CHAPTER 6.1 SUMMARIZING POSSIBLE OUTCOMES AND THEIR PROBABILITIES

1

CHAPTER 6.1SUMMARIZING POSSIBLE OUTCOMES AND THEIR

PROBABILITIES

• DEFINITION: A RANDOM VARIABLE IS A NUMERICAL MEASUREMENT OF THE OUTCOME OF A RANDOM PHENOMENON (EXPERIMENT).

• DEFINITION: A DISCRETE RANDOM VARIABLE X TAKES ITS VALUES FROM A COUNTABLE SET, FOR EXAMPLE, N = {0, 1, 2, 3, 4, 5, 6, 7, . . . }.

• DEFINITION: THE PROBABILITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE IS A FUNCTION SUCH THAT FOR ALL OUTCOMES

x

xpandxp 1)(1)(0

2

MEAN OF A DISCRETE PROBABILITY DISTRIBUTION

• THE MEAN OF A PROBABILITY DISTRIBUTION FOR A DISCRETE RANDOM VARIABLE IS GIVEN BY

IN WORDS, TO GET THE MEAN OF A DISCRETE PROBABILITY DISTRIBUTION, MULTIPLY EACH POSSIBLE VALUE OF THE RANDOM VARIABLE BY ITS PROBABILITY, AND THEN ADD ALL THESE PRODUCTS.

ix

ii xpxXE )()(

3

EXAMPLE: NUMBER OF HOME RUNS IN A GAME FOR BOSTON RED SOX

NUMBER OF HOME RUNS PROBABILITY

0 0.23

1 0.38

2 0.22

3 0.13

4 0.03

5 0.01

6 OR MORE 0.00

SUM 1.00

4

(1) WHAT IS THE EXPECTED (MEAN) NUMBER OF HOME RUNS FOR A BOSTON RED SOX BASEBALL GAME?

(2) INTERPRET WHAT THIS MEAN (EXPECTED VALUE) MEANS.

5

PROBABILITY FOR CONTINUOUS RANDOM VARIABLE

• DEFINITION: A CONTINUOUS RANDOM VARIABLE HAS POSSIBLE VALUES THAT FORM AN INTERVAL, THAT IS, TAKES ITS VALUES FROM AN INTERVAL, FOR EXAMPLE, (2 , 5).

• DEFINITION: THE PROBABILITY DISTRIBUTION OF A CONTINUOUS RANDOM VARIABLE IS SPECIFIED BY A CURVE THAT DETERMINES THE PROBABILITY THAT THE RANDOM VARIABLE FALLS IN ANY PARTICULAR INTERVAL OF VALUES.

6

REMARKS

• EACH INTERVAL HAS PROBABILITY BETWEEN 0 AND 1. THIS IS THE AREA UNDER THE CURVE, ABOVE THAT INTERVAL.

• THE INTERVAL CONTAINING ALL POSSIBLE VALUES HAS PROBABILITY EQUAL TO 1, SO THE TOTAL AREA UNDER THE CURVE EQUALS 1.

• ILLUSTRATIVE PICTURES

7

CHAPTER 6.2 FINDING PROBABILITIES FOR BELL – SHAPED DISTRIBUTIONS – THE NORMAL DISTRIBUTION

• THE NORMAL DISTRIBUTION IS VERY COMMONLY USED FOR CONTINUOUS RANDOM VARIABLES. IT IS CHARACTERIZED BY A PARTICULAR SYMMETRIC, BELL – SHAPED CURVE WITH TWO PARAMETERS – THE MEAN AND STANDARD DEVIATION.

• NOTATION

• ILLUSTRATIVE PICTURES

),( N

8

•THE NORMAL DISTRIBUTION IS ALSO THE MODEL FOR A POPULATION DISTRIBUTION

• THE POPULATION DISTRIBUTION OF A RANDOM VARIABLE X IS OFTEN MODELED BY A BELL – SHAPED CURVE WITH THE PROPERTIES THAT THE PROPORTION OF THE POPULATION FOR WHICH X IS BETWEEN a AND b, IS THE AREA UNDER THE CURVE, AND BETWEEN a AND b.

• ILLUSTRATIVE PICTURE

9

THE EMPIRICAL OR 68 – 95 – 99.7 % RULE

• THE EMPIRICAL RULE STATES THAT FOR AN APPROXIMATELY BELL – SHAPED DISTRIBUTION, ABOUT 68% OF OBSERVATIONS(VALUES) FALL WITHIN ONE STANDARD DEVIATION OF THE MEAN; 95% OF THE VALUES FALL WITHIN TWO STANDARD DEVIATIONS OF THE MEAN; 99.7% OF VALUES FALL WITHIN THREE STANDARD DEVIATIONS OF THE MEAN.

• ILLUSTRATIVE PICTURE

10

FINDING PROBABILITIES FOR CONTINUOUS RANDOM VARIABLES USING THE STANDARD NORMAL

DISTRIBUTION TABLE

• DEFINITION: THE STANDARD NORMAL DISTRIBUTION IS THE NORMAL DISTRIBUTION WITH MEAN = 0 AND STANDARD DEVIATION = 1. IT IS THE DISTRIBUTION OF NORMAL Z – SCORES.

