1 prof. indrajit mukherjee, school of management, iit bombay others convenience stratified judgment...

1Prof. Indrajit Mukherjee, School of Management, IIT Bombay

OthersConvenience

Stratified

Judgment

Non-ProbabilitySamples

Probability Samples

SimpleRandom

Systematic

StratifiedCluster

Samples

Sampling Techniques


TYPE OFSAMPLING

SELECTIONSTRATEGY

PURPOSE

Convenience Select casesbased on theiravailability for

the study.

Saves timetime,

money andeffort; but at the

expense ofinformation and

credibility.


Simple random sampling

Sample method Resulting method

The population is identified uniquely by number. Selection by

random number

Every number of the population has an equal chance of being selected

into the sample


Simulating From Continuous Uniform

( ]

Random numbersUniform [0,1] distribution

Uniform [a, b] distribution

0 r 1

0 a a + r(b-a) b

Shift a Stretch (b - a) b


How to use random number table to select a random samplecorresponds to a number on the list of your population. In the example below, # 08 has been chosen as the starting point and the first student chosen is Carol Chan.

10 09 73 25 33 7637 54 20 48 05 6408 42 26 89 53 1990 01 90 25 29 0912 80 79 99 70 8066 06 57 47 17 3431 06 01 08 05 45

Step 3: Move to the next number, 42 and select the person corresponding to that number intothe sample. #87 – Tan Teck WahStep 4: Continue to the next number that qualifies and select that person into the sample.# 26 -- Jerry Lewis, followed by #89, #53 and #19Step 5: After you have selected the student # 19, go to the next line and choose #90. Continuein the same manner until the full sample is selected. If you encounter a number selectedearlier (e.g., 90, 06 in this example) simply skip over it and choose the next number.

Starting point:move right to the endof the row, then downto the next row row;move left to the endEnd, then down to the next row, and so on.


Systematic sampling (contd.)“Example”

1 26 51 76

2 27 52 77

3 28 53 78

4 29 54 79

5 30 55 80

6 31 56 81

7 32 57 82

8 33 58 83

9 34 59 84

10 35 60 85

11 36 61 86

12 37 62 87

13 38 63 88

14 39 64 89

15 40 65 90

16 41 66 91

17 42 67 92

18 43 68 93

19 44 69 94

20 45 70 95

21 46 71 96

22 47 72 97

23 48 73 98

24 49 74 99

25 50 75 100

Start with #4 and take every 5th unit

N=100

Want n=20

N/n=5

Select a random number from 1-5:Chose 4


Stratified Random Sample: Stratified by Age

20 - 30 years old(homogeneous within)(alike)

30 - 40 years old(within homogeneous) (alike)

40 - 50 yearsold(homogeneous within)(alike)

Heterogeneous(different)between

Heterogeneous(different)between


Sample Spaces and EventsRandom Experiments

Noise variables affect the transformation of inputs to outputs.

Noise variables

Controlled variables

Input OutputSystem


Example

Rotation speed Traverse speedTool type Tool sharpnessShaft material Shaft lengthMaterial removal per cut Part cleanliness Coolant flow Operator Material variation Ambient temperature Coolant age

Machining a shafton a lathe

Outputs (Y’s)DiameterTaperSurface finish


Four Types of Probability

Marginal Union Joint Conditional

The probability of XOccurring P( X )

The probability of Xor Y occurring

The probability of Xand Y occurring

The probability of Xoccurring giventhat Y has occurred P(X|Y)X YX Y

X X Y X Y


P(A and B)(Venn Diagram)

P(A) P(B)

P(A and B)


P(A or B)


Sample Spaces and EventsVenn Diagrams


E4

E1

E2

E3

Venn diagram of four mutually exclusive events


Collectively Exhaustive Events• Events are said to be collectively exhaustive ifthe list of outcomes includes every possibleoutcome: heads and tails as possibleoutcomes of coin flip


Example 3Draw Mutually Collectively

Exclusive Exhaustive

Draw a space and a club Yes Yes

Draw a face card and a Yes Yesnumber cardDraw an ace and a 3 Yes No

Draw a club and a nonclub Yes Yes

Draw a 5 and a diamond No No

Draw a red card and a No Nodiamond


The following circuit operates only if there is a path of functional devices from left to right. The probability that each device function is shown on the graph. Assume that devices fail independently, what is the probability the circuit operates?

