Slide 1

Chap 7-1

Statistics for ManagersUsing Microsoft® Excel5th Edition

Chapter 7

Sampling and Sampling Distributions

Chap 7-2

Learning Objectives

In this chapter, you will learn:

To distinguish between different surveysampling methods

The concept of the sampling distribution

To compute probabilities related to thesample mean and the sample proportion

The importance of the Central LimitTheorem

Chap 7-3

Why Sample?

Selecting a sample is less time-consumingthan selecting every item in the population(census).

Selecting a sample is less costly thanselecting every item in the population.

An analysis of a sample is less cumbersomeand more practical than an analysis of theentire population.

Chap 7-4

Types of Samples

Quota

Samples

Non-ProbabilitySamples

Judgment

Chunk

Probability Samples

Simple

Random

Systematic

Stratified

Cluster

Convenience

Chap 7-5

Types of Samples

In a nonprobability sample, items includedare chosen without regard to their probabilityof occurrence.

In convenience sampling, items are selectedbased only on the fact that they are easy,inexpensive, or convenient to sample.

In a judgment sample, you get the opinions ofpre-selected experts in the subject matter.

Chap 7-6

Types of Samples

In a probability sample, items in the sampleare chosen on the basis of known probabilities.

Probability Samples

Simple

Random

Systematic

Stratified

Cluster

Chap 7-7

Simple Random Sampling

Every individual or item from the frame has anequal chance of being selected

Selection may be with replacement (selectedindividual is returned to frame for possiblereselection) or without replacement (selectedindividual isn’t returned to the frame).

Samples obtained from table of random numbersor computer random number generators.

1 092725 012157 827052 297980 625608 9641342 104460 007903 484595 868313 274221 3671813 676071 388003 266711 323324 044463 7628034 881878 862385 203886 261061 096674 8115485 534500 336348 086585 241740 581286 0084356 094276 615776 242112 985859 075388 0820037 333848 513630 474798 841425 331001 5427408 847886 629263 596457 589243 576797 8009579 942495 695172 523982 264961 771016 11879710 450553 679145 324036 715835 963418 53304811 024670 615375 717260 171144 340939 20871212 932959 205554 113225 704406 263818 633643

Random Number Table

Chap 7-9

Systematic Sampling

Decide on sample size: n

Divide frame of N individuals into groups of kindividuals: k=N/n

Randomly select one individual from the 1st group

Select every kth individual thereafter

For example, suppose you were sampling n = 9individuals from a population of N = 72. So, thepopulation would be divided into k = 72/9 = 8 groups.Randomly select a member from group 1, sayindividual 3. Then, select every 8th individualthereafter (i.e. 3, 11, 19, 27, 35, 43, 51, 59, 67)

Chap 7-10

Stratified Sampling

Divide population into two or more subgroups(called strata) according to some commoncharacteristic.

A simple random sample is selected from eachsubgroup, with sample sizes proportional to stratasizes.

Samples from subgroups are combined into one.

This is a common technique when samplingpopulation of voters, stratifying across racial orsocio-economic lines.

Chap 7-11

Cluster Sampling

Population is divided into several “clusters,” eachrepresentative of the population.

A simple random sample of clusters is selected.

All items in the selected clusters can be used, or itemscan be chosen from a cluster using another probabilitysampling technique.

A common application of cluster sampling involveselection exit polls, where certain election districts areselected and sampled.

Chap 7-12

Comparing Sampling Methods

Simple random sample and Systematic sample

Simple to use

May not be a good representation of the population’sunderlying characteristics

Stratified sample

Ensures representation of individuals across the entirepopulation

Cluster sample

More cost effective

Less efficient (need larger sample to acquire the samelevel of precision)

Chap 7-13

Evaluating Survey Worthiness

What is the purpose of the survey?

Were data collected using a non-probabilitysample or a probability sample?

Coverage error – appropriate frame?

Nonresponse error – follow up

Measurement error – good questions elicit goodresponses

Sampling error – always exists

Chap 7-14

Types of Survey Errors

Coverage error or selection bias

Exists if some groups are excluded from the frameand have no chance of being selected

Non response error or bias

People who do not respond may be different fromthose who do respond

Sampling error

Chance (luck of the draw) variation from sample tosample.