• DEFINITION: THE Z – SCORE FOR A VALUE x OF A RANDOM VARIABLE IS THE NUMBER OF STANDARD DEVIATIONS THAT x FALLS FROM THE MEAN. IT IS CALCULATED AS

•

s

xxzor

xz

11

CLASS EXAMPLE 1

• IN A STANDARD NORMAL MODEL, WHAT PERCENT OF POPULATION IS IN EACH REGION? DRAW A PICTURE IN EACH CASE.

(A) Z < 0.83 (B) Z > 0.83 (C) 0.1 < Z < 0.9

SOLUTION

12

CLASS EXAMPLE 2

• IN A STANDARD NORMAL MODEL, FIND THE VALUE OF Z THAT CUTS OFF

• (A) THE LOWEST 75% OF POPULATION; • (B) THE HIGHEST 20% OF POPULATION (= THE

LOWEST 80%)• SOLUTION

13

CLASS EXAMPLE 3

• SUPPOSE THAT WE MODEL SAT SCORES Y, BY N(500, 100) DISTRIBUTION.

• (A) WHAT PERCENTAGE OF SAT SCORES FALL BETWEEN 450 AND 600?

• (B) FOR WHAT SAT VALUE b, 10% OF SAT SCORES ARE GREATER THAN b?

• SOLUTION

14

CHAPTER 6.3 PROBABILITY MODELS FOR OBSERVATIONS WITH

TWO POSSIBLE OUTCOMES

BERNOULLI TRIALA RANDOM EXPERIMENT WITH TWO

COMPLEMENTARY EVENTS, SUCCESS (S) AND FAILURE (F) IS CALLED A BERNOULLI TRIAL.

P(SUCCESS) = p

P(FAILURE) = q = 1 - p

15

EXAMPLES

• TOSSING A COIN 20 TIMES

SUCCESS = HEADS WITH p = 0.5 AND FAILURE = TAILS WITH q = 1 – p = 0.5

• TAKING A MULTIPLE CHOICE EXAM UNPREPARED.

SUCCESS = CORRECT ANSWER

FAILURE = WRONG ANSWER

p = 0.2; q = 1 – p = 1 – 0.2 = 0.8

16

• PRODUCTS COMING OUT OF A PRODUCTION LINE

SUCCESS = DEFECTIVE ITEMS

FAILURE = NON-DEFECTIVE ITEMS

• ROLLING A DIE 10 TIMES

SUCCESS = GETTING A 6; p = 1/6

FAILURE = NOT GETTING A 6; q = 5/6

17

•AN OFFER FROM A BANK FOR A CREDIT CARD WITH HIGH INTEREST RATE

SUCCESS = DECLINE; FAILURE = ACCEPT

• HAVING HEALTH INSURANCE

SUCCESS = HAVE; FAILLURE = NOT HAVE

• A REFERENDUM WHETHER TO RECALL AN UNFAITHFUL GOVERNOR FROM OFFICE

SUCCESS = VOTE YES; FAILLURE = VOTE NO

18

GEOMETRIC PROBABILITY MODEL

• QUESTION: HOW LONG WILL IT TAKE TO ACHIEVE THE FIRST SUCCESS IN A SERIES OF BERNOULLI TRIALS?

• THE MODEL THAT TELLS US THIS PROBABILITY (THAT IS, THE PROBABILITY UNTIL FIRST SUCCESS) IS CALLED THE GEOMETRIC PROBABILITY MODEL.

19

CONDITIONS

• THE FOLLOWING CONDITIONS MUST HOLD BEFORE USING THE GEOMETRIC PROBABILITY MODEL.

(1) THE TRIALS MUST BE BERNOULLI, THAT IS, THE RANDOM EXPERIMENT MUST HAVE TWO COMPLEMENTARY OUTCOMES – SUCCESS AND FAILURE;

(2) THE TRIALS MUST BE INDEPENDENT OF ONE ANOTHER;

(3) THE PROBABILITY OF SUCCESS IS THE SAME FOR EACH TRIAL.

20

GEOMETRIC PROBABILITY MODEL FOR BERNOULLI TRIALS

• LET p = PROBABILAITY OF SUCCESS

AND q = 1 – p = PROBABILITY OF FAILURE

X = NUMBER OF TRIALS UNTIL FIRST SUCCESS OCCURS

pqxXP x 1)(

2)(

1)(

p

qXSD

pXE

21

EXAMPLE

• ASSUME THAT 13% OF PEOPLE ARE LEFT-HANDED. IF WE SELECT 5 PEOPLE AT RANDOM, FIND THE PROBABILITY OF EACH OUTCOME DESCRIBED BELOW.

• (1) THE FIRST LEFTY IS THE FIFTH PERSON CHOSEN? 0.0745

• (2) THE FIRST LEFTY IS THE SECOND OR THIRD PERSON. 0.211

• (3) IF WE KEEP PICKING PEOPLE UNTIL WE FIND A LEFTY, HOW LONG WILL YOU EXPECT IT WILL TAKE?

