SALEH. M.Soc.ScNAMA : Prof. DR. SAMSUBAR SALEH. M.Soc.Sc PEKERJAAN : DOSEN FEB-UGMPEKERJAAN : DOSEN FEB-UGM JABATAN : GURU BESAR ILMU EKONOMI – IV/cJABATAN : GURU BESAR ILMU EKONOMI – IV/c ALAMAT : JL WELING II / 39 A. DEPOK-SLEMANALAMAT : JL WELING II / 39 A. DEPOK-SLEMAN BIDANG MINAT : EKONOMI-PUBLIKBIDANG MINAT : EKONOMI-PUBLIK STATISTIKA TERAPANSTATISTIKA TERAPAN TEORI EKONOMI MAKRO/MIKROTEORI EKONOMI MAKRO/MIKROPENGALAMAN KERJA : DOSEN FEB-UGMPENGALAMAN KERJA : DOSEN FEB-UGM : ASSISTEN WADEK III & SEK-JURUSAN IE.: ASSISTEN WADEK III & SEK-JURUSAN IE. : PENGELOLA S2/S3 FEB-UGM: PENGELOLA S2/S3 FEB-UGM : SUPERVISOR BSNP DIKNAS: SUPERVISOR BSNP DIKNAS : PENGELOLA PENELITIAN EKONOMI: PENGELOLA PENELITIAN EKONOMI : KONSULTAN PERENC/PEMB DIY: KONSULTAN PERENC/PEMB DIY : KONSULTAN & FASILITATOR KEU-DA: KONSULTAN & FASILITATOR KEU-DAHP: 0811253224.HP: 0811253224.Email :ssamsubar@yahoo.comEmail :ssamsubar@yahoo.com

Text-BooksText-Books

1. Lind & Marchal; Statistics for Business1. Lind & Marchal; Statistics for Business 2. Wonnacott : Statistics2. Wonnacott : Statistics 3. Samsubar Saleh: Statistik Induktif3. Samsubar Saleh: Statistik Induktif

PROBABILITYPROBABILITYPROBABILITYPROBABILITY IS A NUMERICAL MEASURE OF THE IS A NUMERICAL MEASURE OF THE LIKELIHOOD OF AN EVENT OR MORE OCCURRING or LIKELIHOOD OF AN EVENT OR MORE OCCURRING or CHANCE AN EVENT OR MORE WILL OCCUR AS A RESULT CHANCE AN EVENT OR MORE WILL OCCUR AS A RESULT OF AN EXPERIMENT.OF AN EXPERIMENT.THE CONCEPTS OF PROBABILITY ARE SPECIALLY THE CONCEPTS OF PROBABILITY ARE SPECIALLY RELEVANT IN BUSINESS. BUSINESSMEN MAKE DECISIONS RELEVANT IN BUSINESS. BUSINESSMEN MAKE DECISIONS IN THE FACE OF UNCERTAINTY DAY BY DAY. FOR IN THE FACE OF UNCERTAINTY DAY BY DAY. FOR EXAMPLE:EXAMPLE:1. WHEN COMPANY LAUNCHES A NEW PRODUCT.1. WHEN COMPANY LAUNCHES A NEW PRODUCT.2.THE LIKELIHOOD OF A NEW INVESTMENT PROJECT 2.THE LIKELIHOOD OF A NEW INVESTMENT PROJECT BEING PROFITABLE MAY BE UNKNOWN. THEREFORE, BEING PROFITABLE MAY BE UNKNOWN. THEREFORE, DECISION MAKING HAS TO BE BASED ON SOME DECISION MAKING HAS TO BE BASED ON SOME ASSESSMENT OF THE PROBABILITY OF POSSIBLE ASSESSMENT OF THE PROBABILITY OF POSSIBLE OUTCOMES OCCURRING.OUTCOMES OCCURRING.3.ENGINEERING INSTALATION DESIGN.3.ENGINEERING INSTALATION DESIGN.

EXPERIMENT, OUTCOMES AND EVENTEXPERIMENT, OUTCOMES AND EVENT

EACH EXPERIMENT WILL HAVE TWO OR MORE POSSIBLE EACH EXPERIMENT WILL HAVE TWO OR MORE POSSIBLE RESULTS.RESULTS.

OUTCOME; A PARTICULAR RESULT OF AN EXPERIMENT.OUTCOME; A PARTICULAR RESULT OF AN EXPERIMENT. EVENT ; A COLLECTION OF ONE OR MORE OUTCOMES OF AN EVENT ; A COLLECTION OF ONE OR MORE OUTCOMES OF AN

EXPERIMENT.EXPERIMENT.

EXPERIMENT: TOSS 2 COINS ROLL 2 DICEEXPERIMENT: TOSS 2 COINS ROLL 2 DICE HHHHALL POSSIBLE HT THERE ARE 36ALL POSSIBLE HT THERE ARE 36OUTCOMES TH POSSIBLE OUTCOMESOUTCOMES TH POSSIBLE OUTCOMES TTTT SOME POSSIBLE GET THE NUMBER TOTAL NUMBER OF SOME POSSIBLE GET THE NUMBER TOTAL NUMBER OF EVENTS OF TAIL AT LEAST 1 BOTH DICE = 2 OR 10EVENTS OF TAIL AT LEAST 1 BOTH DICE = 2 OR 10

THE MULTIPLICATION RULE OF PROBABILITYTHE MULTIPLICATION RULE OF PROBABILITY

1.1. INDEPENDENT EVENTS: TWO EVENTS ARE INDEPENDENT EVENTS: TWO EVENTS ARE INDEPENDENT WHEN THE OCCURRENCE OF ONE INDEPENDENT WHEN THE OCCURRENCE OF ONE EVENT HAS NO EFFECT ON THE PROBABIEVENT HAS NO EFFECT ON THE PROBABILLITY OF ITY OF OCCURRENCE OF THE OTHER EVENT.OCCURRENCE OF THE OTHER EVENT.

P ( A and B ) = P ( A ) . P ( B )P ( A and B ) = P ( A ) . P ( B )

2. DEPENDENT EVENT: TWO EVENTS ARE DEPENDENT 2. DEPENDENT EVENT: TWO EVENTS ARE DEPENDENT

WHEN THE OCCURRENCE OF ONE EVENT AFFECT WHEN THE OCCURRENCE OF ONE EVENT AFFECT THETHE

PROBABILITY OF OCCURRENCE OTHER EVENT.PROBABILITY OF OCCURRENCE OTHER EVENT.

P ( A and B ) = P ( A ) . P ( B/A )P ( A and B ) = P ( A ) . P ( B/A )

ExampleExample

A BOX CONTAINS OF 10 RED BALLS, 6 A BOX CONTAINS OF 10 RED BALLS, 6 WHITE BALLS AND 4 BLUE BALLS. ONE WHITE BALLS AND 4 BLUE BALLS. ONE BALL IS SELECTED AT RANDOM 3 TIMES. BALL IS SELECTED AT RANDOM 3 TIMES. WHAT IS THE PROBABILITY of OBTAINING:WHAT IS THE PROBABILITY of OBTAINING:

a.a. All balls are white ( with replacement ).All balls are white ( with replacement ).b.b. All balls are white ( without replacement ).All balls are white ( without replacement ).c.c. The 1st selection is red, the 2nd selection is red and the 3rd The 1st selection is red, the 2nd selection is red and the 3rd

selection is blue ( without replacement ).selection is blue ( without replacement ).d.d. There are 2 red balls and 1 There are 2 red balls and 1 blueblue ball ( without replacement ) ball ( without replacement )e.e. There are 1 red, 1 white and 1 blue ball ( Without R).There are 1 red, 1 white and 1 blue ball ( Without R).

THE RULES OF PROBABILITYTHE RULES OF PROBABILITY

THE ADDITION RULE OF PROBABILITY:THE ADDITION RULE OF PROBABILITY:

1. 1. FOR MUTUALLY EXCLUSIVE EVENTS.FOR MUTUALLY EXCLUSIVE EVENTS.

P ( A or B ) = P ( A ) + P ( B ) P ( A or B ) = P ( A ) + P ( B )

EVENTS A & B CANEVENTS A & B CAN’’T OCCUR SIMULTANEOUSLYT OCCUR SIMULTANEOUSLY

OR AT THE SAME TIME.OR AT THE SAME TIME.

2. 2. FOR NON-MUTUALLY EXCLUSIVE EVENTS:FOR NON-MUTUALLY EXCLUSIVE EVENTS:

P ( A or B ) = P ( A ) + P ( B ) – P ( A and B )P ( A or B ) = P ( A ) + P ( B ) – P ( A and B )

A and B CAN OCCUR SIMULTANEOUSLY.A and B CAN OCCUR SIMULTANEOUSLY.

