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GM – 03
QUANTATIVE TECHNIQUES FOR
MANAGERS
11-2
WHAT DOES STATISTICS ACHIEVE
Making Decisions
Data, Information, Knowledge
1.
2.
3.
Data: specific observations of measured
numbers.
Information: processed and summarized
data yielding facts and ideas.
Knowledge: selected and organized
information that provides understanding,
recommendations, and the basis for decisions.
Making Decisions
BRANCHES OF STATISTICS
Descriptive and Inferential Statistics
Descriptive Statistics include graphical and
numerical procedures that summarize
and process data and are used to
transform data into information.
Making Decisions
Descriptive and Inferential Statistics
Inferential Statistics provide the bases for
predictions, forecasts, and estimates
that are used to transform
information to knowledge.
The Journey to Making Decisions
Decision 
Knowledge
Experience, Theory,
Literature, Inferential
Statistics, Computers
Information
Descriptive Statistics,
Probability, Computers
Begin Here:
Identify the
Problem
Data
Describing Data
©
11-7
Summarizing and Describing Data


Tables and Graphs
Numerical Measures
Frequency Distributions
A frequency distribution is a table used to organize
data. The left column (called classes or groups)
includes numerical intervals on a variable being
studied. The right column is a list of the
frequencies, or number of observations, for each
class. Intervals are normally of equal size, must
cover the range of the sample observations, and be
non-overlapping.
11-9
Example of a Frequency Distribution
A Frequency Distribution for the Shampoo Example
Weights (in mL)
220 less than 225
225 less than 230
230 less than 235
235 less than 240
240 less than 245
245 less than 250
Number of Bottles
1
4
29
34
26
6
Cumulative Frequency Distributions
A cumulative frequency distribution contains the
number of observations whose values are less
than the upper limit of each interval. It is
constructed by adding the frequencies of all
frequency distribution intervals up to and
including the present interval.
Relative Cumulative Frequency
Distributions
A relative cumulative frequency
distribution converts all cumulative
frequencies to cumulative percentages
11-12
Example of a Frequency Distribution
A Cumulative Frequency Distribution for the
Shampoo Example
Weights (in mL)
less than 225
less than 230
less than 235
less than 240
less than 245
less than 250
Number of Bottles
1
5
34
68
94
100
Parameters and Statistics
A statistic is a descriptive measure computed
from a sample of data. A parameter is a
descriptive measure computed from an entire
population of data.
Measures of Central Tendency
- Arithmetic Mean A arithmetic mean is of a set of data is
the sum of the data values divided by
the number of observations.
Sample Mean
If the data set is from a sample, then the
sample mean, X , is:
n
X
x
i 1
n
i
x1  x2    xn

n
Population Mean
If the data set is from a population, then the
population mean,  , is:
N
x
x1  x2    xn


N
N
i 1
i
Measures of Central Tendency
- Median An ordered array is an arrangement of data in either
ascending or descending order. Once the data are
arranged in ascending order, the median is the value
such that 50% of the observations are smaller and
50% of the observations are larger.
If the sample size n is an odd number, the median,
Xm, is the middle observation. If the sample size n
is an even number, the median, Xm, is the average
of the two middle observations. The median will
be located in the 0.50(n+1)th ordered position.
Measures of Central Tendency
- Mode The mode, if one exists, is the most
frequently occurring observation in
the sample or population.
Shape of the Distribution
The shape of the distribution is said to
be symmetric if the observations are
balanced, or evenly distributed, about
the mean. In a symmetric distribution
the mean and median are equal.
Shape of the Distribution
A distribution is skewed if the observations are
not symmetrically distributed above and below
the mean. A positively skewed (or skewed to the
right) distribution has a tail that extends to the
right in the direction of positive values. A
negatively skewed (or skewed to the left)
distribution has a tail that extends to the left in
the direction of negative values.
Shapes of the Distribution
Frequency
Symmetric Distribution
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9
Negatively Skewed Distribution
12
12
10
10
8
8
Frequency
Frequency
Positively Skewed Distribution
6
4
2
6
4
2
0
0
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
Measures of Variability
- TheMEASURES
Range -
OF DISPERSION
The range in a set of data is the
difference between the largest and
smallest observations
Measures of Variability
- Sample
Variance
MOST
IMPORTANT
MEASURE OF DISPERSION
The sample variance, s2, is the sum of the squared
differences between each observation and the
sample mean divided by the sample size minus 1.
n
s 
2
 (x  X )
i 1
i
n 1
2
Measures of Variability
- Short-cut Formulas for Sample Variance Short-cut formulas for the sample variance are:
( xi ) 2
xi 

n
2
i 1
s 
n 1
n
or
s2 
2
2
x

n
X
i
n 1
Measures
of Variability
DISTINGUISH
BETWEEN POPULATION VARIANCE
AND SAMPLE
VARIANCE
(IMPORTANT)
- Population
Variance
The population variance, 2, is the sum of the
squared differences between each observation and
the population mean divided by the population size,
N.
N
 
2
 (x  )
i 1
i
N
2
Measures of Variability
- Sample Standard Deviation The sample standard deviation, s, is the positive
square root of the variance, and is defined as:
n
s s 
2
2
(
x

X
)
 i
i 1
n 1
Measures of Variability
- Population Standard DeviationThe population standard deviation, , is
N
  
2
 (x  )
i 1
i
N
2
11-28
For a set of data with a bell-shaped histogram, the
Empirical Rule is:
•
•
•
approximately 68% of the observations are contained
with a distance of one standard deviation around the
mean;  1
approximately 95% of the observations are contained
with a distance of two standard deviations around the
mean;  2
almost all of the observations are contained with a
distance of three standard deviation around the mean;
 3
Coefficient of Variation
The Coefficient of Variation, CV, is a measure of
relative dispersion that expresses the standard
deviation as a percentage of the mean (provided the
mean is positive).
The sample coefficient of variation is
s
CV  100
X
if X  0

CV  100

if   0
The population coefficient of variation is
Five-Number Summary
The Five-Number Summary refers to the five
descriptive measures: minimum, first quartile,
median, third quartile, and the maximum.
X min imum  Q1  Median  Q3  X max imum
Grouped Data Mean
For a population of N observations the mean
K is

fm
i 1
i
i
N
K
For a sample of n observations,
the mean is
X
fm
i
i 1
i
n
Where the data set contains observation values m1, m2, . . ., mk occurring with
frequencies f1, f2, . . . fK respectively
Grouped Data Variance
For a population of N observations the variance
is K
K
2 

i 1
f i (mi   ) 2
N


i 1
f i m i2
N
 2
K of n observations,
K
For a sample
the variance is
s2 

i 1
f i (mi  X ) 2
n 1


i 1
f i m i2  nX 2
n 1
Where the data set contains observation values m1, m2, . . ., mk occurring with
frequencies f1, f2, . . . fK respectively
11-33
1-8 Methods of Displaying Data

Pie Charts


Bar Graphs


Heights of rectangles represent group
frequencies
Frequency Polygons


Categories represented as percentages
of total
Height of line represents frequency
Ogives

Height of line represents cumulative
11-34
Pie Chart
Figure 1-10: Twentysomethings split on job satisfication
Category
Don't like my job but it is on my career path
Job is OK, but it is not on my career path
Enjoy job, but it is not on my career path
My job just pays the bills
Happy with career
6.0%
Do not like my job, but it is on my career path
Happy with career
19.0%
33.0%
Job OK, but it is not on my career path
19.0%
Enjoy job, but it is not on my career path
23.0%
My job just pays the bills
11-35
Bar Chart
Figure 1-11: SHIFTING GEARS
Quartely net income for General Motors (in billions)
1.5
1.2
0.9
0.6
0.3
0.0
1Q
2003
2Q
3Q
C4
4Q
1Q
2004
11-36
Frequency Polygon and Ogive
Relative Frequency Polygon
0.3
Ogive
1.0
0.2
0.5
0.1
0.0
0.0
0
10
20
Sales
30
40
50
0
10
20
30
40
50
Sales
(Cumulative frequency or
relative frequency graph)
11-37
Time Plot
M o n thly S te e l P ro d uc tio n
Millions of Tons
8.5
7.5
6.5
5.5
Month
J F M A M J J A S ON D J F M A M J J A S ON D J F M A M J J A S O
11-38
Example 1-8: Stem-and-Leaf Display
1
2
3
4
5
6
122355567
0111222346777899
012457
11257
0236
02
Figure 1-17: Task Performance Times
11-39
Box Plot
Elements of a Box Plot
Outlier
o
Smallest data
point not below
inner fence
Largest data point
Suspected
not exceeding
outlier
inner fence
X
Outer
Fence
Inner
Fence
Q1-1.5(IQR)
Q1-3(IQR)
X
Q1
Median
Interquartile Range
Q3
Inner
Fence
Q3+1.5(IQR)
*
Outer
Fence
Q3+3(IQR)
Probability
11-41
2 Probability
Using Statistics
 Basic Definitions: Events, Sample Space, and
Probabilities
 Basic Rules for Probability
 Conditional Probability
 Independence of Events
 Combinatorial Concepts
 The Law of Total Probability and Bayes’
Theorem
 Joint Probability Table
 Using the Computer

11-42
2-1 Probability is:




A quantitative measure of
uncertainty
A measure of the strength of belief
in the occurrence of an uncertain
event
A measure of the degree of chance
or likelihood of occurrence of an
uncertain event
Measured by a number between 0
and 1 (or between 0% and 100%)
11-43
Types of Probability

Objective or Classical Probability





based on equally-likely events
based on long-run relative frequency of
events
not based on personal beliefs
is the same for all observers (objective)
examples: toss a coin, throw a die, pick a
card
11-44
Types of Probability (Continued)

Subjective Probability



based on personal beliefs, experiences,
prejudices, intuition - personal judgment
different for all observers (subjective)
examples: Super Bowl, elections, new
product introduction, snowfall
11-45
2-2 Basic Definitions

Set - a collection of elements or
objects of interest

Empty set (denoted by )


Universal set (denoted by S)


a set containing no elements
a set containing all possible elements
Complement (Not). The complement
A

of A is

a set containing all elements of S not in A
11-46
Complement of a Set
S
A
A
Venn Diagram illustrating the Complement of an event
11-47
Basic Definitions (Continued)

Intersection (And)  A  B
–

a set containing all elements in
both A and B
Union (Or)
–
 A  B
a set containing all elements in A
or B or both
11-48
Sets: A Intersecting with B
S
A
B
A B
11-49
Sets: A Union B
S
A
B
A B
11-50
Basic Definitions (Continued)
•
•
Mutually exclusive or disjoint sets
–sets having no elements in common,
having no intersection, whose
intersection is the empty set
Partition
–a collection of mutually exclusive sets
which together include all possible
elements, whose union is the universal
set
11-51
Mutually Exclusive or Disjoint Sets
Sets have nothing in common
S
A
B
11-52
Experiment
•
Process that leads to one of several possible
outcomes *, e.g.:

Coin toss
•

Throw die
•

•
1, 2, 3, 4, 5, 6
Pick a card

•
Heads, Tails
* Also called a basic outcome, elementary event, or simple event
AH, KH, QH, ...
Introduce a new product
Each trial of an experiment has a single
observed outcome.
The precise outcome of a random experiment
is unknown before a trial.
11-53
Events : Definition

Sample Space or Event Set

Set of all possible outcomes (universal set) for a given
experiment

E.g.: Roll a regular six-sided die


Event

Collection of outcomes having a common
characteristic

E.g.: Even number



S = {1,2,3,4,5,6}
A = {2,4,6}
Event A occurs if an outcome in the set A occurs
Probability of an event

Sum of the probabilities of the outcomes of which it
consists

P(A) = P(2) + P(4) + P(6)
Equally-likely Probabilities
(Hypothetical or Ideal Experiments)
•
For example:

Throw a die
• Six possible outcomes {1,2,3,4,5,6}
• If each is equally-likely, the probability of each is
1/6 = 0.1667 = 16.67%
1
P
(
e
)


n( S )
• Probability of each equally-likely outcome is 1
divided by the number of possible outcomes
 Event A (even number)
• P(A) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 =
1/2
P( A) • P( e)
for e in A
n( A ) 3 1

 
n( S ) 6 2
11-54
11-55
Pick a Card: Sample Space
Union of
Events ‘Heart’
and ‘Ace’
P ( Heart  Ace ) 
n ( Heart  Ace )

n(S )
16
4

52
Hearts
Diamonds
Clubs
A
K
Q
J
10
9
8
7
6
5
4
3
2
A
K
Q
J
10
9
8
7
6
5
4
3
2
A
K
Q
J
10
9
8
7
6
5
4
3
2
Spades
A
K
Q
J
10
9
8
7
6
5
4
3
2
Event ‘Ace’
n ( Ace )
P ( Ace ) 
4

n(S )
1

52
13
13
Event ‘Heart’
n ( Heart )
P ( Heart ) 
13

n(S )
1

52
The intersection of the
events ‘Heart’ and ‘Ace’
comprises the single point
circled twice: the ace of hearts
4
n ( Heart  Ace )
P ( Heart  Ace ) 
1

n(S )
52
11-56
2-3 Basic Rules for Probability

Range of Values for P(A):

0  P( A) 1
Complements - Probability of not A
P( A )  1  P( A)

Intersection - Probability of both A and B
P( A  B)  n( A  B)
n( S )

Mutually exclusive events (A and C) :
P( A  C )  0
Basic Rules for Probability
(Continued)
•
Union - Probability of A or B or both (rule of
unions)
P( A  B)  n( A  B)  P( A)  P( B)  P( A  B)
n( S )
 Mutually exclusive events: If A and B are mutually exclusive,
then
P( A  B)  0 so P( A  B)  P( A)  P( B)
11-57
11-58
Sets: P(A Union B)
S
A
B
P( A  B )
11-59
2-4 Conditional Probability
•
Conditional Probability - Probability of A given B
P( A B) 
P( A  B)
, where P( B)  0
P( B)
 Independent events:
P( A B)  P( A)
P( B A)  P( B)
11-60
Conditional Probability (continued)
Rules of conditional probability:
P( A B)  P( A  B) so P( A  B)  P( A B) P( B)
P( B)
 P( B A) P( A)
If events A and D are statistically independent:
P ( A D)  P ( A)
P ( D A)  P ( D)
so
P( A  D)  P( A) P( D)
11-61
Contingency Table - Example 2-2
Counts
AT& T
IBM
Total
Telecommunication
40
10
50
Computers
20
30
50
Total
60
40
100
Probabilities
AT& T
IBM
Total
Telecommunication
.40
.10
.50
Computers
.20
.30
.50
Total
.60
.40
1.00
Probability that a project
is undertaken by IBM
given it is a
telecommunications
project:
P ( IBM  T )
P (T )
0.10

 0. 2
0.50
P ( IBM T ) 
11-62
2-5 Independence of Events
Conditions for the statistical independence of events A and B:
P( A B)  P( A)
P( B A)  P( B)
and
P( A  B)  P( A) P( B)
P ( Ace  Heart )
P ( Heart )
1
1
 52 
 P ( Ace )
13 13
52
P ( Ace Heart ) 
P ( Heart  Ace )
P ( Ace )
1
1
 52   P ( Heart )
4
4
52
P ( Heart Ace ) 
4 13
1
P( Ace  Heart) 
*

 P( Ace) P( Heart)
52 52 52
Independence of Events –
Example 2-5
Events Television (T) and Billboard (B) are
assumed to be independent.
a)P(T  B)  P(T ) P( B)
 0.04 * 0.06  0.0024
b)P(T  B)  P(T )  P( B)  P(T  B)
 0.04  0.06  0.0024 0.0976
11-63
Product Rules for Independent
Events
11-64
The probability of the intersection of several independent events
is the product of their separate individual probabilities:
P( A  A  A  An )  P( A ) P( A ) P( A ) P( An )
1
2
3
1
2
3
The probability of the union of several independent events
is 1 minus the product of probabilities of their complements:
P( A  A  A  An )  1 P( A ) P( A ) P( A ) P( An )
1
2
3
1
2
3
Example 2-7:
(Q Q Q Q ) 1 P(Q )P(Q )P(Q )P(Q )
1 2 3
10
1
2
3
10
10.9010 10.3487 0.6513
11-65
2-6 Combinatorial Concepts
Consider a pair of six-sided dice. There are six possible outcomes
from throwing the first die {1,2,3,4,5,6} and six possible outcomes
from throwing the second die {1,2,3,4,5,6}. Altogether, there are
6*6 = 36 possible outcomes from throwing the two dice.
In general, if there are n events and the event i can happen in
Ni possible ways, then the number of ways in which the
sequence of n events may occur is N1N2...Nn.