7.69 PEOPLE

22

EXAMPLE

• AN OLYMPIC ARCHER IS ABLE TO HIT THE BULL’S-EYE 80% OF THE TIME. ASSUME EACH SHOT IS INDEPENDENT OF THE OTHERS. IF SHE SHOOTS 6 ARROWS, WHAT’S THE PROBABILITY THAT

• (1) HER FIRST BULL’S-EYE COMES ON THE THIRD ARROW? ANS = 0.032

• (2) HER FIRST BULL’S-EYE COMES ON THE FOURTH OR FIFTH ARROW? ANS = 0.00768

• IF SHE KEEPS SHOOTING ARROWS UNTIL SHE HITS THE BULL’S-EYE, HOW LONG DO YOU EXPECT IT WILL TAKE? ANS = 1.25 SHOTS

23

BINOMIAL PROBABILITY MODEL FOR BERNOULLI TRIALS

• QUESTION: WHAT IS THE NUMBER OF SUCCESSES IN A SPECIFIED NUMBER OF TRIALS?

• THE BINOMIAL PROBABILITY MODEL ANSWERS THIS QUESTION, THAT IS, THE PROBABILITY OF EXACTLY k SUCCESSES IN n TRIALS.

• CONDITIONS: SAME AS THOSE FOR THE GEOMETRIC PROBABILITY MODEL

24

BINOMIAL PROBABILITY MODEL

• LET n = NUMBER OF TRIALS

p = PROBABILITY OF SUCCESS

q = PROBABILITY OF FAILURE

X = NUMBER OF SUCCESSESS IN n TRIALS

)!(!

!

,)(

knk

n

k

n

whereqpk

nkXP knk

25

n! = n(n-1)(n-2)(n-3) … 3.2.1

npqXSD

npXE

)(

)(

26

EXAMPLES

• COMPUTE

(1) 3! (2) 4! (3) 5! (4) 6!• COMPUTE

0

12)3(

7

10)2(

2

5)1(

27

EXAMPLE

• ASSUME THAT 13% OF PEOPLE ARE LEFT-HANDED. IF WE SELECT 5 PEOPLE AT RANDOM, FIND THE PROBABILITY OF EACH OUTCOME BELOW.

• (1) THERE ARE EXACTLY 3 LEFTIES IN THE GROUP.• 0.0166

• (2) THERE ARE AT LEAST 3 LEFTIES IN THE GROUP.• 0.0179

• (3) THERE ARE NO MORE THAN 3 LEFTIES IN THE GROUP. 0.9987

28

EXAMPLE

• AN OLYMPIC ARCHER IS ABLE TO HIT THE BULL’S-EYE 80% OF THE TIME. ASSUME EACH SHOT IS INDEPENDENT OF THE OTHERS. IF SHE SHOOTS 6 ARROWS, WHAT’S THE PROBABILITY THAT

• (1) SHE GETS EXACTLY 4 BULL’S-EYES? 0.246• (2) SHE GETS AT LEAST 4 BULL’S-EYES? 0.901• (3) SHE GETS AT MOST 4 BULL’S-EYES? 0.345• (4) SHE MISSES THE BULL’S-EYE AT LEAST ONCE?

• 0.738• (5) HOW MANY BULL’S-EYES DO YOU EXPECT HER

TO GET? 4.8 BULL’SEYES• (6) WITH WHAT STANDARD DEVIATION? 0.98

29

THE NORMAL MODEL TO THE RESCUE OF BINOMIAL MODEL

• IF n, THE FIXED NUMBER OF TRIALS IS LARGE,

THAT IS,

THEN, THE BINOMIAL CUMULATIVE PROBABILITIES CAN BE APPROXIMATED BY THE NORMAL PROBABILITIES WITH THE SAME MEAN OR EXPECTED VALUE = n*p

AND, THE SAME STANDARD DEVIATION =

= SQRT( n*p*q)

10np 10nq

30

EXAMPLE

• TENNESSEE RED CROSS COLLECTED BLOOD FROM 32,000 DONORS. WHAT IS THE PROBABILITY THAT THEY HAD AT LEAST 1850 DONORS OF THE O-NEGATIVE BLOOD GROUP. THE PROBABILITY OF SOMEONE HAVING A 0-NEGATIVE BLOOD TYPE IS 0.06.

• SOLUTION: LET X BE SOMEONE OF THE O-NEGATIVE BLOOD GROUP. THEN THE QUESTION CAN BE FORMULATED MATHEMATICALLY AS

?)1850( XP

31

CHAPTER 6.4HOW LIKELY ARE THE POSSIBLE VALUES OF A

STATISTICS?

• REMINDER: A STATISTIC IS A NUMERICAL SUMMARY OF A SAMPLE DATA. SOME EXAMPLES ARE: SAMPLE PROPORTION, SAMPLE MEAN.

• DEFINITION: THE SAMPLING DISTRIBUTION OF A STATISTIC IS THE PROBABILITY DISTRIBUTION THAT SPECIFIES PROBABILITIES FOR THE POSSIBLE VALUES THE STATISTIC CAN TAKE.

32

SAMPLING DISTRIBUTION MODELS FOR PROPORTIONS AND MEANS

• SAMPLING DISTRIBUTION MODEL FOR A PROPORTION

PROBLEM FORMULATION: SUPPOSE THAT p IS AN UNKNOWN PROPORTION OF ELEMENTS OF A CERTAIN TYPE S IN A POPULATION.

