Sampling & Sampling Distribution

8/9/2019 Sampling & Sampling Distribution

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Objectives

To beconversant withtheterms The advantages of sampling Types of Sampling Sampling Distributions Sampling Errors To learn Central LimitTheorem


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Why Samplethe Population?

Physical impossibility of checkingall items (census).

Thecost of studying all the items.

Highlytime-consuming.


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The destructive nature of certain

tests.

Thesample results areusually

adequate.


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Probability Sampling

Sampleselected suchthateachitem or person in the populationbeing studied has an equal

likelihood of being selected


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Methods of Probability Sampling

Simple Random Sample:A sample so formulated that each itemor person in the whole population has thesame chance of being selectedeg: lottery , housie.

Systematic Random Sampling:

The items or individuals of the populationarranged in some order.Arandom startingpoint is selected and then every kth memberof the population is selected for the sample.


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Stratified Random Sampling:Heterogeneous population first divided intohomogeneous subgroups, or strata,having

somecommon features.Proportionate subsample isselected fromeachsuchstratum at random and combined toget a truesample.

Cluster Sampling:

A population first divided into primaryunits orclusters, which is asheterogeneous asthepopulation .Then random samples of clustersselected for thesurvey.


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Non Probability Sample

Inclusion in thesamplebased on thejudgment or convenience.

Many items may not get a chance.

Statistical theoriescannotbeused.


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Sampling Error

The differencebetween the value of a samplestatistic and the value of thecorrespondingpopulation parameter .

In thecase of mean, Sampling error = X QIn case of Std deviation,

Sampling error = s- W

Assuming thatthesample is random and nonon sampling error hasbeen made


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Non Sampling Error

Errors dueto deficiencies in thecollection , recording and

tabulation of data.


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Sampling DistributionMany number of samples of size n can be drawnfrom a population of N (N C n).

Highly improbablethat all samples havethe

same mean.They vary from sampleto sample.

The probability distribution of all these values ofmeans of varioussamplestaken from a populationiscalled Sampling distribution of means.

Sampling distribution of std deviations,proportionsetccan also be made.


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The law firm of Hoya and Associates

has five partners.At their weeklypartners meeting each reported thenumber of hours they worked fortheir clients in the last week.

P a r tn e r H o u r s

1 . D u n n 2 2

2 . H a r d y 2 6

3 . K ie r s 3 0

4 . M a l in o w s k i 2 6

5 . T i l lm a n 2 2


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If two partners areselectedrandomly,how many differentsamples are possible?

This isthecombination of 5objectstaken 2 at a time.That is:There are a total of 10 different

samples.

10)!25(!2

!525

!

!C


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P a r t n e r s T o t a l e a n1 , 2 4 8 2 4

1 , 3 5 2 2 6

1 , 4 4 8 2 4

1 , 5 4 4 2 2

2 , 3 5 6 2 8

2, 4 52 26

2, 5 48 24

3, 4 56 28

3, 5 52 26

4, 5 48 24


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Organizethesample means into asampling distribution.

S a m p l e

e a n

F r e q u e n c y R e la t i v e

F r e q u e n c y p r o b a b i l i t y

2 2 1 1 / 1 0

2 4 4 4 / 1 0

2 6 3 3 / 1 0

2 8 2 2 / 1 0


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Computethe mean of thesamplemeans. Compare it withthepopulation mean.

2.2510

)2(28)3(26)4(24)1(22!

!

XQ


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The population mean is also 25.2 hours.

2.255

2226302622!

!Q

Notethatthe mean of thesample means isexactlyequal to the population mean if allpossible samples are taken. But not otherwise.


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Standard Error Standard Error (S.E) of a statistic isthestd deviation of sampling distribution ofthatstatistic.

S.E of Mean , Wx = W /root of n S.E of Std Deviation,Ws = W /root of 2n

S.E of proportion,W p = root of (pq/n)

Std error isused to test if the differencebetween samplestatistic and populationparameter issignificant or not.


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A random sample of 500 oranges wastaken from a largeconsignment and 65were found to be defective. Find theStandard error of the proportion of bad

ones in a sample of thissize.

p= 65/500 =0.13

q= 1-p = 0.87, n = 500 Std Error = root (pq/n) = 0.015


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A simple random sample of size 36 is

drawn from a finite population consistingof 101 units.If the population Standarddeviation is 12.6 find thestandard error

of sample mean when thesample is drawni. with replacement ii. withoutreplacement

Ans: i.W/root n =2.1,

ii. =(W/root n) * root {(N-n)/(N-1)}

= 1.693


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Central Limit Theorem

Thesampling distribution of the means ofn samples generated from a populationwill tend towards normal distribution, as

the value of n increases.