EXAMPLEEXAMPLE

SCORES : SCORES : A B C FA B C FCLASSESCLASSES

X X 10 15 8 2 10 15 8 2 Y Y 13 16 7 4 13 16 7 4 Z Z 8 11 6 5 8 11 6 5 CALCULATE THE PROBABILITY:CALCULATE THE PROBABILITY:1. P ( A or B or F ) 2. P ( X or C ) 3. P ( Y or Z )1. P ( A or B or F ) 2. P ( X or C ) 3. P ( Y or Z )4. P ( Z or A ) 5. P ( C/X ) 6. P ( Z/F )4. P ( Z or A ) 5. P ( C/X ) 6. P ( Z/F )

CONDITIONALCONDITIONAL PROBABILITYPROBABILITY

THE PROBABILITY OF A PARTICULAR THE PROBABILITY OF A PARTICULAR EVENT OCCURING, GIVEN THAT EVENT OCCURING, GIVEN THAT ANOTHER EVENT HAS ALREADY ANOTHER EVENT HAS ALREADY OCCURRED.OCCURRED.

IF, P ( A and B ) = P ( A ) . P ( B/A ), THUS;IF, P ( A and B ) = P ( A ) . P ( B/A ), THUS;

P ( A and B )P ( A and B ) P ( B/A ) = -------------------------P ( B/A ) = -------------------------

P ( A )P ( A )

ExampleExample In Lab: A, there are 10 units Acer computer, 5 units Compact In Lab: A, there are 10 units Acer computer, 5 units Compact

and 15 units Toshiba. In lab: B, there are 15 units Acer and 15 units Toshiba. In lab: B, there are 15 units Acer computers, 10 unit Compact and 10 units Toshiba. In lab: C, computers, 10 unit Compact and 10 units Toshiba. In lab: C, there are 10 units Acer, 10 units Compact and 10 units there are 10 units Acer, 10 units Compact and 10 units Toshiba.Toshiba.

A. If each lab is selected one computer randomly, what is the A. If each lab is selected one computer randomly, what is the probability that all of the computers are Toshiba. probability that all of the computers are Toshiba.

B. If one computer is selected at random from one lab, what is B. If one computer is selected at random from one lab, what is the probability that the computer is Acer or Compact.the probability that the computer is Acer or Compact.

BAYESBAYES’’S RULE FOR CONDITIONAL PROBABILITYS RULE FOR CONDITIONAL PROBABILITY

P ( Ai ) P ( B/Ai )P ( Ai ) P ( B/Ai )

P ( Ai/B ) = ----------------------------------------------------------P ( Ai/B ) = ---------------------------------------------------------- P ( A1) P ( B/A1 ) + -------+ P ( An ) P ( B/An ) P ( A1) P ( B/A1 ) + -------+ P ( An ) P ( B/An ) EXAMPLEEXAMPLE 1 1:: A car repair firm employs three paint sprayers, Tomy, Dicky and A car repair firm employs three paint sprayers, Tomy, Dicky and

Harry. Tomy is responsible for painting 25 % of all the cars Harry. Tomy is responsible for painting 25 % of all the cars produced, Dicky for 35 % and Harry for the remaining 40 %. On produced, Dicky for 35 % and Harry for the remaining 40 %. On the basis of frequent quality inspections it is discovered that, on the basis of frequent quality inspections it is discovered that, on average, 5 % of the car sprayed by Tomy fall below the minimum average, 5 % of the car sprayed by Tomy fall below the minimum standard, while for Dicky is 8 % and for Harry 10 %. If a car is standard, while for Dicky is 8 % and for Harry 10 %. If a car is selected at random is judged to be sub-standard. What is the selected at random is judged to be sub-standard. What is the probability that it was sprayed by Harry?probability that it was sprayed by Harry?

Example:2Example:2

The number of students from class A = 35, class B The number of students from class A = 35, class B = 30, class C = 40 and class D = 25. Based on the = 30, class C = 40 and class D = 25. Based on the examination record, it was found that 10% of examination record, it was found that 10% of students from class A was fail, class B = 5%, class students from class A was fail, class B = 5%, class C = 15 % and class D = 20 %. If one student was C = 15 % and class D = 20 %. If one student was selected at random was indicated that he did well selected at random was indicated that he did well in the exam. What is the probability that he is in the exam. What is the probability that he is from class B or D. from class B or D.

Mathematical ExpectationMathematical Expectation Finding the expected value of the Mean, Variance and Finding the expected value of the Mean, Variance and

Standard Deviation of a probability distribution:Standard Deviation of a probability distribution:

Mean = Mean = ∑ Xi . P ( Xi )∑ Xi . P ( Xi ), where Xi is the value of variable X, and , where Xi is the value of variable X, and P (Xi) is the probability of X.P (Xi) is the probability of X.

Variance = ∑ ( Xi - Variance = ∑ ( Xi - µµ ) )22 P ( Xi) or ∑ Xi P ( Xi) or ∑ Xi22 P ( Xi ) - P ( Xi ) - µµ2 2

Std dev = Std dev = √ variance√ variance

Example Example

A family has five children. Please calculate the A family has five children. Please calculate the expected value and standard deviation of having male expected value and standard deviation of having male children.children.

Step1 : find out the value of Xi ( male children ).Step1 : find out the value of Xi ( male children ). Step 2: find out the probability of having Xi.Step 2: find out the probability of having Xi. Xi : ?Xi : ? P( Xi ): ?P( Xi ): ?

ExampleExample The quantity of X demanded probabilityThe quantity of X demanded probability 200 units 0.15200 units 0.15 225 0.20225 0.20 250 0.30250 0.30 275 0.25275 0.25 300 0.10300 0.10a. What is the mean of quantity of X demanded.a. What is the mean of quantity of X demanded.b. What is the variance of the distribution?b. What is the variance of the distribution?c. When the price of X = $ 10 and AC = $ 8c. When the price of X = $ 10 and AC = $ 8 What is the expected value of the profit?What is the expected value of the profit?

Probability distributionsProbability distributions

1. Discrete probability distribution:1. Discrete probability distribution:

Binomial distribution:Binomial distribution:

a.a. In each trials only two outcomes are possible: In each trials only two outcomes are possible: success or failure, good or defect, pass or fail, girl success or failure, good or defect, pass or fail, girl or boy, odd number or even number.or boy, odd number or even number.

b.b. Each trial is independent to each other.Each trial is independent to each other.

c.c. Probability of an event is assumed to remain Probability of an event is assumed to remain constant over all trials ( n trials ).constant over all trials ( n trials ).

Binomial DistributionBinomial Distribution The probability of obtaining the outcome The probability of obtaining the outcome ““successsuccess”” denoted by denoted by ππ and the outcome and the outcome ““failurfailur

ee”” denoted q or (1 – denoted q or (1 – ππ ). ). If we are interested in the number of If we are interested in the number of ““successsuccess”” or or ““failurefailure”” ( x ) occuring in ( n ) trials, ( x ) occuring in ( n ) trials,

then, the probability binomial distribution is given by the formula:then, the probability binomial distribution is given by the formula:

n n xx n – x n – x

C C ππ q q xx

Example:Example: The four engines of a commercial aircraft are design so that they each operate The four engines of a commercial aircraft are design so that they each operate

independently. Test, carried out over a long period of time, show that there is a one-in-a independently. Test, carried out over a long period of time, show that there is a one-in-a hundred chance of in-flight failure hundred chance of in-flight failure oof a single engine.f a single engine.

What is the probability that on a given flight:What is the probability that on a given flight:a.a. No failures occur?No failures occur?b.b. No more than two failures occur?No more than two failures occur?c.c. At least two failures occurs.At least two failures occurs.

BINOMIAL DISTRIBUTIONBINOMIAL DISTRIBUTION PROBABILITY OF A STUDENT WILL PASS IN THE PROBABILITY OF A STUDENT WILL PASS IN THE

FINAL STATISTICS EXAM = 0.80. IF 6 STUDENTS FINAL STATISTICS EXAM = 0.80. IF 6 STUDENTS ARE SELECTED AT RANDOM, WHAT IS THE ARE SELECTED AT RANDOM, WHAT IS THE PROBABILITY OF OBTAINING:PROBABILITY OF OBTAINING:

a. At most 2 of them are fail in the exam.a. At most 2 of them are fail in the exam. b. At least 4 of them will pass in the exam.b. At least 4 of them will pass in the exam.

ExampleExample

If 7 students were selected at random, what is If 7 students were selected at random, what is the probability that:the probability that:

A. More than 4 students will get score A or C.A. More than 4 students will get score A or C. B. at least 5 students will get F.B. at least 5 students will get F. C. less than 3 students will get B or C.C. less than 3 students will get B or C. D. all of them will get A or B or F.D. all of them will get A or B or F.

