data collection for elementary statistics by Taban Rashid

Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.. Chap 3-1
Introduction of Elementary
Statistics

Learning Objectives
In this chapter you learn:
 Understanding the meaning of Statistics, and branches of statistics
 Describe the uses and role of Statistics in business management
 Basic concepts e.g Population, Sample, Parameter, and Statistic
 Describe data and variable types
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc. Chap 3-2

But what is Statistics?
3
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Statistics
Statistics is the science of:
 Collecting, organizing, presenting, analyzing,
and interpreting data to assist in making more
effective decisions.

INTRODUCTION
 Statistics’ is used to refer to;
 Numerical facts, such as the number of people
living in particular area.
The study of ways of collecting, analyzing
and interpreting the facts
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What is Statistics?
• The science of conducting
studies to collect, organize
, summarize, analyze and
draw conclusions from da
ta
• The systematic collection of n
umerical data and its interpret
ation.
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Basic concepts in statistics
 Population and sample
 Parameter and statistic
 Variable
 A constant
 Data and datum

Chap 1-8
Population vs. Sample
Population Sample
All the items or individuals about
which you want to draw conclusion(s)
A portion of the population of
items or individuals

Population and sample
Business Statistics, 4e, by Ken Black. © 2003 John Wiley &
Sons.
105-

Population vs. Sample
a b c d
ef gh i jk l m n
o p q rs t u v w
x y z
Population Sample
b c
g i n
o r u
y
Measures used to describe a
population are called
parameters
Measures computed from
sample data are called
statistics

Key Definitions
 A population is the entire set of individuals or objects under
consideration or the measurements obtained from all
individuals.
 A sample is a portion, or part of the population of interest.
 A parameter is any numerical measure that describes a
characteristic of a population
 A statistic is any numerical measure that describes a
characteristic of a sample

Key concepts used in statistics
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 A variable
A variable is a measurable quantity which changes
over space or time eg Time, cost of goods sold,
number of suppliers, type of specification, volume
and value of stock, return on net assets

Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 3-14
Types of Data/ variable
Data
Qualitative or
Categorical
Quantitative or
Numerical
Discrete Continuous
Non numerical data
Examples:
 Marital Status
 Political Party
 Eye Color
(Defined categories)
Assume only certain
values result from
counting
Examples:
 Number of Children
 Defects per hour
(Counted items)
Assume any value within a
specific range result from
measuring.
Examples:
 Weight
 Voltage
(Measured characteristics)

Types of Variables
 Categorical (qualitative) variables have values
that can only be placed into categories, such as
“yes” and “no”; major; architectural style; etc.
 Numerical (quantitative) variables have values
that represent quantities.
 Discrete variables arise from a counting process
 Continuous variables arise from a measuring
process
. Chap 1-15

 A constant. This is a characteristic that takes the
same value at every time eg no. of months in a
year
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 The values of the variable are the possible observations of
the variable.
 Data are the observed values of a variable or in simple
terms, these are raw facts.
For example: coursework marks for 10 students.
Data & datum i.e. the mark of one student is a datum.

STAGES INVOLVED IN STATISTICS

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Collection of Data
Presentation of data
Analysis of data
Interpretation of data

SCOPE OF STATISTICS
 Human resource; Useful in the recruitment process
 Economics; Forecasting, Demographics, Testing and validating
economic theories
 Finance; Risk analysis, Useful in portfolio diversification and asset
allocation.
 Marketing ; Useful in market research
 Operations management; Design and Quality management
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• Comparisons
• Condensation.
• Formulation and testing hypotheses
• Forecasting and planning
• Policy Making
• Helps in deriving relationship between variables
• Measures Uncertainty
Application/Importance/ uses/ functions of statistics
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Limitations of statistics
 Does not deal with single value (requires aggregated data)
 Can be misused, misinterpreted
 sampling and non sampling errors
 Requires skilled personnel
 Requires data that can be quantified unlike qualitative data
 Cannot be applied to heterogeneous data
 Requires expert knowledge of statistics

Branches/ Types of Statistics
Descriptive
Collecting,
summarizing, analyzing
& describing data
Inferential
Drawing conclusions
and/or making decisions
concerning a population
based only on sample
data
Statistics

Descriptive Statistics
 Collect data
 e.g., Survey
 Present data
 e.g., Tables and graphs
 Characterize data
 e.g., Sample mean =
i
X
n


Inferential Statistics
 Estimation
 e.g., Estimate the population mean
weight using the sample mean
weight
 Hypothesis testing
 e.g., Test the claim that the
population mean weight is 120
pounds
Drawing conclusions about a population based on sample
results.

Collecting Data
Secondary
Data Compilation
Observation
Experimentation
Print or Electronic
Survey
Primary
Data Collection

Chapter Summary
 Reviewed basic concepts of statistics:
 Population vs. Sample
 Parameter vs. Statistic
 Defined descriptive vs. inferential statistics
 Qualitative vs. Quantitative data
 Discrete vs. Continuous data

Statistics
End of chapter 1

Topic II
DATA

Learning Objectives
In this chapter you learn:
 Meaning of data
 Types of data: Primary and Secondary
 Data collection procedure, tools and techniques.
 Classification of data
 Presentation of data (Frequency distribution, Tabulation,
Graphical, Pictorial)

INTRODUCTION
 The objective of statistics is to extract information
from data.
 There are different types of data.
 To help explain this important principle, there is
need to understand some terms.

DATA COLLECTION
 Statistical investigation is a comprehensive process
 Requires systematic collection of data about some group of
people or objects, describing and organizing the data,
analyzing the data
 The validity and accuracy of final judgment is most crucial
 This depends heavily on how well the data was collected in
the first place.
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 By Nature:
 Quantitative data; Data which can be expresse
d numerically or in terms of numbers eg no. of s
tudents
 Qualitative data, can not be expressed numeric
ally
Types of data
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 By timeframe:
 Cross Section Data- Data values observed at a fixed
point in time
 Time Series Data- Ordered data values observed ov
er time
 Panel Data– Data observed over time from the same
units of observation
Types of data
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 By Source: Primary or Secondary
 Primary Data - data gathered for the first time
by the researcher
 Secondary Data - Data taken by the researcher
from secondary sources, internal or external, Al
ready published records/compilation
Types of data
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Advantages of primary data
 The data is original.
 The information obtained is unbiased.
 It provides accurate information and is more reliable.
 It gives a provision to the researcher to capture the changes
occurring in the course of time.
 It is up to date data, relevant and specific to the required
product
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Disadvantages of primary data
 Time consuming to collect
 It requires skilled researchers in order to be
collected.
 It needs a big sample size in order to be accurate.
 It’s more costly to collect
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Advantages of Secondary data
 It’s economical as it saves expenses and efforts
 It is time saving, since it is more quickly obtainable than
primary data.
 It provides a basis for comparison for data collected by the
researcher.
 It helps to make the collection of primary data more specific
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Disadvantages of Secondary data
 Accuracy of secondary data is not known.
 Data may be outdated.
 It may not fit in the framework of the research factors
for example units used.
 Users of such data may not have as thorough
understanding of the background as the original
researcher.
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Considerations to make before data collection
 Statement of the purpose
 should be clearly stated to avoid confusion
 Only necessary information is collected
 Scope of inquiry
 based on space or time- geographical and time
 Choice of statistical unit
DATA COLLECTION
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Data Sources
Data collection technique
Depends on time available, literacy of the re
spondents, language, availability of the reso
urces, the accuracy required
DATA COLLECTION
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41
Major Techniques for Collecting Data
1. Questionnaires
2. Interviews
3. Observation
4. Experimentation

42
Using these data gathering methods
 Each method has advantages and problems.
 No single method can fully measure the variable
 Examples:
 Questionnaires & surveys are open to self-report biases,
such as respondents’ tendency to give socially desirable
answers rather than honest opinions.
 Observations are susceptible to observer biases, such as
seeing what one wants to see rather than what is actually
there.

43
Use more than one
 Because of the biases inherent in any data-collection
method, it is best to use more than one method when
collecting diagnostic data.
 The data from the different methods can be compared,
and if consistent, it is likely the variables are being validly
measured.

44
Questionnaires:
 Questionnaires are one of the most efficient
ways to collect data.
 They contain fixed-response questions about
various features of an organization.
 These on-line or paper-and pencil measures
can be administered to large numbers of
people simultaneously.

45
Questionnaires:
 They can be analyzed quickly.
 They can be easily be fed back to
employees.
 Questionnaires can be standard based on
common research or they can be customized
to meet the specific data gathering need.

GUIDELINES FOR DRAFTING A
QUESTIONNAIRE
 The size of the questionnaire should be small
 The questions should be clear, brief, unambiguous, non-
offending etc
 The questions should be logically arranged
 The questions should be short, simple and easy to
understand
 Avoid questions of personal nature
 The questionnaire should be made to look attractive
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Sample questionnaire
 Demographic statistics
1. Gender of the household head
2. Marital status of the household head
3. Education level of the household head
Male Female
Never married Married Divorced ???
Never studied Primary secondary Diploma

Sample questionnaire
1. I have understood the types of data
2. Statistics is easily understandable
3. I will pass elements of statistics with flying grades
STRONGLY
DISAGREE
DISAGREE NOT SURE AGREE STRONGLY
AGREE
STRONGLY
DISAGREE
DISAGREE NOT SURE AGREE STRONGLY
AGREE

Questionnaire (Advantages)
 Information on character and environment may
help later to interpret some of the results.
 It provides personal rapport which helps to
overcome reluctance to respond
 Supplementary information on informant’s
personal aspects can be noted.

Questionnaire (Advantages)
 High response rate since answers are obtained on
the spot.
 Permits explanation of questions concerning
difficult subject matter
 The wordings in one or more questions can be
altered to suit any informant.

Questionnaire (limitations/ draw backs)
•It is very costly and time consuming
•It is suitable only for intensive studies and not for extensive
enquiries
•Personal prejudice and bias are greater under this method.
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52
Questionnaires (drawbacks)
 Responses are limited to the questions asked in
the instrument.
 They provide little opportunity to probe for
additional data or ask for points of clarification.
 They tend to be impersonal.
 Often elicit response biases – tend to answer in a
socially acceptable manner.

53
Interviews
 Interviews are probably the most widely used
technique for collecting data
 They permit the interviewer to ask the respondent
direct questions.
 Further probing and clarification is possible as the
interview proceeds.

54
Interviews
 Interviews may be highly structured, resembling
questionnaires, or highly unstructured, starting with
general questions that allow the respondent to lead the
way.
 Interviews are usually conducted one-to-one but can be
carried out in a group.
 Group interviews save time and allow people to build on
other’s responses.
 Group interviews may, however, inhibit respondent’s
answers if trust is an issue.

55
Interviews / Focus Groups
 Another unstructured group meeting conducted by a
manager or a consultant.
 A small group of 10-15 people is selected representing a
larger group of people
 Group discussion is started by asking general questions
and group members are encouraged to discuss their
answers in some depth.
 The richness and validity of this information will depend
on the extent that trust exists.

56
Drawback to interviews
 They can consume a great deal of time if interviewers take full
advantage of the opportunity to hear respondents out and change their
questions accordingly.
 Personal biases can also distort the data.
 The nature of the question and the interactions between the
interviewer and the respondent may discourage or encourage certain
kinds of responses.
 It take considerable skill to gather valid data.

57
Observations
 Observing organizational behaviors in their functional
settings is one of the most direct ways to collect data.
 Observation can range from complete participant
observation, where the OD practitioner becomes a
member of the group under study to a more detached
observation using a casually observing and noting
occurrences of specific kinds of behaviors.

58
Advantages to Observation:
 They are free of the biases inherent in the self-report data.
 They put the practitioner directly in touch with the behaviors in question.
 They involved real-time data, describing behavior occurring in the present
rather than the past.
 They are adapting in that they can be modified depending on what is being
observed.

59
Problems with Observation
 Difficulties interpreting the meaning underlying the observations.
 Observers must decide which people to observe; choose time periods,
territory and events
 Failure to attend to these sampling issues can result in a biased sample of
data.