EXAMPLES• PROPORTION OF LEFT - HANDED PEOPLE;• PROPORTION OF HIGH SCHOOL STUDENTS WHO

ARE FAILING A READING TEST;• PROPORTION OF VOTERS WHO WILL VOTE FOR

MR. X.

33

ESTIMATION OF p

• TO ESTIMATE p, WE SELECT A SIMPLE RANDOM SAMPLE (SRS), OF SIZE SAY, n = 1000, AND COMPUTE THE SAMPLE PROPORTION.

• SUPPOSE THE NUMBER OF THE TYPE WE ARE INTERESTED IN, IN THIS SAMPLE OF n = 1000 IS x = 437. THEN THE SAMPLE PROPORTION

IS COMPUTED USING THE FORMULA

n

xp ˆ

p̂

34

IN THE EXAMPLE ABOVE

%7.431000

437ˆ p

35

WHAT IS THE ERROR OF ESTIMATION?

• THAT IS, WHAT IS

• WHAT MODEL CAN HELP US FIND THE BEST ESTIMATE OF THE TRUE PROPORTION OF p?

• LET’S START THE ANALYSIS BY FIRST ANSWERING THE SECOND QUESTION.

?ˆ pp

36

APPROACH

• SUPPOSE THAT WE TAKE A SECOND SAMPLE OF SIZE 1000 AND COMPUTE P(HAT); CLEARLY, THE NEW ESTIMATE WILL BE DIFFERENT FROM 0.437. NOW, TAKE A THIRD SAMPLE, A FOURTH SAMPLE, UNTIL THE TWO THOUSANDTH (2000 –TH) SAMPLE, EACH OF SIZE 1000. IT IS OBVIOUS THAT WE WILL LIKELY OBTAIN TWO THOUSAND DIFFERENT P(HATS) AS ILLUSTRATED IN THE TABLE BELOW.

37

TABLE OF 2000 SAMPLES OF SIZE EACH n=1000, AND THEIR CORRESPONDING P(HATS)

SAMPLES OF SIZE n P(HATS)

… …

1n 1p̂

2n 2p̂

2000n 2000p̂

38

WHAT DO WE DO WITH THE DATA FOR P(HATS)?

• WE CONSTRUCT A HISTOGRAM OF THESE 2000 P(HATS).# OF SAMPLES

P(HATS)p

39

WHAT WE OBSERVE FROM THE HISTOGRAM

• THE HISTOGRAM ABOVE IS AN EXAMPLE OF WHAT WE WOULD GET IF WE COULD SEE ALL THE PROPORTIONS FROM ALL POSSIBLE SAMPLES. THAT DISTRIBUTION HAS A SPECIAL NAME. IT IS CALLED THE SAMPLING DISTRIBUTION OF THE PROPORTIONS.

• OBSERVE THAT THE HISTOGRAM IS UNIMODAL, ROUGHLY SYMMETRIC, AND IT’S CENTERED AT P WHICH IS THE TRUE PROPORTION

40

WHAT DOES THE SHAPE OF THE HISTOGRAM REMIND US ABOUT A MODEL THAT MAY JUST BE THE RIGHT ONE FOR SAMPLE PROPORTIONS?

• ANSWER: IT IS AMAZING AND FORTUNATE THAT A NORMAL MODEL IS JUST THE RIGHT ONE FOR THE HISTOGRAMS OF SAMPLE PROPORTIONS.

• HOW GOOD IS THE NORMAL MODEL?– IT IS GOOD IF THE FOLLOWING

ASSUMPTIONS AND CONDITIONS HOLD.

41

ASSUMPTIONS AND CONDITIONS

• ASSUMPTIONS• INDEPENDENCE ASSUMPTION: THE

SAMPLED VALUES MUST BE INDEPENDENT OF EACH OTHER.

• SAMPLE SIZE ASSUMPTION: THE SAMPLE SIZE, n, MUST BE LARGE ENOUGH

• REMARK: ASSUMPTIONS ARE HARD – OFTEN IMPOSSIBLE TO CHECK. THAT’S WHY WE ASSUME THEM. GLADLY, SOME CONDITIONS MAY PROVIDE INFORMATION ABOUT THE ASSUMPTIONS.

42

CONDITIONS

• RANDOMIZATION CONDITION: THE DATA VALUES MUST BE SAMPLED RANDOMLY. IF POSSIBLE, USE SIMPLE RANDOM SAMPLING DESIGN TO SAMPLE THE POPULATION OF INTEREST.

• 10% CONDITION: THE SAMPLE SIZE, n, MUST BE NO LARGER THAN 10% OF THE POPULATION OF INTEREST.