Pick 5 cards from a
deck of 52 - with
replacement

52*52*52*52*52=525
380,204,032 different
possible outcomes

Pick 5 cards from a
deck of 52 - without
replacement

52*51*50*49*48 =
311,875,200 different
possible outcomes
11-66
Factorial
How many ways can you order the 3 letters A, B, and C?
There are 3 choices for the first letter, 2 for the second, and 1 for
the last, so there are 3*2*1 = 6 possible ways to order the three
letters A, B, and C.
How many ways are there to order the 6 letters A, B, C, D, E,
and F? (6*5*4*3*2*1 = 720)
Factorial: For any positive integer n, we define n factorial as:
n(n-1)(n-2)...(1). We denote n factorial as n!.
The number n! is the number of ways in which n objects can
be ordered. By definition 1! = 1 and 0! = 1.
11-67
Permutations (Order is important)
What if we chose only 3 out of the 6 letters A, B, C, D, E, and F?
There are 6 ways to choose the first letter, 5 ways to choose the
second letter, and 4 ways to choose the third letter (leaving 3
letters unchosen). That makes 6*5*4=120 possible orderings or
permutations.
Permutations are the possible ordered selections of r objects out
of a total of n objects. The number of permutations of n objects
taken r at a time is denoted by nPr, where
n!
P

n r (n  r )!
Forexam ple:
6!
6! 6 * 5 * 4 * 3 * 2 *1


 6 * 5 * 4  120
6 P3 
(6  3)! 3!
3 * 2 *1
Combinations (Order is not
Important)
11-68
Suppose that when we pick 3 letters out of the 6 letters A, B, C, D, E, and F
we chose BCD, or BDC, or CBD, or CDB, or DBC, or DCB. (These are the
6 (3!) permutations or orderings of the 3 letters B, C, and D.) But these are
orderings of the same combination of 3 letters. How many combinations of 6
different letters, taking 3 at a time, are there?
Combinations are the possible selections of r items from a group of n items  n
regardless of the order of selection. The number of combinations is denoted  r
and is read as n choose r. An alternative notation is nCr. We define the number
of combinations of r out of n elements as:
 n
n!
  n C r 
r! (n  r)!
r
Forexam ple:
 n
6!
6!
6 * 5 * 4 * 3 * 2 *1
6 * 5 * 4 120
 6 C3 




 20
r
3
!
(
6

3
)!
3
!
3
!
(3
*
2
*
1)(3
*
2
*
1)
3
*
2
*
1
6
 
2-7 The Law of Total Probability and
Bayes’ Theorem
11-69
The law of total probability:
P( A)  P( A  B)  P( A  B )
In terms of conditional probabilities:
P( A)  P( A  B)  P( A  B )
 P( A B) P( B)  P( A B ) P( B )
More generally (where Bi make up a partition):
P( A)   P( A  B )
i
  P( AB ) P( B )
i
i
11-70
Bayes’ Theorem
•
•
Bayes’ theorem enables you, knowing just a
little more than the probability of A given B,
to find the probability of B given A.
Based on the definition of conditional
probability and the law of total probability.
P ( A  B)
P ( A)
P ( A  B)

P ( A  B)  P ( A  B )
P ( A B) P ( B)

P ( A B) P ( B)  P ( A B ) P ( B )
P ( B A) 
Applying the law of total
probability to the denominator
Applying the definition of
conditional probability throughout
11-71
11-72
2-8 The Joint Probability Table



A joint probability table is similar to a
contingency table , except that it has
probabilities in place of frequencies.
The joint probability for Example 2-11 is
shown below.
The row totals and column totals are
called marginal probabilities.
11-73
The Joint Probability Table


A joint probability table is similar to
a contingency table , except that it
has probabilities in place of
frequencies.
The row totals and column totals
are called marginal probabilities.
The Joint Probability Table:

11-74
The joint probability table is
summarized below.
GROWTH
High
Medium
Low
Total
Appreciates
( Re)
0.21
0.2
0.04
0.45
Depreciates
0.09
0.3
0.16
0.55
0.30
0.5
0.20
1.00
(Re)
Total
Marginal probabilities are the row totals and the column totals.
Random Variables
11-76
3-1 Using Statistics
Consider the different possible orderings of boy (B) and girl (G) in
four sequential births. There are 2*2*2*2=24 = 16 possibilities, so
the sample space is:
BBBB
BBBG
BBGB
BBGG
BGBB
BGBG
BGGB
BGGG
GBBB
GBBG
GBGB
GBGG
GGBB
GGBG
GGGB
GGGG
If girl and boy are each equally likely [P(G) = P(B) = 1/2], and the
gender of each child is independent of that of the previous child,
then the probability of each of these 16 possibilities is:
(1/2)(1/2)(1/2)(1/2) = 1/16.
11-77
Random Variables (Continued)
0
BBBB
BGBB
GBBB
BBBG
BBGB
GGBB
GBBG
BGBG
BGGB
GBGB
BBGG
BGGG
GBGG
GGGB
GGBG
GGGG
1
X
2
3
4
Sample Space
Points on the
Real Line
11-78
Random Variables (Continued)
Since the random variable X = 3 when any of the four outcomes BGGG, GBGG,
GGBG, or GGGB occurs,
P(X = 3) = P(BGGG) + P(GBGG) + P(GGBG) + P(GGGB) = 4/16
The probability distribution of a random variable is a table that lists the
possible values of the random variables and their associated probabilities.
x
0
1
2
3
4
P(x)
1/16
4/16
6/16
4/16
1/16
16/16=1
The Graphical Display for this
Probability Distribution
is shown on the next Slide.
11-79
Random Variables (Continued)
Probability Distribution of the Number of Girls in Four Births
0.4
6/16
Probability, P(X)
0.3
4/16
4/16
0.2
0.1
1/16
0.0
0
1/16
1
2
Number of Girls, X
3
4
11-80
Example 3-1
Consider the experiment of tossing two six-sided dice. There are 36 possible
outcomes. Let the random variable X represent the sum of the numbers on
the two dice:
x
P(x)*
Probability Distribution of Sum of Two Dice
3
1,3
2,3
3,3
4,3
5,3
6,3
4
1,4
2,4
3,4
4,4
5,4
6,4
5
1,5
2,5
3,5
4,5
5,5
6,5
6
1,6
2,6
3,6
4,6
5,6
6,6
7
8
9
10
11
12
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
1
0.17
0.12
p(x)
1,1
2,1
3,1
4,1
5,1
6,1
2
1,2
2,2
3,2
4,2
5,2
6,2
2
3
4
5
6
7
8
9
10
11
12
0.07
0.02
2
3
4
5
6
7
8
9
10
x
* Note that: P(x)  (6  (7  x) 2 ) / 36
11
12
11-81
Example 3-2
Probability Distribution of the Number of Switches
The Probability Distribution of the Number of Switches
P(x)
0.1
0.2
0.3
0.2
0.1
0.1
1
0.4
0.3
P(x)
x
0
1
2
3
4
5
0.2
0.1
0.0
0
1
2
3
4
5
x
Probability of more than 2 switches:
P(X > 2) = P(3) + P(4) + P(5) = 0.2 + 0.1 + 0.1 = 0.4
Probability of at least 1 switch:
P(X 1) = 1 - P(0) = 1 - 0.1 = .9
Discrete and Continuous Random
Variables
11-82
A discrete random variable:

has a countable number of possible values

has discrete jumps (or gaps) between successive values

has measurable probability associated with individual values

counts
A continuous random variable:

has an uncountably infinite number of possible values

moves continuously from value to value

has no measurable probability associated with each value
 measures (e.g.: height, weight, speed, value, duration, length)
Rules of Discrete Probability
Distributions
11-83
The probability distribution of a discrete random
variable X must satisfy the following two conditions.
1. P(x)  0 for all values of x.
2.
 P(x)  1
all x
Corollary:
0  P( X )  1
11-84
Cumulative Distribution Function
The cumulative distribution function, F(x), of a discrete
random variable X is:
F(x)  P( X  x) 
 P(i)
all i  x
P(x)
0.1
0.2
0.3
0.2
0.1
0.1
1.00
F(x)
0.1
0.3
0.6
0.8
0.9
1.0
Cumulative Probability Distribution of the Number of Switches
1 .0
0 .9
0 .8
0 .7
F(x)
x
0
1
2
3
4
5
0 .6
0 .5
0 .4
0 .3
0 .2
0 .1
0 .0
0
1
2
3
x
4
5
11-85
Cumulative Distribution Function
The probability that at most three switches will occur:
x
0
1
2
3
4
5
P(x)
0.1
0.2
0.3
0.2
0.1
0.1
1
F(x)
0.1
0.3
0.6
0.8
0.9
1.0
Note: P(X < 3) = F(3) = 0.8 = P(0) + P(1) + P(2) + P(3)
Using Cumulative Probability
Distributions (Figure 3-8)
The probability that more than one switch will occur:
x
0
1
2
3
4
5
P(x)
0.1
0.2
0.3
0.2
0.1
0.1
1
F(x)
0.1
0.3
0.6
0.8
0.9
1.0
Note: P(X > 1) = P(X > 2) = 1 – P(X < 1) = 1 – F(1) = 1 – 0.3 = 0.7
11-86
Using Cumulative Probability
Distributions (Figure 3-9)
The probability that anywhere from one to three
switches will occur:
x
0
1
2
3
4
5
P(x)
0.1
0.2
0.3
0.2
0.1
0.1
1
F(x)
0.1
0.3
0.6
0.8
0.9
1.0
Note: P(1 < X < 3) = P(X < 3) – P(X < 0) = F(3) – F(0) = 0.8 – 0.1 = 0.7
11-87
3-2 Expected Values of Discrete
Random Variables
The mean of a probability distribution is a
measure of its centrality or location, as is the
mean or average of a frequency distribution. It is
a weighted average, with the values of the
random variable weighted by their probabilities.
0
1
2
3
4
11-88
5
2.3
The mean is also known as the expected value (or expectation) of a random
variable, because it is the value that is expected to occur, on average.
The expected value of a discrete random
variable X is equal to the sum of each
value of the random variable multiplied by
its probability.
  E ( X )   xP ( x )
all x
x
0
1
2
3
4
5
P(x)
0.1
0.2
0.3
0.2
0.1
0.1
1.0
xP(x)
0.0
0.2
0.6
0.6
0.4
0.5
2.3 = E(X) = 
Expected Value of a Function of a
Discrete Random Variables
11-89
The expected value of a function of a discrete random variable X is:
E [ h ( X )]   h ( x ) P ( x )
all x
Example 3-3: Monthly sales of a certain
product are believed to follow the given
probability distribution. Suppose the
company has a fixed monthly production
cost of $8000 and that each item brings
$2. Find the expected monthly profit
h(X), from product sales.
E [ h ( X )]   h ( x ) P ( x )  5400
all x
Number
of items, x
5000
6000
7000
8000
9000
P(x)
0.2
0.3
0.2
0.2
0.1
1.0
xP(x)
h(x) h(x)P(x)
1000 2000
400
1800 4000
1200
1400 6000
1200
1600 8000
1600
900 10000
1000
6700
5400
Note: h (X) = 2X – 8000 where X = # of items sold
The expected value of a linear function of a random variable is:
E(aX+b)=aE(X)+b
In this case: E(2X-8000)=2E(X)-8000=(2)(6700)-8000=5400
Variance and Standard Deviation of a
Random Variable
11-90
The variance of a random variable is the expected
squared deviation from the mean:

2
 V ( X )  E [( X   ) 2 ] 

(x   ) 2 P(x)
a ll x

 

 E ( X 2 )  [ E ( X )] 2    x 2 P ( x )     xP ( x ) 
 a ll x
  a ll x

2
The standard deviation of a random variable is the
square root of its variance:   SD( X )  V ( X )
Random Variable – using Example 32
11-91
 2  V ( X )  E[( X  )2]
Table 3-8
Number of
Switches, x
0
1
2
3
4
5
P(x) xP(x)
0.1
0.0
0.2
0.2
0.3
0.6
0.2
0.6
0.1
0.4
0.1
0.5
2.3
Recall:  = 2.3.
(x-)
-2.3
-1.3
-0.3
0.7
1.7
2.7
(x-)2 P(x-)2
5.29
0.529
1.69
0.338
0.09
0.027
0.49
0.098
2.89
0.289
7.29
0.729
2.010
x2P(x)
0.0
0.2
1.2
1.8
1.6
2.5
7.3
  ( x  )2 P( x)  2.01
all x
 E( X 2)  [ E( X )]2
2




   x2 P( x)    xP( x)
all x
 all x

 7.3  2.32  2.01
Variance of a Linear Function of a
Random Variable
The variance of a linear function of a random variable is:
V(aX b)  a2V( X)  a22
Example 33:
Number
of items, x P(x)
5000
0.2
6000
0.3
7000
0.2
8000
0.2
9000
0.1
1.0
2 V(X)
xP(x)
1000
1800
1400
1600
900
6700
x2 P(x)
5000000
10800000
9800000
12800000
8100000
46500000
 E ( X 2 )  [ E ( X )]2
2

 

2
   x P( x )     xP( x ) 
all x
 all x

 46500000  ( 67002 )  1610000
  SD( X )  1610000  1268.86
V (2 X  8000)  (2 2 )V ( X )
 ( 4)(1610000)  6440000
 ( 2 x  8000)  SD(2 x  8000)
 2 x  (2)(1268.86)  2537.72
11-92
Some Properties of Means and
Variances of Random Variables
The mean or expected value of the sum of random variables
is the sum of their means or expected values:
( XY)  E( X Y)  E( X)  E(Y)  X  Y
For example: E(X) = $350 and E(Y) = $200
E(X+Y) = $350 + $200 = $550
The variance of the sum of mutually independent random
variables is the sum of their variances:
 2 ( X Y )  V ( X  Y)  V ( X ) V (Y)   2 X   2 Y
if and only if X and Y are independent.
For example: V(X) = 84 and V(Y) = 60
V(X+Y) = 144
11-93
Some Properties of Means and
Variances of Random Variables
NOTE:
E( X  X ... X )  E( X )  E( X ) ... E( X )
1 2
1
2
k
k
E(a X  a X ... a X )  a E( X )  a E( X ) ... a E( X )
1 1 2 2
1 2
2
k k 1
k
k
The variance of the sum of k mutually independent random
variables is the sum of their variances:
V ( X  X ... X ) V ( X ) V ( X ) ...V ( X )
1 2
1
2
k
k
and
V (a X  a X ... a X )  a2V ( X )  a2V ( X ) ... a2V ( X )
1 1 2 2
1 2
2
k k 1
k
k
11-94
Chebyshev’s Theorem Applied to
Probability Distributions
Chebyshev’s Theorem applies to probability distributions just
as it applies to frequency distributions.
For a random variable X with mean ,standard deviation ,
and for any number k > 1:
1
P( X    k)  1 2
k
1
1
1 3