Data collection methods
• Direct interview
• Many researchers believe that the best way to survey people is by
means of personal interviews
• Involves an interviewer soliciting information from a respond
ent by asking prepared questions.
• Observation
• This is data collected through direct observation.
• Experimentation
• A more expensive but better way to produce data is through exper
imentation.
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Tables and Charts for
Numerical Data
Numerical Data
Stem-and-Leaf
Display
Histogram Polygon Ogive
Frequency Distributions
and
Cumulative Distributions

Tabulating Numerical Data:
Frequency Distributions
What is a Frequency Distribution?
 A frequency distribution is a list or a table …
 containing class groupings (ranges within which
the data fall) ...
 and the corresponding frequencies with which
data fall within each grouping or category

Why Use a Frequency Distribution?
 It is a way to summarize numerical data
 It condenses the raw data into a more
useful form...
 It allows for a quick visual interpretation of
the data

Class Intervals
and Class Boundaries
 Each class grouping has the same width
 Determine the width of each interval by
 Usually at least 5 but no more than 15
groupings
 Class boundaries never overlap
 Round up the interval width to get desirable
endpoints
groupings
class
desired
of
number
range
interval
of
Width 

Frequency Distribution Example
Example: A manufacturer of insulation randomly
selects 20 winter days and records the daily
high temperature
24, 35, 17, 21, 24, 37, 26, 46, 58, 30,
32, 13, 12, 38, 41, 43, 44, 27, 53, 27

 Sort raw data in ascending order:
12, 13, 17, 21, 24, 24, 26, 27, 27, 30, 32, 35, 37, 38, 41, 43, 44, 46, 53, 58
 Find range: 58 - 12 = 46
 Select number of classes: 5 (usually between 5 and 15)
 Compute class interval (width): 10 (46/5 then round up)
 Determine class boundaries (limits): 10, 20, 30, 40, 50, 60
 Compute class midpoints: 15, 25, 35, 45, 55
 Count observations & assign to classes
(continued)

Class Frequency
10 up to 20 3 .15 15
20 up to 30 6 .30 30
30 up to 40 5 .25 25
40 up to 50 4 .20 20
50 up to 60 2 .10 10
Total 20 1.00 100
Relative
Frequency Percentage
Data in ordered array:
12, 13, 17, 21, 24, 24, 26, 27, 27, 30, 32, 35, 37, 38, 41, 43, 44, 46, 53, 58
(continued)

Tabulating Numerical Data:
Cumulative Frequency
Class
10 up to 20 3 15 3 15
20 up to 30 6 30 9 45
30 up to 40 5 25 14 70
40 up to 50 4 20 18 90
50 up to 60 2 10 20 100
Total 20 100
Percentage
Cumulative
Percentage
12, 13, 17, 21, 24, 24, 26, 27, 27, 30, 32, 35, 37, 38, 41, 43, 44, 46, 53, 58
Frequency
Cumulative
Frequency

Graphing Numerical Data:
The Histogram
 A graph of the data in a frequency distribution
is called a histogram
 The class boundaries (or class midpoints)
are shown on the horizontal axis
 the vertical axis is either frequency, relative
frequency, or percentage
 Bars of the appropriate heights are used to
represent the number of observations within
each class

Histogram Example
Histogram: Daily High Temperature
0
1
2
3
4
5
6
7
5 15 25 35 45 55 65
Frequency
Class Midpoints
(No gaps
between
bars)
Class
10 up to 20 15 3
20 up to 30 25 6
30 up to 40 35 5
40 up to 50 45 4
50 up to 60 55 2
Frequency
Class
Midpoint

Graphing Numerical Data:
The Frequency Polygon
Frequency Polygon: Daily High Temperature
0
1
2
3
4
5
6
7
5 15 25 35 45 55 65
Frequency
Class Midpoints
Class
10 up to 20 15 3
20 up to 30 25 6
30 up to 40 35 5
40 up to 50 45 4
50 up to 60 55 2
Frequency
Class
Midpoint
(In a percentage
polygon the vertical axis
would be defined to
show the percentage of
observations per class)

Graphing Cumulative Frequencies:
The Ogive (Cumulative % Polygon)
Ogive: Daily High Temperature
0
20
40
60
80
100
10 20 30 40 50 60
Cumulative
Percentage
Class Boundaries (Not Midpoints)
Class
Less than 10 10 0
10 up to 20 20 15
20 up to 30 30 45
30 up to 40 40 70
40 up to 50 50 90
50 up to 60 60 100
Cumulative
Percentage
upper
class
boundary
10 20 30 40 50 60

Stem-and-Leaf Diagram
 A simple way to see distribution details in a
data set
METHOD: Separate the sorted data series
into leading digits (the stem) and
the trailing digits (the leaves)

Example
 Here, use the 10’s digit for the stem unit:
21, 24, 24, 26, 27, 27, 30, 32, 38, 41
 21 is shown as
 38 is shown as
 41 is shown as
Stem Leaf
2 1
3 8
4 1

Example
 Completed stem-and-leaf diagram:
Stem Leaves
2 1 4 4 6 7 7
3 0 2 8
4 1
(continued)
21, 24, 24, 26, 27, 27, 30, 32, 38, 41

Using other stem units
 Using the 100’s digit as the stem:
 Round off the 10’s digit to form the leaves
 613 would become 6 1
 776 would become 7 8
 . . .
 1224 becomes 12 2
Stem Leaf

Using other stem units
 Using the 100’s digit as the stem:
 The completed stem-and-leaf display:
Stem Leaves
(continued)
6 1 3 6
7 2 2 5 8
8 3 4 6 6 9 9
9 1 3 3 6 8
10 3 5 6
11 4 7
12 2
Data:
613, 632, 658, 717,
722, 750, 776, 827,
841, 859, 863, 891,
894, 906, 928, 933,
955, 982, 1034,
1047,1056, 1140,
1169, 1224

Tables and Charts for Categorical
Data
Categorical
Data
Graphing Data
Pie
Charts
Pareto
Diagram
Bar
Charts
Tabulating Data
Summary
Table

The Summary Table
Example: Current Investment Portfolio
Investment Amount Percentage
Type (in thousands $) (%)
Stocks 46.5 42.27
Bonds 32.0 29.09
CD 15.5 14.09
Savings 16.0 14.55
Total 110.0 100.0
(Variables are
Categorical)
Summarize data by category

Bar and Pie Charts
 Bar charts and Pie charts are often used
for qualitative data (categories or nominal
scale)
 Height of bar or size of pie slice shows the
frequency or percentage for each
category

Bar Chart Example
Investor's Portfolio
0 10 20 30 40 50
Stocks
Bonds
CD
Savings
Amount in $1000's
Stocks 46.5 42.27
Bonds 32.0 29.09
CD 15.5 14.09
Savings 16.0 14.55
Total 110.0 100.0
Current Investment Portfolio

Pie Chart Example
Percentages
are rounded to
the nearest
percent
Savings
15%
CD
14%
Bonds
29%
Stocks
42%
Stocks 46.5 42.27
Bonds 32.0 29.09
CD 15.5 14.09
Savings 16.0 14.55
Total 110.0 100.0

Pareto Diagram
 Used to portray categorical data (nominal scale)
 A bar chart, where categories are shown in descending order of
frequency
 A cumulative polygon is often shown in the same graph
 Used to separate the “vital few” from the “trivial many”

Pareto Diagram Example
cumulative
%
invested
(line
graph)
%
invested
in
each
category
(bar
graph)
0%
5%
10%
15%
20%
25%
30%
35%
40%
45%
Stocks Bonds Savings CD
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%

Tabulating and Graphing
Multivariate Categorical Data
 Contingency Table for Investment Choices ($1000’s)
Investment Investor A Investor B Investor C Total
Category
Stocks 46.5 55 27.5 129
Bonds 32.0 44 19.0 95
CD 15.5 20 13.5 49
Savings 16.0 28 7.0 51
Total 110.0 147 67.0 324
(Individual values could also be expressed as percentages of the overall total,
percentages of the row totals, or percentages of the column totals)

Tabulating and Graphing
Multivariate Categorical Data
 Side-by-side bar charts
(continued)
Comparing Investors
0 10 20 30 40 50 60
S toc k s
B onds
CD
S avings
Inves tor A Inves tor B Inves tor C

Side-by-Side Chart Example
 Sales by quarter for three sales territories:
0
10
20
30
40
50
60
1st Qtr 2nd Qtr 3rd Qtr 4th Qtr
East
West
North
1st Qtr 2nd Qtr 3rd Qtr 4th Qtr
East 20.4 27.4 59 20.4
West 30.6 38.6 34.6 31.6
North 45.9 46.9 45 43.9

 Scatter Diagrams are used to
examine possible relationships
between two numerical variables
 The Scatter Diagram:
 one variable is measured on the vertical
axis and the other variable is measured
on the horizontal axis
Scatter Diagrams

Scatter Diagram Example
Cost per Day vs. Production Volume
0
50
100
150
200
250
0 10 20 30 40 50 60 70
Volume per Day
Cost
per
Day
Volume
per day
Cost per
day
23 131
24 120
26 140
29 151
33 160
38 167
41 185
42 170
50 188
55 195
60 200

 A Time Series Plot is used to study
patterns in the values of a variable
over time
 The Time Series Plot:
 one variable is measured on the vertical
axis and the time period is measured on
the horizontal axis
Time Series Plot

Scatter Diagram Example
Number of Franchises, 1996-2004
0
20
40
60
80
100
120
1994 1996 1998 2000 2002 2004 2006
Year
Number
of
Franchises
Year
Number of
Franchises
1996 43
1997 54
1998 60
1999 73
2000 82
2001 95
2002 107
2003 99
2004 95

Misusing Graphs and Ethical Issues
Guidelines for good graphs:
 Do not distort the data
 Avoid unnecessary adornments (no “chart junk”)
 Use a scale for each axis on a two-dimensional
graph
 The vertical axis scale should begin at zero
 Properly label all axes
 The graph should contain a title
 Use the simplest graph for a given set of data

Chapter Summary
 Data in raw form are usually not easy to use for
decision making -- Some type of organization is
needed:
 Table  Graph
 Techniques reviewed in this chapter:
 Frequency distributions, histograms and polygons
 Cumulative distributions and ogives
 Stem-and-leaf display
 Bar charts, pie charts, and Pareto diagrams
 Contingency tables and side-by-side bar charts
 Scatter diagrams and time series plots

Chapters 3 & 4
Describing Data:
Numerical Measures, Displaying &
Exploring Data
Basic Statistics for
Business & Economics

Learning Objectives
In this chapter, you learn:
 To describe the properties of central tendency,
variation, and shape in numerical data
 To calculate descriptive summary measures for a
population
 To calculate the coefficient of variation and Z-
statistic
 To construct and interpret a box-and-whisker plot

Chapter Topics
 Measures of central tendency, variation,
location, and shape
 Mean, median, mode, geometric mean
 Range, interquartile range, variance and standard
deviation, coefficient of variation
 Quartiles, Z-statistic
 Symmetric and skewed distributions
 Population summary measures
 Mean, variance, and standard deviation
 The empirical rule and Chebyshev rule
 Five number summary and box-and-whisker
plot

Summary Measures
Arithmetic Mean
Median
Mode
Describing Data Numerically
Variance
Standard Deviation
Coefficient of Variation
Range
Interquartile Range
Geometric Mean
Skewness
Central Tendency Variation Shape
Location
Fractiles
Z - Statistic
Weighted Mean

Measures of Central Tendency
Central Tendency
Arithmetic Mean Median Mode Geometric Mean
n
X
X
n
i
i


 1
n
/
1
n
2
1
G )
X
X
X
(
X 


 
Midpoint of
ranked
values
Most
frequently
observed
value

Arithmetic Mean
 The arithmetic mean (mean) is the most
common measure of central tendency
 For a sample of size n:
Sample size
n
X
X
X
n
X
X n
2
1
n
1
i
i






 
Observed values

Arithmetic Mean
 The most common measure of central tendency
 Mean = sum of values divided by the number of values
 Affected by extreme values (outliers)
(continued)
0 1 2 3 4 5 6 7 8 9 10
Mean = 3
0 1 2 3 4 5 6 7 8 9 10
Mean = 4
3
5
15
5
5
4
3
2
1






4
5
20
5
10
4
3
2
1







Median
 In an ordered array, the median is the “middle”
number (50% above, 50% below)
 Not affected by extreme values
0 1 2 3 4 5 6 7 8 9 10
Median = 3
0 1 2 3 4 5 6 7 8 9 10
Median = 3

Finding the Median
 The location of the median:
 If the number of values is odd, the median is the middle number
 If the number of values is even, the median is the average of the two middle numbers
 Note that is not the value of the median, only the position of the
median in the ranked data
data
ordered
the
in
position
2
1
n
position
Median


2
1
n 

Mode
 A measure of central tendency
 Value that occurs most often
 Not affected by extreme values
 Used for either numerical or categorical
(nominal) data
 There may may be no mode
 There may be several modes
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Mode = 9
0 1 2 3 4 5 6
No Mode

Review Example
 Five houses on a hill by the beach
$2,000 K
$500 K
$300 K
$100 K
$100 K
House Prices:
$2,000,000
500,000
300,000
100,000
100,000

Review Example:
Summary Statistics
 Mean: ($3,000,000/5)
= $600,000
 Median: middle value of ranked data
= $300,000
 Mode: most frequent value
= $100,000
House Prices:
$2,000,000
500,000
300,000
100,000
100,000
Sum $3,000,000

Which measure of location
is the “best”?
 Mean is generally used, unless
extreme values (outliers) exist
 Then median is often used, since
the median is not sensitive to
extreme values.
 Example: Median home prices may be
reported for a region – less sensitive to
outliers

Geometric Mean
 Geometric mean
 Used to measure the rate of change of a variable
over time
 Geometric mean rate of return
 Measures the status of an investment over time
 Where Ri is the rate of return in time period i
n
/
1
n
2
1
G )
X
X
X
(
X 


 
1
)]
R
1
(
)
R
1
(
)
R
1
[(
R n
/
1
n
2
1
G 






 

Example
An investment of $100,000 declined to $50,000 at the
end of year one and rebounded to $100,000 at end
of year two:
000
,
100
$
X
000
,
50
$
X
000
,
100
$
X 3
2
1 


50% decrease 100% increase
The overall two-year return is zero, since it started and
ended at the same level.