• SUCCESS/FAILURE CONDITION: THE SAMPLE SIZE HAS TO BE BIG ENOUGH SO THAT WE EXPECT AT LEAST 10 SUCCESSES AND AT LEAST 10 FAILLURES. THAT IS,

)(10

)(10

FAILLUREnq

SUCCESSnp

43

THE CENTRAL LIMIT THEOREM FOR THE SAMPLING DISTRIBUTION OF A PROPORTION

• FOR A LARGE SAMPLE SIZE n, THE SAMPLING DISTRIBUTION OF P(HAT) IS APPROXIMATELY

THAT IS, P(HAT) IS NORMAL WITH

q

ppN ,

n

pqpDEVIATIONSTANDARD

ppEMEAN

)ˆ(

)ˆ(

44

EXAMPLE 1

• ASSUME THAT 30% OF STUDENTS AT A UNIVERSITY WEAR CONTACT LENSES

• (A) WE RANDOMLY PICK 100 STUDENTS. LET P(HAT) REPRESENT THE PROPORTION OF STUDENTS IN THIS SAMPLE WHO WEAR CONTACTS. WHAT’S THE APPROPRIATE MODEL FOR THE DISTRIBUTION OF P(HAT)? SPECIFY THE NAME OF THE DISTRIBUTION, THE MEAN, AND THE STANDARD DEVIATION. BE SURE TO VERIFY THAT THE CONDITIONS ARE MET.

• (B) WHAT’S THE APPROXIMATE PROBABILITY THAT MORE THAN ONE THIRD OF THIS SAMPLE WEAR CONTACTS?

45

SOLUTION TO EXAMPLE 1

46

EXAMPLE 2

• INFORMATION ON A PACKET OF SEEDS CLAIMS THAT THE GERMINATION RATE IS 92%. WHAT’S THE PROBABILITY THAT MORE THAN 95% OF THE 160 SEEDS IN THE PACKET WILL GERMINATE? BE SURE TO DISCUSS YOUR ASSUMPTIONS AND CHECK THE CONDITIONS THAT SUPPORT YOUR MODEL.

• SOLUTION

47

CHAPTER 6.5 – 6.6 SAMPLING DISTRIBUTION OF THE SAMPLE

MEAN

APPROACH FOR ESTIMATING

SAME AS FOR SAMPLING DISTRIBUTION FOR PROPORTIONS ILLUSTRATED ABOVE

X

n

xxxxTHATRECALL n

...21

X

48

ASSUMPTIONS AND CONDITIONS

• ASSUMPTIONS• INDEPENDENCE ASSUMPTION: THE SAMPLED

VALUES MUST BE INDEPENDENT OF EACH OTHER

• SAMPLE SIZE ASSUMPTION: THE SAMPLE SIZE MUST BE SUFFICIENTLY LARGE.

• REMARK: WE CANNOT CHECK THESE DIRECTLY, BUT WE CAN THINK ABOUT WHETHER THE INDEPENDENCE ASSUMPTION IS PLAUSIBLE.

49

CONDITIONS

• RANDOMIZATION CONDITION: THE DATA VALUES MUST BE SAMPLED RANDOMLY, OR THE CONCEPT OF A SAMPLING DISTRIBUTION MAKES NO SENSE. IF POSSIBLE, USE SIMPLE RANDOM SAMPLING DESIGN TO ABTAIN THE SAMPLE.

• 10% CONDITION: WHEN THE SAMPLE IS DRAWN WITHOUT REPLACEMENT (AS IS USUALLY THE CASE), THE SAMPLE SIZE, n, SHOULD BE NO MORE THAN 10% OF THE POPULATION.

• LARGE ENOUGH SAMPLE CONDITION: IF THE POPULATION IS UNIMODAL AND SYMMETRIC, EVEN A FAIRLY SMALL SAMPLE IS OKAY. IF THE POPULATION IS STRONGLY SKEWED, IT CAN TAKE A PRETTY LARGE SAMPLE TO ALLOW USE OF A NORMAL MODEL TO DESCRIBE THE DISTRIBUTION OF SAMPLE MEANS

50

CENTRAL LIMIT THEOREM FOR THE SAMPLING DISTRIBUTION FOR MEANS

• FOR A LARGE ENOUGH SAMPLE SIZE, n, THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN IS APPROXIMATELY

• THAT IS, NORMAL WITH

X

n

N,

deviationdardspopulation

nxDEVIATIONSTANDARD

meanpopulationxEMEAN

tan

)(

)(

51

EXAMPLE 3

• SUPPOSE THE MEAN ADULT WEIGHT, , IS 175 POUNDS WITH STANDARD DEVIATION, , OF 25 POUNDS. AN ELEVATOR HAS A WEIGHT LIMIT OF 10 PERSONS OR 2000 POUNDS. WHAT IS THE PROBABILITY THAT 10 PEOPLE WHO GET ON THE ELEVATOR OVERLOAD ITS WEIGHT LIMIT?

• SOLUTION

52

EXAMPLE 4

• STATISTICS FROM CORNELL’S NORTHEAST REGIONAL CLIMATE CENTER INDICATE THAT ITHACA, NY, GETS AN AVERAGE OF 35.4 INCHES OF RAIN EACH YEAR, WITH A STANDARD DEVIATION OF 4.2 INCHES. ASSUME THAT A NORMAL MODEL APPLIES.

• (A) DURING WHAT PERCENTAGE OF YEARS DOES ITHACA GET MORE THAN 40 INCHES OF RAIN?

• (B) LESS THAN HOW MUCH RAIN FALLS IN THE DRIEST 20% OF ALL YEARS?