Example : 2Example : 2

Suppose that a family has 5 children. What is the Suppose that a family has 5 children. What is the probability of having:probability of having:

a.a. At least 3 of them are male.At least 3 of them are male.

b.b. All of them are female.All of them are female.

c.c. At most there is one male.At most there is one male.

d.d. the number of female is more than 4.the number of female is more than 4.

Example : 3Example : 3

Two dice is thrown 5 times, what is the Two dice is thrown 5 times, what is the probability of obtaining:probability of obtaining:

a.a. Total number of both dice is 10 appears at Total number of both dice is 10 appears at most twice.most twice.

b.b. Total number of both dice is 5 appears at Total number of both dice is 5 appears at least 4 times.least 4 times.

c.c. Total number of both dice is 8 appears Total number of both dice is 8 appears exactly 3 times.exactly 3 times.

Continuous Probability DistributionContinuous Probability Distribution

The normal probability distribution has the following The normal probability distribution has the following major characteristics :major characteristics :

1. It is bell-shaped and has a single peak.1. It is bell-shaped and has a single peak. 2. The mean, median and mode are equal.2. The mean, median and mode are equal. 3. The location of normal distribution is3. The location of normal distribution is

determined by the determined by the µµ and and σσ.. 4. It falls off smoothly in either direction from the4. It falls off smoothly in either direction from the

µµ and the curve gets closer and closer to X and the curve gets closer and closer to X

axis but never actually touches it.axis but never actually touches it.

NORMAL DISTRIBUTIONNORMAL DISTRIBUTION

1. Xi 1. Xi ~ N ( µ, ~ N ( µ, σσ ) )

2. Zi ~ N ( 0, 1 )2. Zi ~ N ( 0, 1 )

What is the difference in meaning between the What is the difference in meaning between the first and the second terms.first and the second terms.

STANDARD NORMAL VALUESTANDARD NORMAL VALUE ( Z ) ( Z )

XXii - - µµ ZZii = --------------- = ---------------

σσ ZZii MEASURESMEASURES THE DISTANCE OF ANY THE DISTANCE OF ANY

PARTICULAR VALUE OF X FROM THE PARTICULAR VALUE OF X FROM THE µµ, , MEASURED IN UNITS OF THE STANDARD MEASURED IN UNITS OF THE STANDARD DEVIATION.DEVIATION.

THE SIGN OF Z MIGHT BE POSITIVE, NEGATIVE THE SIGN OF Z MIGHT BE POSITIVE, NEGATIVE OR ZERO. IT DEPENDS ON THE DIFFERENCE OR ZERO. IT DEPENDS ON THE DIFFERENCE BETWEEN XBETWEEN Xii VALUE AND ITS MEAN ( VALUE AND ITS MEAN ( µµ ). ).

FINDING AREAS UNDER THE NORMAL CURVEFINDING AREAS UNDER THE NORMAL CURVE

ZZii-value PROBABILITY( AREAS )-value PROBABILITY( AREAS )

1.65 0.45051.65 0.4505

1.96 0.47501.96 0.4750

1.28 0.3991.28 0.39977

0.50 0.19150.50 0.1915

1.00 0.34131.00 0.3413Starting point of z value is always from where ZStarting point of z value is always from where Zii = 0 up = 0 up

to the point of another zto the point of another zii value ( the difference between value ( the difference between

mean and a selected value of Xmean and a selected value of Xii ). ).

Example: ( ZExample: ( Zii value :appendix D value :appendix D,, p:720) p:720)

The scores of 250 students in mathematics class follow the The scores of 250 students in mathematics class follow the normal distribution, with a mean of 69 and standard normal distribution, with a mean of 69 and standard deviation of 10. What is the probability or areas under this deviation of 10. What is the probability or areas under this normal curve for the scores :normal curve for the scores :

a.a. Between 60 and 65.Between 60 and 65.

b.b. ≥ ≥ 8080

c.c. Between 64 and 80.Between 64 and 80.

d.d. If the minimum passing grade is 55. how many students fail If the minimum passing grade is 55. how many students fail in this exam.in this exam.

e.e. What is the minimum scores for 5 % the best.What is the minimum scores for 5 % the best.

f.f. What is the maximum scores for 37.5% above the average.What is the maximum scores for 37.5% above the average.

Normal DistributionNormal Distribution

An electrical firmAn electrical firm manufactures light bulbs that have a manufactures light bulbs that have a length of life that is normally distributed with mean 800 length of life that is normally distributed with mean 800 hours and variance of 1600 hours. Find the probability:hours and variance of 1600 hours. Find the probability:

a. That a bulb burns between 778 and 834 hours.a. That a bulb burns between 778 and 834 hours.

b. That a bulb burns b. That a bulb burns ≥ 850 hours or ≤ 775 hours.≥ 850 hours or ≤ 775 hours.

c. That a bulb burns between 720 and 780 hours.c. That a bulb burns between 720 and 780 hours.

d. The minimum length of life for 12.5% the longest.d. The minimum length of life for 12.5% the longest.

e. The minimum length of life for 22% below the average. e. The minimum length of life for 22% below the average.

MID-EXAM QUESTIONSMID-EXAM QUESTIONS

1. MEASURES OF CENTRAL TENDENCY:1. MEASURES OF CENTRAL TENDENCY:

MEAN, MEDIAN AND MODE AND MEAN, MEDIAN AND MODE AND COEFFICIENT OF VARIATION.COEFFICIENT OF VARIATION.

2. MEASURES OF LOCALITY:2. MEASURES OF LOCALITY:

QUARTILE, DECILE AND PERCENTILES.QUARTILE, DECILE AND PERCENTILES.

3. DISPERSION: VARIANCE, STANDARD 3. DISPERSION: VARIANCE, STANDARD DEVIATION AND COEFFICIENT OF DEVIATION AND COEFFICIENT OF VARIATION FOR UNGROUPED DATA.VARIATION FOR UNGROUPED DATA.

MID EXAM QUESTIONSMID EXAM QUESTIONS

4. PROBABILITY ( 5 QUESTIONS ):4. PROBABILITY ( 5 QUESTIONS ):

1.1. INDEPENDENT TO CONDITIONAL PROBINDEPENDENT TO CONDITIONAL PROB

2.2. BAYES THEOREMBAYES THEOREM

3.3. BINOMIAL DISTRIBUTIONBINOMIAL DISTRIBUTION

4.4. NORMAL DISTRIBUTIONNORMAL DISTRIBUTION

5.5. BINOMIAL TO NORMALBINOMIAL TO NORMAL

MID EXAM QUESTIONSMID EXAM QUESTIONS BINOMIAL DISTRIBUTION ( 2 QUEST)BINOMIAL DISTRIBUTION ( 2 QUEST) NORMAL DISTRIBUTION ( 2 QUEST )NORMAL DISTRIBUTION ( 2 QUEST )

YOU MAY SELECT ONLY 4 QUESTIONS.YOU MAY SELECT ONLY 4 QUESTIONS.

DURATION = 120 MINUTESDURATION = 120 MINUTES

OPEN-BOOK AND NO CHEATING OK?OPEN-BOOK AND NO CHEATING OK?

THERE IS NO MORE MAKE-UP EXAM!THERE IS NO MORE MAKE-UP EXAM!

REWARD SYSTEM: CHOCOLATE BAR.REWARD SYSTEM: CHOCOLATE BAR.

WHEN U FIND ANY DIFFICULTIES PLEASE WHEN U FIND ANY DIFFICULTIES PLEASE DISCUSS WITH ME OR CONTACT TO DISCUSS WITH ME OR CONTACT TO 0811253224.0811253224.

The Normal approximation to the BinomialThe Normal approximation to the Binomial

Binomial distribution is effective if the number of Binomial distribution is effective if the number of trials is relatively small. But, generating a binomial trials is relatively small. But, generating a binomial distribution for a large number of trials would be distribution for a large number of trials would be very time consuming. A more efficient approach is to very time consuming. A more efficient approach is to apply the normal approximation to the Binomial:apply the normal approximation to the Binomial:

The approach is determined as follows:The approach is determined as follows:

µµ = n = n ππ Xi – n Xi – n ππ

σσ = = √√ n n ππ q then, Z =------------- q then, Z =-------------

√ √ n n ππ q q

Example Example

If 10 % of all business executives fill out a given If 10 % of all business executives fill out a given marketing survey questionnaire, what is the marketing survey questionnaire, what is the probability of getting:probability of getting:

a.a. At least 15 questionnaires At least 15 questionnaires will be backwill be back out of 200 out of 200 distributed to executives?distributed to executives?

b.b. What is the probability that between 23 and 3What is the probability that between 23 and 300 questionnaires will be filled out?questionnaires will be filled out?

c.c. What is the probability that 17 questionnaires or What is the probability that 17 questionnaires or less will be filled out.less will be filled out.