Example
Use the 1-year returns to compute the arithmetic
mean and the geometric mean:
%
0
1
1
1
)]
2
(
)
50
[(.
1
%))]
100
(
1
(
%))
50
(
1
[(
1
)]
R
1
(
)
R
1
(
)
R
1
[(
R
2
/
1
2
/
1
2
/
1
n
/
1
n
2
1
G



















 
%
25
2
%)
100
(
%)
50
(
X 



Arithmetic
mean rate
of return:
Geometric
mean rate
of return:
Misleading result
More
accurate
result
(continued)

Measures of Variation
Same center,
different variation
Variation
Variance Standard
Deviation
Coefficient
of Variation
Range Interquartile
Range
 Measures of variation give
information on the spread
or variability of the data
values.

Range
 Simplest measure of variation
 Difference between the largest and the smallest
values in a set of data:
Range = Xlargest – Xsmallest
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Range = 14 - 1 = 13
Example:

Disadvantages of the Range
 Ignores the way in which data are distributed
 Sensitive to outliers
7 8 9 10 11 12
Range = 12 - 7 = 5
7 8 9 10 11 12
Range = 12 - 7 = 5
1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,5
1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,120
Range = 5 - 1 = 4
Range = 120 - 1 = 119

Variance
 Average (approximately) of squared deviations of values from
the mean
 Sample variance:
1
-
n
)
X
(X
S
n
1
i
2
i
2




Where = mean
n = sample size
Xi = ith value of the variable X
X

Standard Deviation
 Most commonly used measure of variation
 Shows variation about the mean
 Is the square root of the variance
 Has the same units as the original data
 Sample standard deviation:
1
-
n
)
X
(X
S
n
1
i
2
i





Calculation Example:
Sample Standard Deviation
Sample
Data (Xi) : 10 12 14 15 17 18 18 24
n = 8 Mean = X = 16
4.3095
7
130
1
8
16)
(24
16)
(14
16)
(12
16)
(10
1
n
)
X
(24
)
X
(14
)
X
(12
)
X
(10
S
2
2
2
2
2
2
2
2
























A measure of the “average”
scatter around the mean

Measuring variation
Small standard deviation
Large standard deviation

Comparing Standard Deviations
Mean = 15.5
S = 3.338
11 12 13 14 15 16 17 18 19 20 21
11 12 13 14 15 16 17 18 19 20 21
Data B
Data A
Mean = 15.5
S = 0.926
11 12 13 14 15 16 17 18 19 20 21
Mean = 15.5
S = 4.567
Data C

Advantages of Variance and
Standard Deviation
 Each value in the data set is used in the
calculation
 Values far from the mean are given extra
weight
(because deviations from the mean are squared)

Coefficient of Variation
 Measures relative variation
 Always in percentage (%)
 Shows variation relative to mean
 Can be used to compare two or more sets of
data measured in different units
100%
X
S
CV 










Comparing Coefficient
of Variation
 Stock A:
 Average price last year = $50
 Standard deviation = $5
 Stock B:
 Average price last year = $100
 Standard deviation = $5
Both stocks
have the same
standard
deviation, but
stock B is less
variable relative
to its price
10%
100%
$50
$5
100%
X
S
CVA 












5%
100%
$100
$5
100%
X
S
CVB 













Numerical Measures
for a Population
 Population summary measures are called parameters
 The population mean is the sum of the values in the
population divided by the population size, N
N
X
X
X
N
X
N
2
1
N
1
i
i







 
μ = population mean
N = population size
Where

Population Variance
 Average of squared deviations of values from the mean
 Population variance:
N
μ)
(X
σ
N
1
i
2
i
2




Where μ = population mean
N = population size

Population Standard Deviation
 Most commonly used measure of variation
 Shows variation about the mean
 Is the square root of the population variance
 Has the same units as the original data
 Population standard deviation:
N
μ)
(X
σ
N
1
i
2
i





 If the data distribution is approximately
bell-shaped, then the interval:
 contains about 68% of the values in
the population or the sample
The Empirical Rule
1σ
μ 
μ
68%
1σ
μ 

 contains about 95% of the values in
the population or the sample
 contains about 99.7% of the values
in the population or the sample
The Empirical Rule
2σ
μ 
3σ
μ 
3σ
μ 
99.7%
95%
2σ
μ 

 Regardless of how the data are distributed,
at least (1 - 1/k2) x 100% of the values will
fall within k standard deviations of the mean
(for k > 1)
 Examples:
(1 - 1/22) x 100% = 75% …........ k=2 (μ ± 2σ)
(1 - 1/32) x 100% = 89% ………. k=3 (μ ± 3σ)
(1 - 1/42) x 100% = 93.8% ……..k=4 (μ ± 4σ)
Chebyshev Rule
within
At least

Approximating the Mean from a
Frequency Distribution
 Sometimes only a frequency distribution is available, not
the raw data
 Use the midpoint of a class interval to approximate the
values in that class
 Where n = number of values or sample size
c = number of classes in the frequency distribution
mj = midpoint of the jth class
fj = number of values in the jth class
n
f
m
X
c
1
j
j
j




Approximating the Standard Deviation from
a Frequency Distribution
 Assume that all values within each class interval
are located at the midpoint of the class
 Approximation for the standard deviation from a
frequency distribution:
1
-
n
f
)
X
(m
S
c
1
j
j
2
j





Measures of Location
Location
Z - Statistic
Fractiles
 Measures of location give
information on the relative
position of the data
values.
 Fractiles partition
ranked data into parts
that are approximately
equal.

Quartiles
 Quartiles split the ranked data into 4 segments with
an equal number of values per segment
25% 25% 25% 25%
 The first quartile, Q1, is the value for which 25% of the
observations are smaller and 75% are larger
 Q2 is the same as the median (50% are smaller, 50% are
larger)
 Only 25% of the observations are greater than the third
quartile
Q1 Q2 Q3

Quartile Formulas
Find a quartile by determining the value in the
appropriate position in the ranked data, where
First quartile position: Q1 = (n+1)/4
Second quartile position: Q2 = (n+1)/2 (the median position)
Third quartile position: Q3 = 3(n+1)/4
where n is the number of observed values

Quartiles
(n = 9)
Q1 is in the (9+1)/4 = 2.5 position of the ranked data
so use the value half way between the 2nd and 3rd values,
so Q1 = 12.5
Sample Data in Ordered Array: 11 12 13 16 16 17 18 21 22
 Example: Find the first quartile
Q1 and Q3 are measures of noncentral location
Q2 = median, a measure of central tendency

Quartiles
(n = 9)
Q1 is in the (9+1)/4 = 2.5 position of the ranked data,
so Q1 = 12.5
Q2 is in the (9+1)/2 = 5th position of the ranked data,
so Q2 = median = 16
Q3 is in the 3(9+1)/4 = 7.5 position of the ranked data,
so Q3 = 19.5
Sample Data in Ordered Array: 11 12 13 16 16 17 18 21 22
 Example:
(continued)

Interquartile Range
 Can eliminate some outlier problems by using
the interquartile range
 Eliminate some high- and low-valued
observations and calculate the range from the
remaining values
 Interquartile range = 3rd quartile – 1st quartile
= Q3 – Q1

Interquartile Range
Median
(Q2)
X
maximum
X
minimum Q1 Q3
Example:
25% 25% 25% 25%
12 30 45 57 70
Interquartile range
= 57 – 30 = 27

Z-Statistic
 A measure of distance from the mean (for example, a
Z-statistic of 2.0 means that a value is 2.0 standard
deviations from the mean)
 The difference between a value and the mean, divided
by the standard deviation
 A Z statistic above 3.0 or below -3.0 is considered an
outlier
S
X
X
Z



Z-statistic
Example:
 If the mean is 14.0 and the standard deviation is 3.0,
what is the Z statistic for the value 18.5?
 The value 18.5 is 1.5 standard deviations above the
mean
 (A negative Z-statistic would mean that a value is less
than the mean)
1.5
3.0
14.0
18.5
S
X
X
Z 




(continued)

Shape of a Distribution
 Describes how data are distributed
 Measures of shape
 Symmetric or skewed
Mean = Median
Mean < Median Median < Mean
Right-Skewed
Left-Skewed Symmetric

Pearson’s Coefficient of Skewness
 SK ranges from -3 up to 3
 A value of 0 indicates a symmetric distribution and will occur when the
mean and median are equal.
 A value near -3 indicates considerable negative skewness; the mean and
median are to the left of the mode.
 A value near 1.5 indicated moderate positive skewness; the mean and
median are to the right of the mode.
 
s
median
X
Sk


3

Exploratory Data Analysis
 Box-and-Whisker Plot: A Graphical display of data using 5-
number summary:
Minimum -- Q1 -- Median -- Q3 -- Maximum
Example:
Minimum 1st Median 3rd Maximum
Quartile Quartile
Minimum 1st Median 3rd Maximum
Quartile Quartile
25% 25% 25% 25%

Shape of Box-and-Whisker Plots
 The Box and central line are centered between the
endpoints if data are symmetric around the median
 A Box-and-Whisker plot can be shown in either vertical
or horizontal format
Min Q1 Median Q3 Max

Distribution Shape and
Box-and-Whisker Plot
Positively-Skewed
Negatively-Skewed Symmetric
Q1 Q2 Q3 Q1 Q2 Q3 Q1 Q2 Q3

Box-and-Whisker Plot Example
 Below is a Box-and-Whisker plot for the following
data:
0 2 2 2 3 3 4 5 5 10 27
 The data positively skewed, as the plot depicts
0 2 3 5 27
0 2 3 5 27
Min Q1 Q2 Q3 Max

Chapter Summary
 Described measures of central tendency
 Mean, weighted mean, median, mode, geometric mean
 Described measures of variation
 Range, interquartile range, variance and standard deviation, coefficient of variation,
 Described measures of location
 Fractile, Z-statistic
 Illustrated shape of distribution
 Symmetric, skewed, box-and-whisker plots

Chapter 5
A Survey of Probability
Concepts
Business and Statistics

Learning Objectives
 Basic probability concepts and definitions
 Marginal and Joint probability
 Conditional probability
 Various counting rules

Important Terms
 Experiment – A process that leads to the
occurrence of one and only one of several
possible observations.
 Outcome – A particular result of an experiment
that can be described by a single characteristic
 Event – A collection of one or more outcomes of
an experiment.
 Sample Space – the collection of all possible
events

Sample Space
The Sample Space is the collection of all
possible events
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:

Events
 Outcome
 An outcome from a sample space with one
characteristic
 e.g., A red card from a deck of cards
 Complement of an event A (denoted A’)
 All outcomes that are not part of event A
 e.g., All cards that are not diamonds
 Joint event
 Involves two or more characteristics simultaneously
 e.g., An ace that is also red from a deck of cards

Mutually Exclusive Events
 Mutually exclusive events
 Events that cannot occur together
example:
A = queen of diamonds; B = queen of clubs
 Events A and B are mutually exclusive