1

  75%
2
4
4
2
At 1  12  1  1  8  89%
9 9
3
least
1
1
1 15

1


 94%
2
16
16
4
2
Lie
within
3
4
Standard
deviations
of the mean
11-95
11-96
3-3 Bernoulli Random Variable
• If an experiment consists of a single trial and the outcome of the
trial can only be either a success* or a failure, then the trial is
called a Bernoulli trial.
• The number of success X in one Bernoulli trial, which can be 1 or
0, is a Bernoulli random variable.
• Note: If p is the probability of success in a Bernoulli experiment,
the E(X) = p and V(X) = p(1 – p).
* The
terms success and failure are simply statistical terms, and do not have
positive or negative implications. In a production setting, finding a defective
product may be termed a “success,” although it is not a positive result.
11-97
3-4 The Binomial Random Variable
Consider a Bernoulli Process in which we have a sequence of n identical
trials satisfying the following conditions:
1. Each trial has two possible outcomes, called success *and failure.
The two outcomes are mutually exclusive and exhaustive.
2. The probability of success, denoted by p, remains constant from trial
to trial. The probability of failure is denoted by q, where q = 1-p.
3. The n trials are independent. That is, the outcome of any trial does
not affect the outcomes of the other trials.
A random variable, X, that counts the number of successes in n Bernoulli
trials, where p is the probability of success* in any given trial, is said to
follow the binomial probability distribution with parameters n (number
of trials) and p (probability of success). We call X the binomial random
variable.
* The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a
production setting, finding a defective product may be termed a “success,” although it is not a positive result.
11-98
Binomial Probabilities (Introduction)
Suppose we toss a single fair and balanced coin five times in succession,
and let X represent the number of heads.
There are 25 = 32 possible sequences of H and T (S and F) in the sample space for this
experiment. Of these, there are 10 in which there are exactly 2 heads (X=2):
HHTTT HTHTH HTTHT HTTTH THHTT THTHT THTTH TTHHT TTHTH TTTHH
The probability of each of these 10 outcomes is p3q3 = (1/2)3(1/2)2=(1/32), so the
probability of 2 heads in 5 tosses of a fair and balanced coin is:
P(X = 2) = 10 * (1/32) = (10/32) = 0.3125
10
(1/32)
Number of outcomes
with 2 heads
Probability of each
outcome with 2 heads
11-99
Binomial Probabilities (continued)
P(X=2) = 10 * (1/32) = (10/32) = .3125
Notice that this probability has two parts:
10
(1/32)
Number of outcomes
with 2 heads
Probability of each
outcome with 2 heads
In general:
1. The probability of a given sequence
of x successes out of n trials with
probability of success p and
probability of failure q is equal to:
pxq(n-x)
2. The number of different sequences of n trials that
result in exactly x successes is equal to the number
of choices of x elements out of a total of n elements.
This number is denoted:
n!
 n
nCx    
 x x!( n  x)!
11-100
The Binomial Probability Distribution
The binomial probability distribution:
n!
 n x ( n  x )
P( x)    p q

p x q ( n x)
x!( n  x)!
 x
where :
p is the probability of success in a single trial,
q = 1-p,
n is the number of trials, and
x is the number of successes.
N u m b er o f
su ccesses, x
0
1
2
3

n
P ro b ab ility P (x )
n!
p 0 q (n 0)
0 !( n  0 ) !
n!
p 1 q ( n  1)
1 !( n  1 ) !
n!
p 2 q (n 2)
2 !( n  2 ) !
n!
p 3 q (n 3)
3 !( n  3 ) !

n!
p n q (n n)
n !( n  n ) !
1 .0 0
The Normal Distribution
11-102
4-1 Introduction
As n increases, the binomial distribution approaches a ...
n=6
n = 10
Binomial Distribution: n=10, p=.5
Binomial Distribution: n=14, p=.5
0.3
0.3
0.2
0.2
0.2
0.1
P(x)
0.3
P(x)
P(x)
Binomial Distribution: n=6, p=.5
n = 14
0.1
0.0
0.1
0.0
0
1
2
3
4
5
6
0.0
0
x
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
x
x
Normal Probability Density Function:
1
0.4




0.3
for
  x
2p 2
where e  2 . 7182818 ... and p  3 . 14159265 ...
f(x)
f ( x) 
 2
x

e 2 2




Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
x
5
11-103
The Normal Probability Distribution
The normal probability density function:
 2
x

e 2  2 for
f (x) 
1
0.4




0.3
 x
2 p 2
where e  2 .7182818 ... and p  3.14159265 ...
f(x)




Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
x
5
4-2 Properties of the Normal
Distribution
• The normal is a family of
Bell-shaped and symmetric
distributions. because the distribution is
symmetric, one-half (.50 or 50%) lies
on either side of the mean.
Each is characterized by a different pair
of mean, , and variance, . That is:
[X~N(,)].
Each is asymptotic to the horizontal
axis.
The area under any normal probability
density function within k of  is the
same for any normal distribution,
regardless of the mean and variance.
11-104
4-2 Properties of the Normal
Distribution (continued)
•
•
•
If several independent random
variables are normally distributed
then their sum will also be normally
distributed.
The mean of the sum will be the
sum of all the individual means.
The variance of the sum will be the
sum of all the individual variances
(by virtue of the independence).
11-105
4-2 Properties of the Normal
Distribution (continued)
•
If X1, X2, …, Xn are independent
normal random variable, then their
sum S will also be normally
distributed with
• E(S) = E(X1) + E(X2) + … + E(Xn)
• V(S) = V(X1) + V(X2) + … + V(Xn)
• Note: It is the variances that can be
added above and not the standard
deviations.
11-106
4-2 Properties of the Normal
Distribution – Example 4-1
Example 4.1: Let X1, X2, and X3 be
independent random variables that are
normally distributed with means and
variances as shown.
X1
X2
X3
Mean
Variance
10
1
20
2
30
3
Let S = X1 + X2 + X3. Then E(S) = 10 + 20 + 30 = 60 and
V(S) = 1 + 2 + 3 = 6. The standard deviation of S is 6
= 2.45.
11-107
4-2 Properties of the Normal
Distribution (continued)
•
If X1, X2, …, Xn are independent
normal random variable, then the
random variable Q defined as Q =
a1X1 + a2X2 + … + anXn + b will also
be normally distributed with
• E(Q) = a1E(X1) + a2E(X2) + … +
anE(Xn) + b
• V(Q) = a12 V(X1) + a22 V(X2) + … +
an2 V(Xn)
•
Note: It is the variances that can be
added above and not the standard
deviations.
11-108
4-2 Properties of the Normal
Distribution – Example 4-3
Example 4.3: Let X1 , X2 , X3 and X4 be independent
random variables that are normally distributed
with means and variances as shown. Find the
mean and variance of Q = X1 - 2X2 + 3X2 - 4X4 + 5
Mean
Variance
X1
12
4
X2
-5
2
X3
8
5
X4
10
1
E(Q) = 12 – 2(-5) + 3(8) – 4(10) + 5 = 11
V(Q) = 4 + (-2)2(2) + 32(5) + (-4)2(1) = 73
SD(Q) = 73  8.544
11-109
Computing the Mean, Variance and Standard
11-110
Deviation for the Sum of Independent Random
Variables Using the Template
11-111
Normal Probability Distributions
All of these are normal probability density functions, though each has a different mean and variance.
Normal Distribution:  =40, =1
Normal Distribution:  =30, =5
0.4
Normal Distribution:  =50, =3
0.2
0.2
0.2
f(y)
f(x)
f(w)
0.3
0.1
0.1
0.1
0.0
0.0
35
40
45
0.0
0
10
20
30
w
40
x
W~N(40,1)
X~N(30,25)
50
60
35
45
50
55
y
Y~N(50,9)
Normal Distribution:  =0, =1
Consider:
0.4
f(z)
0.3
0.2
0.1
0.0
-5
0
z
Z~N(0,1)
5
P(39  W  41)
P(25  X  35)
P(47  Y  53)
P(-1  Z  1)
The probability in each
case is an area under a
normal probability density
function.
65
4-4 The Standard Normal
Distribution
11-112
The standard normal random variable, Z, is the normal random
variable with mean  = 0 and standard deviation  = 1: Z~N(0,12).
Standard Normal Distribution
0 .4
=1
{
f(z)
0 .3
0 .2
0 .1
0 .0
-5
-4
-3
-2
-1
0
=0
Z
1
2
3
4
5
Finding Probabilities of the Standard
Normal Distribution: P(0 < Z < 1.56)
11-113
Standard Normal Probabilities
Standard Normal Distribution
0.4
f(z)
0.3
0.2
0.1
{
1.56
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
Look in row labeled 1.5
and column labeled .06 to
find P(0  z  1.56) =
0.4406
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
Finding Probabilities of the Standard
Normal Distribution: P(Z < -2.47)
11-114
To find P(Z<-2.47):
z ...
.
.
P(0 < Z < 2.47) = .4932
.
< -2.47) = .5 - P(0 < Z2.3< ...2.47)
0.4909
2.4 ... 0.4931
= .5 - .4932 = 0.0068
2.5 ... 0.4948
.
.
.
Find table area for 2.47
P(Z
.06
.
.
.
0.4911
0.4932
0.4949
.07
.
.
.
0.4913
0.4934
0.4951
Standard Normal Distribution
Area to the left of -2.47
P(Z < -2.47) = .5 - 0.4932
= 0.0068
0.4
Table area for 2.47
P(0 < Z < 2.47) = 0.4932
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
.08
.
.
.
Finding Probabilities of the Standard
Normal Distribution: P(1< Z < 2)
11-115
To find P(1  Z  2):
1. Find table area for 2.00
F(2) = P(Z  2.00) = .5 + .4772 =.9772
2. Find table area for 1.00
F(1) = P(Z  1.00) = .5 + .3413 = .8413
3. P(1  Z  2.00) = P(Z  2.00) - P(Z  1.00)
z
.
.
.
0.9
1.0
1.1
.
.
.
1.9
2.0
2.1
.
.
.
= .9772 - .8413 = 0.1359
Standard Normal Distribution
0.4
Area between 1 and 2
P(1  Z  2) = .9772 - .8413 = 0.1359
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
.00
.
.
.
0.3159
0.3413
0.3643
.
.
.
0.4713
0.4772
0.4821
.
.
.
...
...
...
...
...
...
...
Finding Values of the Standard Normal
Random Variable: P(0 < Z < z) = 0.40
To find z such that
P(0  Z  z) = .40:
1. Find a probability as close as
possible to .40 in the table of
standard normal probabilities.
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
.
.
.
.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
.
.
.
.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
.
.
.
2. Then determine the value of z
from the corresponding row
and column.
Area to the left of 0 = .50
Also, since P(Z  0) = .50
.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
.
.
.
.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
.
.
.
.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
.
.
.
.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
.
.
.
.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
.
.
.
.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
.
.
.
Standard Normal Distribution
0.4
P(z  0) = .50
Area = .40 (.3997)
0.3
f(z)
P(0  Z  1.28)  .40
.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
.
.
.
0.2
0.1
0.0
P(Z  1.28)  .90
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
Z = 1.28
11-116
.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
.
.
.
99% Interval around the Mean
To have .99 in the center of the distribution, there
should be (1/2)(1-.99) = (1/2)(.01) = .005 in each
tail of the distribution, and (1/2)(.99) = .495 in
each half of the .99 interval. That is:
P(0  Z  z.005) = .495
z
.
.
.
2.4 ...
2.5 ...
2.6 ...
.
.
.
.04
.
.
.
0.4927
0.4945
0.4959
.
.
.
.05
.
.
.
0.4929
0.4946
0.4960
.
.
.
Look to the table of standard normal probabilities
Area in center left = .495
to find that:
.06
.
.
.
0.4931
0.4948
0.4961
.
.
.
.07
.
.
.
0.4932
0.4949
0.4962
.
.
.
11-117
.08
.
.
.
0.4934
0.4951
0.4963
.
.
.
.09
.
.
.
0.4936
0.4952
0.4964
.
.
.
Total area in center = .99
0.4
 z.005  
z.005  
P(-.2575 Z  ) = .99
Area in center right = .495
f(z)
0.3
0.2
Area in right tail = .005
Area in left tail = .005
0.1
0.0
-5
-4
-3
-2
-z.005
-2.575
-1
0
Z
1
2
3
z.005
2.575
4
5
4-5 The Transformation of Normal
Random Variables
11-118
The area within k of the mean is the same for all normal random variables. So an area
under any normal distribution is equivalent to an area under the standard normal. In this
example: P(40  X  P(-1  Z     since and 
The transformation of X to Z:
X   x
Z 

Normal Distribution:  =50, =10
x
0.07
0.06
Transformation
f(x)
(1) Subtraction: (X - x)
0.05
0.04
0.03
=10
{
0.02
Standard Normal Distribution
0.01
0.00
0.4
0
20
30
40
50
60
70
80
90 100
X
0.3
0.2
(2) Division by x)
{
f(z)
10
1.0
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
The inverse transformation of Z to X:
X  x  Z x
11-119
Using the Normal Transformation
Example 4-9
Example 4-10
X~N(160,302)
X~N(127,222)
P (100  X  180)
 100    X    180   
 P
P( X  150)
X   150   

 P




 
 
100  160
180  160




P
Z

30
30
 P 2  Z .6667
 0.4772  0.2475  0.7247


 
 
150  127

 P Z 


22 

 P Z  1.045
 0.5  0.3520  0.8520
Using the Normal Transformation Example 4-11
11-120
Normal Distribution:  = 383,  = 12
Example 4-11
0.05
0.04
X~N(383,122)
0.03


 P 0.9166  Z  1.333
 0.4088  0.3203  0.0885
Template solution
0.02
0.01
Standard Normal Distribution
0.00
340
0.4
390
X
0.3
f(z)

 

 
 394  383   399  383

P
Z
 12

12
f(X)
P ( 394  X  399)
394   X   399   

 P



0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
440
The Transformation of Normal
Random Variables
The transformation of X to Z:
Z 
X  x
x
The inverse transformation of Z to X:
X  
 Z
x
x
The transformation of X to Z, where a and b are numbers::
a  





P ( X a ) P Z

 
b  





P ( X b) P Z

 
b  
a







P (a X b ) P
Z
 
 
11-121
11-122
Normal Probabilities (Empirical Rule)
• The probability that a normal random
S ta n d a rd N o rm a l D is trib u tio n
variable will be within 1 standard
deviation from its mean (on either
side) is 0.6826, or approximately 0.68.
variable will be within 2 standard
deviations from its mean is 0.9544, or
approximately 0.95.
• The probability that a normal random
variable will be within 3 standard
deviation from its mean is 0.9974.
0.3
f(z)
• The probability that a normal random
0.4
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
11-123
4-6 The Inverse Transformation
The area within k of the mean is the same for all normal random variables. To find a
probability associated with any interval of values for any normal random variable, all that
is needed is to express the interval in terms of numbers of standard deviations from the
mean. That is the purpose of the standard normal transformation. If X~N(50,102),
 x   70   