Let T and B denote the events that the top and bottom devices operate, Respectively. There is a path if at least on device operates. The probability that the circuit operates is

P(T or B) =1-[P(T or B)’]=1-P(T’ and B’)

A simple formula for the solution can be derived from the complements T and B’. From the independence assumption.

P(T’ and B’)=P(T’) P(B’)=(1-0.95)2 =0.052

P(T or B)=1- 0.052 0.9975

0.95

0.95

a b


Probability(D|F)

P(D|F) = P(DF)/P(F)

/

P(D) P(DF) P(F)


Random Variables (Numeric)

Experiment Outcome Random Variable Range of RandomVariable

Stock 50Xmas trees

Number oftrees sold

X = number oftrees sold

0,1,2,, 50

Inspect 600items

Number acceptable

Y = number acceptable

0,1,2,…,600

Send out5,000 sales letters

Number ofPeople responding

Z = number ofpeople responding

0,1,2,…,5,000

Build anapartment

building

%completedafter 4months

R = %completedafter 4 months

0≤R ≤ 100

Test thelifetime of a

light bulb(minutes)

Time bulblasts - up to

80,000minutes

S = time bulbburns

0 ≤ S ≤ 80,000


105 221 183 186 121 181 180 143

97 154 153 174 120 168 167 141

245 228 174 199 181 158 176 110

163 131 154 115 160 208 158 133

207 180 190 193 194 133 156 123

134 178 76 167 184 135 229 146

218 157 101 171 165 172 158 169

199 151 142 163 145 171 148 158

160 175 149 87 160 237 150 135

196 201 200 176 150 170 118 149

Comressive strength (in psi) of 80 aluminum-lithium alloy specimens


Frequency Distributions and Histograms

Histogram of compressive strength for 80 aluminum-lithium alloy specimens.

0

0.0625

0.1250

0.1895

0.2500

0.3125 25

0

5

10

15

20

70 90 110 130 150 170 190 210 230 250

Fre

qu

en

cy

compressive strength (psi)


438 450 487 451 452 441 444 461 432 471

413 450 430 437 465 444 471 453 431 458

444 450 446 444 466 458 471 452 455 445

468 459 450 453 473 454 458 438 447 463

445 466 456 434 471 437 459 445 454 423

472 470 433 454 464 443 449 435 435 451

474 457 455 448 478 465 462 454 425 440

454 441 459 435 446 435 460 428 449 442

455 450 423 432 459 444 445 454 449 441

449 445 455 441 464 457 437 434 452 439

Histograms – Useful for large data sets

Group values of the variable into bins, then count the number ofobservations that fall into each binPlot frequency (or relative frequency) versus the values of thevariable


30

0

10

20

405 415 425 435 445 455 465 475 485 495

Minitab histogram for the metal layer thickness data in table

Metal thickness

Frequency


Histogram ExampleData in ordered array:12, 13, 17, 21, 24, 24, 26, 27, 27, 30, 32, 35, 37, 38, 41, 43, 44, 46, 53, 58

No gapsbetweenbars, sincecontinuousData

10

0

5

5 15 25 36 45 55

00

3

65

4

2

Histogram

Class midpoints

Frequency


How Many Class Intervals?

• Many (Narrow classintervals)• may yield a very jagged distributionwith gaps from empty classes• Can give a poor indication of howfrequency varies across classes

• Few (Wide class intervals)• may compress variation too muchand yield a blocky distribution• can obscure important patterns ofvariation.