Measurement error

Due to weaknesses in question design, respondenterror, and interviewer’s impact on the respondent

Chap 7-15

Measurement Error

A person with one watch always knows whattime it is;

A person with two watches always searches toidentify the correct one;

A person with ten watches is always remindedof the difficulty in measuring time

Chap 7-16

Sampling Distributions

A sampling distribution is a distribution of all of thepossible values of a statistic for a given size sampleselected from a population.

For example, suppose you sample 50 students fromyour college regarding their mean GPA. If youobtained many different samples of 50, you willcompute a different mean for each sample. We areinterested in the distribution of all potential meanGPA we might calculate for any given sample of 50students.

Population

(Size of population = N)

Sample number 1

Sample number 2

Sample number 3

Sample number NCn

Each sample size = n

Point Estimators

Sample Population

Chap 7-19

Sampling DistributionsSample Mean Example

Suppose your population (simplified) wasfour people at your institution.

Population size N = 4

Random variable, x, is age of individuals

Values of x: 18, 20, 22, 24 (years)

Chap 7-20

Sampling DistributionsSample Mean Example

Summary Measures for the Population Distribution:

18 20 22 24

A B C D

P(x)

Uniform Distribution

Chap 7-21

Sampling DistributionsSample Mean Example

1stObs.

2nd Observation

18,18

18,20

18,22

18,24

20,18

29,20

20,22

20,24

22,18

22,20

22,22

22,24

24,18

24,20

24,22

24,24

Now consider all possible samples of size n=2

1stObs.

2nd Observation

16 SampleMeans

16 possible samples(sampling withreplacement)

Chap 7-22

Sampling DistributionsSample Mean Example

Sampling Distribution of All Sample Means

1stObs

2nd Observation

16 SampleMeans

18 19 20 21 22 23 24

P(X)

(no longer uniform)

Sample MeansDistribution

Chap 7-23

Sampling DistributionsSample Mean Example

Summary Measures of this Sampling Distribution:

Chap 7-24

Sampling DistributionsSample Mean Example

Population

N = 4

Sample Means Distribution

n = 2

18 20 22 24

A B C D

P(X)

18 19 20 21 22 23 24

P(X)

Example:

Let N = {2,6,10,11,21}

Find µ, median and σ

µ = 10

median = 10

σ = 6.36

How many samples of size 3 are

possible?

ave

med

2,6,10

3.26

2,6,11

6.33

4.5

3.68

2,6,21

9.67

10.01

8.17

2,10,11

7.67

4.93

4.03

2,10,21

9.53

7.79

2,11,21

11.33

9.50

7.76

6,10,11

2.64

2.16

6,10,21

12.33

7.77

6.34

6,11,21

12.67

7.63

6.24

10,11,21

6.08

4.97

5C3 = 10

med 6 10 11

P(x) .3 .4 .3

Chap 7-26

Sampling DistributionsStandard Error

Different samples of the same size from the same populationwill yield different sample means.

A measure of the variability in the mean from sample tosample is given by the Standard Error of the Mean:

Note that the standard error of the mean decreases as thesample size increases.

Chap 7-27

Sampling DistributionsStandard Error: Normal Pop.

If a population is normal with mean μ and standarddeviation σ, the sampling distribution of the mean is alsonormally distributed with

and

(This assumes that sampling is with replacement or samplingis without replacement from an infinite population)

Chap 7-28

Sampling DistributionsZ Value: Normal Pop.

Z-value for the sampling distribution of the sample mean:

where:= sample mean

= population mean

= population standard deviation

n = sample size

Chap 7-29

Sampling DistributionsProperties: Normal Pop.

(i.e. isunbiased )

Normal PopulationDistribution

Normal SamplingDistribution

(has the same mean)

Chap 7-30

Sampling DistributionsProperties: Normal Pop.

For sampling with replacement:

 As n increases,

 decreases

Larger samplesize

Smaller samplesize

Chap 7-31

Sampling DistributionsNon-Normal Population

The Central Limit Theorem states that as the samplesize (that is, the number of values in each sample)gets large enough, the sampling distribution of themean is approximately normally distributed. This istrue regardless of the shape of the distribution of theindividual values in the population.