70  50 

  P Z 
  P( Z  2)
P( X  70)  P
 

 
10 
That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean
of X: 70 =  + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal
distribution.
Normal Distribution:  = 124,  = 12
Example 4-12
X~N(124,122)
P(X > x) = 0.10 and P(Z > 1.28) 0.10
x =  + z = 124 + (1.28)(12) = 139.36
0.04
.
.
.
...
...
...
.
.
.
.07
.
.
.
0.3790
0.3980
0.4147
.
.
.
.08
.
.
.
0.3810
0.3997
0.4162
.
.
.
.09
.
.
.
0.3830
0.4015
0.4177
.
.
.
f(x)
0.03
z
.
.
.
1.1
1.2
1.3
.
.
.
0.02
0.01
0.01
0.00
80
130
X
139.36
180
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
0.0012
.
0.0010
.
f(x)
0.0008
.
0.0006
.
0.0004
.
0.0002
.
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
0.3
f(z)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
11-124
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
.4750
.4750
0.3
f(z)
2. Shade the area
corresponding to
the desired
probability.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
11-125
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
3. From the table
of the standard
normal
distribution,
find the z value
or values.
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
2. Shade the area
corresponding
to the desired
probability.
S tand ard Norm al D istrib utio n
0.4
.4750
.
.
.
...
...
...
.
.
.05
.
.
.
0.4678
0.4744
0.4798
.
.
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.4750
0.3
f(z)
z
.
.
.
1.8
1.9
2.0
.
.
.07
.
.
.
0.4693
0.4756
0.4808
.
.
11-126
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
Z
-1.96
1.96
3
4
5
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
3. From the table
of the standard
normal
distribution,
find the z value
or values.
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
11-127
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
2. Shade the area
corresponding
to the desired
probability.
0.4
.4750
.
.
.
...
...
...
.
.
.05
.
.
.
0.4678
0.4744
0.4798
.
.
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.4750
0.3
f(z)
z
.
.
.
1.8
1.9
2.0
.
.
4. Use the
transformation
from z to x to get
value(s) of the
original random
variable.
S tand ard Norm al D istrib utio n
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
Z
-1.96
1.96
3
4
5
x =   z = 2450 ± (1.96)(400)
= 2450 ±784=(1666,3234)
Finding Values of a Normal Random
Variable, Given a Probability
The normal distribution with  = 3.5 and  = 1.323 is a close
approximation to the binomial with n = 7 and p = 0.50.
P(x<4.5) = 0.7749
Normal Distribution:  = 3.5,  = 1.323
Binomial Distribution: n = 7, p = 0.50
0.3
0.3
P( x 4) = 0.7734
0.2
f(x)
P(x)
0.2
0.1
0.1
0.0
0.0
0
5
10
0
1
2
3
X
4
5
6
7
X
MTB > cdf 4.5;
SUBC> normal 3.5 1.323.
Cumulative Distribution Function
MTB > cdf 4;
SUBC> binomial 7,.5.
Cumulative Distribution Function
Normal with mean = 3.50000 and standard deviation = 1.32300
Binomial with n = 7 and p = 0.500000
x P( X <= x)
4.5000
0.7751
x P( X <= x)
4.00
0.7734
11-128
11-129
FOR ANY RESEARCH WE ARE ALWAYS INTERESTED
TO UNDERSTAND THE POPULATION PARAMETER
SO THAT DECISIONS CAN BE MADE BASED ON
INFORMATION.
EX: A MARKETER MAY BE INTERESTED TO KNOW
AVERAGE CONSUMPTION OF SUGAR PER
HOUSEHOLD PER MONTH IN THE CITY OF
DELHI.
THIS INFORMATION IS THE POPULATION PARAMETER
WHERE THE WHOLE OF CITY DELHI HOUSEHOLD IS
THE POPULATION AND THE AVERAGE CONSUMPTION
OF SUGAR IS THE PARAMETER REPRESENTED BY ‘µ’
11-130
HOWEVER, FINDING THIS PARAMETER IS DIFFICULT
AS IT WILL BE VIRTUALLY IMPRACTICAL TO CONTACT
ALL THE HOUSEHOLD OF DELHI ( OR TIME TAKEN
WOULD BE VERY LARGE) AND THE PURPOSE OF
THE STUDY ITSELF MAY BE TIME BARED.
HENCE WE MUST RESORT TO COLLECTING THE
INFORMATION FROM ONLY A SUBSET OF THE
POPULATION WHICH IS CALLED THE SAMPLE. THIS
SAMPLE INFORMATION FOR THE SAME VARIABLE
IS REFERRED TO AS THE STATISTIC
( x with a bar on the top )
11-131
HOWEVER SAMPLE MEAN IS NOT EQUAL TO
POPULATION TO MEAN AND THE DIFFERENCE
IN THE SAME IS THE ERROR IN ESTIMATING
THE PARAMETER ( KNOWN AS TOTAL ERROR)
THIS ERROR OCCURS FOR SEVERAL REASONS.
11-132
Sample vs. Census
Type of Study
Conditions Favoring the Use of
Sample
Census
1. Budget
Small
Large
2. Time available
Short
Long
3. Population size
Large
Small
4. Variance in the characteristic
Small
Large
5. Cost of sampling errors
Low
High
6. Cost of nonsampling errors
High
Low
11-133
THUS IT IS CLEAR THAT SAMPLING IS REQUIRED
AND IF SAMPLE SIZE IS PROPERLY CHOSEN THEN
THE ERROR IS ALSO CAN BE KEPT AT A MINIMUM
LEVEL.
11-134
SAMPLING DISTRIBUTION
IF THE TARGET SEGMENT ( POPULATION ) CONTAINS
‘N’ ELEMENTS AND FROM THIS POPULATION WE PICK
RANDOMLY ‘n’ ELEMENTS.
IN HOW MANY POSSIBLE WAYS CAN WE PICK UP
THESE ‘n’ ELEMENTS?
N^n ways if done with replacement
NCn ways if done without replacement
FOR EACH OF THESE SAMPLES THERE WILL BE A
SAMPLE MEAN. THE WAY THESE SAMPLE MEANS ARE
SPREAD IS KNOWN AS SAMPLING DISTRIBUTION.
11-135
SAMPLING DISTRIBUTION
Let us illustrate the concept of Sampling Distribution:
Consider a population consisting of only three members
( A, B and C). If a question is asked to them as to how
Many chocolates do they eat in a day, the answer is
A= 1 per day, B = 2 per day and C = 3 per day. Hence
The variable is number of chocolates which is
{ 1, 2, 3 } . This gives the population average (µ = 2)
And a variance ( σ^2) = 2/3.
If sampling of size is 2 is taken with replacement let
Us list all possible samples along with its sample means
11-136
SAMPLING DISTRIBUTION
Possible freq prob
Possible sample are
Sample
sample
mean
Mean
( 1,1)
1
( 1,2)
1.5
1
1
1/9
( 1,3)
2
1.5
2
2/9
( 2,1)
1.5
2
3
3/9
( 2,2)
2
2.5
2
2/9
( 2,3)
2.5
3
1
1/9
( 3,1)
2
( 3,2)
2.5
( 3,3)
3
Expected value of sample mean = 2 = population mean
Expected variance of sample means = σ^2/n
11-137
SAMPLING DISTRIBUTION
P
R
O
B
A
B
I
L
I
T
Y
Does this appear to be normally
Distributed?
Yes indeed!
1 1.5 2 2.5 3
Sample mean
11-138
SAMPLING DISTRIBUTION
Thus the Central Limit Theorem says that
The distribution of the sample mean is always
Normally distributed as long as sample size is
large
Such that :
Expected value of sample Mean = population mean
and standard deviation of sample mean
= population standard deviation/n
This is true irrespective of the distribution of the
Population.
Properties of the Sampling Distribution of the
Sample Mean
is more bell-shaped and
symmetric.
 Both have the same
center.
 The sampling distribution
of the mean is more
compact, with a smaller
variance.
Uniform Distribution (1,8)
0.2
P(X)
Comparing the population
distribution and the
sampling distribution of the
mean:
 The sampling distribution
0.1
0.0
1
2
3
4
5
6
7
8
X
Sampling Distribution of the Mean
0.10
P(X)
•
11-139
0.05
0.00
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0
X
11-140
Relationships between Population Parameters and the
Sampling Distribution of the Sample Mean
The expected value of the sample mean is equal to the population mean:
E( X )    
X
X
The variance of the sample mean is equal to the population variance divided by
the sample size:
V(X) 
2
X


2
X
n
The standard deviation of the sample mean, known as the standard error of
the mean, is equal to the population standard deviation divided by the square
root of the sample size:
SD( X )   
X

X
n
11-141
Sampling from a Normal Population
When sampling from a normal population with mean  and standard
deviation , the sample mean, X, has a normal sampling distribution:
This means that, as the
sample size increases, the
sampling distribution of the
sample mean remains
centered on the population
mean, but becomes more
compactly distributed around
that population mean
n
2
)
Sampling Distribution of the Sample Mean
0.4
Sampling Distribution: n =16
0.3
Sampling Distribution: n = 4
f(X)
X ~ N (,

0.2
Sampling Distribution: n = 2
0.1
Normal population
Normal population
0.0

11-142
The Central Limit Theorem
0.25
P(X)
0.20
0.15
0.10
0.05
0.00
X
n = 20
P(X)
0.2
0.1
0.0
X
When sampling from a population
with mean  and finite standard
deviation , the sampling
distribution of the sample mean will
tend to a normal distribution with

mean  and standard deviation n as
the sample size becomes large
(n >30).
n=5
Large n
0.4
0.2
0.1
0.0
-

X
For “large enough” n: X ~ N ( , / n)
2
f(X)
0.3
The Central Limit Theorem Applies to Sampling
Distributions from Any Population
Normal
Uniform
Skewed
11-143
General
Population
n=2
n = 30

X

X

X

X
11-144
SAMPLING DISTRIBUTION - EXAMPLE
Let us assume that we are interested in understanding
What will be the average consumption of sugar per
Household per month in a given target population?
What this means is we are interested to get the
Information ( µ = average sugar consumed /month)
We can only estimate the same based on sample
Information. i.e. based on sample mean . This can be
Done as follows.
11-145
SAMPLING DISTRIBUTION - EXAMPLE
For the example let us assume that we sampled randomly
100 household and got the information that the sample
Mean was 1890 grams per household per month.
Let us also assume that the population standard deviation
Was known as 230 grams .
We use the fact that the sample mean obtained was one
Among the different sample means possible and that
That the sample means would be normally distributed.
Hence an interval estimate can be obtained as
µ = x-bar ± Z 
where Z = std. normal deviate
n
11-146
SAMPLING DISTRIBUTION - EXAMPLE
For the example let us assume that we sampled randomly
100 household and got the information that the sample
Mean was 1890 grams per household per month.
Let us also assume that the population standard deviation
Was known as 230 grams .
Substituting the values we get
µ = 1890 ± Z x (230 / √100 )
for a 90% confidence Z = 1.28 ( refer Z table )
= 1890 ± 1.28 x (230 / √100 )
= 1890 ± 29.44
there is a 90% chance that the actual (µ ) will be
contained within 1860.56 to 1919.44 grams.
11-147
FROM THE EXAMPLE JUST EXPLAINED YOU CAN
SEE THAT
(sample mean(x-bar) – µ ) = Z
Error in estimating µ
Hence

n

n
Is a function of 
n
is often referred to as ‘standard error’
Hence if error is known then the sample size can be
Determined ( this is based on sampling error alone)
11-148
Sampling without replacement
When we sample without replacement and from a finite
Population the standard deviation of sample means
( also known as standard error ) incorporates a
Finite population multiplier and is as follows:

n
√ ( N-n)/(N-1)
N = population size
n = sample size
Finite population
Multiplier ( always ≤ 1)
It can be noted that as N goes to ∞ the multiplier
Becomes = 1 and hence the standard error is the same
As if the sampling is done with replacement.
11-149
Sampling distribution for proportion
Consider an example :
The number of times a hotel is unable to accommodate
Their customer with rooms because the hotel is full.
This can only be expressed in terms of proportion
i.e. 10% and so on.
In this case also the sampling distribution of proportion
If the sample size is large behaves like a normal
Distribution with expected value of sample proportion
Equal to population population and the standard error
Equal to √(pq/n) , where ‘q’ = 100-p if ‘p’ in percentage
11-150
Sampling distribution for proportion
Similarly interval estimation for proportion can be
Found. Example if a sample of 1000 voters selected
And 400 of them decided to vote for a political party (X)
Then the proportion of the population that is expected
To vote for the party (x) would be
P = 40% ±Z √(40*60/1000)
= 40% ±3.038 ( Z = 1.96 for 95% confidence level)
hence the interval would be 36.92% to 43.04%
11-151
Sampling distribution of difference of two means
Consider a group of male employees and a group of
Female employees in IT industry at a given level. It is
Desired to understand what is the level of difference
In their salaries ( given there is discrimination )
11-152
Sampling distribution of difference of two means
In this situation there could be many possible samples
that can be drawn of size (n1) from males and similarly
many samples that can be drawn of size ( n2) from
female employees.
For each sample taken from each group the sample
mean can be substracted which will give us the level
of difference in the salary.
This is difference of two sample means and the distribution
would also behave like a normal distribution for large sample
11-153
Sampling distribution of difference of two means
Thus the sampling distribution for
Difference of two means is normally distributed
With expected value (x (male) – x (female) )
=0
and variance for the difference of two means
=
s12 s22
+
Standard
n1 n2
s
Error
11-154
Sampling distribution for small sample
In our earlier discussion we have always emphasized
The need for a large sample for the sample mean to
Be distributed as a Normal Distribution
What is meant by LARGE sample ?
11-155
Sampling distribution for small sample
Gossett was working on samples which were
Considered as small such as 10, 15, 25 etc and he
Found that the sample mean distribution was not
Exactly Normal distribution but was nevertheless
Symmetric but with large variance. He denoted
His distribution as
Student ‘t’ distribution.
Thus the mean value of ‘t’ = 0 and the probability
Density function was not only a function of the
Mean and variance but also dependent on what he
Called as “degree of freedom”
Confidence Interval for a
Mean () with Unknown 

11-156
Degrees of Freedom
•
•
Degrees of Freedom (d.f.) is a parameter
based on the sample size that is used to
determine the value of the t statistic.
Degrees of freedom tell how many
observations are used to calculate , less
the number of intermediate estimates
used in the calculation.
n=n-1
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for a
Mean () with Unknown 

11-157
Degrees of Freedom
•
•
McGraw-Hill/Irwin
As n increases, the t distribution
approaches the shape of the normal
distribution.
For a given confidence level, t is always
larger than z, so a confidence interval
based on t is always wider than if z were
used.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-158
Degree of freedom
To understand the degree of freedom let us
Consider the numbers
1
2
3
total
6
We will have the freedom to change any two out
Of these three numbers without change in the total
Thus the degree of freedom would be 2
Thus degree of freedom would be (n-1)
Where n = sample size.
11-159
Sampling distribution for small sample
Thus for a small sample and whenever the population
Variance is unknown, the distribution of the sample
Means behaves like a ‘t’ distribution .
This ‘t’ distribution becomes very close to Normal
Distribution when the degree of freedom is 29 and
Above.
Hence the definition for a large sample in statistics
Is when the sample size is 30 or more.
For smaller than 30 the distribution needed would be
‘t’ .
Confidence Interval for a
Mean () with Unknown 

11-160
Student’s t Distribution
•
•
t distributions are symmetric and shaped
like the standard normal distribution.
The t distribution is dependent on the
size of the sample.
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-161
Sampling distribution for small sample
Thus for all calculations with small samples
Z value will be substituted with ‘t’ values.
Usually small samples are not used when
Proportions are involved.
11-162
Confidence Interval for a Mean () with Unknown 
Student’s t Distribution
Use the Student’s t distribution instead of the
normal distribution when the population is
normal but the standard deviation  is
unknown and the sample size is small.
s
x+t
n
The confidence interval for  (unknown )
is
s
s
x-t
x+t
n
McGraw-Hill/Irwin
<<
n
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for a
Mean () with Unknown 
11-163
Student’s t Distribution
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for a
Mean () with Unknown 