DonDon’’t forget to apply the correction factor!t forget to apply the correction factor!

Extra-BonusExtra-BonusA manufacture produces 10.000 units of product A manufacture produces 10.000 units of product

daily. On the basis of frequent inspections it is daily. On the basis of frequent inspections it is dicovered, 500 unit of product is classified as dicovered, 500 unit of product is classified as defective. What is the probability that:defective. What is the probability that:

a.a. More than 525 units of product is defective.More than 525 units of product is defective.

b.b. At least 9450 units of product is good.At least 9450 units of product is good.

c.c. Less than 475 units of product is defective.Less than 475 units of product is defective.

d.d. The number of good product is between 9450 units The number of good product is between 9450 units and 9550 units.and 9550 units.

The last quiz before mid examThe last quiz before mid exam

1. Binomial distribution1. Binomial distribution 2. Normal distribution2. Normal distribution 3. From Binomial to Normal.3. From Binomial to Normal.

The quiz will be given on Tuesday next The quiz will be given on Tuesday next week. ( don’t forget to bring your calculator week. ( don’t forget to bring your calculator and statistics tables. and statistics tables.

SAMPLING METHODS & ESTIMATION SAMPLING METHODS & ESTIMATION MODELSMODELS

Sampling MethodsSampling Methods REASONS TO SAMPLEREASONS TO SAMPLE

1. TO OBSERVE THE WHOLE POPULATION WO1. TO OBSERVE THE WHOLE POPULATION WOUULD LD BE TIME CONSUMING AND VERY EXPENSIVE.BE TIME CONSUMING AND VERY EXPENSIVE.

2. THE PHYSICAL IMPOSSIBILITY OF CHECKING ALL 2. THE PHYSICAL IMPOSSIBILITY OF CHECKING ALL ITEMS IN THE POPULATION.ITEMS IN THE POPULATION.

3. THE DESTRUCTIVE NATURE OF SOME TESTS.3. THE DESTRUCTIVE NATURE OF SOME TESTS.

4. THE SAMPLE RESULTS ARE ADEQUATE TO INFER 4. THE SAMPLE RESULTS ARE ADEQUATE TO INFER THE CHARACTERISTICS OF POPULATION.THE CHARACTERISTICS OF POPULATION.

ERROR: sampling error VS non-sampling errorERROR: sampling error VS non-sampling error

SAMPLING METHODSSAMPLING METHODS 1. SIMPLE RANDOM SAMPLE: A SAMPLE SELECTED SO THAT 1. SIMPLE RANDOM SAMPLE: A SAMPLE SELECTED SO THAT

EACH ITEM IN THE POPULATION HAS THE SAME EACH ITEM IN THE POPULATION HAS THE SAME PROBABILITY OF BEING INCLUDED.PROBABILITY OF BEING INCLUDED.

2. SYSTEMATIC RANDOM SAMPLE: A RANDOM STARTING 2. SYSTEMATIC RANDOM SAMPLE: A RANDOM STARTING POINT k IS SELECTEPOINT k IS SELECTEDD, AND THEN EVERY , AND THEN EVERY kth kth MEMBER OF MEMBER OF THE POPULATION IS SELECTED.THE POPULATION IS SELECTED.

3. STRATIFIED RANDOM SAMPLE : A POPULATION IS 3. STRATIFIED RANDOM SAMPLE : A POPULATION IS DIVIDED INTO SUB-GROUPS, CALLED STRATA, AND A DIVIDED INTO SUB-GROUPS, CALLED STRATA, AND A SAMPLE IS RANDOMLY SELECTED FROMSAMPLE IS RANDOMLY SELECTED FROM EACH STRATUM.EACH STRATUM.

4.CLUSTER SAMPLING: A POPULATION IS DIVIDED INTO 4.CLUSTER SAMPLING: A POPULATION IS DIVIDED INTO CLUSTERS USING NATURALLY OCCURING GEOGRAPHIC OR CLUSTERS USING NATURALLY OCCURING GEOGRAPHIC OR OTHER BOUNDARIES. THEN, CLUSTER ARE RANDOMLY OTHER BOUNDARIES. THEN, CLUSTER ARE RANDOMLY SELECTED AND A SAMPLE IS COLLECTED BY RANDOMLY SELECTED AND A SAMPLE IS COLLECTED BY RANDOMLY SELECTING FROM EACH CLUSTER. SELECTING FROM EACH CLUSTER.

Sampling MethodsSampling Methods

Purposive random samplingPurposive random sampling Proportional random samplingProportional random sampling

Non-Random samplingNon-Random sampling Convenience samplingConvenience sampling Judgment samplingJudgment sampling Snowball samplingSnowball sampling

HOW CAN SAMPLES BE USED TO MAKE ESTIMATES?HOW CAN SAMPLES BE USED TO MAKE ESTIMATES?

A SAMPLE IS A TOOL TO INFER SOMETHING ABOUT A SAMPLE IS A TOOL TO INFER SOMETHING ABOUT POPULATION OR STATISTICS CAN BE USED TO FIND POPULATION OR STATISTICS CAN BE USED TO FIND SOMETHING ABOUT A CHARACTERISTIC OF SOMETHING ABOUT A CHARACTERISTIC OF POPULATION OR A PARAMETER.POPULATION OR A PARAMETER.

EXAMPLE: EXAMPLE: X CAN BE USED TO ESTIMATE X CAN BE USED TO ESTIMATE µµ..

Central limit theorem :Central limit theorem :If all samples of a particular size are selected from any If all samples of a particular size are selected from any

population, the sampling distribution of the sample mean population, the sampling distribution of the sample mean is approximately a normal distribution. This is approximately a normal distribution. This approximation improves with larger samples.approximation improves with larger samples.

SAMPLING DISTRIBUTION OF THE SAMPLE MEANSAMPLING DISTRIBUTION OF THE SAMPLE MEAN

SAMPLING DISTRIBUTION OF THE SAMPLE MEAN IS A SAMPLING DISTRIBUTION OF THE SAMPLE MEAN IS A PROBABILITY DISTRIBUTION OF ALL POSSIBLE PROBABILITY DISTRIBUTION OF ALL POSSIBLE SAMPLE MEANS OF A GIVEN SAMPLE SIZE ( n ). SAMPLE MEANS OF A GIVEN SAMPLE SIZE ( n ).

THE FOLLOWING EXAMPLE ILLUSTRATES THE THE FOLLOWING EXAMPLE ILLUSTRATES THE CONTRUCTION OF A SAMPLING DISTRIBUTION OF CONTRUCTION OF A SAMPLING DISTRIBUTION OF SAMPLE MEAN. SAMPLE MEAN.

LET US ASSUME THAT A POPULATION CONSISTS OF LET US ASSUME THAT A POPULATION CONSISTS OF FOUR ELEMENTS X1 =1, X2 = 2, X3 = 3 AND X4 = 4. FOUR ELEMENTS X1 =1, X2 = 2, X3 = 3 AND X4 = 4. CONSIDER ALL THE SAMPLES OF A GIVEN SIZE n = 2 CONSIDER ALL THE SAMPLES OF A GIVEN SIZE n = 2 THAT COULD BE DRAWN FROM THIS POPULATION. THAT COULD BE DRAWN FROM THIS POPULATION. (CONSIDER CAREFULLY WHETHER ONE IS SAMPLING (CONSIDER CAREFULLY WHETHER ONE IS SAMPLING WITH OR WITHOUT REPLACEMENT).WITH OR WITHOUT REPLACEMENT).

Sampling with ReplacementSampling with Replacement All possible All possibleAll possible All possible

samples sample means samples sample meanssamples sample means samples sample means

1. 1.1 1.0 9. 3.1 2.01. 1.1 1.0 9. 3.1 2.0

2. 1.2 1.5 10. 3.2 2.52. 1.2 1.5 10. 3.2 2.5

3. 1.3 2.0 11. 3.3 3.03. 1.3 2.0 11. 3.3 3.0

4. 1.4 2.5 12. 3.4 3.54. 1.4 2.5 12. 3.4 3.5

5. 2.1 1.5 13. 4.1 2.55. 2.1 1.5 13. 4.1 2.5

6. 2.2 2.0 14. 4.2 3.06. 2.2 2.0 14. 4.2 3.0

7. 2.3 2.5 15. 4.3 3.57. 2.3 2.5 15. 4.3 3.5

8 2.4 3.0 16. 4.4 4.08 2.4 3.0 16. 4.4 4.0

µµ = ( 1 + 2 + 3 + 4 ) : 4 = 2.5 ( population mean ) = ( 1 + 2 + 3 + 4 ) : 4 = 2.5 ( population mean )

µµx = ( 1 + 1.5 + 2.0 +------------- + 4.0 ) : 16 = 2.5 The mean of all the x = ( 1 + 1.5 + 2.0 +------------- + 4.0 ) : 16 = 2.5 The mean of all the means is equal to the population mean or means is equal to the population mean or µµ..