• (C) A CORNELL UNIVERSITY STUDENT IS IN ITHACA FOR 4 YEARS. LET y (bar) REPRESENT THE MEAN AMOUNT OF RAIN FOR THOSE 4 YEARS. DESCRIBE THE SAMPLING DISTRIBUTION MODEL OF THIS SAMPLE MEAN, y (bar).

• (D) WHAT’S THE PROBABILITY THAT THOSE 4 YEARS AVERAGE LESS THAN 30 INCHES OF RAIN?

53

SOLUTION TO EXAMPLE 4

54

CHAPTER 7.1 – 7.2

CONFIDENCE INTERVALS FOR PROPORTIONS

ESTIMATIONPOINT ESTIMATION PRODUCES A NUMBER

(AN ESTIMATE) WHICH IS BELIEVED TO BE CLOSE TO THE VALUE OF UNKNOWN PARAMETER.

FOR EXAMPLE: A CONCLUSION MAYBE THAT “PROPORTION P OF LEFT-HANDED

STUDENTS IN MSU IS APPROXIMATELY O.46”

55

SOME POINT ESTIMATORS

PARAMETER ESTIMATOR

PROPORTION P

MEAN

STANDARD DEVIATION

S

P̂

X

56

INTERVAL ESTIMATION

• PRODUCES AN INTERVAL THAT CONTAINS THE ESTIMATED PARAMETER WITH A PRESCRIBED CONFIDENCE.

• A CONFIDENCE INTERVAL OFTEN HAS THE FORM:

)(MEERROROFMARGINESTIMATEPOINT

57

DEFINITION

• GIVEN A CONFIDENCE LEVEL C%, THE CRITICAL VALUE IS THE NUMBER SO THAT THE AREA UNDER THE PROPER CURVE AND BETWEEN IS C (IN DECIMALS).

*C

** CANDC

58

SOME CRITICAL VALUES FOR STANDARD NORMAL DISTRIBUTION

C % CONFIDENCE LEVEL

CRITICAL VALUE

80% 1.282

90% 1.645

95% 1.960

98% 2.326

99% 2.576

*Z

59

WHAT DOES C% CONFIDENCE REALLY MEAN?

• FORMALLY, WHAT WE MEAN IS THAT C% OF SAMPLES OF THIS SIZE WILL PRODUCE CONFIDENCE INTERVALS THAT CAPTURE THE TRUE PROPORTION.

• C% CONFIDENCE MEANS THAT ON AVERAGE, IN C OUT OF 100 ESTIMATIONS, THE INTERVAL WILL CONTAIN THE TRUE ESTIMATED PARAMETER.

• E.G. A 95% CONFIDENCE MEANS THAT ON THE AVERAGE, IN 95 OUT OF 100 ESTIMATIONS, THE INTERVAL WILL CONTAIN THE TRUE ESTIMATED PARAMETER.

60

CONFIDENCE INTERVAL FOR PROPORTION P [ONE-PROPORTION Z-INTERVAL]

ASSUMPTIONS AND CONDITIONS• RANDOMIZATION CONDITION

• 10% CONDITION

• SAMPLE SIZE ASSUMPTION OR SUCCESS/FAILURE CONDITION

• INDEPENDENCE ASSUMPTION• NOTE: PROPER RANDOMIZATION CAN HELP

ENSURE INDEPENDENCE.

61

CONSTRUCTING CONFIDENCE INTERVALS

ESTIMATOR SAMPLE PROPORTION

STANDARD ERROR

C% MARGIN OF ERROR

C% CONFIDENCE INTERVAL

P̂

n

qpPSE

ˆˆ)ˆ(

)ˆ()ˆ( * pSEzpME

)ˆ(ˆ pMEp

62

SAMPLE SIZE NEEDED TO PRODUCE A CONFIDENCE INTERVAL WITH A GIVEN MARGIN OF ERROR, ME

SOLVING FOR n GIVES

WHERE IS A REASONABLE GUESS. IF WE CANNOT MAKE A GUESS, WE TAKE

n

qpzpME

ˆˆ)ˆ( *

2

2*

)(

ˆˆ)(

ME

qpzn

qANDp ˆˆ5.0ˆˆ qp

63

EXAMPLE 1A MAY 2002 GALLUP POLL FOUND THAT ONLY 8% OF A

RANDOM SAMPLE OF 1012 ADULTS APPROVED OF ATTEMPTS TO CLONE A HUMAN.

(A) FIND THE MARGIN OF ERROR FOR THIS POLL IF WE WANT 95% CONFIDENCE IN OUR ESTIMATE OF THE PERCENT OF AMERICAN ADULTS WHO APPROVE OF CLONING HUMANS.

(B) EXPLAIN WHAT THAT MARGIN OF ERROR MEANS.

(C) IF WE ONLY NEED TO BE 90% CONFIDENT, WILL THE MARGIN OF ERROR BE LARGER OR SMALLER? EXPLAIN.

(D) FIND THAT MARDIN OF ERROR.

(E) IN GENERAL, IF ALL OTHER ASPECTS OF THE SITUATION REMAIN THE SAME, WOULD SMALLER SAMPLES PRODUCE SMALLER OR LARGER MARGINS OF ERROR?