Calculation of Grouped Mean

Class Interval Frequency Class Midpoint fM

20-under 30 6 25 15030-under 40 18 35 63040-under 50 11 45 49550-under 60 11 55 60560-under 70 3 65 19570-under 80 1 75 75

50 2150

215043.0

50

fm

f


Mode of Grouped Data

• Midpoint of the modal class• Modal class has the greatest frequency

Class Interval Frequency

20-under 30 3530-under 40 1840-under 50 1150-under 60 1160-under 70 370-under 80 1

30+40Mode= 35

2


6 1 5 7 8 6 0 2 4 25 2 4 4 1 4 1 7 2 34 3 3 3 6 3 2 3 4 55 2 3 4 4 4 2 3 5 75 4 5 5 4 5 3 3 3 12


Frequency Distribution:Discrete Data

• Discrete data: possible values are countable

Example: Anadvertiser asks200 customershow many daysper week theyread the dailynewspaper.

Number of days read Frequency

0 441 242 183 164 205 226 267 30

total 200


Relative FrequencyRelative Frequency: What proportion is in each

22% of thepeople in thesample reportthat they read theNewspaper days per week

Number of days read Frequency

Relative frequency

0 44 0.221 24 0.122 18 0.093 16 0.084 20 0.105 22 0.116 26 0.137 30 0.15

total 200 1.00


Relative Frequency Plot andProbability Distributions

Histogram approximates a probability density function.

F(x)

X


Interpretations of Probability

Relative frequency of corrupted pulses sent over acommunications channel.

Relative frequency of corrupted pulse=2/10

Corrupted pulse

Time

Volt

age


Interpretations of Probability

P(E)=30(0.01)=0.30Probability of the event E is the sum of the probabilities of the outcomes in E

Diodes

E

S


Random Variables

Question Random Variable x Type

Familysize

x = Number of dependents in family reported on tax return

Discrete

Distance fromhome to store

x = Distance in miles fromhome to the store site

Continuous

Own dogor cat

x = 1 if own no pet;= 2 if own dog(s) only;= 3 if own cat(s) only;= 4 if own dog(s) and cat(s)

Discrete


Using past data on TV sales, …a tabular representation of the probabilitydistribution for TV sales was developed.

Unit soldNumber of days read

0 801 502 403 104 20

total 200

x f(x)0 .401 .252 .203 .054 .10

total 1.00


Graphical Representation of the ProbabilityDistribution

.50

.10

.20

.30

.40

0 1 2 3 4

Values of random variable x (TV sales)

Pro

babili

ty


• Uniform Probability Distribution• Normal Probability Distribution• Exponential Probability Distribution

Uniform

Normal Exponential

F(x)

X X X

F(x) F(x)


x1 x2 x3 x4 x5

F(x)

X

F(x)

X

p(x3)

p(x4)

p(x5)p(x1)

p(x2)

a b

Sometimes called aprobability mass function

Sometimes called a probabilitydensity function

Probability distributions (a)Discrete case (b)continuous case


Throwing a Dice

1 2 3 4 5 6

1/6

Distribution of X

P(X

)

X


Example 2aOutcome Probabilityof Roll = 5

Die 1 Die 21 4 1/362 3 1/363 2 1/364 1 1/36

Rolling two diceresults in a total offive spots showing. There are a total of 36possibleoutcomes


sample Sample sample1,1 1 3,1 2 5,1 31,2 1.5 3,2 2.5 5,2 3.51,3 2 3,3 3 5,3 41,4 2.5 3,4 3.5 5,4 4.51,5 3 3,5 4 5,5 51,6 3.5 3,6 4.5 5,6 5.52,1 1.5 4,1 2.5 6,1 3.52,2 2 4,2 3 6,2 42,3 2.5 4,3 3.5 6,3 4.52,4 3 4,4 4 6,4 52,5 3.5 4,5 4.5 6,5 5.52,6 4 4,6 5 6,6 6

X X X

All Samples of subgroup size 2 from a Population


1 1/361.5 2/362 3/36

2.5 4/363 5/36

3.5 6/364 5/36

4.5 4/365 3/36

5.5 2/366 1/36

X P X

SamplingDistribution of X


1 2 3 4 5 6

(b) Sampling distribution of

6/36

4/36

2/36

X

p X

X


0

.6

.5

.4

.3

.2

.1

61 3.5Sampling Distribution of for n = 5

X

X


Sampling Distributions of MeansFigure Distributions of averagescores from throwing dice.