Collectively Exhaustive Events
 Collectively exhaustive events
 One of the events must occur
 The set of events covers the entire sample space
example:
A = aces; B = black cards;
C = diamonds; D = hearts
 Events A, B, C and D are collectively exhaustive
(but not mutually exclusive – an ace may also be
a heart)
 Events B, C and D are collectively exhaustive and
also mutually exclusive

Visualizing Events
 Venn Diagrams
 Let A = aces
 Let B = red cards
A
B
A ∩ B = ace and red
A U B = ace or red

Visualizing Events
 Contingency Tables
 Tree Diagrams
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Full Deck
of 52 Cards
Sample
Space
Sample
Space
2
24
2
24

Probability
 Probability is the numerical measure
of the likelihood that an event will
occur
 The probability of any event must be
between 0 and 1, inclusively
 The sum of the probabilities of all
mutually exclusive and collectively
exhaustive events is 1
Certain
Impossible
0.5
1
0
0 ≤ P(A) ≤ 1 For any event A
1
P(C)
P(B)
P(A) 


If A, B, and C are mutually exclusive and
collectively exhaustive

Assessing Probability
 There are three approaches to assessing the probability
of an uncertain event:
1. a priori classical probability: Assigning probabilities based on
the assumption of equally likely outcomes.
2. empirical probability: Assigning probabilities based on
experimentation or historical data.
3. subjective probability: Assigning probability based on judgement
an individual judgment or opinion about the probability of occurrence
outcomes
elementary
of
number
total
occur
can
event
the
ways
of
number
T
X
occurrence
of
y
probabilit 

observed
outcomes
of
number
total
observed
outcomes
favorable
of
number
occurrence
of
y
probabilit 

Computing Joint and
Marginal Probabilities
 The probability of a joint event, A and B:
 Computing a marginal (or simple) probability:
 Where B1, B2, …, Bk are k mutually exclusive and collectively
exhaustive events
outcomes
elementary
of
number
total
B
and
A
satisfying
outcomes
of
number
B
and
A
P 
)
(
)
B
d
an
P(A
)
B
and
P(A
)
B
and
P(A
P(A) k
2
1 


 

Joint Probability Example
P(Red and Ace)
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
2
cards
of
number
total
ace
and
red
are
that
cards
of
number



Marginal Probability Example
P(Ace)
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
4
52
2
52
2
)
(
)
Re
( 



 Black
and
Ace
P
d
and
Ace
P

Joint Probabilities Using Contingency
Table
P(A1 and B2) P(A1)
Total
Event
P(A2 and B1)
P(A1 and B1)
Event
Total 1
Joint Probabilities Marginal (Simple) Probabilities
A1
A2
B1 B2
P(B1) P(B2)
P(A2 and B2) P(A2)

General Addition Rule
P(A or B) = P(A) + P(B) - P(A and B)
General Addition Rule:
If A and B are mutually exclusive, then
P(A and B) = 0, so the rule can be simplified:
P(A or B) = P(A) + P(B)
For mutually exclusive events A and B

General Addition Rule Example
P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)
= 26/52 + 4/52 - 2/52 = 28/52
Don’t count
the two red
aces twice!
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52

Computing Conditional Probabilities
 A conditional probability is the probability of one
event, given that another event has occurred:
P(B)
B)
and
P(A
B)
|
P(A 
P(A)
B)
and
P(A
A)
|
P(B 
Where P(A and B) = joint probability of A and B
P(A) = marginal probability of A
P(B) = marginal probability of B
The conditional
probability of A given
that B has occurred
The conditional
probability of B given
that A has occurred

Conditional Probability Example
 What is the probability that a car has a CD
player, given that it has AC ?
i.e., we want to find P(CD | AC)
 Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a CD player
(CD). 20% of the cars have both.

No CD
CD Total
AC 0.2 0.5 0.7
No AC 0.2 0.1 0.3
Total 0.4 0.6 1.0
 Of the cars on a used car lot, 70% have air conditioning
(AC) and 40% have a CD player (CD).
20% of the cars have both.
0.2857
0.7
0.2
P(AC)
AC)
and
P(CD
AC)
|
P(CD 


(continued)

No CD
CD Total
AC 0.2 0.5 0.7
No AC 0.2 0.1 0.3
Total 0.4 0.6 1.0
 Given AC, we only consider the top row (70% of the cars). Of these,
20% have a CD player. 20% of 70% is about 28.57%.
0.2857
0.7
0.2
P(AC)
AC)
and
P(CD
AC)
|
P(CD 


(continued)

Using Decision Trees
P(AC and CD) = 0.2
P(AC and CD’) = 0.5
P(AC’ and CD’) = 0.1
P(AC’ and CD) = 0.2
7
.
5
.
3
.
2
.
3
.
1
.
All
Cars
7
.
2
.
Given AC or
no AC:
Joint Probability
Conditional
Probability

Using Decision Trees
P(CD and AC) = 0.2
P(CD and AC’) = 0.2
P(CD’ and AC’) = 0.1
P(CD’ and AC) = 0.5
4
.
2
.
6
.
5
.
6
.
1
.
All
Cars
4
.
2
.
Given CD or
no CD:
Conditional
Probability
Joint Probability

Statistical Independence
 Two events are independent if and only
if:
 Events A and B are independent when the probability
of one event is not affected by the other event
P(A)
B)
|
P(A 

Multiplication Rules
 Multiplication rule for two events A and B:
P(B)
B)
|
P(A
B)
and
P(A 
P(A)
B)
|
P(A 
Note: If A and B are independent, then
and the multiplication rule simplifies to
P(B)
P(A)
B)
and
P(A 

Counting Rules
 Rules for counting the number of possible outcomes
 Counting Rule 1:
 If any one of k different mutually exclusive and collectively exhaustive
events can occur on each of n trials, the number of possible outcomes
is equal to
kn

Counting Rules
 If there are k1 events on the first trial, k2 events on
the second trial, … and kn events on the nth trial, the
number of possible outcomes is
 Example:
 You want to go to a park, eat at a restaurant, and see a
movie. There are 3 parks, 4 restaurants, and 6 movie
choices. How many different possible combinations are
there?
 Answer: (3)(4)(6) = 72 different possibilities
(k1)(k2)…(kn)
(continued)

Counting Rules
 The number of ways that n items can be arranged in order is
 Example:
 Your restaurant has five menu choices for lunch. How many ways can you order
them on your menu?
 Answer: 5! = (5)(4)(3)(2)(1) = 120 different possibilities
n! = (n)(n – 1)…(1)
(continued)

Counting Rules
 Permutations: The number of ways of arranging X
objects selected from n objects in order is
 Example:
 Your restaurant has five menu choices, and three are
selected for daily specials. How many different ways can
the specials menu be ordered?
 Answer: different possibilities
(continued)
X)!
(n
n!
Px
n


60
2
120
3)!
(5
5!
X)!
(n
n!
nPx 






Counting Rules
 Combinations: The number of ways of selecting X objects from n objects,
irrespective of order, is
 Example:
 Your restaurant has five menu choices, and three are selected for daily specials.
How many different special combinations are there, ignoring the order in which they
are selected?
 Answer: different possibilities
(continued)
X)!
(n
X!
n!
Cx
n


10
(6)(2)
120
3)!
(5
3!
5!
X)!
(n
X!
n!
Cx
n 






Chapter Summary
 Discussed basic probability concepts
 Sample spaces and events, contingency tables, simple
probability, marginal probability, and joint probability
 Examined basic probability rules
 General addition rule, addition rule for mutually exclusive events,
rule for collectively exhaustive events
 Defined conditional probability
 Statistical independence,decision trees, and the multiplication
rule
 Examined counting rules

Chapter 6
Discrete Probability
Distributions
Business and Economics

Learning Objectives
 The properties of a probability distribution
 To calculate the expected value and variance of a
probability distribution
 To calculate probabilities from binomial,
hypergeometric, and Poisson distributions
 How to use the binomial, hypergeometric, and
Poisson distributions to solve business problems

Introduction to Probability
Distributions
 Random Variable
 Represents a possible numerical value from
an uncertain event
Random
Variables
Discrete
Random Variable
Continuous
Random Variable
Ch. 5 Ch. 6

Discrete Random Variables
 Can only assume a countable number of values
Examples:
 Roll a die twice
Let X be the number of times 4 comes up
(then X could be 0, 1, or 2 times)
 Toss a coin 5 times.
Let X be the number of heads
(then X = 0, 1, 2, 3, 4, or 5)

Discrete Probability Distribution
X Value Probability
0 1/4 = 0.25
1 2/4 = 0.50
2 1/4 = 0.25
Experiment: Toss 2 Coins. Let X = # heads.
T
T
4 possible outcomes
T
T
H
H
H H
Probability Distribution
0 1 2 X
0.50
0.25
Probability

Discrete Random Variable
Summary Measures
 Expected Value (or mean) of a discrete
distribution (Weighted Average)
 Example: Toss 2 coins,
X = # of heads,
compute expected value of X:
E(X) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25)
= 1.0
X P(X)
0 0.25
1 0.50
2 0.25





N
1
i
i
i )
X
(
P
X
E(X)

Summary Measures
 Variance of a discrete random variable
 Standard Deviation of a discrete random variable
where:
E(X) = Expected value of the discrete random variable X
Xi = the ith outcome of X
P(Xi) = Probability of the ith occurrence of X




N
1
i
i
2
i
2
)
P(X
E(X)]
[X
σ
(continued)





N
1
i
i
2
i
2
)
P(X
E(X)]
[X
σ
σ

Summary Measures
 Example: Toss 2 coins, X = # heads,
compute standard deviation (recall E(X) = 1)
)
P(X
E(X)]
[X
σ i
2
i

 
0.707
0.50
(0.25)
1)
(2
(0.50)
1)
(1
(0.25)
1)
(0
σ 2
2
2








(continued)
Possible number of heads
= 0, 1, or 2

Probability Distributions
Continuous
Probability
Distributions
Binomial
Hypergeometric
Poisson
Probability
Distributions
Discrete
Probability
Distributions
Normal
Ch. 5 Ch. 6

The Binomial Distribution
Binomial
Hypergeometric
Poisson
Probability
Distributions
Discrete
Probability
Distributions

Binomial Probability Distribution
 A fixed number of observations, n
 e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse
 Two mutually exclusive and collectively exhaustive
categories
 e.g., head or tail in each toss of a coin; defective or not defective
light bulb
 Generally called “success” and “failure”
 Probability of success is p, probability of failure is 1 – 
 Constant probability for each observation
 e.g., Probability of getting a tail is the same each time we toss
the coin

Binomial Probability Distribution
(continued)
 Observations are independent
 The outcome of one observation does not affect the
outcome of the other
 Two sampling methods
 Infinite population without replacement
 Finite population with replacement

Possible Binomial Distribution
Settings
 A manufacturing plant labels items as
either defective or acceptable
 A firm bidding for contracts will either get a
contract or not
 A marketing research firm receives survey
responses of “yes I will buy” or “no I will
not”
 New job applicants either accept the offer
or reject it

Rule of Combinations
 The number of combinations of selecting X
objects out of n objects is
X)!
(n
X!
n!
Cx
n


where:
n! =(n)(n - 1)(n - 2) . . . (2)(1)
X! = (X)(X - 1)(X - 2) . . . (2)(1)
0! = 1 (by definition)

Binomial Distribution Formula
P(X) = probability of X successes in n trials,
with probability of success  on each trial
X = number of ‘successes’ in sample,
(X = 0, 1, 2, ..., n)
n = sample size (number of trials or observations)
 = probability of “success”
P(X)
n
X ! n X
 (1- )
X n X
!
( )!