64

SOLUTION

65

EXAMPLE 2

DIRECT MAIL ADVERTISERS SEND SOLICITATIONS (a.k.a. “junk mail”) TO THOUSANDS OF POTENTIAL CUSTOMERS IN THE HOPE THAT SOME WILL BUY THE COMPANY’S PRODUCT. THE RESPONSE RATE IS USUALLY QUITE LOW. SUPPOSE A COMPANY WANTS TO TEST THE RESPONSE TO A NEW FLYER, AND SENDS IT TO 1000 PEOPLE RANDOMLY SELECTED FROM THEIR MAILING LIST OF OVER 200,000 PEOPLE. THEY GET ORDERS FROM 123 OF THE RECIPIENTS.

(A) CREATE A 90% CONFIDENCE INTERVAL FOR THE PERCENTAGE OF PEOPLE THE COMPANY CONTACTS WHO MAY BUY SOMETHING.

(B) EXPLAIN WHAT THIS INTERVAL MEANS.(C) EXPLAIN WHAT “90% CONFIDENCE” MEANS.(D) THE COMPANY MUST DECIDE WHETHER TO NOW DO A

MASS MAILING. THE MAILING WON’T BE COST-EFFECTIVE UNLESS IT PRODUCES AT LEAST A 5% RETURN. WHAT DOES YOUR CONFIDENCE INTERVAL SUGGEST? EXPLAIN.

66

SOLUTION

67

EXAMPLE 3

IN 1998 A SAN DIEGO REPRODUCTIVE CLINIC REPORTED 49 BIRTHS TO 207 WOMEN UNDER THE AGE OF 40 WHO HAD PREVIOUSLY BEEN UNABLE TO CONCEIVE.

(A) FIND A 90% CONFIDENCE INTERVAL FOR THE SUCCESS RATE AT THIS CLINIC.

(B) INTERPRET YOUR INTERVAL IN THIS CONTEXT.

(C) EXPLAIN WHAT “90 CONFIDENCE” MEANS.

(D) WOULD IT BE MISLEADING FOR THE CLINIC TO ADVERTISE A 25% SUCCESS RATE? EXPLAIN.

(E) THE CLINIC WANTS TO CUT THE STATED MARGIN OF ERROR IN HALF. HOW MANY PATIENTS’ RESULTS MUST BE USED?

(F) DO YOU HAVE ANY CONCERNS ABOUT THIS SAMPLE? EXPLAIN.

68

SOLUTION

69

CHAPTER 7.3 – 7.4CONFIDENCE INTERVALS TO ESTIMATE A

POPULATION MEAN

• NOTES TO BE TAKEN IN CLASS

70

CHAPTER 8TESTING HYPOTHESES ABOUT

PROPORTIONS

• PROBLEM• SUPPOSE WE TOSSED A COIN 100 TIMES

AND WE OBTAINED 38 HEADS AND 62 TAILS. IS THE COIN BIASED?

• THERE IS NO WAY TO SAY YES OR NO WITH 100% CERTAINTY. BUT WE MAY EVALUATE THE STRENGTH OF SUPPORT TO THE HYPOTHESIS THAT “THE COIN IS BIASED.”

71

TESTING

• HYPOTHESESNULL HYPOTHESIS – ESTABLISHED FACT;– A STATEMENT THAT WE EXPECT DATA TO

CONTRADICT;– NO CHANGE OF PARAMETERS.ALTERNATIVE HYPOTHESIS – NEW CONJECTURE;– YOUR CLAIM;– A STATEMENT THAT NEEDS A STRONG

SUPPORT FROM DATA TO CLAIM IT;– CHANGE OF PARAMETERS

0H

AH

72

IN OUR PROBLEM

.""

5.0;:

5.0;:0

HEADSTURNSCOINTHE

THATYPROBABILITTHEISpWHERE

pBIASEDISCOINH

pFAIRISCOINH

A

73

EXAMPLE

• WRITE THE NULL AND ALTERNATIVE HYPOTHESES YOU WOULD USE TO TEST EACH OF THE FOLLOWING SITUATIONS.

• (A) IN THE 1950s ONLY ABOUT 40% OF HIGH SCHOOL GRADUATES WENT ON TO COLLEGE. HAS THE PERCENTAGE CHANGED?

• (B) 20% OF CARS OF A CERTAIN MODEL HAVE NEEDED COSTLY TRANSMISSION WORK AFTER BEING DRIVEN BETWEEN 50,000 AND 100,000 MILES. THE MANUFACTURER HOPES THAT REDESIGN OF A TRANSMISSION COMPONENT HAS SOLVED THIS PROBLEM.

• (C) WE FIELD TEST A NEW FLAVOR SOFT DRINK, PLANNING TO MARKET IT ONLY IF WE ARE SURE THAT OVER 60% OF THE PEOPLE LIKE THE FLAVOR.

74

ATTITUDE

• ASSUME THAT THE NULL HYPOTHESIS

IS TRUE AND UPHOLD IT,

UNLESS DATA STRONGLY SPEAKS AGAINST IT.

0H

75

TEST MECHANIC

• FROM DATA, COMPUTE THE VALUE OF A PROPER TEST STATISTICS, THAT IS, THE Z-STATISTICS.