Variance and Standard DeviationVariance and Standard Deviationσσ22= ( 1 – 2.5 )= ( 1 – 2.5 )22+( 2 – 2.5 )+( 2 – 2.5 )2 2 +( 3 – 2.5 )+( 3 – 2.5 )22 + ( 4 – 2.5 ) + ( 4 – 2.5 )22 : 4: 4

= = ( 2.25 + 0.25 + 0.25 + 2.25): 4 = 5/4( 2.25 + 0.25 + 0.25 + 2.25): 4 = 5/4 or or σσ22 = 5/4 = 1.25 = 5/4 = 1.25

σσ = √ 1.25 = 1.118 ( std deviation of population) = √ 1.25 = 1.118 ( std deviation of population)The standard deviation of the sample means is formally The standard deviation of the sample means is formally

referred as standard error of the mean.referred as standard error of the mean.σσxx22==( 1 -2.5 )( 1 -2.5 )22 + ( 1.5 – 2.5 ) + ( 1.5 – 2.5 )22 +------------------------- + ( 4 – 2.5 ) +------------------------- + ( 4 – 2.5 )22 : 16 : 16 = 10 : 16 = 0.625 and = 10 : 16 = 0.625 and σσx = √ 0.625 = 0.79 orx = √ 0.625 = 0.79 or σσ 1.118 1.118 σσx = ----- = ----------- = 0.79x = ----- = ----------- = 0.79 √ √ n √ 2n √ 2Note that when we increase n, the Note that when we increase n, the σσx will decrease.x will decrease.

Without ReplacementWithout Replacement

All possible samples Sample means All possible samples Sample means

1.2 1.51.2 1.5

1.3 2.01.3 2.0

1.4 2.51.4 2.5

2.3 2.52.3 2.5

2.4 3.02.4 3.0

3.4 3.53.4 3.5

µµx = ( 1.5 + 2.0 + ------- + 3.5 ) : 6 = 2.5x = ( 1.5 + 2.0 + ------- + 3.5 ) : 6 = 2.5

σσx2 = ( 1.5 – 2.5 )2 + ( 2.0 – 2.5 )2 +----- + ( 3.5 – 2.5)2 : 6x2 = ( 1.5 – 2.5 )2 + ( 2.0 – 2.5 )2 +----- + ( 3.5 – 2.5)2 : 6

= (1 + 0.25 + 0 + 0 + 0.25 + 1) = 2.5 : 6 = 0.417= (1 + 0.25 + 0 + 0 + 0.25 + 1) = 2.5 : 6 = 0.417

σσ

σσx = √ 0.417 = 0.645 or x = √ 0.417 = 0.645 or σσx = --- ( √ ( N – n ) : ( N – 1 )x = --- ( √ ( N – n ) : ( N – 1 )

√ √ n n

Example: Example: Sampling Distribution of The MeanSampling Distribution of The Mean

A population consists of three elements:A population consists of three elements:

X1 = 25, X2= 30 and X3 = 35.X1 = 25, X2= 30 and X3 = 35.

Calculate:Calculate:

a.a. Sampling distribution of the mean if the Sampling distribution of the mean if the sample size = 2 ( with replacement and sample size = 2 ( with replacement and without replacement ).without replacement ).

b. Variance and std deviation of the sample b. Variance and std deviation of the sample means.means.

1. 1. Confidence interval for the population mean ( Confidence interval for the population mean ( µµ ) ) when n ≥ 30 ( Z value )when n ≥ 30 ( Z value )

Confidence interval for Confidence interval for µµ is estimated by: is estimated by:

X X ±± a sampling error or X a sampling error or X ±± z value SE z value SE

Z value depends on confidence level.Z value depends on confidence level.

SE = standard error of the meanSE = standard error of the mean

SE = SE = σσ/√n, if std deviation of population is known./√n, if std deviation of population is known.

But, if the std deviation of population is unknown,But, if the std deviation of population is unknown,

SE = s/√n , s = std deviation of sample. SE = s/√n , s = std deviation of sample.

Sampling Error and Confidence Interval EstimateSampling Error and Confidence Interval Estimate Sampling Error : The diffrence between a sample Sampling Error : The diffrence between a sample

statistic and its corresponding population parameter. statistic and its corresponding population parameter. This error will decrease as the sample size increase.This error will decrease as the sample size increase.

Confidence interval estimate: A range of values Confidence interval estimate: A range of values constructed from sample data so that the population constructed from sample data so that the population parameter is likely to within that range occur at parameter is likely to within that range occur at specified probability. The specified probability is specified probability. The specified probability is called the level of confidence.called the level of confidence.

EXAMPLE: EXAMPLE: Confidence Interval for Confidence Interval for µµ ( n≥30 ( n≥30 ) )

The mean and std deviation for the quality grade The mean and std deviation for the quality grade point averages of a random samples of 36 college point averages of a random samples of 36 college seniors are calculated to be 2.9 and 0.3 seniors are calculated to be 2.9 and 0.3 respectively. respectively.

a. Find the 92 % confidence interval for the mean of a. Find the 92 % confidence interval for the mean of the entire senior class.the entire senior class.

b. How large the sample size is required if we want b. How large the sample size is required if we want to reduce the sampling error not more than 0.05.to reduce the sampling error not more than 0.05.

Confidence Interval for Confidence Interval for µµ ( n < 30 ) ( n < 30 )

The weights of 7 similar boxes of cereal are 9.8, 10.2, 10.4, The weights of 7 similar boxes of cereal are 9.8, 10.2, 10.4, 9.8, 10.0, 10.2 and 9.6 ounces.9.8, 10.0, 10.2 and 9.6 ounces.

Find a 90% confidence interval for he mean of all such boxes Find a 90% confidence interval for he mean of all such boxes of cereal. ( use t – table ).of cereal. ( use t – table ).

StepStep11: calculate the mean and std deviation of sample.: calculate the mean and std deviation of sample.

2: use t table to find critical value for 90% C.L ( tv).2: use t table to find critical value for 90% C.L ( tv).

3: substitute of all information from step 1 and 23: substitute of all information from step 1 and 2

into the formula:into the formula:

X - tv ( sd/X - tv ( sd/√n) ≤ u ≤ X + tv ( sd/√n )√n) ≤ u ≤ X + tv ( sd/√n )

Level of ConfidenceLevel of ConfidenceRequired level value of Required level value of αα value of Z table value of Z table

of confidenceof confidence

90 % 10 % 90 % 10 % ±± 1.65 1.65

95 % 5 % 95 % 5 % ±± 1.96 1.96

99 % 1 % 99 % 1 % ±± 2.58 2.58

Example : The prices at which certain type of instant coffee was Example : The prices at which certain type of instant coffee was being sold on a given day were collected from a random being sold on a given day were collected from a random sample of 45 shops around the country. The mean price was $ sample of 45 shops around the country. The mean price was $ 1.95 with a standard deviation of $ 0.27. Compute a 80 % 1.95 with a standard deviation of $ 0.27. Compute a 80 % confidence interval for the population mean. confidence interval for the population mean.

2. 2. A CONFIDENCE INTERVAL FOR A A CONFIDENCE INTERVAL FOR A

PROPORTION OF POPULATION ( PROPORTION OF POPULATION ( ππ ) )

A PROPORTION OF POPULATION ( A PROPORTION OF POPULATION ( ππ ) CAN BE ) CAN BE ESTIMATED BY THE FORMULA:ESTIMATED BY THE FORMULA:

P = X/nP = X/nP = THE FRACTION OR PERCENT INDICATING THE PART P = THE FRACTION OR PERCENT INDICATING THE PART

OF THE SAMPLE HAVING A PARTICULAR TRAIT OF OF THE SAMPLE HAVING A PARTICULAR TRAIT OF INTEREST.INTEREST.

n = SAMPLE SIZE, X IS A PART OF SAMPLE SIZE WHICH n = SAMPLE SIZE, X IS A PART OF SAMPLE SIZE WHICH HAVE A PARTICULAR TRAIT OF INTEREST.HAVE A PARTICULAR TRAIT OF INTEREST.