Example: Flip a coin four
times, let x = # heads:
n = 4
 = 0.5
1 -  = (1 - 0.5) = 0.5
X = 0, 1, 2, 3, 4
The number of observations with
exactly x successes among n trials
Probability of x successes among n
trials for any 1 particular order

Example:
Calculating a Binomial Probability
What is the probability of one success in five
observations if the probability of success is .1?
X = 1, n = 5, and  = 0.1
0.32805
.9)
(5)(0.1)(0
0.1)
(1
(0.1)
1)!
(5
1!
5!
)
(1
X)!
(n
X!
n!
1)
P(X
4
1
5
1
X
n
X














Binomial Distribution
 The shape of the binomial distribution depends on the
values of  and n
n = 5  = 0.1
n = 5  = 0.5
Mean
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
 Here, n = 5 and  = 0.1
 Here, n = 5 and  = 0.5

Binomial Distribution
Characteristics
 Mean
 Variance and Standard Deviation

n
E(x)
μ 

)
-
(1
n
σ2



)
-
(1
n
σ 


Where n = sample size
 = probability of success
(1 – ) = probability of failure

Binomial Characteristics
n = 5  = 0.1
n = 5  = 0.5
Mean
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
0.5
(5)(0.1)
n
μ 

 
0.6708
0.1)
(5)(0.1)(1
)
-
(1
n
σ



 

2.5
(5)(0.5)
n
μ 

 
1.118
0.5)
(5)(0.5)(1
)
-
(1
n
σ



 

Examples

Using Binomial Tables
n = 10
x …  =.20  =.25  =.30  =.35  =.40  =.45  =.50
0
1
2
3
4
5
6
7
8
9
10
…
…
…
…
…
…
…
…
…
…
…
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0001
0.0000
0.0000
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0000
0.0000
0.0282
0.1211
0.2335
0.2668
0.2001
0.1029
0.0368
0.0090
0.0014
0.0001
0.0000
0.0135
0.0725
0.1757
0.2522
0.2377
0.1536
0.0689
0.0212
0.0043
0.0005
0.0000
0.0060
0.0403
0.1209
0.2150
0.2508
0.2007
0.1115
0.0425
0.0106
0.0016
0.0001
0.0025
0.0207
0.0763
0.1665
0.2384
0.2340
0.1596
0.0746
0.0229
0.0042
0.0003
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
10
9
8
7
6
5
4
3
2
1
0
…  =.80  =.75  =.70  =.65  =.60  =.55  =.50 x
Examples:
n = 10,  = 0.35, x = 3: P(x = 3|n =10,  = 0.35) = 0.2522
n = 10,  = 0.75, x = 2: P(x = 2|n =10,  = 0.75) = 0.0004

The Hypergeometric
Distribution
Binomial
Poisson
Probability
Distributions
Discrete
Probability
Distributions
Hypergeometric

The Hypergeometric Distribution
 “n” trials in a sample taken from a population of size N
 Two sampling methods
 finite population without replacement
 sample size is more than 5% of the population Sample taken
from finite population without replacement or the
 Outcomes of trials are dependent
 Concerned with finding the probability of “X” successes
in the sample where there are “S” successes in the
population

Hypergeometric Distribution
Formula



























 

n
N
X
n
S
N
X
S
C
]
C
][
C
[
P(X)
n
N
X
n
S
N
X
S
Where
N = population size
S = number of successes in the population
N – S = number of failures in the population
n = sample size
X = number of successes in the sample
n – X = number of failures in the sample

Properties of the
 The mean of the hypergeometric distribution is
 The standard deviation is
Where is called the “Finite Population Correction Factor”
from sampling without replacement from a
finite population
N
nS
E(x)
μ 

1
-
N
n
-
N
N
S)
-
nS(N
σ 2


1
-
N
n
-
N

Using the
■ Example: 3 different computers are checked from 10 in
the department. 4 of the 10 computers have illegal
software loaded. What is the probability that 2 of the 3
selected computers have illegal software loaded?
N = 10 n = 3
S = 4 X = 2
0.3
120
(6)(6)
3
10
1
6
2
4
n
N
X
n
S
N
X
S
2)
P(X 






















































The probability that 2 of the 3 selected computers have illegal
software loaded is 0.30, or 30%.

The Poisson Distribution
Binomial
Hypergeometric
Poisson
Probability
Distributions
Discrete
Probability
Distributions

The Poisson Distribution
 Apply the Poisson Distribution when:
 You wish to count the number of times an event
occurs in a given area of opportunity
 The probability that an event occurs in one area of
opportunity is the same for all areas of opportunity
 The number of events that occur in one area of
opportunity is independent of the number of events
that occur in the other areas of opportunity
 The probability that two or more events occur in an
area of opportunity approaches zero as the area of
opportunity becomes smaller
 The average number of events per unit is  (lambda)

Poisson Distribution Formula
where:
X = number of events in an area of opportunity
 = expected number of events
e = base of the natural logarithm system (2.71828...)
!
)
(
X
e
X
P
x





Poisson Distribution
Characteristics
 Mean
 Variance and Standard Deviation
λ
μ 
λ
σ2

λ
σ 
where  = expected number of events

Using Poisson Tables
X

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90
0
1
2
3
4
5
6
7
0.9048
0.0905
0.0045
0.0002
0.0000
0.0000
0.0000
0.0000
0.8187
0.1637
0.0164
0.0011
0.0001
0.0000
0.0000
0.0000
0.7408
0.2222
0.0333
0.0033
0.0003
0.0000
0.0000
0.0000
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
0.0000
0.0000
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.5488
0.3293
0.0988
0.0198
0.0030
0.0004
0.0000
0.0000
0.4966
0.3476
0.1217
0.0284
0.0050
0.0007
0.0001
0.0000
0.4493
0.3595
0.1438
0.0383
0.0077
0.0012
0.0002
0.0000
0.4066
0.3659
0.1647
0.0494
0.0111
0.0020
0.0003
0.0000
Example: Find P(X = 2) if  = 0.50
0.0758
2!
(0.50)
e
X!
e
2)
P(X
2
0.50
X
λ






λ

Graph of Poisson Probabilities
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3 4 5 6 7
x
P(x)
X
 =
0.50
0
1
2
3
4
5
6
7
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
P(X = 2) = 0.0758
Graphically:
 = 0.50

Poisson Distribution Shape
 The shape of the Poisson Distribution depends on
the parameter  : As  becomes larger, the Poisson
distribution becomes more symmetrical.
0.00
0.05
0.10
0.15
0.20
0.25
1 2 3 4 5 6 7 8 9 10 11 12
x
P(x)
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3 4 5 6 7
x
P(x)
 = 0.50  = 3.00

Chapter Summary
 Addressed the probability of a discrete random variable
 Discussed the Binomial distribution
 Discussed the Hypergeometric distribution
 Discussed the Poisson distribution

Chapter 7
The Normal Probability
Distribution

Learning Objectives
 To compute probabilities from the normal
distribution
 To compute probabilities from the normal
distribution to approximate probabilities from the
binomial distribution
 To use the normal probability plot to determine
whether a set of data is approximately normally
distributed

Probability Distributions
Continuous
Probability
Distributions
Binomial
Hypergeometric
Poisson
Probability
Distributions
Discrete
Probability
Distributions
Normal
Ch. 5 Ch. 6

Continuous Probability Distributions
 A continuous random variable is a variable that
can assume any value on a continuum (can
assume an uncountable number of values)
 thickness of an item
 time required to complete a task
 temperature of a solution
 height, in inches
 These can potentially take on any value,
depending only on the ability to measure
accurately.

The Normal Distribution
Probability
Distributions
Normal
Continuous
Probability
Distributions

 ‘Bell Shaped’
 Symmetrical
 Mean, Median and Mode
are Equal
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+  to  
Mean
= Median
= Mode
X
P(X)
μ
σ

By varying the parameters μ and σ, we obtain
different normal distributions
Many Normal Distributions

Shape
X
f(X)
μ
σ
Changing μ shifts the
distribution left or right.
Changing σ increases
or decreases the
spread.

The Normal Probability
Density Function
 The formula for the normal probability density
function is
Where e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
σ = the population standard deviation
X = any value of the continuous variable
2
μ)/σ]
(1/2)[(X
e
2π
1
P(X) 




The Standardized Normal
 Any normal distribution (with any mean and
standard deviation combination) can be
transformed into the standardized normal
distribution (Z)
 Need to transform X units into Z units

Translation to the Standardized
Normal Distribution
 Translate from X to the standardized normal
(the “Z” distribution) by subtracting the mean
of X and dividing by its standard deviation:
σ
μ
X
Z


The Z distribution always has mean = 0 and
standard deviation = 1

The Standardized Normal
Probability Density Function
 The formula for the standardized normal
probability density function is
Where e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
Z = any value of the standardized normal distribution
2
(1/2)Z
e
2π
1
P(Z) 


The Standardized
Normal Distribution
 Also known as the “Z” distribution
 Mean is 0
 Standard Deviation is 1
Z
P(Z)
0
1
Values above the mean have positive Z-values,
values below the mean have negative Z-values

Example
 If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
 This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
2.0
50
100
200
σ
μ
X
Z 





Comparing X and Z units
Z
100
2.0
0
200 X
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (X) or in standardized units (Z)
(μ = 100, σ = 50)
(μ = 0, σ = 1)

Finding Normal Probabilities
Probability is the
area under the
curve!
a b X
P(X) P a X b
( )
≤
Probability is measured by the area
under the curve
≤
P a X b
( )
<
<
=
(Note that the
probability of any
individual value is zero)

Probability as
Area Under the Curve
P(X)
X
μ
0.5
0.5
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
1.0
)
X
P( 




0.5
)
X
P(μ 



0.5
μ)
X
P( 




Empirical Rules
μ ± 1σ encloses about
68% of X’s
P(X)
X
μ μ+1σ
μ-1σ
What can we say about the distribution of values
around the mean? There are some general rules:
σ
σ
68.26%

The Empirical Rule
 μ ± 2σ covers about 95% of X’s
 μ ± 3σ covers about 99.7% of X’s
x
μ
2σ 2σ
x
μ
3σ 3σ
95.44% 99.73%
(continued)

The Standardized Normal Table
 The Standardized Normal table in the
textbook (Appendix D) gives the probability
between zero and a desired value for Z
Scale of Z
0 2.00
0.4772
Example:
P(0 < Z < 2.00) = 0.4772

The Standardized Normal Table
The value within the
table gives the
probability from Z = 0 up
to the desired Z value
.4772
2.0
P(0 < Z < 2.00) = 0.4772
The row shows
the value of Z
to the first
decimal point
The column gives the value of
Z to the second decimal point
2.0
.
.
.
(continued)
Z 0.00 0.01 0.02 …
0.0
0.1

General Procedure for
Finding Probabilities
 Draw the normal curve for the problem in
terms of X
 Translate X-values to Z-values
 Use the Standardized Normal Table
To find P(a < X < b) when X is
distributed normally:

 Suppose X is normal with mean 8.0 and
standard deviation 5.0
 Find P(X < 8.6)
X
8.6
8.0

standard deviation 5.0. Find P(X < 8.6)
Z
0.12
0
X
8.6
8
μ = 8
σ = 10
μ = 0
σ = 1
(continued)
0.12
5.0
8.0
8.6
σ
μ
X
Z 




P(X < 8.6) P(Z < 0.12)

Solution: Finding P(Z < 0.12)
Z
0.12
Z .00 .01
0.0 .0000 .0040 .0080
.0398 .0438
0.2 .0793 .0832 .0871
0.3 .1179 .1217 .1255
.5478
.02
0.1 .0478
Standardized Normal Probability
Table (Portion)
0.00
= P(Z < 0.12)
P(X < 8.6)
= 0.5 + 0.0478

Upper Tail Probabilities
standard deviation 5.0.
 Now Find P(X > 8.6)
X
8.6
8.0

Upper Tail Probabilities
 Now Find P(X > 8.6)…
(continued)
Z
0.12
0
Z
0.12
0.0478
0
0.500 0.5 - 0.0478
= 0.4522
P(X > 8.6) = P(Z > 0.12) = 0.5 - P(Z ≤ 0.12)
= 0.5 - 0.0478 = 0.4522

Probability Between
Two Values
standard deviation 5.0. Find P(8 < X < 8.6)
P(8 < X < 8.6)
= P(0 < Z < 0.12)
Z
0.12
0
X
8.6
8
0
5
8
8
σ
μ
X
Z 




0.12
5
8
8.6
σ
μ
X
Z 




Calculate Z-values:

Solution: Finding P(0 < Z < 0.12)
Z
0.12
0.0478
0.00
= P(0 < Z < 0.12)
P(8 < X < 8.6)
= P(Z < 0.12) – P(Z ≤ 0)
= 0.5478 - .5000 = 0.0478
0.5000
Z .00 .01
0.0 .0000 .0040 .0080
.0398 .0438
0.2 .0793 .0832 .0871
0.3 .1179 .1217 .1255
.02
0.1 .0478
Table (Portion)

Probabilities in the Lower Tail
 Now Find P(7.4 < X < 8)
X
7.4
8.0

Probabilities in the Lower Tail
Now Find P(7.4 < X < 8)…
X
7.4 8.0
P(7.4 < X < 8)
= P(-0.12 < Z < 0)
= P(Z < 0) – P(Z ≤ -0.12)
= 0.5000 - 0.4522 = 0.0478
(continued)
0.0478
0.4522
Z
-0.12 0
The Normal distribution is
symmetric, so this probability
is the same as P(0 < Z < 0.12)