Measures of the sampling distribution:

Chap 7-32

Sampling DistributionsNon-Normal Population

Population Distribution

Sampling Distribution

(becomes normal as n increases)

Largersamplesize

Smaller sample size

Chap 7-33

Sampling DistributionsNon-Normal Population

For most distributions, n > 30 will give asampling distribution that is nearly normal

For fairly symmetric distributions, n > 15 willgive a sampling distribution that is nearlynormal

For normal population distributions, thesampling distribution of the mean is alwaysnormally distributed

Chap 7-34

Sampling DistributionsExample

Suppose a population has mean μ = 8 and standarddeviation σ = 3. Suppose a random sample of size n = 36is selected.

What is the probability that the sample mean is between7.75 and 8.25?

Even if the population is not normally distributed, thecentral limit theorem can be used (n > 30).

So, the distribution of the sample mean is approximatelynormal with

Chap 7-35

Sampling DistributionsExample

= 2(.5000-.3085)

= 2(.1915)

= 0.3830

-0.5 0.5

Standardized NormalDistribution

7.75 8.25

SamplingDistribution

Sample

PopulationDistribution

Chap 7-36

Sampling DistributionsExample

First, compute Z values for both 7.75 and 8.25.

Now, use the cumulative normal table to computethe correct probability.

As the sample size increases,the sampling distribution ofsample means approaches anormal distribution.

Example: Given the population of women has normallydistributed weights with a mean of 143 lb and a standarddeviation of 29 lb,a.) if one woman is randomly selected, find the probabilitythat her weight is greater than 150 lb.b.) if 36 different women are randomly selected, find theprobability that their mean weight is greater than 150 lb.

a.) if one woman is randomly selected, find the probabilitythat her weight is greater than 150 lb.

 = 143

150

= 29

0.24

0.0948

0.5 - 0.0948 = 0.4052

z = 150-143 = 0.24

a.) if one woman is randomly selected, the probability thather weight is greater than 150 lb. is 0.4052.

 = 143

150

= 29

0.24

0.0948

0.5 - 0.0948 = 0.4052

Example: Given the population of women has normallydistributed weights with a mean of 143 lb and a standarddeviation of 29 lb,b.) if 36 different women are randomly selected, find theprobability that their mean weight is greater than 150 lb.

b.) if 36 different women are randomlyselected, find the probability that their meanweight is greater than 150 lb.

x = 143

150

x= 29 = 4.83333

b.) if 36 different women are randomlyselected, find the probability that their meanweight is greater than 150 lb.

x = 143

150

x= 4.83333

1.45

0.4265

z = 150-143 = 1.45

b.) if 36 different women are randomlyselected, find the probability that their meanweight is greater than 150 lb.

x = 143

150

x= 4.83333

1.45

0.4265

0.5 - 0.4265 = 0.0735

z = 150-143 = 1.45

b.) if 36 different women are randomlyselected, the probability that their meanweight is greater than 150 lb is 0.0735.

x = 143

150

x= 4.83333

1.45

0.4265

0.5 - 0.4265 = 0.0735

z = 150-143 = 1.45

Example: Given the population of women has normallydistributed weights with a mean of 143 lb and a standarddeviation of 29 lb,

a.) if one woman is randomly selected, find the probability

that her weight is greater than 150 lb.P(x > 150) = 0.4052

Example: Given the population of women has normallydistributed weights with a mean of 143 lb and a standarddeviation of 29 lb,

a.) if one woman is randomly selected, find the probability

that her weight is greater than 150 lb.P(x > 150) = 0.4052

b.) if 36 different women are randomly selected, their

mean weight is greater than 150 lb. P(x> 150) = 0.0735

Given the population of women has normallydistributed weights with a mean of 143 lb anda standard deviation of 29 lb,

a.) if one woman is randomly selected, find the

probability that her weight is greater than 150 lb.P(x > 150) = 0.4052

b.) if 36 different women are randomly selected, their

mean weight is greater than 150 lb. P(x > 150) = 0.0735It is much easier for an individual to deviate from themean than it is for a group of 36 to deviate from the mean.