11-164
Comparison of z and t
•
•
•
•
McGraw-Hill/Irwin
For very small samples, t-values differ
substantially from the normal.
As degrees of freedom increase, the tvalues approach the normal z-values.
For example, for n = 31, the degrees of
freedom are: n = 31 – 1 = 30
What would the t-value be for a 90%
confidence interval?
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for a
Mean () with Unknown 

11-165
Comparison of z and t
For n = 30, the corresponding z-value is 1.645.
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for the Difference of
Two Means, small sample 1 – 2
•
11-166
The procedure for constructing a
confidence interval for 1 – 2 depends on
our assumption about the unknown
variances.
Assuming equal variances:
(x1 – x2) + t
(n1 – 1)s12 + (n2 – 2)s22
n1 + n2 - 2
1 1
+
n1 n2
with n = (n1 – 1) + (n2 – 1) degrees of freedom
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for the Difference of
Two Means, small sample 1 – 2
11-167
Assuming equal variances:
(x1 – x2) + t
(n1 – 1)s12 + (n2 – 2)s22
n1 + n2 - 2
with n = (n1 – 1) + (n2 – 1)
degrees of freedom
McGraw-Hill/Irwin
1 1
+
n1 n2
Pooled standard
deviation
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Confidence Interval for the Difference of
Two Means, small sample 1 – 2
11-168
Assuming equal variances:
(x1 – x2) + t
(n1 – 1)s12 + (n2 – 2)s22
n1 + n2 - 2
with n = (n1 – 1) + (n2 – 1)
degrees of freedom
McGraw-Hill/Irwin
1 1
+
n1 n2
Standard
Error for
Differences
Of means
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-169
F- Distribution
F – distribution ( Fisher’s ) is the ratio of the variations.
If two samples are drawn and we wish to know
Whether the samples are drawn from a single population
Or from two separate population, then an F- Statistic
Is calculated.
This F- Statistic = ratio of samples variances of the two
samples. ( S12 / S22 )
11-170
F – distribution curve
F – Distribution is a probability density function
whose shape of the curve is as follows:
y
F- Statistic
11-171
F- distribution
We will have more occasions to talk about
This F- statistic later while discussing
Hypothesis testing.
11-172
11-173
Chi-Square - Distribution
Chi-Square Distributed (
) is a distribution
When we wish to estimate the population variance
From a known sample variance.
Similarly there are many non parametric tests
Where we would use a Chi-Square tests.
The shape of the Chi-square distribution varies with
The degree of freedom
11-174
Chi-Square distribution
y
Chi-square statistic
11-175
Chi-square distribution
Square of the Z distribution behaves like a
Chi-Square distribution.
Similarly a sum of the square of several Normal
Distribution also behaves like a Chi-Square
Distribution.
We will have more to talk about this distribution
When we look at hypothesis testing.
11-176
11-177
Visual Displays and Correlation Analysis

Visual Displays
•
•
•
Begin the analysis of bivariate data (i.e.,
two variables) with a scatter plot.
A scatter plot
- displays each observed data pair (xi, yi)
as a dot on an x-y grid
indicates visually the strength of the
relationshi between the two variables
.
11-178
Visual Displays and Correlation Analysis

Visual Displays
cost of maintenance per month
maintenance cost
25000
20000
15000
10000
5000
0
0
5
10
15
20
25
30
35
hrs driven per week
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-179
Visual Displays and Correlation Analysis

Correlation Analysis
Weak Positive
Correlation
Strong Positive
Correlation
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-180
Visual Displays and Correlation Analysis

Correlation Analysis
Strong Negative
Correlation
Weak Negative
Correlation
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-181
Visual Displays and Correlation Analysis

Correlation Analysis
Nonlinear Relation
No Correlation
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-182
Visual Displays and Correlation Analysis

Correlation Analysis
•
The sample correlation coefficient (r)
measures the degree of linearity in the
relationship between X and Y.
-1 < r < +1
Strong negative relationship
•
Strong positive relationship
r = 0 indicates no linear relationship
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-183
Visual Displays and Correlation Analysis

Correlation Analysis
Cov ( x,y)
= ------------s(x) s(y)
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-184
Correlation coefficient can also be found
As follows:
r=
(n(∑xy) - ( ∑x ∑y)
√{(n∑x2 ) – (∑x)2 } x {(n∑y2) –(∑y)2}
11-185
11-186
Use of Excel for finding the correlation
11-187
Use of Excel for finding the correlation
11-188
Use of Excel for finding the correlation
11-189
Use of Excel for finding the correlation
11-190
Correlation coefficient can also be found
As follows:
r=
(n(∑xy) - ( ∑x ∑y)
√{(n∑x2 ) – (∑x)2 } x {(n∑y2) –(∑y)2}
0.9322
Hence r^2 = 0.869
11-191
Properties of correlation coefficient
1. The value of ‘r’ always varies between -1 to +1
2. The change of origin and scale does not effect
the value of the coefficient
( what this means is as follows)
11-192
Change of origin and scale means
11-193
Properties of correlation coefficient
1. The value of ‘r’ always varies between -1 to +1
2. The change of origin and scale does not effect
the value of the coefficient
3. If ‘x’ and ‘y’ are interchanged the coefficient is
not effected. i.e. it remains unaltered.
( we usually refer to ‘x’ as independent variable
‘y’ as dependent variable
4. the fourth property can be explained after we
explain regression ( hence hold till such time )
11-194
Bivariate Regression

What is Bivariate Regression?
•
•
•
McGraw-Hill/Irwin
Bivariate Regression analyzes the
relationship between two variables.
It specifies one dependent
(response) variable and one
independent (predictor) variable.
This hypothesized relationship may
be linear, quadratic, or whatever.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-195
Bivariate Regression
Chart Title
cost of maintenance
25000
20000
15000
10000
5000
0
-5000
0
5
10
15
20
25
30
35
hrs of vehicle driven
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-196
Bivariate Regression
In the equation y= a + bx how to find the value of
‘a’ and ‘b’ which are the intercept and slope.
Chart Title
cost of maintenance
25000
20000
15000
10000
5000
0
-5000
0
5
10
15
20
25
30
35
hrs of vehicle driven
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-197
Bivariate Regression
In the equation y= a + bx how to find the value of
‘a’ and ‘b’ which are the intercept and slope.
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-198
How to develop a Regression line
11-199
How to develop a Regression line
11-200
Normal Equations
∑ Y = na + b∑X
∑XY = a∑X + b∑X2
Which when simplified becomes
(n∑ XY) – (∑X ∑Y)
b=
----------------------(n∑X^2 ) - (∑X)^2
a= Y(bar) – b X ( bar )
11-201
Normal Equations
For the problem considered earlier
b = 777.32 ( y – dependent variable
x – independent variable)
And
a= -6115.9
11-202
Normal Equations
If we had desired the normal equations for the
Situation when ‘x’ = dependent variable
and ‘y’ = independent variable
The normal equations would simply change so that
Where there ‘x’ replace with ‘y’ and replace ‘y’ with ‘x’
You will find the numerator would remain unchanged
But the denominator would be
(n∑X^2 ) - (∑X)^2
11-203
Normal Equations
Hence the new a’ = 9.393
b’ = 0.001118
when x = dependent variable
y= independent variable
11-204
Properties of correlation coefficient
1. The value of ‘r’ always varies between -1 to +1
2. The change of origin and scale does not effect
the value of the coefficient
3. If ‘x’ and ‘y’ are interchanged the coefficient is
not effected. i.e. it remains unaltered.
( we usually refer to ‘x’ as independent variable
‘y’ as dependent variable
4. b x b’ = r^2 which is 777.32 x 0.001118 = 0.869
11-205
11-206
From here it follows:
1. Both the regression coefficients must have the
same sign ( either + or - )
2. If one regression coefficient is greater than 1
then the other regression coefficient must be
<1.
3. If one regression coefficient is < 1 then the other
regression coefficient may be > or < than 1.
11-207
Regression Terminology

Fitting a Regression on a Scatter
Plot in Excel
•
McGraw-Hill/Irwin
Step 1:
- Highlight the data columns.
- Click on the Chart Wizard and choose Scatter
Plot
- In the completed graph, click once on the
points
in the scatter plot to select the data
- Right-click and choose Add Trendline
- Choose Options and check Display Equation
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-208

Fitting a Regression on a Scatter
Plot in Excel
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-209

Fitting a Regression on a Scatter
Plot in Excel
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-210