Finding the X value for a Known
Probability
 Steps to find the X value for a known probability:
1. Find the Z value for the known probability
2. Convert to X units using the formula:
Zσ
μ
X 


Finding the X value for a Known
Probability
Example:
 Now find the X value so that only 20% of all
values are below this X
X
? 8.0
0.2000
Z
? 0
(continued)

Find the Z value for
20% in the Lower Tail
 20% area in the lower
tail is consistent with a
Z value of -0.84
Z .03
0.7 .2673 .2704
.2967
0.9 .3238 .3264
.04
0.8 .2995
Table (Portion)
.05
.2734
.3023
.3289
…
…
…
…
1. Find the Z value for the known probability
X
? 8.0
0.3000
0.2000
Z
-0.84 0

Finding the X value
2. Convert to X units using the formula:
80
.
3
0
.
5
)
84
.
0
(
0
.
8
Zσ
μ
X






So 20% of the values from a distribution
with mean 8.0 and standard deviation
5.0 are less than 3.80

Normal Approximation to the Binomial
Distribution
 The binomial distribution is a discrete distribution, but the normal
is continuous
 To use the normal to approximate the binomial, accuracy is
improved if you use a correction for continuity adjustment
 Example:
 X is discrete in a binomial distribution, so P(X = 4) can be approximated
with a continuous normal distribution by finding
P(3.5 < X < 4.5)

Distribution
 The closer  is to 0.5, the better the normal approximation to the
binomial
 The larger the sample size n, the better the normal
approximation to the binomial
 General rule:
 The normal distribution can be used to approximate the binomial
distribution if
n ≥ 5
and
n(1 – ) ≥ 5
(continued)

Distribution
 The mean and standard deviation of the
binomial distribution are
μ = n 
 Transform binomial to normal using the formula:
(continued)
)
(1
n
n
X
σ
μ
X
Z








)
(1
n
σ 
 


Using the Normal Approximation
to the Binomial Distribution
 If n = 1000 and  = 0.2, what is P(X ≤ 180)?
 Approximate P(X ≤ 180) using a continuity correction
adjustment:
P(X ≤ 180.5)
 Transform to standardized normal:
 So P(Z ≤ -1.54) = 0.0618
1.54
0.2)
)(1
(1000)(0.2
)
(1000)(0.2
180.5
)
(1
n
n
X
Z 










X
180.5 200
-1.54 0 Z

Evaluating Normality
 Not all continuous random variables are normally distributed
 It is important to evaluate how well the data set is approximated
by a normal distribution

Evaluating Normality
 Construct charts or graphs
 For small- or moderate-sized data sets, do stem-and-
leaf display and box-and-whisker plot look
symmetric?
 For large data sets, does the histogram or polygon
appear bell-shaped?
 Compute descriptive summary measures
 Do the mean, median and mode have similar values?
 Is the interquartile range approximately 1.33 σ?
 Is the range approximately 6 σ?
(continued)

Assessing Normality
 Observe the distribution of the data set
 Do approximately 2/3 of the observations lie within
mean 1 standard deviation?
 Do approximately 80% of the observations lie within
mean 1.28 standard deviations?
 Do approximately 95% of the observations lie within
mean 2 standard deviations?
 Evaluate normal probability plot
 Is the normal probability plot approximately linear
with positive slope?
(continued)




The Normal Probability Plot
 Normal probability plot
 Arrange data into ordered array
 Find corresponding standardized normal quantile values
 Plot the pairs of points with observed data values on the vertical axis and
the standardized normal quantile values on the horizontal axis
 Evaluate the plot for evidence of linearity

The Normal Probability Plot
A normal probability plot for data
from a normal distribution will be
approximately linear:
30
60
90
-2 -1 0 1 2 Z
X
(continued)

Normal Probability Plot
Left-Skewed Right-Skewed
Rectangular
30
60
90
-2 -1 0 1 2 Z
X
(continued)
30
60
90
-2 -1 0 1 2 Z
X
30
60
90
-2 -1 0 1 2 Z
X Nonlinear plots
indicate a deviation
from normality

Chapter 8
Sampling Methods and the
Central Limit Theorem

Learning Objectives
 To distinguish between different survey
sampling methods
 The concept of the sampling distribution
 The importance of the Central Limit Theorem
 To compute probabilities related to the sample
mean

Reasons for Drawing a Sample
 Less time consuming than a census
 Less costly to administer than a census
 Less cumbersome and more practical to
administer than a census of the targeted
population

Types of Samples Used
 Nonprobability Sample
 Items included are chosen without regard to
their probability of occurrence
 Probability Sample
 Items in the sample are chosen on the basis
of known probabilities

Types of Samples Used
Quota
Samples
Non-Probability
Samples
Judgement Chunk
Probability Samples
Simple
Random
Systematic
Stratified
Cluster
Convenience
(continued)

Probability Sampling
 Items in the sample are chosen based on known probabilities
Probability Samples
Simple
Random
Systematic Stratified Cluster

Simple Random Samples
 Every individual or item from the frame has an equal chance of
being selected
 Selection may be with replacement or without replacement
 Samples obtained from table of random numbers or computer
random number generators

Systematic Samples
 Decide on sample size: n
 Divide frame of N individuals into groups of k
individuals: k=N/n
 Randomly select one individual from the 1st
group
 Select every kth individual thereafter
N = 64
n = 8
k = 8
First Group

Stratified Samples
 Divide population into two or more subgroups (called
strata) according to some common characteristic
 A simple random sample is selected from each subgroup,
with sample sizes proportional to strata sizes
 Samples from subgroups are combined into one
Population
Divided
into 4
strata
Sample

Cluster Samples
 Population is divided into several “clusters,”
each representative of the population
 A simple random sample of clusters is selected
 All items in the selected clusters can be used, or items can be
chosen from a cluster using another probability sampling
technique
Population
divided into
16 clusters. Randomly selected
clusters for sample

Advantages and Disadvantages
 Simple random sample and systematic sample
 Simple to use
 May not be a good representation of the population’s
underlying characteristics
 Stratified sample
 Ensures representation of individuals across the
entire population
 Cluster sample
 More cost effective
 Less efficient (need larger sample to acquire the
same level of precision)

Sampling Distributions
 A sampling distribution is a
distribution of all of the possible
values of a statistic for a given size
sample selected from a population.

Sampling Distribution of the
Sample mean
 A sampling distribution of the
sample mean is a distribution of all
of the possible sample means for a
given size sample selected from a
population.

Developing a
Sampling Distribution
 Assume there is a population …
 Population size N=4
 Random variable, X,
is age of individuals
 Values of X: 18, 20,
22, 24 (years)
A B C D

Developing a
.3
.2
.1
0
18 20 22 24
A B C D
Uniform Distribution
P(x)
x
(continued)
Summary Measures for the Population Distribution:
21
4
24
22
20
18
N
X
μ i







2.236
N
μ)
(X
σ
2
i





Now consider all possible samples of size n=2
16 possible samples
(sampling with
replacement)
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
(continued)
Developing a
16 Sample
Means
1st
Obs
2nd Observation
18 20 22 24
18 18,18 18,20 18,22 18,24
20 20,18 20,20 20,22 20,24
22 22,18 22,20 22,22 22,24
24 24,18 24,20 24,22 24,24

Sampling Distribution of All Sample Means
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24 18 19 20 21 22 23 24
0
.1
.2
.3
P(X)
X
Sample Means
Distribution
16 Sample Means
_
Developing a
(continued)
(no longer uniform)
_

Summary Measures of this Sampling Distribution:
Developing a
(continued)
21
16
24
21
19
18
N
X
μ i
X







 
1.58
16
21)
-
(24
21)
-
(19
21)
-
(18
N
)
μ
X
(
σ
2
2
2
2
X
i
X










Comparing the Population with its
18 19 20 21 22 23 24
0
.1
.2
.3
P(X)
X
18 20 22 24
A B C D
0
.1
.2
.3
Population
N = 4
P(X)
X _
1.58
σ
21
μ X
X


2.236
σ
21
μ 

Sample Means Distribution
n = 2
_

Standard Error of the Mean
 Different samples of the same size from the same
population will yield different sample means
 A measure of the variability in the mean from sample to
sample is given by the Standard Error of the Mean:
(This assumes that sampling is with replacement or
sampling is without replacement from an infinite population)
 Note that the standard error of the mean decreases as
the sample size increases
n
σ
σX


If the Population is Normal
 If a population is normal with mean μ and
standard deviation σ, the sampling distribution
of is also normally distributed with
and
X
μ
μX

n
σ
σX


Z-value for Sampling Distribution
of the Mean
 Z-value for the sampling distribution of :
where: = sample mean
= population mean
= population standard deviation
n = sample size
X
μ
σ
n
σ
μ)
X
(
σ
)
μ
X
(
Z
X
X 



X

Sampling Distribution Properties

(i.e. is unbiased )
Normal Population
Distribution
Normal Sampling
Distribution
(has the same mean)
x
x
x
μ
μx 
μ
x
μ

Sampling Distribution Properties
As n increases,
decreases
Larger
sample size
Smaller
sample size
x
(continued)
x
σ
μ

If the Population is not Normal
 We can apply the Central Limit Theorem:
 Even if the population is not normal,
 …sample means from the population will be
approximately normal as long as the sample size is
large enough.
Properties of the sampling distribution:
and
μ
μx 
n
σ
σx 

Central Limit Theorem
n↑
As the
sample
size gets
large
enough…
the sampling
distribution
becomes
almost normal
regardless of
shape of
population
x

If the Population is not Normal
Population Distribution
(becomes normal as n increases)
Central Tendency
Variation
x
x
Larger
sample
size
Smaller
sample size
(continued)
Sampling distribution
properties:
μ
μx 
n
σ
σx 
x
μ
μ

How Large is Large Enough?
 For most distributions, n > 30 will give a
sampling distribution that is nearly normal
 For fairly symmetric distributions, n > 15
 For normal population distributions, the
sampling distribution of the mean is always
normally distributed

Example
 Suppose a population has mean μ = 8 and
standard deviation σ = 3. Suppose a random
sample of size n = 36 is selected.
 What is the probability that the sample mean is
between 7.8 and 8.2?

Example
Solution:
 Even if the population is not normally
distributed, the central limit theorem can be
used (n > 30)
 … so the sampling distribution of is
approximately normal
 … with mean = 8
 …and standard deviation
(continued)
x
x
μ
0.5
36
3
n
σ
σx 



Example
Solution (continued):
(continued)
0.3108
0.4)
Z
P(-0.4
36
3
8
-
8.2
n
σ
μ
-
X
36
3
8
-
7.8
P
8.2)
X
P(7.8



















Z
7.8 8.2 -0.4 0.4
Sampling
Distribution
Standard Normal
Distribution .1554
+.1554
Population
Distribution
?
?
?
?
?
?
?
?
?
?
?
?
Sample Standardize
8
μ  8
μX
 0
μz 
x
X

Chapter Summary
 Described different types of samples and sampling
techniques
 Introduced sampling distributions
 Described the sampling distribution of the mean
 For normal populations
 Using the Central Limit Theorem
 Calculated probabilities related to the sample mean

Chaptere 9
Estimation and Confidence
Intervals
Business Statistics for

Learning Objectives
 To construct and interpret confidence interval estimates
for the mean and the proportion
 How to determine the sample size necessary to
develop a confidence interval for the mean or
proportion

Confidence Intervals
Content of this chapter
 Confidence Intervals for the Population
Mean, μ
 when Population Standard Deviation σ is Known
 when Population Standard Deviation σ is Unknown
 Confidence Intervals for the Population
Proportion, p
 Determining the Required Sample Size

Point and Interval Estimates
 A point estimate is a single number,
 a confidence interval provides additional
information about variability
Point Estimate
Lower
Confidence
Limit
Upper
Confidence
Limit
Width of
confidence interval

Point Estimates
We can estimate a
Population Parameter …
with a Sample
Statistic
(a Point Estimate)
Mean
Proportion p
π
X
μ

 How much uncertainty is associated with a
point estimate of a population parameter?
 An interval estimate provides more
information about a population characteristic
than does a point estimate
 Such interval estimates are called confidence
intervals

Confidence Interval Estimate
 An interval gives a range of values:
 Takes into consideration variation in sample
statistics from sample to sample
 Based on observations from 1 sample
 Gives information about closeness to
unknown population parameters
 Stated in terms of level of confidence
 Can never be 100% confident

Estimation Process
(mean, μ, is
unknown)
Population
Random Sample
Mean
X = 50
Sample
I am 95%
confident that
μ is between
40 & 60.