Given the population of women has normallydistributed weights with a mean of 143 lb and astandard deviation of 29 lb,

Find an interval that will include 90% of the

sample means



x= 4.83333

.45

Central Limit Theorem

1. Rainfall in the Northeast has a mean pH level of 4.8and a standard deviation of .5. What is the probabilitythat the average pH level of 36 randomly selected futurerainstorms in the area will have a pH value greater than5.0?

x = 4.8

n = 36

 = .5

Central Limit Theorem

2. Average amount spent by a customer at a particularstore averages $54.70. with a standard deviation of$17.10. If 50 customers are chosen at random, find thefollowing probabilities,

a. The average amount spent exceeds $60.00.

x = 54.70

n = 50

 = 17.10

Central Limit Theorem

2. Average amount spent by a customer at a particular storeaverages $54.70. with a standard deviation of $17.10. If 50customers are chosen at random, find the followingprobabilities,

b. The average amount spent is between $50.00 and $60.00.

x = 54.70

n = 50

 = 17.10

Central Limit Theorem

3. A production for steel rods have an average of 6meters and a variance of 64 cm. These rods are tiedinto bundles of 40.

a.What is the probability that the average length of arandomly selected bundle is less than 598 cm?

x = 6

n = 40

 = 8

Central Limit Theorem

3. A production for steel rods have an average of 6meters and a variance of 64 cm. These rods are tiedinto bundles of 40.

b. What is the probability that the average length of abundle is less than 598 cm or more than 601 cm?

x = 6

n = 40

 = 8

Central Limit Theorem

3. A production for steel rods have an average of 6 metersand a variance of 64 cm. These rods are tied into bundlesof 40.

c. Suppose that the bundle size is 25. What is the probabilityof selecting a bundle that has an average length less than598 cm?

x = 6

n = 25

 = 8

Sample Proportion

Standard Error of the Proportion

Finding Z for the Sampling Distribution of

the Proportion

Chap 7-58

Sampling DistributionsThe Proportion

 The proportion of the population having somecharacteristic is denoted π.

Sample proportion ( p ) provides an estimate of π:

0 ≤ p ≤ 1

p has a binomial distribution

(assuming sampling with replacement from a finite population or withoutreplacement from an infinite population)

Chap 7-59

Sampling DistributionsThe Proportion

Standard error for the proportion:

Z value for the proportion:

Chap 7-60

Sampling DistributionsThe Proportion: Example

If the true proportion of voters who supportProposition A is π = .4, what is the probability thata sample of size 200 yields a sample proportionbetween .40 and .45?

In other words, if π = .4 and n = 200, what is

P(.40 ≤ p ≤ .45) ?

Chap 7-61

Sampling DistributionsThe Proportion: Example

Find :

Convert tostandardizednormal:

Chap 7-62

Sampling DistributionsThe Proportion: Example

Use cumulative normal table:

P(0 ≤ Z ≤ 1.44) = P(Z ≤ 1.44) – 0.5 = .4251

.45

1.44

.4251

Standardize

Sampling Distribution

StandardizedNormal Distribution

.40

Central Limit Theorem

1. If 100 consumers are randomly selected regardingtheir brand preference, what is the probability ofobserving a sample proportion preferring Brand A assmall as .15 or less, if in fact, 25% of the populationprefer Brand A?

Central Limit Theorem

2. In a random survey, 40% of all U.S. employeesparticipate in self-insurance health plans.

a. In a random sample of 100 employees, what is theprobability that at least half of these employees participatein such a plan?

Central Limit Theorem

2. In a random survey, 40% of all U.S. employeesparticipate in self-insurance health plans.

b. What is the probability that out of these 100employees that between 20% and 30% participate?

Central Limit Theorem

3. Tests show that 25% of the population eats breakfast.In a random sample of 500 people, what is theprobability that less than 15% chosen eat breakfast?

Central Limit Theorem

3. Tests show that 25% of the population eats breakfast.In a random sample of 500 people, what is theprobability that between 20% and 50% eat breakfast?

Chap 7-68

Chapter Summary

Described different types of samples

Examined survey worthiness and types of surveyerrors

Introduced sampling distributions

Described the sampling distribution of the mean

For normal populations

Using the Central Limit Theorem

Described the sampling distribution of a proportion

Calculated probabilities using sampling distributions

In this chapter, we have