CONFIDENCE INTERVAL FOR POPULATION PROPORTION CONFIDENCE INTERVAL FOR POPULATION PROPORTION ( ( ππ ) )

P P ±± Z Zv v √√ P( 1- P ) / nP( 1- P ) / n page:300 no:18 page:300 no:18

ApplicationApplication

The distribution of household that favor a certain bath soap in The distribution of household that favor a certain bath soap in West Java Province (use a 90% confidence level).West Java Province (use a 90% confidence level).

Brand Number of householdsBrand Number of households Palmolive 1500Palmolive 1500 Lux 1800Lux 1800 Zest 1500Zest 1500 Beauty 1000Beauty 1000 Minty 1200Minty 1200 Other brands 2000Other brands 2000 Calculate confidence interval for the proportion of households Calculate confidence interval for the proportion of households

that favor LUX bath soap for their families.that favor LUX bath soap for their families.

4. 4. Confidence interval for the population mean ( Confidence interval for the population mean ( µµ ) when the ) when the sample size ( n < 30 ) t–table ( p:722 )sample size ( n < 30 ) t–table ( p:722 )

Confidence interval for the Confidence interval for the µµ is estimated by: is estimated by:

X X ±± a sampling error or X a sampling error or X ±± t table.SE t table.SE

SE = S/√n ( std deviation of population is unknown).SE = S/√n ( std deviation of population is unknown).Example :Example :

The operating life of rechargeable cordless screwdrivers The operating life of rechargeable cordless screwdrivers produced by a firm is assumed to be normally distributed. A produced by a firm is assumed to be normally distributed. A sample of 15 screwdrivers is tested and the mean life is found sample of 15 screwdrivers is tested and the mean life is found to be 8900 hours, with a sample std deviation of 500 hours. to be 8900 hours, with a sample std deviation of 500 hours. Construct a 90% confidence interval estimate for the Construct a 90% confidence interval estimate for the population mean.population mean.

Finding t - valueFinding t - value

Example : n = 10, confidence level = 90%.Example : n = 10, confidence level = 90%. n = 24, confidence level = 95%n = 24, confidence level = 95% Degree of freedom of t- value = df (n-1 )Degree of freedom of t- value = df (n-1 )

df Two-tailed test,df Two-tailed test,αα t-value t-value

9 9 10% 10% ±± 1.833 1.833

23 5% 23 5% ±± 2.069 2.069

3. 3. A confidence Interval for The diffrence between Two A confidence Interval for The diffrence between Two Population Means ( Population Means ( µµ1 - 1 - µµ2 ), n ≥ 302 ), n ≥ 30

Confidence interval for the diffrence between Confidence interval for the diffrence between µµ1 - 1 - µµ2 is:2 is:

( X1 – X2 ) ( X1 – X2 ) ±± Zv Zv σσxx

σσx = pooled standard deviation.x = pooled standard deviation.

X1 = the sample mean of X1X1 = the sample mean of X1

X2 = the sample mean of X2X2 = the sample mean of X2

n1 = the sample size of X1n1 = the sample size of X1

n2 = the sample size of x2n2 = the sample size of x2

ExampleExample

A study was made to estimate the difference in salaries A study was made to estimate the difference in salaries of college professors in the private and state colleges of college professors in the private and state colleges of Virginia. A random sample of 100 professor in the of Virginia. A random sample of 100 professor in the private colleges showed an average of $ 15.000 per private colleges showed an average of $ 15.000 per month with a standard deviation of $ 1200. A random month with a standard deviation of $ 1200. A random sample of 200 professors in state colleges showed an sample of 200 professors in state colleges showed an average salary of $ 16.000 with a standard deviation average salary of $ 16.000 with a standard deviation of $ of $ 11400. Find a 90% confidence interval for the 400. Find a 90% confidence interval for the difference between the average salaries of professors difference between the average salaries of professors teaching in state and private colleges in Virginia.teaching in state and private colleges in Virginia.

5. 5. Estimating the difference between two population Estimating the difference between two population means ( means ( µµ1- 1- µµ2 ), n< 302 ), n< 30

(( X1 X1 -- X2 X2)) ±± t tvv S SPP

A taxi company is trying to decide whether to purchase brand A or brand B tires for its A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleets of taxis. To estimate the difference in the two brands, an experiment is conducted fleets of taxis. To estimate the difference in the two brands, an experiment is conducted using using 112 tires of brand A and 2 tires of brand A and 114 tires of brand B. The results are:4 tires of brand B. The results are:

Brand A Brand BBrand A Brand BAverage wear-out 22.500 miles 23.600 milesAverage wear-out 22.500 miles 23.600 milesStd deviation 100 miles Std deviation 100 miles 550 miles0 miles

Compute 95% confidence interval for the difference between the average quality for the Compute 95% confidence interval for the difference between the average quality for the two brands. two brands.

EXAMPLE: POPULATION PROPORTIONEXAMPLE: POPULATION PROPORTION

A new program for a youth club is planned in a small city. To A new program for a youth club is planned in a small city. To determine whether or not the program will get the city determine whether or not the program will get the city governmentgovernment’’s support. It is necessary to estimate the proportion s support. It is necessary to estimate the proportion of young people who plan to use the clubof young people who plan to use the club’’s facilities. A survey of s facilities. A survey of 100 randomly selected young people has shown that 22 will use 100 randomly selected young people has shown that 22 will use the facilities if they become available. Construct a 90% the facilities if they become available. Construct a 90% confidence interval for the true proportion.confidence interval for the true proportion.

n = 100, X = 22, P = 22/100 = 0.22 and ( 1 – P ) = 0.78n = 100, X = 22, P = 22/100 = 0.22 and ( 1 – P ) = 0.78 Z value for CL = 90 % = Z value for CL = 90 % = ±± 1.65. 1.65.

Example:Example:

The performance of stocks between Cement The performance of stocks between Cement Industries and Oil industries in 2007 ( in % ).Industries and Oil industries in 2007 ( in % ).

Cement Industries Oil IndustriesCement Industries Oil Industries Average yields 22 18Average yields 22 18 Variance 3 2Variance 3 2 Sample size 35 40Sample size 35 40

Calculate interval estimate the difference in average Calculate interval estimate the difference in average yields between cement industries and oil industries yields between cement industries and oil industries ( CL = 85 % ).( CL = 85 % ).

Quiz 4 : Regression and CorrelationQuiz 4 : Regression and CorrelationMidterm grades Final gradesMidterm grades Final grades

77 77 82 82

50 50 66 66

71 7871 78

72 8072 80

9292 95 95

a. Construct the regression line by using OLS.a. Construct the regression line by using OLS.

b. Calculate R and R square.b. Calculate R and R square.

c. When you got score 75 in mid-exam, what is your prediction c. When you got score 75 in mid-exam, what is your prediction

score in final-exam?score in final-exam?

TESTING HYPOTHESES TESTING HYPOTHESES ( CH: 10 P 116-521) ( CH: 10 P 116-521)

HYPOTHESIS TESTING: A PROCEDURE BASED ON HYPOTHESIS TESTING: A PROCEDURE BASED ON SAMPLE EVIDENCE TO DETERMINE WHETHER THE SAMPLE EVIDENCE TO DETERMINE WHETHER THE HYPOTHESIS IS A REASONABLE STATEMENT.HYPOTHESIS IS A REASONABLE STATEMENT.

HYPOTHESIS: A STATEMENT ABOUT A POPULATION HYPOTHESIS: A STATEMENT ABOUT A POPULATION DEVELOPED FOR THE PURPOSE OF TESTING.DEVELOPED FOR THE PURPOSE OF TESTING.

NULL HYPOTHESIS ( Ho ): SPECIFIES THE VALUE OF NULL HYPOTHESIS ( Ho ): SPECIFIES THE VALUE OF THE PARAMETER TO BE TESTED. e.g: THE PARAMETER TO BE TESTED. e.g: µ, µ, ππ, ( µ1- µ2) OR , ( µ1- µ2) OR HYPOTHESES THAT WE FORMULATE WITH THE HYPOTHESES THAT WE FORMULATE WITH THE HOPE OF REJECTING.HOPE OF REJECTING.

ALTERNATIVE HYPOTHESIS ( Ha ): HYPOTHESES ALTERNATIVE HYPOTHESIS ( Ha ): HYPOTHESES THAT WE FORMULATE WITH THE HOPE OF THAT WE FORMULATE WITH THE HOPE OF ACCEPTING.ACCEPTING.

HYPOTHESIS TESTING PROCEDUREHYPOTHESIS TESTING PROCEDURE

TWO-TAILED TEST: BASICALLY BE USED TO TWO-TAILED TEST: BASICALLY BE USED TO TEST IF THE QUESTION ASK YOU ABOUT: TEST IF THE QUESTION ASK YOU ABOUT: THE DIFFERENCE, CHANGE OR EFFECT.THE DIFFERENCE, CHANGE OR EFFECT.