General Formula
 The general formula for all
confidence intervals is:
Point Estimate ± (Critical Value)(Standard Error)

Confidence Level
 Confidence Level
 Confidence for which the interval
will contain the unknown
population parameter
 A percentage (less than 100%)

Confidence Level, (1-)
 Suppose confidence level = 95%
 Also written (1 - ) = 0.95
 A relative frequency interpretation:
 In the long run, 95% of all the confidence
intervals that can be constructed will contain the
unknown true parameter
 A specific interval either will contain or will
not contain the true parameter
 No probability involved in a specific interval
(continued)

Population
Mean
σ Unknown
Confidence
Intervals
Population
Proportion
σ Known

Confidence Interval for μ
(σ Known)
 Assumptions
 Population standard deviation σ is known
 Population is normally distributed
 If population is not normal, use large sample
 Confidence interval estimate:
where is the point estimate
Z is the normal distribution critical value for a probability of /2 in each tail
is the standard error
n
σ
Z
X 
X
n
σ/

Finding the Critical Value, Z
 Consider a 95% confidence interval:
Z= -1.96 Z= 1.96
0.95
1 

0.025
2

α
0.025
2

α
Point Estimate
Lower
Confidence
Limit
Upper
Confidence
Limit
Z units:
X units: Point Estimate
0
1.96
Z 


Common Levels of Confidence
 Commonly used confidence levels are 90%,
95%, and 99%
Confidence
Level
Confidence
Coefficient, Z value
1.28
1.645
1.96
2.33
2.58
3.08
3.27
0.80
0.90
0.95
0.98
0.99
0.998
0.999
80%
90%
95%
98%
99%
99.8%
99.9%


1

Intervals and Level of Confidence
μ
μx

Intervals
extend from
to
(1-)x100%
of intervals
constructed
contain μ;
()x100% do
not.
Sampling Distribution of the Mean
n
σ
Z
X 
n
σ
Z
X 
x
x1
x2
/2
 /2



1

Example
 A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
 Determine a 95% confidence interval for the
true mean resistance of the population.

Example
 A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
 Solution:
2.4068
1.9932
0.2068
2.20
)
11
(0.35/
1.96
2.20
n
σ
Z
X








(continued)

Interpretation
 We are 95% confident that the true mean
resistance is between 1.9932 and 2.4068
ohms
 Although the true mean may or may not be
in this interval, 95% of intervals formed in
this manner will contain the true mean

Population
Mean
σ Unknown
Confidence
Intervals
Population
Proportion
σ Known

(σ Unknown)
 If the population standard deviation σ is
unknown, we can substitute the sample
standard deviation, S
 This introduces extra uncertainty, since S
is variable from sample to sample
 So we use the t distribution instead of the
normal distribution

(σ Unknown)
 Assumptions
 Population standard deviation is unknown
 Population is normally distributed
 If population is not normal, use large sample
 Use Student’s t Distribution
 Confidence Interval Estimate:
(where t is the critical value of the t distribution with n -1 degrees of
freedom and an area of α/2 in each tail)
n
S
t
X 1
-
n

(continued)

Student’s t Distribution
 The t is a family of distributions
 The t value depends on degrees of
freedom (d.f.)
 Number of observations that are free to vary after
sample mean has been calculated
d.f. = n - 1

Degrees of Freedom (df)
Idea: Number of observations that are free to vary
after sample mean has been calculated
Example: Suppose the mean of 3 numbers is 8.0
Let X1 = 7
Let X2 = 8
What is X3?
If the mean of these three
values is 8.0,
then X3 must be 9
(i.e., X3 is not free to vary)
Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2
(2 values can be any numbers, but the third is not free to vary
for a given mean)

Student’s t Distribution
t
0
t (df = 5)
t (df = 13)
t-distributions are bell-
shaped and symmetric, but
have ‘fatter’ tails than the
normal
Standard
Normal
(t with df = ∞)
Note: t Z as n increases

Student’s t Table
Upper Tail Area
df .25 .10 .05
1 1.000 3.078 6.314
2 0.817 1.886 2.920
3 0.765 1.638 2.353
t
0 2.920
The body of the table
contains t values, not
probabilities
Let: n = 3
df = n - 1 = 2
 = 0.10
/2 = 0.05
/2 = 0.05

t distribution values
With comparison to the Z value
Confidence t t t Z
Level (10 d.f.) (20 d.f.) (30 d.f.) ____
0.80 1.372 1.325 1.310 1.28
0.90 1.812 1.725 1.697 1.645
0.95 2.228 2.086 2.042 1.96
0.99 3.169 2.845 2.750 2.58
Note: t Z as n increases

Example
A random sample of n = 25 has X = 50 and
S = 8. Form a 95% confidence interval for μ
 d.f. = n – 1 = 24, so
The confidence interval is
2.0639
t0.025,24
1
n
,
/2 



t
25
8
(2.0639)
50
n
S
t
X 1
-
n
/2, 

 
46.698 ≤ μ ≤ 53.302

Population
Mean
σ Unknown
Confidence
Intervals
Population
Proportion
σ Known

Confidence Intervals for the
Population Proportion, π
 An interval estimate for the population
proportion ( π ) can be calculated by
adding an allowance for uncertainty to
the sample proportion ( p )

Confidence Intervals for the
Population Proportion, π
 Recall that the distribution of the sample
proportion is approximately normal if the
sample size is large, with standard deviation
 We will estimate this with sample data:
(continued)
n
p)
p(1
n
)
(1
σp

 


Confidence Interval Endpoints
 Upper and lower confidence limits for the
population proportion are calculated with the
formula
 where
 Z is the standard normal value for the level of confidence desired
 p is the sample proportion
 n is the sample size
n
p)
p(1
Z
p



Example
 A random sample of 100 people
shows that 25 are left-handed.
 Form a 95% confidence interval for
the true proportion of left-handers

Example
 A random sample of 100 people shows
that 25 are left-handed. Form a 95%
confidence interval for the true proportion
of left-handers.
/100
0.25(0.75)
1.96
25/100
p)/n
p(1
Z
p




0.3349
0.1651
(0.0433)
1.96
0.25





(continued)

Interpretation
 We are 95% confident that the true
percentage of left-handers in the population
is between
16.51% and 33.49%.
 Although the interval from 0.1651 to 0.3349
may or may not contain the true proportion,
95% of intervals formed from samples of
size 100 in this manner will contain the true
proportion.

Determining Sample Size
For the
Mean
Determining
Sample Size
For the
Proportion

Sampling Error
 The required sample size can be found to reach
a desired margin of error (e) with a specified
level of confidence (1 - )
 The margin of error is also called sampling error
 the amount of imprecision in the estimate of the
population parameter
 the amount added and subtracted to the point
estimate to form the confidence interval

For the
Mean
Determining
Sample Size
n
σ
Z
X 
n
σ
Z
e 
Sampling error
(margin of error)

For the
Mean
Determining
Sample Size
n
σ
Z
e 
(continued)
2
2
2
e
σ
Z
n 
Now solve
for n to get

 To determine the required sample size for the
mean, you must know:
 The desired level of confidence (1 - ), which
determines the critical Z value
 The acceptable sampling error, e
 The standard deviation, σ
(continued)

Required Sample Size Example
If  = 45, what sample size is needed to
estimate the mean within ± 5 with 90%
confidence?
(Always round up)
219.19
5
(45)
(1.645)
e
σ
Z
n 2
2
2
2
2
2



So the required sample size is n = 220

If σ is unknown
 If unknown, σ can be estimated when
using the required sample size formula
 Use a value for σ that is expected to be
at least as large as the true σ
 Select a pilot sample and estimate σ with
the sample standard deviation, S

Determining
Sample Size
For the
Proportion
2
2
e
)
(1
Z
n
π
π 

Now solve
for n to get
n
)
(1
Z
e
π
π 

(continued)

 To determine the required sample size for the
proportion, you must know:
 The desired level of confidence (1 - ), which
determines the critical Z value
 The acceptable sampling error, e
 The true proportion of “successes”, π
 π can be estimated with a pilot sample, if
necessary (or conservatively use π = 0.5)
(continued)

How large a sample would be necessary
to estimate the true proportion defective in
a large population within ±3%, with 95%
confidence?
(Assume a pilot sample yields p = 0.12)

Solution:
For 95% confidence, use Z = 1.96
e = 0.03
p = 0.12, so use this to estimate π
So use n = 451
450.74
(0.03)
0.12)
(0.12)(1
(1.96)
e
)
(1
Z
n 2
2
2
2





π
π
(continued)

Chapter Summary
 Introduced the concept of confidence intervals
 Discussed point estimates
 Developed confidence interval estimates
 Created confidence interval estimates for the mean
(σ known)
 Determined confidence interval estimates for the
mean (σ unknown)
 Created confidence interval estimates for the
proportion
 Determined required sample size for mean and
proportion settings

Chapter 10
One-Sample Tests of a
Hypothesis
Business Statistics for

Learning Objectives
 The basic principles of hypothesis testing
 How to use hypothesis testing to test a mean or
proportion
 The assumptions of each hypothesis-testing
procedure, how to evaluate them, and the
consequences if they are seriously violated
 How to avoid the pitfalls involved in hypothesis testing
 The ethical issues involved in hypothesis testing

What is a Hypothesis?
 A hypothesis is a claim
(assumption) about a
population parameter:
 population mean
 population proportion
Example: The mean monthly cell phone bill
of this city is μ = $42
Example: The proportion of adults in this
city with cell phones is π = 0.68

The Null Hypothesis, H0
 States the claim or assertion to be tested
Example: The average number of TV sets in
U.S. Homes is equal to three ( )
 Is always about a population parameter,
not about a sample statistic
3
μ
:
H0 
3
μ
:
H0  3
X
:
H0 

The Null Hypothesis, H0
 Begin with the assumption that the null
hypothesis is true
 Similar to the notion of innocent until
proven guilty
 Refers to the status quo
 Always contains “=” , “≤” or “” sign
 May or may not be rejected
(continued)

The Alternative Hypothesis, H1
 Is the opposite of the null hypothesis
 e.g., The average number of TV sets in U.S.
homes is not equal to 3 ( H1: μ ≠ 3 )
 Challenges the status quo
 Never contains the “=” , “≤” or “” sign
 May or may not be proven
 Is generally the hypothesis that the
researcher is trying to prove

Population
Claim: the
population
mean age is 50.
(Null Hypothesis:
REJECT
Suppose
the sample
mean age
is 20: X = 20
Sample
Null Hypothesis
20 likely if μ = 50?