Fitting a Regression on a Scatter
Plot in Excel
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-211
Regression Terminology
McGraw-Hill/Irwin
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11-212
11-213
WHAT IS A HYPOTHESIS
A hypothesis is a conjectural statement about the a
Certain characteristic in the whole population or target
Segment.
Ex: What is the average expenditure per month incurred
on vehicle maintenance. If it is suggested that this
average is Rs.1200 per month, then this will be a
hypothesis.
What this implied is that if we take all the people who drive their
Vehicle and find the expenditure of everyone and average the results
It would be Rs. 1200 per month.
11-214
WHAT IS A HYPOTHESIS
A hypothesis is a conjectural statement about the a
Certain characteristic in the whole population or target
Segment.
Ex: A car dealer claims that on an average the mileage
of a car ( given model ) gives at least 15 kms to
a litre of petrol.
This implied that for the given model, if we take the average of
All the vehicles and find its mileage per litre of petrol, it would be
At least 15 kms to a litre of petrol.
11-215
WHAT IS A HYPOTHESIS
A hypothesis is a conjectural statement about the a
Certain characteristic in the whole population or target
Segment.
Ex: An exporter claims that the proportion of defects
in his consignment will be at most 2% .
This means that if we take all his consignments and find the proportion
Of defects, the average defectives will not exceed 2%.
11-216
WHAT IS A HYPOTHESIS
A hypothesis is a conjectural statement about the a
Certain characteristic in the whole population or target
Segment.
Ex: A refill manufacturer for ball point pens claims that
the length of the refill on an average is 140 mm.
This would imply that while each of the refill’s may ( or cannot)
Be exactly 140 mm but on an average the length of refills
would be 140 mm.
11-217
What do all these hypothesis show?
1. They are all conjectures about the population
parameter.
2. They talk always about the population parameter
3. They are statements made only on the basis of
the research question in hand and not on the basis
of the data collected.
11-218
Why do we need to do hypothesis testing
While we want to verify the statement specified in the
Hypothesis, it would be impossible to do so without doing
A census. Hence if a census is carried out then
Hypothesis testing is not essential.
However we know that a census is not practical and also
Need not be accurate. Hence we need to comment on the
Hypothesis on the basis of sample information only.
Hence we draw inference about the population parameter
Based on sample information, this drawing of inference is
Called hypothesis testing.
11-219
Characteristics of a good hypothesis
1. hypothesis should be based on sound previous research
2. look for realistic explanations
3. state the variables clearly
4. it should be easily amenable to test
5. measure the variables in the correct scale
11-220
Basics of hypothesis testing
As we said before the hypothesis is about the characteristics
About the population. We usually call it the ‘parameter’
Parameter can only be obtained by doing a census
Which is not possible or not practical.
Hence our inference about the parameter is based on the
Sample information.
Hence based on a sample information we may reject a true
Hypothesis and conversely we may accept a hypothesis
When it is actually not true.
Both of these are errors in making the inference.
11-221
Basics of hypothesis testing
Consider the problem:
A car dealer claims that on an average the mileage of a
car ( given model ) gives at least 15 kms to a litre of petrol
The problem here is that if the average is more than 15
km/litre then we are satisfied but if it gives less than than
What is claimed by the dealer then we have difficulty in
Believing the claim of the dealer. Hence this is what we
Wish to verify or infer. This inference must be based on
The information available from a single sample of size ‘n’
This ‘n’ can be either 30 cars or 40 cars or even about 20
cars.
11-222
Basics of hypothesis testing
Consider the problem:
A car dealer claims that on an average the mileage of a
car ( given model ) gives at least 15 kms to a litre of petrol
Hence we can write the hypothesis as follows:
Null Hypothesis :
Generally we not disagree with the dealer to begin with
unless there is sufficient evidence to disagree.
Hence we write Null hypothesis: ( Ho ): µ ≥ 15
Alternative hypothesis:
( Ha ): µ < 15
11-223
Basics of hypothesis testing
Consider the problem:
A car dealer claims that on an average the mileage of a
car ( given model ) gives at least 15 kms to a litre of petrol
( Ho ): µ ≥ 15
( Ha ): µ < 15
11-224
Approaches for hypothesis testing
11-225
Errors in hypothesis testing
Hypothesis
true
Hypothesis
false
Accept
hypothesis
No error
Type II error
or ß- error
Reject
hypothesis
Type I error or No error
alpha error
11-226
Steps in Hypothesis Testing
1. Based on the research question develop Null and
Alternative hypothesis
2. Decide on the level of Type I Error or alpha error.
3. Decide whether the test is a single tail or a two tail
11-227
Steps in Hypothesis Testing
4. Decide on the appropriate Test Statistic which will
be used ( Z or t or any other )
5. Calculate the test statistic.
11-228
Steps in Hypothesis Testing
6. Read the test statistic for the level of type I error from
The table of Z or ‘t’ etc.
7. Compare the calculated test statistic with that of the
table value.
8. Make a conclusion.
11-229
Worked Example -1:
A manufacturing firm has been averaging shipping of a product within 30 days
of receiving the order. Of late it is believed that the average shipping time has
increased. To test this a sample of size 49 is drawn randomly from the
Shipments made during a given period of time. The sample shows an average
shipping time of 36 days. The population standard deviation is believed to be 7 days.
Is there sufficient evidence to believe that the shipping is getting delayed.
A 5% level of significance test is thought to be good.
Step 1: Formulate the null and alternative hypothesis:
Ho: µ ≤ 30
Ha: µ > 30
11-230
Worked Example -1 contd:
A manufacturing firm has been averaging shipping of a product within 30 days
of receiving the order. Of late it is believed that the average shipping time has
increased. To test this a sample of size 49 is drawn randomly from the
Shipments made during a given period of time. The sample shows an average
shipping time of 36 days. The population standard deviation is believed to be 7 days.
Is there sufficient evidence to believe that the shipping is getting delayed.
A 5% level of significance test is thought to be good.
Step 2: Decide on the level of significance or type I error
This is given in the problem as 5%
Step 3: Looking at the hypothesis it is clear that it is
single tail and the problem area is to the right
hence right tail.
11-231
Worked Example -1 contd:
A manufacturing firm has been averaging shipping of a product within 30 days
of receiving the order. Of late it is believed that the average shipping time has
increased. To test this a sample of size 49 is drawn randomly from the
Shipments made during a given period of time. The sample shows an average
shipping time of 36 days. The population standard deviation is believed to be 7 days.
Is there sufficient evidence to believe that the shipping is getting delayed.
A 5% level of significance test is thought to be good.
Step 4: decide on the test statistic. Since sample size is > 30
we can go for the Z statistic.
Step 5: Calculate Z statistic = (x-bar - µ )/ S.E. ( standard
error )
= (36 – 30 )/ (7 / √49) = 6
11-232
Worked Example -1 contd:
A manufacturing firm has been averaging shipping of a product within 30 days
of receiving the order. Of late it is believed that the average shipping time has
increased. To test this a sample of size 49 is drawn randomly from the
Shipments made during a given period of time. The sample shows an average
shipping time of 36 days. The population standard deviation is believed to be 7 days.
Is there sufficient evidence to believe that the shipping is getting delayed.
A 5% level of significance test is thought to be good.
Step 5: Calculate Z statistic = (x-bar - µ )/ S.E. ( standard
error )
= (36 – 30 )/ (7 / √49) = 6
Step 6: The Z ( table value at 5% level of alpha), single
tail = 1.645.
Step 7. Since the Z (cal) > Z ( table value ) hence reject
or unable to accept the null hypothesis
11-233
Worked Example -1 contd: ( single mean)
A manufacturing firm has been averaging shipping of a product within 30 days
of receiving the order. Of late it is believed that the average shipping time has
increased. To test this a sample of size 49 is drawn randomly from the
Shipments made during a given period of time. The sample shows an average
shipping time of 36 days. The population standard deviation is believed to be 7 days.
Is there sufficient evidence to believe that the shipping is getting delayed.
A 5% level of significance test is thought to be good.
Step 7. Since the Z (cal) > Z ( table value ) hence reject
or unable to accept the null hypothesis
Step 8. Conclusion: There appears to be a delay in the
shipping time in recent times.
11-234
Worked example :-2 ( single mean)
Let us consider the car example: The dealer claims that a particular
model gives at least 15Km/litre of fuel. A random sample of 36 cars
gives a mean of 14.6 km/litre and a population standard deviation
is assumed known as 0.75 km/litre. Assume 5% level of significance
Step 1.
Ho: µ ≥ 15
Ha: µ < 15
Step 2: The significance level is given as 5% .
Step 3: This is also a single tail test but the direction
is towards the left.
Step 4: Since the sample size is large ( 36) we can use
the Z – test.
11-235
Worked example :-2 ( single mean)
Let us consider the car example: The dealer claims that a particular
model gives at least 15Km/litre of fuel. A random sample of 36 cars
gives a mean of 14.6 km/litre and a population standard deviation
is assumed known as 0.75 km/litre. Assume 5% level of significance
Step 4: Since the sample size is large ( 36) we can use
the Z – test.
Step 5: Calculate the Z – statistic
Z = (14.6 – 15 ) / ( 0.75/ √36)
= -0.4 / (0.75/6) = -3.2
Step 6: Evaluate the value of Z at 5% level of
significance. (remember this is now to the
left side, hence ‘Z’ should be negative.
hence it is Z = -1.645.
11-236
Worked example :-2 ( single mean)
Let us consider the car example: The dealer claims that a particular
model gives at least 15Km/litre of fuel. A random sample of 36 cars
gives a mean of 14.6 km/litre and a population standard deviation
is assumed known as 0.75 km/litre. Assume 5% level of significance
Step 5: Calculate the Z – statistic
Z = (14.6 – 15 ) / ( 0.75/ √36)
= -0.4 / (0.75/6) = -3.2
Step 6: Evaluate the value of Z at 5% level of
significance. (remember this is now to the
left side, hence ‘Z’ should be negative.
hence it is Z = -1.645.
Step 7. Compare Z calculated with Z table value
for a left tail test the rule is
if Z calculated ≤ Z table value  reject Ho
else accept Ho.
11-237
Worked example :-2 ( single mean)
Let us consider the car example: The dealer claims that a particular
model gives at least 15Km/litre of fuel. A random sample of 36 cars
gives a mean of 14.6 km/litre and a population standard deviation
is assumed known as 0.75 km/litre. Assume 5% level of significance
Step 6: Evaluate the value of Z at 5% level of
significance. (remember this is now to the
left side, hence ‘Z’ should be negative.
hence it is Z = -1.645.
Step 7. Compare Z calculated with Z table value
for a left tail test the rule is
if Z calculated ≤ Z table value  reject Ho
else accept Ho.
In this Z(cal) = - 3.2 < Z ( table value) -1.645
hence reject Ho.
Step 8. Conclusion is that the dealer claim cannot be accepted
11-238
Worked Example -3: ( single mean)
A refill manufacturer claims that the refill length for a ball Point pen is 140 mm
long. A sample of size 100 is selected and finds that the mean length of refills
is 141.77mm with a standard deviation of 5.88 mm. At 5% level of significance
Can it be concluded that the refills are of poor quality.
Step1: Null hypothesis and alternative hypothesis:
Ho: µ = 140 mm
Ha: µ ≠ 140 mm
Step 2: Alpha level is given at 5% level.
Step 3. In this case it is a two tail test as both sides are
not acceptable because the refill will not fit in the
pen and hence poor quality.
11-239
Worked Example -3 contd: ( single mean)
A refill manufacturer claims that the refill length for a ball Point pen is 140 mm
long. A sample of size 100 is selected and finds that the mean length of refills
is 141.77mm with a standard deviation of 5.88 mm. At 5% level of significance
Can it be concluded that the refills are of poor quality.
Step 4. Since the sample size is 100 which is large hence
Z test can be done.
Step 5. Calculate the Z statistic :
Z= (141.77 – 140) / ( 5.88 / √100)
= 1.77 / 0.588
= 3.01
Step 6. Read the Z statistic from table at 5% two tail
= 1.96
11-240
Worked Example -3 contd: ( single mean)
A refill manufacturer claims that the refill length for a ball Point pen is 140 mm
long. A sample of size 100 is selected and finds that the mean length of refills
is 141.77mm with a standard deviation of 5.88 mm. At 5% level of significance
Can it be concluded that the refills are of poor quality.
Step 5. Calculate the Z statistic :
Z= (141.77 – 140) / ( 5.88 / √100)
= 1.77 / 0.588
= 3.01
Step 6. Read the Z statistic from table at 5% two tail
= 1.96
Step 7: Z ( cal ) > Z ( table value) => hence reject Null
hypothesis.
Step 8. Conclusion: The refills produced are or poor quality
11-241
Rule for rejecting a Null hypothesis
For right tail test ( single tail ):
If Z ( cal ) ≥ Z ( table value )  Reject Ho
For a left tail test ( single tail) :
If Z ( cal ) ≤ Z ( table value )  Reject Ho
For a two tail test :
If |Z ( cal )| ≥ |Z (table value)|  Reject Ho
It can be observed that even for a single tail
Test if we consider the modulus value of Z then
The same rule as that for two tail test can be
Used for rejecting Ho ( use caution here )
11-242
When population σ is unknown
While conducting hypothesis testing, usually population
σ is unknown. Under these situations the sample
standard deviation ( s ) is used instead of σ and
Hence the standard error would be = ( s/√ n).
It must be made sure that sample standard deviation
Must be calculated with ( n-1) in the denominator
As stated in the earlier subject DRM 01 as only then
It becomes unbiased estimator of ‘σ’.
Further it must be ensured that the sample size should
Be large ( definition of large was n > 30)
11-243
When sample size is small <30
When the sample size is < 30, it is considered as a
Small sample and hence ‘t’ distribution should be used
Instead of ‘z’ . Further if the population standard deviation
Is unknown, then also it is recommended that ‘t’
Distribution is used. Hence the only change is :
(Z) statistic = (x-bar - µ )/ S.E. ( standard error)
Replace with ‘t’
11-244
Worked example – single proportion
Insurance companies have recently created difficulty in settling medical
Claims directly to the hospitals. One reason can be attributed to false
Billing by individuals who have taken medical insurance. A company
Believes that of recent there has been an increase in the number of
False medical claims which has gone up to 5%. A random sample of
100 customers indicated that 7 customers had falsified their claim.
Is there any reason to believe that false medical claims have gone up?
Use 5% level of significance.
Step 1: Ho : p ≤ 5%
Ha : p > 5%
Step 2: level of significance given as 5%
Step 3: this is a single tail ( right tail ) test.
11-245
Worked example – single proportion
Insurance companies have recently created difficulty in settling medical Claims directly
to the hospitals. One reason can be attributed to false Billing by individuals who have
taken medical insurance. A company Believes that of recent there has been an increase
in the number of False medical claims which has gone up to 5%. A random sample of
100 customers indicated that 7 customers had falsified their claim. Is there any reason
to believe that false medical claims have gone up? Use 5% level of significance.
Step 1: Ho : p ≤ 5%
Ha : p > 5%
Step 2: level of significance given as 5%
Step 3: this is a single tail ( right tail ) test.
Step 4. Z statistic will be used as sample size is large
Step 5. Z = (ṕ - p )/ standard error
(ṕ - p )/ (√pq/n)
= (7%-5%)/√(5%x95%)/100)
11-246
Worked example – single proportion
Insurance companies have recently created difficulty in settling medical Claims directly
to the hospitals. One reason can be attributed to false Billing by individuals who have
taken medical insurance. A company Believes that of recent there has been an increase
in the number of False medical claims which has gone up to 5%. A random sample of
100 customers indicated that 7 customers had falsified their claim. Is there any reason
to believe that false medical claims have gone up? Use 5% level of significance.
Step 1: Ho : p ≤ 5%
Ha : p > 5%
Step 2: level of significance given as 5%
Step 3: this is a single tail ( right tail ) test.
Step 4. Z statistic will be used as sample size is large
Step 5. Z = (ṕ - p )/ standard error
(ṕ - p )/ (√pq/n)
= (7%-5%)/√(5%x95%)/100)
= 2/ (2.179) = 0.91
Step 6. Z table value at 5% single tail = 1.645
Step 7 : Compare Z cal with Z table value
11-247
Worked example – single proportion
Insurance companies have recently created difficulty in settling medical Claims directly
to the hospitals. One reason can be attributed to false Billing by individuals who have
taken medical insurance. A company Believes that of recent there has been an increase
in the number of False medical claims which has gone up to 5%. A random sample of
100 customers indicated that 7 customers had falsified their claim. Is there any reason
to believe that false medical claims have gone up? Use 5% level of significance.
Step 5. Z = (ṕ - p )/ standard error
(ṕ - p )/ (√pq/n)
= (7%-5%)/√(5%x95%)/100) = 2/ (2.179) = 0.91
Step 6. Z table value at 5% single tail = 1.645
Step 7 : Compare Z cal with Z table value hence Z cal < Z table value
Hence accept Ho
Step 8 : We cannot conclude that the number of false
claims have gone beyond 5%. Even though sample
shows a 7%.
11-248
Hypothesis test for difference of two means
Let us consider the following situations:
Case-1: Does going to VLCC help in reducing weight ?
Case -2: Does the company always assess the rent for
the residential quarter less than the employee
himself?
Case-3: Is their gender discrimination among employers
in a given industry for the same level of job?
Case -4: Is the new drug more effective in treatment
of a disease than the existing drug?
11-249
In all cases we are taking about the difference of
Two mean.
Case-1: The mean before joining VLCC and the mean
after joining VLCC (µbefore -µ after )
Case 2: The mean value of residential quarter assessed
by the company and mean value of residential
quarter assessed by the employee for whom
it is meant. (µ(company) - µ ( employee)
Case 3: The mean wage given to women employees and
the mean wage given to men employees
(µ (women) - µ ( men )
Case 4: The mean time to recover with new drug and
mean time to recover with existing drug.
µ (new drug ) - µ ( old drug )
11-250
Despite each of these cases being a difference of
Two means; there is one essential difference:Case 1 & 2:
In both these cases we are talking about the same
Sample .
Case 1: - Same sample weight before joining VLCC
same sample weight after joining VLCC
Case 2: Same house assessed by Company
Same house assessed by employee
11-251
Despite each of these cases being a difference of
Two means; there is one essential difference:In Cases 3 & 4:Case 3. mean wages for a group of females
mean wages for a group of men
each sample is independently drawn
Case 4: mean time of recovery using new drug
mean time of recovery using existing drug
each sample is independently drawn.
( obviously the same patient cannot be
both the drugs )
11-252
Difference of two means
Difference of two means
Dependent
Sample
Samples independently
drawn
Both situations are treated differently.
11-253
Difference of two means – dependent samples
Ex: It was intended to understand whether there is any difference
in the productivity of a worker immediately after a weekly off
or immediately before a weekly off. If the weekly off was on a
Sunday, it was desired to find if the productivity is different
on Saturday or on a Monday. Hence productivity was measured
on Saturday’s and Monday’s for the same set of workers. The
data are as follows: ( use 5% level of significance)
Worker Id
Productivity
Sat
Mon
1
25
28
2
32
29
3
20
29
4
26
36
5
29
35
6
21
30
7
18
32
8
17
24
9
27
25
11-254
Dependent sample case:
Step 1: To develop a Null and alternative hypothesis
Ho: µ (sat) = µ (mon)
Ha: µ (sat) ≠ µ (mon)
the problem does not suggest that productivity
is more on Saturday’s than on Monday’s or vice-versa
Hence it could either way. Hence ≠ symbol in the
Alternative hypothesis:
This also implies:
Ho: µ (sat) - µ (mon) = 0 or difference=0
Ha: µ (sat) - µ (mon) ≠ 0 or difference ≠0
11-255
Difference of two means – dependent samples
Worker Id
1
2
3
4
5
6
7
8
9
Productivity
Sat
Mon
25
28
32
29
20
29
26
36
29
35
21
30
18
32
17
24
27
25
difference
-3
3
-9
-10
-6
-9
-14
-7
2
average difference = - 5.88
sample standard deviation(s) = 5.622
11-256
Dependent sample case:
Step 2: Level of alpha is given as 5%
Step 3: This is a two tail test
Step 4: The test statistic will be a ‘t’ distribution as
the sample size is small and also the ‘σ’ is
unknown.
t (cal) = diff – 0 / ( s/√ n)
= (- 5.88 – 0) / (5.622/ √9 )
= -5.88 / 1.874
= -3.13 or modulus = 3.13
Step 5: find the table value of ‘t’ for 5% two tail
with a degree of freedom of 8 ( n-1)
= 2.306.
11-257
Dependent sample case:
Step 2: Level of alpha is given as 5%
Step 3: This is a two tail test
Step 4: The test statistic will be a ‘t’ distribution as
the sample size is small and also the ‘σ’ is
unknown.
t (cal) = diff – 0 / ( s/√ n)
= (- 5.88 – 0) / (5.622/ √9 )
= -5.88 / 1.874
= -3.13 or modulus = 3.13
Step 5: find the table value of ‘t’ for 5% two tail
with a degree of freedom of 8 ( n-1)
= - 2.306.
Step 6: Compare t(cal) with t(Table value)
If |t(cal)|> |t(table value)  reject Ho
hence 3.13 > 2.306
Step 7: Reject Ho
Step 8 Conclusion : There is a change in the productivity between
before a weekend and after a weekend.
11-258
Dependent sample case:
Hence it is clear that in the case of a difference
Of two means – dependent sample case
Is treated as if it is a single mean case;
Further data is always obtained in pairs.
And the sample sizes are usually less than 30
Hence a ‘t’ is normally used.
This test is also called Paired ‘t’ test
11-259
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Consider case 3 and 4 discussed earlier. Reproducing
Below both the cases for ready reference
Case 3. mean wages for a group of females
mean wages for a group of men
each sample is independently drawn
Case 4: mean time of recovery using new drug
mean time of recovery using existing drug
each sample is independently drawn.
( obviously the same patient cannot be
both the drugs )
11-260
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Case 3. mean wages for a group of females
mean wages for a group of men
each sample is independently drawn
In this case Null hypothesis: Ho : µ(women) = µ(men)
Ha: µ(women) ≠ µ(men)
Or depending on the problem it could have been a single
Tail test. One tail or two tail depends on the Research
Question being addressed.
What is the implication of acceptance of the Null Hypothesis?
It means that both groups belong to the same population
i.e. there is only population and hence one mean and variance
11-261
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Case 3. mean wages for a group of females
mean wages for a group of men
each sample is independently drawn
In this case Null hypothesis: Ho : µ(women) = µ(men)
Ha: µ(women) ≠ µ(men)
Or depending on the problem it could have been a single
Tail test. One tail or two tail depends on the Research
Question being addressed.
What if Null hypothesis is rejected or not accepted?
This would imply that all men belong to a population and all women
Belong to a different population and since there are two population;
There are also two different means and the variance may be either
Equal or unequal.
11-262
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Case 3. mean wages for a group of females
mean wages for a group of men
each sample is independently drawn
In this case Null hypothesis: Ho : µ(women) = µ(men)
Ha: µ(women) ≠ µ(men)
Above implies Ho: µ(women) - µ(men) = 0
Ha: µ(women) - µ(men) ≠ 0
Writing the above hypothesis is Step 1.
Now we collect samples of the two groups and find out
Their wages. Average wage for females and for men are
Separately calculated and also sample variances.
11-263
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
In this case Null hypothesis: Ho : µ(women) = µ(men)
Ha: µ(women) ≠ µ(men)
Above implies Ho: µ(women) - µ(men) = 0
Ha: µ(women) - µ(men) ≠ 0
Writing the above hypothesis is Step 1.
Now we collect samples of the two groups and find out
Their wages. Average wage for females and for men are
Separately calculated and also sample variances.
Step2:
Step3:
Step4:
Step5:
Decide on the level of significance
Decide whether single tail or two tail test
Decide on the test statistic. Z for large sample and ‘t’ for small sample.
Calculate Z statistic
= {x-bar(women)-x-bar(men) - µ(women) - µ(men) } / S.Error
Recall that in the course on sampling methods we have indicated the
standard error for the difference of two means- independent case.
11-264
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
In this case Null hypothesis: Ho : µ(women) = µ(men)
Ha: µ(women) ≠ µ(men)
Above implies Ho: µ(women) - µ(men) = 0
Ha: µ(women) - µ(men) ≠ 0
Writing
Step2:
Step3:
Step4:
Step5:
the above hypothesis is Step 1.
Decide on the level of significance
Decide whether single tail or two tail test
Decide on the test statistic. Z for large sample and ‘t’ for small sample.
Calculate Z statistic
= {x-bar(women)- x-bar(men) - µ(women) - µ(men) } / S.Error
Recall that in the course on sampling methods we have indicated the
standard error for the difference of two means- independent case.
Step 6: Find Z for alpha level of significance from table
Step 7: Compare Z (cal) with Z(alpha) : If Z (cal) ≥ Z(alpha) reject Ho
Step 8: Conclude your result.
11-265
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Example: A firm is interested to understand whether there is any
difference in the stress level of employees working in
the HR department and in the Marketing department.
A random sample of 30 HR employees were considered
and their stress level measured as 5.36 in a scale of 10
and from a random sample of 40 marketing personnel
showed a stress level of 6.23 in a scale of 10.
At 5% level of significance can be conclude that the
stress levels are different for the different groups.
The variance in the stress levels for HR was 2.3 and that
of marketing was 1.87.
Step 1: Ho : µ(mktg) = µ(HR) µ(mktg) - µ(HR) = 0
Ha: µ(mktg) ≠ µ(HR) µ(mktg) - µ(HR) ≠ 0
11-266
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Example: A firm is interested to understand whether there is any
difference in the stress level of employees working in
the HR department and in the Marketing department.
A random sample of 30 HR employees were considered
and their stress level measured as 5.36 in a scale of 10
and from a random sample of 40 marketing personnel
showed a stress level of 6.23 in a scale of 10.
At 5% level of significance can be conclude that the
stress levels are different for the different groups.
The variance in the stress levels for HR was 2.3 and that
of marketing was 1.87.
Step 1: Ho : µ(mktg) = µ(HR) µ(mktg) - µ(HR) = 0
Ha: µ(mktg) ≠ µ(HR) µ(mktg) - µ(HR) ≠ 0
Step2: Alpha level is specified as 5%
Step 3: This is a two tail test
11-267
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Example: A firm is interested to understand whether there is any
difference in the stress level of employees working in
the HR department and in the Marketing department.
A random sample of 30 HR employees were considered
and their stress level measured as 5.36 in a scale of 10
and from a random sample of 40 marketing personnel
showed a stress level of 6.23 in a scale of 10.
At 5% level of significance can be conclude that the
stress levels are different for the different groups.
The variance in the stress levels for HR was 2.3 and that
of marketing was 1.87.
Step 6: Calculate Z = (6.23-5.66)/ Standard error
11-268
Recall that Standard error for difference of two means
Independent sample case is
 12
n1