( ALPHA IS DIVIDED BY 2 OR ( ALPHA IS DIVIDED BY 2 OR ± ± 1/21/2αα ) ) ONE TAILED TEST : BASICALLY BE USED TO ONE TAILED TEST : BASICALLY BE USED TO

TEST IF THE QUESTION ASK YOU ABOUT: TEST IF THE QUESTION ASK YOU ABOUT: GREATER THAN, LESS THAN, INCREASE, GREATER THAN, LESS THAN, INCREASE, DECREASE, OR MORE THAN. THE CRITICAL DECREASE, OR MORE THAN. THE CRITICAL STATISTIC TABLE MAY BE IN THE LEFT OR STATISTIC TABLE MAY BE IN THE LEFT OR IN THE RIGHT TAIL OF THE CURVE. IN THE RIGHT TAIL OF THE CURVE.

THE GENERAL PROCEDURE FOR A HYPOTHESIS TESTTHE GENERAL PROCEDURE FOR A HYPOTHESIS TEST

START State the Ho and Ha ( two or one tailed test )START State the Ho and Ha ( two or one tailed test )

Select the desired CL or alphaSelect the desired CL or alpha

Determine the critical table ( Z, t, F or xDetermine the critical table ( Z, t, F or x2 2 ))

Determine the rejection and acceptance areas of Ho.Determine the rejection and acceptance areas of Ho.

Compute the critical ratio by using data sample ( test-statistic )Compute the critical ratio by using data sample ( test-statistic )

Compare the critical ratio and table ( accept or reject Ho )Compare the critical ratio and table ( accept or reject Ho )

Conclusion Conclusion

1.Testing the Population Mean ( 1.Testing the Population Mean ( µ ), n ≥ 30µ ), n ≥ 30

Example: Mr X, a mouthwash distributor, has stated that the Example: Mr X, a mouthwash distributor, has stated that the average cost to process a sales order is $ 13,25. Miss Y, cost average cost to process a sales order is $ 13,25. Miss Y, cost controller, fears that the average cost of processing is more controller, fears that the average cost of processing is more than $13.25. She is interested in taking action if cost are high, than $13.25. She is interested in taking action if cost are high, but she does not care if the actual average cost is below the but she does not care if the actual average cost is below the assumed value. A random sample of 100 orders had a sample assumed value. A random sample of 100 orders had a sample mean of $13.35, assuming the std deviation is $ 0.50. mean of $13.35, assuming the std deviation is $ 0.50. Conduct Conduct a test at 5 % level of significance and can you conclude that a test at 5 % level of significance and can you conclude that the average cost of a sales order is more than $13.25?the average cost of a sales order is more than $13.25?

TEST STATISTIC FOR TEST STATISTIC FOR µ ( n ≥ 30)µ ( n ≥ 30) Ho=Ho= Ha=Ha= CRITICAL VALUES ( Z TABLE )CRITICAL VALUES ( Z TABLE ) ACCEPTANCE & REJECTION AREASACCEPTANCE & REJECTION AREAS TEST STATISTIC: X - TEST STATISTIC: X - µµ Zh = ---------Zh = --------- SD/√nSD/√n. TEST STATISTIC IS FOUND FROM SAMPLE . TEST STATISTIC IS FOUND FROM SAMPLE

INFORMATION.INFORMATION.. DECISION: ACCEPT OR REJECT Ho.. DECISION: ACCEPT OR REJECT Ho.. CONCLUSION:. CONCLUSION:

2. TEST STATISTIC FOR 2. TEST STATISTIC FOR ππ (POPULATION PROPORTION) (POPULATION PROPORTION)

TEST STATISTIC:TEST STATISTIC:

P – P – ππ Zh = ---------------Zh = --------------- ππ ( 1 – ( 1 –ππ )/n )/n

TESTING FOR A PROPORTION TESTING FOR A PROPORTION POPULATIONPOPULATION

RESEARCH AT THE UGM INDICATES THAT 50% RESEARCH AT THE UGM INDICATES THAT 50% OF THE STUDENTS CHANGE THEIR MAJOR OF OF THE STUDENTS CHANGE THEIR MAJOR OF STUDY AFTER THEIR FIRST YEAR IN A STUDY AFTER THEIR FIRST YEAR IN A PROGRAM. A RANDOM SAMPLE OF 100 PROGRAM. A RANDOM SAMPLE OF 100 STUDENTS IN THE BUSINESS PROGRAM STUDENTS IN THE BUSINESS PROGRAM REVEALED THAT 48% HAD CHANGED THEIR REVEALED THAT 48% HAD CHANGED THEIR MAJOR AREA OF STUDY AFTER THEIR FIRST MAJOR AREA OF STUDY AFTER THEIR FIRST YEAR OF THE PROGRAM. YEAR OF THE PROGRAM. Has there been a Has there been a significant decrease in the proportion of students who significant decrease in the proportion of students who change their major after the first year in this program? change their major after the first year in this program? Test at the 0.05 level of significance.Test at the 0.05 level of significance.

Example:2Example:2Brand of bath soap Number of HouseholdBrand of bath soap Number of Household

Palmolive 1000Palmolive 1000

Lux 1500Lux 1500

Zest 1000Zest 1000

Beauty 750Beauty 750

Minty 1250Minty 1250

Maya 1000Maya 1000

Other Brands 3500Other Brands 3500

a.a. Can you conclude that the proportion of HH that like Lux Can you conclude that the proportion of HH that like Lux is different from 14%.is different from 14%.

b.b. Can you conclude that the proportion of HH that like Lux Can you conclude that the proportion of HH that like Lux is higher than 14 %. ( Alpha = 15 % ).is higher than 14 %. ( Alpha = 15 % ).

5. Testing The Population Mean (5. Testing The Population Mean (µ ), n < 30µ ), n < 30( Use t-table )( Use t-table ) p: 785 p: 785

Test Statistic:Test Statistic: X - X - µµ

th = ------------th = ------------

SD / nSD / n

ExampleExample

The distribution of gasoline consumption of 6 sample The distribution of gasoline consumption of 6 sample Honda Astrea can be reported as follow:Honda Astrea can be reported as follow:Sample RangeSample Range

1 54 km/l1 54 km/l

2 532 53

3 563 56

4 524 52

5 505 50

6 55 6 55

The manufacture claimed that the average of gasoline The manufacture claimed that the average of gasoline

Consumption was 55 km/l. Can you conclude that this claim is Consumption was 55 km/l. Can you conclude that this claim is

overestimate? ( use alpha = 5 % ). overestimate? ( use alpha = 5 % ).

TESTING THE DIFFERENCE BETWEEN TWO TESTING THE DIFFERENCE BETWEEN TWO POPULATION MEANS POPULATION MEANS

PARAMETRIC STATISTICS:PARAMETRIC STATISTICS: 1. SPSS: PAIRED SAMPLES1. SPSS: PAIRED SAMPLES 2. EVIEWS: UNPAIRED SAMPLES: 2. EVIEWS: UNPAIRED SAMPLES: EQUALITY OF VARIANCE TESTEQUALITY OF VARIANCE TEST 3. EXCEL: DATA ANALYSIS3. EXCEL: DATA ANALYSIS

NON PARAMETRIC STATISTICS:NON PARAMETRIC STATISTICS: SPSS : RELATED SAMPLES, WILCOXON SPSS : RELATED SAMPLES, WILCOXON

SIGN RANK TEST.SIGN RANK TEST.

3.TEST STATISTIC FOR (3.TEST STATISTIC FOR (µ1 -µ2), n ≥ 30µ1 -µ2), n ≥ 30(Independent-samples test ) (Independent-samples test )

Ho:Ho: Ha:Ha: TEST STATISTIC:TEST STATISTIC: X1 – X2X1 – X2 Zh = ---------------------------Zh = --------------------------- var1/n1 + var2/n2var1/n1 + var2/n2

EXAMPLEEXAMPLE THE PRODUCTION OF ELECTRONIC ITEMS DAILY THE PRODUCTION OF ELECTRONIC ITEMS DAILY

IN COMPANY ABC:IN COMPANY ABC:

MALE WOKERS FEMALE WORKERSMALE WOKERS FEMALE WORKERS

AVERAGE PROD 450 UNITS 460 UNITSAVERAGE PROD 450 UNITS 460 UNITS

VARIANCE 90 UNITS 110 UNITSVARIANCE 90 UNITS 110 UNITS

SAMPLE SIZE 40 WORKERS 50 WORKERSSAMPLE SIZE 40 WORKERS 50 WORKERS

CAN YOU CONCLUDE THAT FEMALE WORKERSCAN YOU CONCLUDE THAT FEMALE WORKERS

PRODUCTIVITY IS HIGHER THAN MALE WORKERS?PRODUCTIVITY IS HIGHER THAN MALE WORKERS?