Is
Hypothesis Testing Process
If not likely,
Now select a
random sample
H0: μ = 50 )
X

Reason for Rejecting H0
Sampling Distribution of X
μ = 50
If H0 is true
If it is unlikely that
we would get a
sample mean of
this value ...
... then we
reject the null
hypothesis that
μ = 50.
20
... if in fact this were
the population mean…
X

Level of Significance, 
 Defines the unlikely values of the sample
statistic if the null hypothesis is true
 Defines rejection region of the sampling
distribution
 Is designated by  , (level of significance)
 Typical values are 0.01, 0.05, or 0.10
 Is selected by the researcher at the beginning
 Provides the critical value(s) of the test

Level of Significance
and the Rejection Region
H0: μ ≥ 3
H1: μ < 3
0
H0: μ ≤ 3
H1: μ > 3


Represents
critical value
Lower-tail test
Level of significance = 
0
Upper-tail test
Two-tail test
Rejection
region is
shaded
/2
0

/2

H0: μ = 3
H1: μ ≠ 3

Errors in Making Decisions
 Type I Error
 Reject a true null hypothesis
 Considered a serious type of error
The probability of Type I Error is 
 Called level of significance of the test
 Set by the researcher in advance

Errors in Making Decisions
 Type II Error
 Fail to reject a false null hypothesis
The probability of Type II Error is β
(continued)

Outcomes and Probabilities
Actual
Situation
Decision
Do Not
Reject
H0
No error
(1 - )

Type II Error
( β )
Reject
H0
Type I Error
( )

Possible Hypothesis Test Outcomes
H0 False
H0 True
Key:
Outcome
(Probability) No Error
( 1 - β )

Type I & II Error Relationship
 Type I and Type II errors cannot happen at
the same time
 Type I error can only occur if H0 is true
 Type II error can only occur if H0 is false
If Type I error probability (  ) , then
Type II error probability ( β )

Factors Affecting Type II Error
 All else equal,
 β when the difference between
hypothesized parameter and its true value
 β when 
 β when σ
 β when n

Hypothesis Tests for the Mean
 Known  Unknown
Hypothesis
Tests for 
(Z test) (t test)

Z Test of Hypothesis for the Mean
(σ Known)
 Convert sample statistic ( ) to a Z test statistic
X
The test statistic is:
n
σ
μ
X
Z


σ Known σ Unknown
Hypothesis
Tests for 
(Z test) (t test)

Critical Value
Approach to Testing
 For a two-tail test for the mean, σ known:
 Convert sample statistic ( ) to test statistic (Z
statistic )
 Determine the critical Z values for a specified
level of significance  from a table or
computer
 Decision Rule: If the test statistic falls in the
rejection region, reject H0 ; otherwise do not
reject H0
X

Two-Tail Tests
Do not reject H0 Reject H0
Reject H0
 There are two
cutoff values
(critical values),
defining the
regions of
rejection
/2
-Z 0
H0: μ = 3
H1: μ  3
+Z
/2
Lower
critical
value
Upper
critical
value
3
Z
X

6 Steps in
Hypothesis Testing
1. State the null hypothesis, H0 and the
alternative hypothesis, H1
2. Choose the level of significance, , and the
sample size, n
3. Determine the appropriate test statistic and
sampling distribution
4. Determine the critical values that divide the
rejection and nonrejection regions

6 Steps in
Hypothesis Testing
5. Collect data and compute the value of the test
statistic
6. Make the statistical decision and state the
managerial conclusion. If the test statistic falls
into the nonrejection region, do not reject the
null hypothesis H0. If the test statistic falls into
the rejection region, reject the null hypothesis.
Express the managerial conclusion in the
context of the problem
(continued)

Hypothesis Testing Example
Test the claim that the true mean # of TV
sets in US homes is equal to 3.
(Assume σ = 0.8)
1. State the appropriate null and alternative
hypotheses
 H0: μ = 3 H1: μ ≠ 3 (This is a two-tail test)
2. Specify the desired level of significance and the
sample size
 Suppose that  = 0.05 and n = 100 are chosen
for this test

2.0
.08
.16
100
0.8
3
2.84
n
σ
μ
X
Z 







3. Determine the appropriate technique
 σ is known so this is a Z test.
4. Determine the critical values
 For  = 0.05 the critical Z values are ±1.96
5. Collect the data and compute the test statistic
 Suppose the sample results are
n = 100, X = 2.84 (σ = 0.8 is assumed known)
So the test statistic is:
(continued)

 6. Is the test statistic in the rejection region?
Reject H0 Do not reject H0
 = 0.05/2
-Z= -1.96 0
Reject H0 if
Z < -1.96 or
Z > 1.96;
otherwise
do not
reject H0
(continued)
 = 0.05/2
Reject H0
+Z= +1.96
Here, Z = -2.0 < -1.96, so the
test statistic is in the rejection
region

6(continued). Reach a decision and interpret the result
-2.0
Since Z = -2.0 < -1.96, we reject the null hypothesis
and conclude that there is sufficient evidence that the
mean number of TVs in US homes is not equal to 3
(continued)
 = 0.05/2
-Z= -1.96 0
 = 0.05/2
Reject H0
+Z= +1.96

p-Value Approach to Testing
 p-value: Probability of obtaining a test
statistic more extreme ( ≤ or  ) than the
observed sample value given H0 is true
 Also called observed level of significance
 Smallest value of  for which H0 can be
rejected

p-Value Approach to Testing
 Convert Sample Statistic (e.g., ) to Test
Statistic (e.g., Z statistic )
 Obtain the p-value from a table or computer
 Compare the p-value with 
 If p-value <  , reject H0
 If p-value   , do not reject H0
X
(continued)

p-Value Example
 Example: How likely is it to see a sample mean of
2.84 (or something further from the mean, in either
direction) if the true mean is  = 3.0?
0.0228
/2 = 0.025
-1.96 0
-2.0
0.0228
2.0)
P(Z
0.0228
2.0)
P(Z





Z
1.96
2.0
X = 2.84 is translated
to a Z score of Z = -2.0
p-value
= 0.0228 + 0.0228 = 0.0456
0.0228
/2 = 0.025

p-Value Example
 Compare the p-value with 
 If p-value <  , reject H0
 If p-value   , do not reject H0
Here: p-value = 0.0456
 = 0.05
Since 0.0456 < 0.05,
we reject the null
hypothesis
(continued)
0.0228
/2 = 0.025
-1.96 0
-2.0
Z
1.96
2.0
0.0228
/2 = 0.025

Connection to Confidence Intervals
 For X = 2.84, σ = 0.8 and n = 100, the 95%
confidence interval is:
2.6832 ≤ μ ≤ 2.9968
 Since this interval does not contain the hypothesized
mean (3.0), we reject the null hypothesis at  = 0.05
100
0.8
(1.96)
2.84
to
100
0.8
(1.96)
-
2.84 

One-Tail Tests
 In many cases, the alternative hypothesis
focuses on a particular direction
H0: μ ≥ 3
H1: μ < 3
H0: μ ≤ 3
H1: μ > 3
This is a lower-tail test since the
alternative hypothesis is focused on
the lower tail below the mean of 3
This is an upper-tail test since the
alternative hypothesis is focused on
the upper tail above the mean of 3

Lower-Tail Tests
 There is only one
critical value, since
the rejection area is
in only one tail 
-Z 0
μ
H0: μ ≥ 3
H1: μ < 3
Z
X
Critical value

Upper-Tail Tests
Reject H0
Do not reject H0

Zα
0
μ
H0: μ ≤ 3
H1: μ > 3
 There is only one
critical value, since
the rejection area is
in only one tail
Critical value
Z
X
_

Example: Upper-Tail Z Test
for Mean ( Known)
A phone industry manager thinks that
customer monthly cell phone bills have
increased, and now average over $52 per
month. The company wishes to test this
claim. (Assume  = 10 is known)
H0: μ ≤ 52 the average is not over $52 per month
H1: μ > 52 the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)
Form hypothesis test:

 Suppose that  = 0.10 is chosen for this test
Find the rejection region:
Reject H0
Do not reject H0
 = 0.10
1.28
0
Reject H0
Reject H0 if Z > 1.28
Example: Find Rejection Region
(continued)

Review:
One-Tail Critical Value
Z .07 .09
1.1 .8790 .8810 .8830
1.2 .8980 .9015
1.3 .9147 .9162 .9177
z 0 1.28
.08
Standardized Normal
Distribution Table (Portion)
What is Z given  = 0.10?
 = 0.10
Critical Value
= 1.28
0.90
.3997
0.10
0.90

Example: Test Statistic
Obtain sample and compute the test statistic
Suppose a sample is taken with the following
results: n = 64, X = 53.1 (=10 was assumed known)
 Then the test statistic is:
0.88
64
10
52
53.1
n
σ
μ
X
Z 




(continued)

Example: Decision
Reach a decision and interpret the result:
Reject H0
Do not reject H0
 = 0.10
1.28
0
Reject H0
Do not reject H0 since Z = 0.88 ≤ 1.28
i.e.: there is not sufficient evidence that the
mean bill is over $52
Z = 0.88
(continued)

p -Value Solution
Calculate the p-value and compare to 
(assuming that μ = 52.0)
Reject H0
 = 0.10
Do not reject H0
1.28
0
Reject H0
Z = 0.88
(continued)
0.1894
0.8106
1
0.88)
P(Z
64
10/
52.0
53.1
Z
P
53.1)
X
P(










 



p-value = 0.1894
Do not reject H0 since p-value = 0.1894 >  = 0.10

t Test of Hypothesis for the Mean
(σ Unknown)
 Convert sample statistic ( ) to a t test statistic
X
The test statistic is:
n
S
μ
X
t 1
-
n


Hypothesis
Tests for 
σ Known σ Unknown
(Z test) (t test)

Example: Two-Tail Test
( Unknown)
The average cost of a
hotel room in New York
is said to be $168 per
night. A random sample
of 25 hotels resulted in
X = $172.50 and
S = $15.40. Test at the
 = 0.05 level.
(Assume the population distribution is normal)
H0: μ = 168
H1: μ  168

Example Solution:
Two-Tail Test
  = 0.05
 n = 25
  is unknown, so
use a t statistic
 Critical Value:
t24 = ± 2.0639
Do not reject H0: not sufficient evidence that
true mean cost is different than $168
Reject H0
Reject H0
/2=.025
-t n-1,α/2
Do not reject H0
0
/2=.025
-2.0639 2.0639
1.46
25
15.40
168
172.50
n
S
μ
X
t 1
n 





1.46
H0: μ = 168
H1: μ  168
t n-1,α/2

Connection to Confidence Intervals
 For X = 172.5, S = 15.40 and n = 25, the 95%
confidence interval is:
172.5 - (2.0639) 15.4/ 25 to 172.5 + (2.0639) 15.4/ 25
166.14 ≤ μ ≤ 178.86
 Since this interval contains the Hypothesized mean (168),
we do not reject the null hypothesis at  = 0.05

Hypothesis Tests for Proportions
 Involves categorical variables
 Two possible outcomes
 “Success” (possesses a certain characteristic)
 “Failure” (does not possesses that characteristic)
 Fraction or proportion of the population in the
“success” category is denoted by π

Proportions
 Sample proportion in the success category is
denoted by p

 When both nπ and n(1-π) are at least 5, p can
be approximated by a normal distribution with
mean and standard deviation

size
sample
sample
in
successes
of
number
n
X
p 



p
μ
n
)
(1
σ

 

p
(continued)

Hypothesis Tests for Proportions
 The sampling
distribution of p is
approximately
normal, so the test
statistic is a Z
value:
n
)
(1
p
Z
π
π
π



nπ  5
and
n(1-π)  5
Hypothesis
Tests for p
nπ < 5
or
n(1-π) < 5
Not discussed
in this chapter

Z Test for Proportion
in Terms of Number of Successes
 An equivalent form
to the last slide,
but in terms of the
number of
successes, X:
)
(1
n
n
X
Z






X  5
and
n-X  5
Hypothesis
Tests for X
X < 5
or
n-X < 5
Not discussed
in this chapter

Example: Z Test for Proportion
A marketing company
claims that it receives
8% responses from its
mailing. To test this
claim, a random sample
of 500 were surveyed
with 25 responses. Test
at the  = 0.05
significance level.
Check:
nπ = (500)(.08) = 40
n(1-π) = (500)(.92) = 460


Z Test for Proportion: Solution
 = 0.05
n = 500, p = 0.05
Reject H0 at  = 0.05
H0: π = 0.08
H1: π  0.08
Critical Values: ± 1.96
Test Statistic:
Decision:
Conclusion:
z
0
Reject Reject
.025
.025
1.96
-2.47
There is sufficient
evidence to reject the
company’s claim of 8%
response rate.
2.47
500
.08)
.08(1
.08
.05
n
)
(1
p
Z 










-1.96

p-Value Solution
Calculate the p-value and compare to 
(For a two-tail test the p-value is always two-tail)
Do not reject H0
Reject H0
Reject H0
/2 = .025
1.96
0
Z = -2.47
(continued)
0.0136
2(0.0068)
2.47)
P(Z
2.47)
P(Z






p-value = 0.0136:
Reject H0 since p-value = 0.0136 <  = 0.05
Z = 2.47
-1.96
/2 = .025
0.0068
0.0068

Potential Pitfalls and
Ethical Considerations
 Use randomly collected data to reduce selection biases
 Do not use human subjects without informed consent
 Choose the level of significance, α, and the type of test
(one-tail or two-tail) before data collection
 Do not employ “data snooping” to choose between one-
tail and two-tail test, or to determine the level of
significance
 Do not practice “data cleansing” to hide observations
that do not support a stated hypothesis
 Report all pertinent findings

Chapter Summary
 Addressed hypothesis testing methodology
 Performed Z Test for the mean (σ known)
 Discussed critical value and p–value
approaches to hypothesis testing
 Performed one-tail and two-tail tests

Chapter Summary
 Performed t test for the mean (σ
unknown)
 Performed Z test for the proportion
 Discussed pitfalls and ethical issues
(continued)

data collection for elementary statistics by Taban Rashid

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