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6


SamplingDistributionBecomesAlmostNormalRegardlessof Shape ofPopulation

As SampleSize GetsLargeEnough

Central Limit Theorem

X


Probability DistributionsOutcome X Number responding p(X)

SA 5 10 0.1

A 4 20 0.2

N 3 30 0.3

D 2 40 0.3

SD 1 50 0.1

1 2 3 4 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Chart Title


X P(X=x) X P(X=x)0 0.205891 6 0.0019391 0.343152 7 0.0002772 0.266896 8 0.0000313 0.128505 9 0.0000034 0.042835 10 05 0.010471


Probability Distributions andProbability Density Functions

Density function of a loading on a long, thin beam.

X

Loadin

g



Probability determined from the area under f(x).

P(a<X<b)

X

F(x)

a b


Continuous Uniform Random Variable

1/(b-a)

X

F(x)

a bContinuous uniform probability density function.


Uniform Distribution or rectangularprobability distribution

1/(b-a)

X

F(x)

a b

1, where f x a x b

b a

area = width x height = (b – a) x1

b a

1

b a


ExampleThe amount of gasoline sold daily at a service station isuniformly distributed with a minimum of 2,000 gallons and amaximum of 5,000 gallons.

X

F(x)

1

5,000 2,000 2,000 5,000

Find the probability that daily sales will fall betweenand 3,000 gallons.2,500 Algebraically: what is P(2,500 ≤ X ≤ 3,000) ?


Example

X

F(x)

1

3,000 2,000 5,000

12,500 3,000 3,000 2,500 0.1667

3,000P X

“there is about a 17% chance that between 2,500 and 3,000gallons of gas will be sold on a given day”


Cumulative Distribution Functions

X

F(x)

0 12.5

1

Cumulative Distribution Functions



X

F(x)

12.5 12.6

Probability density function


Normal Probability DistributionCharacteristicsThe distribution is symmetric, and is bell-shaped.

x


Normal Probability Distribution

Characteristics

Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and .5 to the right).

x

0.5 0.5


The mean is not necessarily the 50th percentile of the distribution (that’s the median)The mean is not necessarily the most likely value of the random variable (that’s the mode)

Two probability distributions with same mean but different standard deviations

µ Median Mode

The mean of a distribution

Two probability distributions with different means

µ µ Mode

µ=20µ=10 µ=10

σ=2

σ=4


µ-1σ µ+1σ µ+2σ µ+3σµ-2σµ-3σ µ

99.73%

68.26%

95.46%

µ

σ2

F(x)

x

Areas under normal distributionThe normal distribution


Normal Distribution

Standardizing a normal random variable.


A normal distribution whose mean is zero and standarddeviation is one is called the standard normal distribution.

σ=1µ=0

As we shall see shortly, any normal distribution can be converted to astandard normal distribution with simple algebra. This makes calculations much easier.

Standard Normal Distribution…

2

1 0

2 11

1 2

x

f x e x


Normal DistributionExample

0 1.5 z


Areas under Standardized Normal Distribution


Using Excel to ComputeStandard Normal ProbabilitiesFormula Worksheet

A B

1 Probabilities: standard normal distribution

2 P (z < 1 00) =NORMSDIST(1)

3 P (0.00 < z < 1.00) =NORMSDIST(1)-NORMSDIST(0)

4 P (0.00 < z < 1.25) =NORMSDIST(1.25)-NORMSDIST(0)

5 P (-1.00 < z < 1.00) =NORMSDIST(1)-NORMSDIST(-1)

6 P (z > 1.58) =1-NORMSDIST(1.58)

7 P (z < -0.50) =NORMSDIST(-0.5)

8

1 prof. indrajit mukherjee, school of management, iit bombay others convenience stratified judgment...

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