( USE ALPHA = 10 % )( USE ALPHA = 10 % )

4:Two –Sample tests about Proportion ( 4:Two –Sample tests about Proportion ( ππ1 – 1 – ππ2 )2 )

A random sample of 100 young women revealed 19 liked A random sample of 100 young women revealed 19 liked the Heavenly frgrance well enough to purchase it. the Heavenly frgrance well enough to purchase it. Similarly, a sample of 200 older women revealed 62 liked Similarly, a sample of 200 older women revealed 62 liked the fragrance well enough to make a purchase. Can you the fragrance well enough to make a purchase. Can you conclude that there is a difference in the proportion of conclude that there is a difference in the proportion of younger and older women who like the Heavenly younger and older women who like the Heavenly fragrance? ( alpha = 5%).fragrance? ( alpha = 5%).

Pooled sample proportion Pc = (X1 + X2) / (n1 + n2 )Pooled sample proportion Pc = (X1 + X2) / (n1 + n2 )

Test Statistic: Zh = P1 –P2 / SdTest Statistic: Zh = P1 –P2 / Sd

Sd = Pc ( 1 – Pc ) / n1 + Pc ( 1 – Pc ) / n2Sd = Pc ( 1 – Pc ) / n1 + Pc ( 1 – Pc ) / n2

6.TEST STATISTIC FOR (6.TEST STATISTIC FOR (µ1 - µ2), n < 30µ1 - µ2), n < 30( USE- t TABLE )( USE- t TABLE )

TEST STATISTIC:TEST STATISTIC: X1 - X2X1 - X2 th = -------------------------------------------th = ------------------------------------------- ( n1-1) var1+ (n2-1) var2 1 + 1( n1-1) var1+ (n2-1) var2 1 + 1 (n1 + n2 – 2) n1 n2(n1 + n2 – 2) n1 n2

Testing (Testing (µ1 - µ2 ), n < 30µ1 - µ2 ), n < 30

Dr. Dony, a psychologist, administered IQ tests to Dr. Dony, a psychologist, administered IQ tests to determine if female FEB students were as smart as determine if female FEB students were as smart as male students. The random sample of 15 females male students. The random sample of 15 females had a mean score of 131 with std deviation of 17. had a mean score of 131 with std deviation of 17. The random sample of 13 male students had a The random sample of 13 male students had a mean of 126 and a std deviation of 14. At 0.01 level mean of 126 and a std deviation of 14. At 0.01 level of significance:of significance:

a. is there a significant difference in their IQ?a. is there a significant difference in their IQ? b. Can you conclude that females IQ is greater b. Can you conclude that females IQ is greater

than males IQ?than males IQ?

6. Dependent Samples testSample Productivity of Workers After Training Before Training d ( d-d ) ( d – d )2

1 235 units 228 units 7 2.4 5.762 210 205 5 0.4 0.163 231 219 12 7.4 54.764 242 240 2 -2.6 6.765 205 198 7 2.4 5.766 230 223 7 2.4 5.767 231 227 4 -0.6 0.368 210 215 -5 -9.6 92.169 225 222 3 -1.6 2.5610 249 245 4 -0.6 0.36 Can you conclude that the training program increase the

productivity of the employees? ( alpha = 5% ).

∑ ∑ d = 46 d = 46/10 = 4.6d = 46 d = 46/10 = 4.6

SD = ∑ ( d – d )2SD = ∑ ( d – d )2 ( n – 1 )( n – 1 ) Test statistic: dTest statistic: d SD / nSD / n

Example:2Example:2Sample of customer Score of: New Menu Old menuSample of customer Score of: New Menu Old menu

1 36 351 36 35

2 48 462 48 46

3 50 513 50 51

4 76 744 76 74

5 55 565 55 56

6 60 596 60 59

7 71 727 71 72

8 66 648 66 64

Can you conclude that the new menu is more delicious than Can you conclude that the new menu is more delicious than the old one? ( alpha = 1% ).the old one? ( alpha = 1% ).

7: CHY SQUARE TEST7: CHY SQUARE TESTTable: p: 787Table: p: 787

Chy- square tests are used in a procedure that Chy- square tests are used in a procedure that involves the comparison of the differences between involves the comparison of the differences between the sample frequencies of the occurrence ( Oij ) and the sample frequencies of the occurrence ( Oij ) and the hypothetical or theoretical population frequencies the hypothetical or theoretical population frequencies ( Eij ) ( Expected value ). ( Goodness of Fit ).( Eij ) ( Expected value ). ( Goodness of Fit ).

It can also be used to test relationship between It can also be used to test relationship between variables ( independency test ).variables ( independency test ).

Critical value of xCritical value of x22 depends on the number of rows depends on the number of rows and columnsand columns

Chy square tableChy square table

Chy- square table is always one tailed in the Chy- square table is always one tailed in the right side of the curve.right side of the curve.

Critical table is XCritical table is X22 άά df ( r -1 )( c - 1 ). df ( r -1 )( c - 1 ). Ho is accepted if test statistics calculated is Ho is accepted if test statistics calculated is

less than or equal to its critical table.less than or equal to its critical table. Ho is rejected if test statistics calculated is Ho is rejected if test statistics calculated is

greater than its critical table. greater than its critical table.

Chy-Square testChy-Square test

Test statistic:Test statistic: ( Oij – Eij )( Oij – Eij )22

XXhh2 2 ==ΣΣ -------------- --------------

EijEij Oij = observed frequencies in iOij = observed frequencies in i thth row & j row & jthth column column

Eij = expected frequencies in iEij = expected frequencies in i thth row & j row & jthth column column

EXAMPLEEXAMPLE A GARMENT COMPANY IN CAKUNG RECORDS A GARMENT COMPANY IN CAKUNG RECORDS

THE PERFORMANCE OF ITS LABOUR THE PERFORMANCE OF ITS LABOUR PRODUCTIVITY RANDOMLY.PRODUCTIVITY RANDOMLY.

THE LEVEL OF PRODUCTIVITYTHE LEVEL OF PRODUCTIVITY LOW MODERATE HIGHLOW MODERATE HIGH WORK-SHIFTWORK-SHIFT MORNING (I) 40 45 50MORNING (I) 40 45 50 AFTERNOON(II) 60 55 60AFTERNOON(II) 60 55 60 NIGHT(III) 40 30 25NIGHT(III) 40 30 25 FROM THE DATA ABOVE, WHAT IS YOUR FROM THE DATA ABOVE, WHAT IS YOUR

CONCLUSION? ( ALPHA = 5 % ).CONCLUSION? ( ALPHA = 5 % ).

CHY SQUARE: INDEPENDENCY TESTCHY SQUARE: INDEPENDENCY TEST

DISTRIBUTION OF DATA COLLECTED BY UNILEVER DISTRIBUTION OF DATA COLLECTED BY UNILEVER

INDONESIA WAS RECORDED AS FOLLOWINDONESIA WAS RECORDED AS FOLLOW:: Color of bath soap: Pink White YellowColor of bath soap: Pink White Yellow Gender: Man 10 10 20Gender: Man 10 10 20 Woman 25 20 5Woman 25 20 5

What is your conclusion? ( UseAlpha = 5 % )What is your conclusion? ( UseAlpha = 5 % )

Exercises: check your book p : 661-662-663.Exercises: check your book p : 661-662-663.

ANALYSIS of VARIANCEANALYSIS of VARIANCE This method can be used to test the difference This method can be used to test the difference

among population means ( the number of among population means ( the number of specific sample is more than two or k > 2.specific sample is more than two or k > 2.

The steps of testing the hypothesis.The steps of testing the hypothesis. 1. Calculate the variance within samples ( 1. Calculate the variance within samples ( σσ22

ww).). 2. Calculate variance between or among the 2. Calculate variance between or among the

samples (samples (σσ22bb).).

3. Calculate F table.3. Calculate F table. 4. Calculate F statistic. 4. Calculate F statistic.

ANALYSIS OF VARIANCEANALYSIS OF VARIANCE

FERTILIZER: A B C DFERTILIZER: A B C D

SAMPLESAMPLE

11 10 kgs 12 kgs 11 kgs 9 kgs10 kgs 12 kgs 11 kgs 9 kgs

22 12 11 10 1012 11 10 10

33 13 10 9 813 10 9 8

44 11 10 10 811 10 10 8

55 14 12 10 1014 12 10 10

Can you conclude that the average productivity of thoseCan you conclude that the average productivity of those

fertilizers are significantly different? ( alpha = 5% ). fertilizers are significantly different? ( alpha = 5% ).