 22
n2
If population variance ‘σ’ is unknown use
Unbiased estimator ‘s’ – sample variance.
11-269
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Example: A firm is interested to understand whether there is any
difference in the stress level of employees working in
the HR department and in the Marketing department.
A random sample of 30 HR employees were considered
and their stress level measured as 5.36 in a scale of 10
and from a random sample of 40 marketing personnel
showed a stress level of 6.23 in a scale of 10.
At 5% level of significance can be conclude that the
stress levels are different for the different groups.
The variance in the stress levels for HR was 2.3 and that
of marketing was 1.87.
Step 6: Calculate Z = (6.23-5.36)/ Standard error
= 0.87/ √{(1.87/40) + (2.3/30)}
= 0.87 / 0.35 = 2.486
11-270
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Example: A firm is interested to understand whether there is any
difference in the stress level of employees working in
the HR department and in the Marketing department.
A random sample of 30 HR employees were considered
and their stress level measured as 5.36 in a scale of 10
and from a random sample of 40 marketing personnel
showed a stress level of 6.23 in a scale of 10.
At 5% level of significance can be conclude that the
stress levels are different for the different groups.
The variance in the stress levels for HR was 2.3 and that
of marketing was 1.87.
Step 6: Calculate Z = (6.23-5.36)/ Standard error
= 0.87/ √{(1.87/40) + (2.3/30)}
= 0.87 / 0.35 = 2.486
Step 7: Z table value ( two tail ) 5% alpha = 1.96
Hence Z ( cal) > Z ( table value) Reject Ho.
2.486 > 1.96  reject Ho
11-271
DIFFERENCE OF TWO MEANS – INDEPENDENT SAMPLE
Example: A firm is interested to understand whether there is any
difference in the stress level of employees working in
the HR department and in the Marketing department.
A random sample of 30 HR employees were considered
and their stress level measured as 5.36 in a scale of 10
and from a random sample of 40 marketing personnel
showed a stress level of 6.23 in a scale of 10.
At 5% level of significance can be conclude that the
stress levels are different for the different groups.
The variance in the stress levels for HR was 2.3 and that
of marketing was 1.87.
Step 6: Calculate Z = (6.23-5.36)/ Standard error
= 0.87/ √{(1.87/40) + (2.3/30)}
= 0.87 / 0.35 = 2.486
Step 7: Z table value ( two tail ) 5% alpha = 1.96
Hence Z ( cal) > Z ( table value) Reject Ho.
2.486 > 1.96  reject Ho
Step 8: Conclusion: Stress levels are different for marketing and HR.
11-272
11-273
Difference of two means – independent sample
small sample
If the null hypothesis is accepted, it would imply
that both group sample came from the same population
and hence for the one population there can be only
one mean and one variance.
However, if the null hypothesis is rejected it implies that
Both group sample belongs to different population and
Therefore for each of the population there will be
Different mean. But we assume that the two groups
Has the same variance. That is homogeneity of variances
Is assumed. When such assumption is made the standard
Error can be recalled as :
11-274
Difference of two means – independent sample
small sample
( homogeneity of variance)
Recall that in the case of small sample for independent
Samples case; the standard error was calculated
Using the pooled estimates as follows:
standard error = s(pooled)√{1/n1+1/n2)
Where s^2(pooled) = {(n1-1)s1^2} +{(n2-1)s2^2}
(n1+n2-2)
Or s(pooled) = √ s^2 (pooled)
11-275
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
Step -1:
Ho: µ(a) = µ(b)
Ha: µ(a) ≠ µ(b)
Step 2: Level of significance is known as 5%
11-276
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
Step -1:
Ho: µ(a) = µ(b)
Ha: µ(a) ≠ µ(b)
Step 2: Level of significance is known as 5%
Step 3: This is a two tail test as the question is
only asking if the life of the two brands
of batteries are different.
11-277
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
Step 4: Choosing the test statistic. Since the sample
size is 8 and 11 respectively ( small ) a
t-statistic is used to infer the hypothesis.
Step 5: Calculate the ‘t’ statistic:
t= (mean for B/A – mean for B/B)- 0
standard error
11-278
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
Step 5..contd: mean life for brand A = 39.625
mean life for brand B = 42.18182
variance for brand A = 7.125
variance for brand B = 11.3636
pooled variance = 9.6183
t= (39.625-42.18182)-0 / 3.1013√(1/8+1/11)
= -2.55682 / 1.44 = -1.7742
11-279
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
Step 5
t= (39.625-42.18182)-0 / 3.1013√(1/8+1/11)
= -2.55682 / 1.44 = -1.7742
absolute value = 1.7742
Step 6: tabulated value of ‘t’ for 5% alpha at 17 df = 2.1098
Step 7: Compare absolute values: ‘t’ (cal) < ‘t’(tablulated)
hence unable to reject the null hypothesis
11-280
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
Step 5 t= (39.625-42.18182)-0 / 3.1013√(1/8+1/11)
= -2.55682 / 1.44 = -1.7742 , absolute value = 1.7742
Step 6: tabulated value of ‘t’ for 5% alpha at 17 df = 2.1098
Step 7: Compare absolute values: ‘t’ (cal) < ‘t’(tablulated)
hence unable to reject the null hypothesis
Step 8: Conclusion: Hence the mean life of batteries of both the
brands are similar and hence both vendors
can be considered for selection based on
other considerations such as price, delivery etc.
11-281
Worked example: independent small sample ‘t’ - test
A car manufacturer is intending to procure batteries for its given
Model from two different vendors. However before procuring they
Wish to know if the life of the two batteries would be similar. A
Sample of batteries from both the manufacturers are selected
Radomly and the life ( in months) was found as follows:
Brand A: 38, 37, 42, 44, 36, 39, 40, 41
Brand B: 42, 41, 37, 39, 40, 43, 44, 45, 46, 48, 39
Is there reason to believe that the life of batteries of the two
brands are different ? Use 5% level of significance.
One of the assumptions made to solve this problem
Is that the population variances are equal even if
The alternative hypothesis is accepted. However, we
Have not checked this aspect. Hence it is necessary
To do this check this aspect which we shall take up
Now.
11-282
CHECKING FOR HOMOGENEITY OF POPULATION VARIANCE
HOMOGENEITY OF POPULATION VARAINCE IS CHECKED BY
CARRYING A HYPOTHESIS TEST WHICH AS GIVEN BELOW:
STEP 1: Ho: σ1^2 = σ2^2
Ha: σ1^2 ≠ σ2^2
Step 2: Decide the level of significance : assume 5%
Step 3: this is a two tail test based on the alternative hypothesis
Step 4: Decide on the test statistic:
For this test which is a ratio of the two sample variances
is the ‘F’ test also known as Fisher’s Test:
Step 5: Calculate ‘F’ statistic = s1^2/ S2^2
for the previous problem it is
= 7.125 / 11.3636 = 0.627
11-283
CHECKING FOR HOMOGENEITY OF POPULATION VARIANCE
HOMOGENEITY OF POPULATION VARAINCE IS CHECKED BY
CARRYING A HYPOTHESIS TEST WHICH AS GIVEN BELOW:
STEP 1: Ho: σ1^2 = σ2^2
Ha: σ1^2 ≠ σ2^2
Step 2: Decide the level of significance : assume 5%
Step 3: this is a two tail test based on the alternative hypothesis
Step 4: Decide on the test statistic:
For this test which is a ratio of the two sample variances
is the ‘F’ test also known as Fisher’s Test:
Step 5: Calculate ‘F’ statistic = s1^2/ S2^2
for the previous problem it is
= 7.125 / 11.3636 = 0.627
Step 6: Read the table value for ‘F’ from the table . This requires
degree of freedom for the numerator and denominator
which is (n1-1) and (n2-1) i.e. 7 and 10 respectively.
11-284
F- table – how to read.
The value of F when the blue shaded portion
Is 0.975 , we take the reciprocal of F value
Of 0.025 with degree of freedom interchanged
Hence F ( 0.025) with 10,7 df = 4.76
And 1/4.76 = 0.21.
11-285
11-286
CHECKING FOR HOMOGENEITY OF POPULATION VARIANCE
HOMOGENEITY OF POPULATION VARAINCE IS CHECKED BY
CARRYING A HYPOTHESIS TEST WHICH AS GIVEN BELOW:
STEP 1: Ho: σ1^2 = σ2^2
Ha: σ1^2 ≠ σ2^2
Step 2: Decide the level of significance : assume 5%
Step 3: this is a two tail test based on the alternative hypothesis
Step 4: Decide on the test statistic:
For this test which is a ratio of the two sample variances
is the ‘F’ test also known as Fisher’s Test:
Step 5: Calculate ‘F’ statistic = s1^2/ S2^2
for the previous problem it is
= 7.125 / 11.3636 = 0.627
Step 6: Read the table value for ‘F’ from the table . This requires
degree of freedom for the numerator and denominator
which is (n1-1) and (n2-1) i.e. 7 and 10 respectively.
Step 7: Now we can see that ‘F’ statistic calculated is in between
table value of 0.21 and 3.95. Accept Ho.
Step 8 : hence homogeneity of variance is established.
11-287
11-288
Difference of two proportions
A candidate who was interested in filing his nomination papers for an
Election wanted to understand whether his popularity in two adjacent
Constituency was equally population or he had more popularity in a
Particular constituency. He then availed the services of a research agency
To check his popularity in the two constituency.
In this problem we would not be able to check the
Difference of two means but difference of two proportions.
Two independent samples will be drawn from the two
Constituency and find out how many support his candidature.
We can take a worked example to explain this test.
11-289
Difference of two proportions
A candidate who was interested in filing his nomination papers for an
Election wanted to understand whether his popularity in two adjacent
Constituency was equally population or he had more popularity in a
Particular constituency. He then availed the services of a research agency
To check his popularity in the two constituency. Random samples were
Drawn from each constituency A and B and preference for his candidature
was measured by a survey. In constituency A – sample size 800 and 390
Favored him and in Constituency B – sample size 900 and 490 favored him
Can we say that Constituency B is favorable for this candidate. Use 5%
Level of significance.
Step 1: Ho: p(b) = p(a)
Ha: p(b) > p(a)
Step 2: level of alpha is given as 5%
Step 3: this is a single tail test which is based on the question
Step 4: Choose the test statistic: Large sample and hence Z can be used
11-290
Difference of two proportions
A candidate who was interested in filing his nomination papers for an Election
wanted to understand whether his popularity in two adjacent Constituency was
equally population or he had more popularity in a Particular constituency. He
then availed the services of a research agencyTo check his popularity in the two
constituency. Random samples were Drawn from each constituency A and B and
preference for his candidature was measured by a survey. In constituency A –
sample size 800 and 390 Favored him and in Constituency B – sample size 900
and 490 favored him Can we say that Constituency B is favorable for this
candidate. Use 5% Level of significance.
Step 1: Ho: p(b) = p(a)
Ha: p(b) > p(a)
Step 2: level of alpha is given as 5%
Step 3: this is a single tail test which is based on the question
Step 4: Choose the test statistic: Large sample and hence Z can be used
Step 5: Calculate Z = {p(b)-p(a) }-0 / standard error
for difference of two
proportion
11-291
Difference of two proportions
That the standard error for the difference of two proportions
Is given by
p1(1 - p1) + p2(1 - p2)
n1
n2
Hence in this problem p(a) = 390 / 800 = 0.4875
or 48.75%
p(b) = 490/900 = 54.44%
Hence standard error
= √{48.75x51.25/800} x {(54.44x45.56/900}
= 2.4246
11-292
Difference of two proportions
A candidate who was interested in filing his nomination papers for an Election
wanted to understand whether his popularity in two adjacent Constituency was
equally population or he had more popularity in a Particular constituency. He
then availed the services of a research agencyTo check his popularity in the two
constituency. Random samples were Drawn from each constituency A and B and
preference for his candidature was measured by a survey. In constituency A –
sample size 800 and 390 Favored him and in Constituency B – sample size 900
and 490 favored him Can we say that Constituency B is favorable for this
candidate. Use 5% Level of significance.
Step 1: Ho: p(b) = p(a)
Ha: p(b) > p(a)
Step 2: level of alpha is given as 5%
Step 3: this is a single tail test which is based on the question
Step 4: Choose the test statistic: Large sample and hence Z can be used
Step 5: Calculate Z = {p(b)-p(a) }-0 / standard error for porportion
= (54.44-48.75) -0 / 2.4246
= 2.347
Step 6: Read the table value of Z at 5% alpha, single tail = 1.645
Step 7: Compare Z ( cal) with Z ( table value ) 2.347> 1.645 reject Ho.
11-293
Difference of two proportions
A candidate who was interested in filing his nomination papers for an Election
wanted to understand whether his popularity in two adjacent Constituency was
equally population or he had more popularity in a Particular constituency. He
then availed the services of a research agencyTo check his popularity in the two
constituency. Random samples were Drawn from each constituency A and B and
preference for his candidature was measured by a survey. In constituency A –
sample size 800 and 390 Favored him and in Constituency B – sample size 900
and 490 favored him Can we say that Constituency B is favorable for this
candidate. Use 5% Level of significance.
Step 1: Ho: p(b) = p(a)