• IF IT IS FAR FROM WHAT IS EXPECTED UNDER THE NULL HYPOTHESIS ASSUMPTION, THEN WE REJECT THE NULL HYPOTHESIS.

76

COMPUTATION OF THE Z – STATISTICS OR PROPER TEST STATISTICS

n

qppSD

wherepSD

ppz

oo

o

.)ˆ(

,)ˆ(

ˆ

77

CONSIDERING THE EXAMPLE AT THE BEGINNING:

4.205.0

50.038.0

05.0100

)5.0(5.0)ˆ(,5.0,38.0ˆ

o

O

zAND

PSDPP

78

THE P – VALUE AND ITS COMPUTATION

• THE PROBABILITY THAT IF THE NULL HYPOTHESIS IS CORRECT, THE TEST STATISTIC TAKES THE OBSERVED OR MORE EXTREME VALUE.

• P – VALUE MEASURES THE STRENGTH OF EVIDENCE AGAINST THE NULL HYPOTHESIS. THE SMALLER THE P – VALUE, THE STRONGER THE EVIDENCE AGAINST THE NULL HYPOTHESIS.

79

THE WAY THE ALTERNATIVE HYPOTHESIS IS WRITTEN IS HELPFUL IN COMPUTING THE P - VALUE

NORMAL CURVEAH

oA ppH :

oA ppH :

)( ozzP

)( ozzP

)(2 ozzP oA ppH :

valuep

80

IN OUR EXAMPLE,

• P – VALUE = P( z < - 2.4) = 0.0082

• INTERPRETATION: IF THE COIN IS FAIR, THEN THE PROBABILITY OF OBSERVING 38 OR FEWER HEADS IN 100 TOSSES IS 0.0082

81

CONCLUSION: GIVEN SIGNIFICANCE LEVEL = 0.05

• WE REJECT THE NULL HYPOTHESIS IF THE P – VALUE IS LESS THAN THE SIGNIFICANCE LEVEL OR ALPHA LEVEL.

• WE FAIL TO REJECT THE NULL HYPOTHESIS (I.E. WE RETAIN THE NULL HYPOTHESIS) IF THE P – VALUE IS GREATER THAN THE SIGNIFICANCE LEVEL OR ALPHA LEVEL.

82

ASSUMPTIONS AND CONDITIONS

• RANDOMIZATION

• INDEPENDENT OBSERVATIONS

• 10% CONDITION

• SUCCESS/FAILURE CONDITION

83

EXAMPLE 1

• THE NATIONAL CENTER FOR EDUCATION STATISTICS MONITORS MANY ASPECTS OF ELEMENTARY AND SECONDARY EDUCATION NATIONWIDE. THEIR 1996 NUMBERS ARE OFTEN USED AS A BASELINE TO ASSESS CHANGES. IN 1996, 31% OF STUDENTS REPORTED THAT THEIR MOTHERS HAD GRADUATED FROM COLLEGE. IN 2000, RESPONSES FROM 8368 STUDENTS FOUND THAT THIS FIGURE HAD GROWN TO 32%. IS THIS EVIDENCE OF A CHANGE IN EDUCATION LEVEL AMONG MOTHERS?

84

EXAMPLE 1 CONT’D

• (A) WRITE APPROPRIATE HYPOTHESES.

• (B) CHECK THE ASSUMPTIONS AND CONDITIONS.

• (C) PERFORM THE TEST AND FIND THE P – VALUE.

• (D) STATE YOUR CONCLUSION.

• (E) DO YOU THINK THIS DIFFERENCE IS MEANINGFUL? EXPLAIN.

85

SOLUTION

86

EXAMPLE 2

• IN THE 1980s IT WAS GENERALLY BELIEVED THAT CONGENITAL ABNORMALITIES AFFECTED ABOUT 5% OF THE NATION’S CHILDREN. SOME PEOPLE BELIEVE THAT THE INCREASE IN THE NUMBER OF CHEMICALS IN THE ENVIRONMENT HAS LED TO AN INCREASE IN THE INCIDENCE OF ABNORMALITIES. A RECENT STUDY EXAMINED 384 CHILDREN AND FOUND THAT 46 OF THEM SHOWED SIGNS OF AN ABNORMALITY. IS THIS STRONG EVIDENCE THAT THE RISK HAS INCREASED? ( WE CONSIDER A P – VALUE OF AROUND 5% TO REPRESENT STRONG EVIDENCE.)

87

EXAMPLE 2 CONT’D

• (A) WRITE APPROPRIATE HYPOTHESES.• (B) CHECK THE NECESSARY ASSUMPTIONS.

• (C) PERFORM THE MECHANICS OF THE TEST. WHAT IS THE P – VALUE?

• (D) EXPLAIN CAREFULLY WHAT THE P – VALUE MEANS IN THIS CONTEXT.

• (E) WHAT’S YOUR CONCLUSION?• (F) DO ENVIRONMENTAL CHEMICALS CAUSE

CONGENITAL ABNORMALITIES?

88

SOLUTION

89

CHAPTER 8 CONT’DTESTING HYPOTHESES ABOUT MEANS

• NOTES TO BE TAKEN IN CLASS