This is irrespective of the distribution ofthe population itself.

The mean of thesampling distribution ofmeans will equal the population mean.


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Normal Uniform Skewed

Population

n = 2

n = 30

XQXQXQXQ

General

The Central LimitTheorem AppliestoSampling Distributions from AnyAny Population


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The mean of thesampling distributionof thesample means, X tendsto Qand itsstandard deviation Wx tends toW / root n,

For largesamples, (n > 30),

X can betaken as = Q

And Wx as = W / root n.


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Purpose of sampling The main purpose of sampling isto infer the

parameter from thestatistic.eg:exit poll

Itcan beused to identifytheunknownpopulation from which a given samplecan beexpected to belong.eg :Oil spill, blood stain

Itcan beused to assesthe probability of aone or moresamples belonging to thesamepopulation.eg:twin sisters

Central LimitTheorem permitsusto usenormal distribution to testthese, if thesample size is largeenough.


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Example 1

The weight of persons using a lift isnormally distributed with a mean of 70 kg

and a standard deviation of 9 kg. The lifthas a maximum capacity of 300 kg.Iffour persons enter the lift, what is theprobability of the load exceeding the

limit? Hint similar to marks / first class case

Sample mean how far away from mean of mean


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Population mean Q = 70 kg

Sample mean = 300/4 = 75 kg

Thesampling error of sampling distribution

of means from a population is = W/root n

= 9/root 4 = 4.5kg

P( x> 75 ) isto be found out.

z = x-Q/Wx = (75-70)/4.5 = 1.11 From normal table,corresponding

P (z >1.11) = 0.5 -.3665 = 0.1335


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Example 2

From past experience it is knownthat the std deviation of thenumber of days of absence ofworkers in an industry is 15.In arandom sample of 100 workers, whatis the probability for the mean

value to differ from the actual meanby more than 3 days ?


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Using normal distribution,

Standard error of sampling distribution

Wx = W/root n =15/ root

100 =

1.53 days = 2 Wx

>3 days (ie x) = >2 (ie z) from the mean

Hence P (x > 3 or < -3) = P (z>2 or < -2).

From table, this is =(.5-.4772)+(.5-.4772)=0.0456


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Example 3

In a normal distribution with mean 375 andstd deviation 48, how big a sample shouldbe taken so that the probability for the

sample mean to fall between 370 and 380will be at least 0.95? Ans :

Probability on either side of mean = 0.475.

From table, corresponding value of z =1.96

1.96 = (380-375)/(48/root n) . n =354.3

So sample size to be at least 355.


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An oil refinery has back up monitors to keeptrack of the oil flow to prevent disruption inprocess.A monitor has an average life of4300 hrs and a std deviation of 730 hrs.Ithas two standby identical units which willautomatically come into operation, if anymonitor fails. What is the probability that aset of monitors will last for

i.13000 hrs at least? ii.12630hrs at the most?

Hint: Treat it as a sample 3 monitors.

Ans: i. 0.4685 ii. 0.4154

Example 4 / HW


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Treat it as a sample 3 monitors. Average life of population Q = 4300hrs. W = 730 hrs. S.E = 730/root 3 = 421.5i.To get a life of 13000 hrs with 3 monitors

working one after other,the mean of sampleshould be 13000/3 hrs = 4333.33 hrs.z = (4333.33 - 4300)/421.5 = 0.079

Probability (z > 0.079) = 0.5-.0315 = 0.4685ii. z = (4210 4300)/421.5 = - 0.2135Probability (z< -0.2135)= 0.5-0.0846= 0.4154


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Example HW

An underwater salvageteam is planning toexplorethesite where 45 Spanishshipscarrying gold sank. From records, it appears

that itcan generate a revenue of 225,000$per ship, and thestd deviation isexpected tobe 39,000$.Theteams financier isskeptic andopined that if theexpedition expenses of 2.1

m$ is not recovered in the first 9 ships,he willcancel the reminder of theexploration.Whatisthe probability for theexploration tocontinuebeyond 9 ships? Ans : 0.2393


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Sara Gorden is. 6.46 LEVIN

Example HW


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Assignment- Sampling

Sampling-page 311-6.3,19,21 Levin Page 323- 6.39,

6-46, 6-54, 6-62,6-64

Page 339 for objective

Sampling & Sampling Distribution

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