Ha: p(b) > p(a)
Step 2: level of alpha is given as 5%
Step 3: this is a single tail test which is based on the question
Step 4: Choose the test statistic: Large sample and hence Z can be used
Step 5: Calculate Z = {p(b)-p(a) }-0 / standard error for porportion
= (54.44-48.75) -0 / 2.4246
= 2.347
Step 6: Read the table value of Z at 5% alpha, single tail = 1.645
Step 7: Compare Z ( cal) with Z ( table value ) 2.347> 1.645 reject Ho.
Step 8: Conclusion: Constituency B is more popular than A for this candidate.
11-294
Analysis of variance
Consider the following research question:
Three different types of seeds are sown in a exactly similar
Types of soil and the same type of fertilizer is added for
Each of the type of plants. The yield for the product is
As follows:
Yield ( million tons)
seed A
seed B
Seed C
plot-1
8
12
9
plot-2
9
10
10
plot-3
10
10
9
plot-4
9
13
8
plot-5
9
11
8
If all the seed varieties are similar they should given
Similar yield and if some of them are superior then
One or more type would give a larger yield.
11-295
Analysis of variance
Consider the following research question:
Three different types of seeds are sown in a exactly similar Types of soil and the
same type of fertilizer is added for Each of the type of plants. The yield for the
product is As follows: Yield ( million tons)
seed A
seed B
Seed C
plot-1
8
12
9
plot-2
9
10
10
plot-3
10
10
9
plot-4
9
13
8
plot-5
9
11
8
Step 1: Ho: µa = µb = µc
Ha: at least two means are unequal
In this case a direct comparison is not feasible as which
two can be compared. You need 3 different comparisons
A with B , A with C and B with C.
In this case the type I error would be very large . Hence
We must adopt another method.
11-296
Basics of ANOVA
11-297
Variance calculation
A) Calculate the correction factor (c/f)= GT^2/Total sample
B) Calculate the total sum of squares ( TSS)
= square each value and sum it up – c/f
C) Calculate sum of squares between samples (SSB)
= (total for seedA)^2/n(a) + (total for seedB)^2/n(B)+
(total for seedC)^2/n(c) – c/f
In the problem stated above the values for each of these
Are as follows:
c/f = (145^2/15)= 1401.66
TSS= 1431-1401.66= 29.34
SSB= 1419.4-1401.66 = 17.74
Now we can construct the ANOVA table
11-298
ANOVA table ( step 2)
Source of Sum of
variation squares
Degree Mean
of
square
freedom
F (cal)
Between SSB
treatment
K-1
MSB/
MSW
Within
SSW
treatment
n(a)+n( MSW =
b)+n(c)- SSW/df
k
Total
n(a) +
n(b) +
n(c)-1
TSS=
SSB+SSW
MSB=
SSB/df
F ( table
value)
11-299
ANOVA table ( filled in table )
Source of Sum of
variation squares
Degree Mean
of
square
freedom
F (cal)
F ( table
value)
Between 17.74
treatment
2
8.866
9.178
6.93
Within
11.6
treatment
12
0.966
Total
14
29.34
Mean Square is the variance .
Hence MSB =variance between seeds
MSW= variance within seeds
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ANOVA
Step 2: Level of alpha is to be decided
Step 3: This is always a single tail test.
Step 4: Since ratio of variances are being considered
hence it is an ‘F’ – test of Fisher’s test.
Step 5: Calculate the F( cal) as stated earlier
Step 6: Read the F – table value for alpha level and
df for between and within
Step 7: Compare F(cal) with F(table value)
If F(cal) ≥ F(table value )  Reject Ho
Step 8: Conclusion.
11-301
ANOVA table ( filled in table )
Source of
variation
Sum of
squares
Degree of
freedom
Mean
square
F (cal)
F ( table
value)
Between
treatment
17.74
2
8.866
9.178
6.93
Within
treatment
11.6
12
0.966
Total
29.34
14
Hence F(cal) > F( table value)  Reject Ho
Conclusion: The three types of seeds do not give the
same yield.
11-302
Significance testing for correlation coefficient
Let us recall from the previous course on Sampling methods
Where we had calculated the correlation coefficient based
On sample information.
The sample information contained only a few values for
The independent variable and a corresponding few values
For the dependent variable. Hence if the correlation exists
For these values, then how can we be sure that if all the
Values in the population are known, then a correlation will
Exist. To answer this question, it is necessary to test
The significance of the correlation coefficient.
The procedure is discussed below:
11-303
Significance testing for correlation coefficient
Step 1. Define the Null and Alternative hypothesis:
Ho: ρ = 0
Ha: ρ ≠ 0
ρ = population correlation
Step 2: Decide on the level of alpha ( type I error)
let us say it is 5%
Step 3: This is a two tail test based on the sign of the
alternative hypothesis.
Step 4: Decide on the test statistic:
Since the sample is usually small we use a
‘t’ – test.
11-304
Significance testing for correlation coefficient
Step 1. Define the Null and Alternative hypothesis:
Ho: ρ = 0
Ha: ρ ≠ 0
ρ = population correlation
Step 2: Decide on the level of alpha ( type I error) let us say it is 5%
Step 3: This is a two tail test based on the sign of the alternative hypothesis.
Step 4: Decide on the test statistic: Since the sample is usually small we use
a ‘t’ – test.
Step5: t= (r-ρ) / standard error for correlation
standard error = √{(1-r^2)/(n-2)}
where ‘n’ = number of pairs (x,y) of sample data
Step6: Read table value of ‘t’ for significance level, and
degree of freedom ( n-2)
11-305
Significance testing for correlation coefficient
Step 1. Define the Null and Alternative hypothesis:
Ho: ρ = 0
Ha: ρ ≠ 0
ρ = population correlation
Step 2: Decide on the level of alpha ( type I error) let us say it is 5%
Step 3: This is a two tail test based on the sign of the alternative hypothesis.
Step 4: Decide on the test statistic: Since the sample is usually small we use
a ‘t’ – test.
Step5: t= (r-ρ) / standard error for correlation
standard error = √{(1-r^2)/(n-2)}
where ‘n’ = number of pairs (x,y) of sample data
Step6: Read table value of ‘t’ for significance level, and df ( n-2)
Step 7: Compare ‘t’(cal) with ‘t’ (table value)
If t(cal) ≥ t(table value) Reject Ho
Step 8: Conclude whether the correlation exists in the
population
11-306
Worked example: test for correlation
Consider a sample data collected on the number of
Hours study done by student and the marks obtained
by student. Data is as follows:
Student
hours of study /day
marks obtained(%)
a
12
63
b
10
68
c
8
53
d
9
60
e
15
75
f
14
80
g
11
68
h
13
53
11-307
Worked example: test for correlation
Consider a sample data collected on the number of
Hours study done by student and the marks obtained by student. Data is as follows:
Student
hours of study /day
marks obtained(%)
a
12
63
b
10
68
c
8
53
d
9
60
e
15
75
f
14
80
g
11
68
h
13
53
Correlation coefficient ‘r’ ( sample) = 0.613
You can refer to the lectures on Sampling methods for
Getting the details of how to calculate this value.
11-308
Worked example: test for correlation
Step 1: Ho: ρ = 0
Ha: ρ ≠ 0
Step 2: Assume alpha level is 5%
Step 3: This is a two tail test
Step 4: Decide test statistic which is ‘t’ in this case
Step 5: calculate ‘t’ = (0.613-0)/√(1-0.613^2)/6
= 0.613/0.3225 = 1.90
Step 6: ‘t’ ( table value) at 5% two tail , df=6
= 2.447
Step7: Compare ‘t’ ( cal) with ‘t’ (table value)
1.90 < 2.447  Accept Ho
Step 8: Conclusion: There is no significant correlation
between number of hours of study
and marks obtained.
11-309
Worked example: test for correlation
Step 1: Ho: ρ = 0
Ha: ρ ≠ 0
Step 2: Assume alpha level is 5%
Step 3: This is a two tail test
Step 4: Decide test statistic which is ‘t’ in this case
Step 5: calculate ‘t’ = (0.613-0)/√(1-0.613^2)/6 = 0.613/0.3225 = 1.90
Step 6: ‘t’ ( table value) at 5% two tail , df=6 = 2.447
Step7: Compare ‘t’ ( cal) with ‘t’ (table value) 1.90 < 2.447  Accept Ho
Step 8: Conclusion: There is no significant correlation between number of hours of
study and marks obtained.
The above example clearly brings out that even though
There is non zero correlation in the sample data but we
Cannot conclude in the population that a correlation exists.
If the correlation had been a larger value or if the sample
Size had been larger and then for the same correlation
Coefficient we may have concluded that a correlation exists
In the population. This is important to understand
11-310
Testing for the Regression coefficient
Recall that we had developed a regression equation to
Estimate the dependent variable (y) if we know the value
Of the independent variable (x) .
y= a + bx
where y= dependent variable
x = independent variable
a = intercept
b = regression coefficient (also known as slope)
Just like correlation, the regression coefficient is based on
Sample data only and in order to use the same to estimate
The regression coefficient in the population we need to test
Its significance. Population equation would be given as
Y = ßo + ß1(X) where ß1= regression coefficient
in the population
11-311
Testing for the Regression coefficient
Step1: Null and hypothesis:
Ho: ß1 = 0
Ha: ß1 ≠ 0
Step2: Decide on the level of alpha
Step3: This is a two tail test based on the sign of Ha
Step4: Decide on the test statistic . Usually ‘t’ because
the sample size would be small
Step5: Calculate ‘t’ statistic = (b1-ß1)/standard error(b1)
Step6: Read the table value of ‘t’ for alpha level and df
df = n-2
Step7: Compare ‘t’ (cal) with ‘t’ (table value)
If ‘t’ (cal) ≥ ‘t’ (table value)  Reject Ho
Step8: Conclusion.
11-312
Worked example for testing of regression coefficient
An icecream vendor wants to determine the sale of his product
based on the maximum temperature during the day. He collects
data which are as follows:
(y) sales(kgs) 223, 252, 230, 195, 185, 170, 272, 222, 215, 235
(x) Temp(c) 27, 30, 31, 28, 26, 23, 32, 29, 28, 30
The equation developed for this
y= a + b(x)
where a = -76.57 you can use the formula’s
b = 10.44 given in the subject
on sampling methods
11-313
Understanding the regression equation
We first need to understand what is the regression equation
That we have developed.
Getting a regression equation only means that we have
minimized the error in estimating the value of ‘y’ but not
made it zero.
Therefore for the problem stated earlier we can calculate what
Would be the error that would be made if we use the equation
Developed and what would be the error if we did not know
The equation. If we did not know the equation we would
Have used the mean value of ‘y’ to estimate ‘y’
Then the error would have been ∑{y-y(bar)}^2
If we know the equation then the error would be
∑{(y(actual) – y(est)}^2
let us calculate both these values for the problem just given
11-314
Understanding the regression equation
Total error =∑{y-y(bar)}^2
= 8440.9
If we know the equation then the error would be
Error if equation is known
∑{(y(actual) – y(est)}^2
= 1640.86
This means that an error of Total error – Error still there
8440.9-1640.86 = 6800.04
has been explained because of the regression equation.
11-315
Understanding the regression equation
Source of
error
Sum of
squares of
error
df
Mean
square
F(cal)
F( table
value)
Due to
regression
6800.04
1
6800.04
33.15
11.26
Error still
remaining
(residual
error)
1640.86
8
205.10
Total Error
8440.9
Mean
square
error
9
Error variance
11-316
Understanding the regression equation
Square root of the mean square error which is
Gives the standard error of the estimate (Se)
Se = √ 205.1 = 14.32.
Standard error for (b1) = Se / √{∑(x-x(bar))^2}
∑(x-x(bar))^2} = 62.4
√ 62.4 = 7.89
Standard error (b1) = 14.32 / 7.89 = 1.814
11-317
Worked example for testing of regression coefficient
An icecream vendor wants to determine the sale of his product
based on the maximum temperature during the day. He collects
data which are as follows:
(y) sales(kgs) 223, 252, 230, 195, 185, 170, 272, 222, 215, 235
(x) Temp(c) 27, 30, 31, 28, 26, 23, 32, 29, 28, 30
Step 1: Ho: =0
Ha: ß1≠0
Step2: Assume an alpha level of 5%
Step3: This is a two tail test.
Step4: Decide on the test statistic :- ‘t’ in this case
Step5: Calculate ‘t’ = (b-ß1) / standard error (b1)
(10.44-0) / 1.814 = 5.755
Step 6: Read table value ‘t’ , 5% alpha, df= 8
= 2.306
11-318
Worked example for testing of regression coefficient
An icecream vendor wants to determine the sale of his product
based on the maximum temperature during the day. He collects
data which are as follows:
(y) sales(kgs) 223, 252, 230, 195, 185, 170, 272, 222, 215, 235
(x) Temp(c) 27, 30, 31, 28, 26, 23, 32, 29, 28, 30
Step 1: Ho: =0
Ha: ß1≠0
Step2: Assume an alpha level of 5%
Step3: This is a two tail test.
Step4: Decide on the test statistic :- ‘t’ in this case
Step5: Calculate ‘t’ = (b-ß1) / standard error (b1)
=(10.44-0) / 1.814 = 5.755
Step 6: Read table value ‘t’ , 5% alpha, df= 8 = 2.306
Step7: ‘t’(cal) > ‘t’(table value)  Reject Ho
Step8: The regression coefficient calculated in the sample is
significant and can be used to find the interval estimate in for the
population parameter.
11-319
References used for this subject
1. Statistics for Business and Economics, 5th edition,
Paul Newbold, William Carlson & Betty Thorne
Prentice Hall Publication.
2. General Statistics : Warren Chase and Fred Bown
John Wiley Publication
3. Marketing Research, 5th edition, Naresh Malhotra,
Pearson Education Publication
4. Statistics for Business and Economics, 8th edition,
Anderson, Sweeney & Williams.
Thomson South-Western Publication
5. Complete Business Statistics, 6th edition,
Azcel & Sounderpandian
Tata McGraw Hill publication.
6. Applied Statistics in Business & Economics,
David P. Doane & Lori E. Seward : Tata McGraw Hill