د حاتم البيطار استشاري وجراح الفم والاسنان 01005684344 اتصل للحجز بالعيادة Dr.Hatem EL-Bitar

INTENSE PULSED LIGHT
AND ROSACEA

 Rosacea is a chronic acneiform disorder
affecting both the skin and eye, it is a
syndrome of undetermined etiology
characterized by both vascular and
papulopustular component involving the face
and occasionally the neck and upper trunk.

 Presentation
– Facial erythema.
– Telangectasia.
– Papules.
– Pustules.
– Sebaceous gland hypertrophy (Rhinophyma).
– Ocular lesions.

 IPL is an effective treatment for the signs
and symptoms of rosacea and
represents a new category of therapeutic
options for the rosacea patients.

 Flash lamp emit every wave length of
light in the visible spectrum and a little
into the band of IR radiation up to 1200
nm.

 Pulsed dye laser (Disadv.)
–Small spot size.
–Purpura.
Time consuming.
Period of down time.

Why IPL
 Wave length in not fixed (wide range).
 Large spot size (ttt of entire face).
 Purpura is uncommon (no down time
procedure).
 Minimal risk of scarring.

 The light energy passes through the
epidermis and is absorbed by oxy
hge. by selective photo-thermolysis.

Fungal Infections
General Properties of Fungi

Structural-functional
relationships
The fungi are more evolutionarily
advanced forms of
microorganisms.
Fungi can be divided into two basic
morphological forms, yeasts and
hyphae.

 Yeasts are unicellular fungi which
reproduce asexually by blastoconidia
formation (budding) or fission.
 Hyphae are multi-cellular fungi which
reproduce asexually and/or sexually.

 Dimorphism is the condition where by a
fungus can exhibit either the yeast form or
the hyphal form, depending on growth
conditions. Very few fungi exhibit
dimorphism. Most fungi occur in the hyphae
form as branching, threadlike tubular
filaments.

 These filamentous structures either lack
cross walls (coenocytic) or have cross
walls (septate) depending on the species. In
some cases septate hyphae develop clamp
connections at the septa which connect the
hyphal elements.

A.Yeast cells reproducing by blastoconidia formation;
B. Yeast dividing by fission;
C. Pseudohyphal development;
D. Coenocytic hyphae;
E. Septate hyphae;
F. Septate hyphae with clamp connections

 The presence/absence of conidia and their
size, shape and location are major features
used in the laboratory to identify the species
of fungus in clinical specimens.

Metabolism
 All fungi are free living, i.e., they are not
obligate intracellular parasites.

For medical purposes the important
aspects of fungal metabolism are:
1. The synthesis of chitin and other compounds,
for use in forming the cell wall. These
induce immune hypersensitivity.

 2. The synthesis of ergosterol for incorporation
 into the plasma membrane.
 This makes the plasma membrane sensitive to
those
antimicrobial agents which either block the
 synthesis of ergosterol or prevent its incorporation
 into the membrane or bind to it, e.g. amphotericin
B.

 3. The synthesis of toxins such
as Aflatoxins.
 - these are carcinogens produced by
Aspergillus flavus when growing on grain.
When these grains are eaten by humans or
when they are fed to dairy cattle and they
get into the milk supply, they affect humans.

 4. The synthesis of proteins on ribosomes
that are different from those found in
bacteria.
 This makes the fungi immune to those
antimicrobial agents that are directed
against the bacterial ribosome,
 e.g. chloramphenicol.

Classification of Fungal Infections
1. Superficial mycoses
2. Cutaneous mycoses
3. Subcutaneous mycoses
4. Systemic mycoses
5. Opportunistic mycoses

 In general, humans have a high level of
innate immunity to fungi and most of the
infections they cause are mild and self-
limiting.


 This resistance is due to:
 Fatty acid content of the skin
pH of the skin, mucosal surfaces and body
fluids
Epithelial turnover
Normal flora
Transferrin
Cilia of respiratory tract

Response of bone to inflammation

Bone as an organ is composed of:•
1- Bone tissue: (Described in microscopic terms and defined by the relation of
its collagen fibers and mineral structure to the bone cells)
2- Cartilage.
3-Fat marrow elements.
4- Vessels.
5-Nerves.
6- Fibrous tissue.

• Macroscopically two types of bone are recognized to:
1- Cortical bone: This is dense compact bone
whose outer shell defines the shape of the bone.
2- Coarse cancellous bone: This is also termed: spongy, trabecular
or marrow bone.
Generally, found at the ends of long bones
within the medullary canal (medullary space).
Note:
Changes in the rate of bone turnover (resorption / deposition and formation)
are manifested principally in cancellous bone
because the cancellous bone has a high surface-to-volume ratio
which means it contains many more bone cells per unit volume than does the
cortical bone.

All bones contain both cancellous and cortical elements but their proportions differ.•
For example:
The skull is formed by outer and inner tables of compact bone, with only
a small amount of cancellous bone within the marrow space called
diploë.

Periosteum:
It is a specialized connective tissue that covers the outer surface of all bones
and is capable of forming bone.
Endosteum:
It lines the bone marrow cavity and it is made of one layer of osteogenic cell.

The bone matrix
It is composed of:
1- Mineralized (inorganic) phase:
Mineralized matrix is composed of hydroxyapatite crystals and other ions as
carbonate, citrate, fluoride, chloride, sodium, magnesium, potassium and strontium.
2- Organic matrix:
Type I collagen and other proteins as:
a- Osteocalcin:
It is protein produced by osteoblast.
Blood levels of this protein serve as a useful marker of bone formation.
b- Osteopontin and Sialoprotein:
They are other bone matrix proteins that help anchor cells to the bone matrix.

3- Cells:
They are Osteoprogenitor, Osteoblast, Osteocyte and Osteoclast.
a- Osteoprogenitor cells:
can differentiate into Osteoblasts and Osteocytes. They
They give rise to osteoblasts.
They are found in the inner layer of periosteum and endosteum.
b- Osteoblasts:
They are bone forming cells as their cytoplasm contains alkaline phosphatase
which is responsible for bone deposition.
They are found in the inner layer of periosteum and in endosteum.

c- Osteocyte:
It is the mature bone cell i.e. It is the osteoblast which is completely embedded in
bone matrix and isolated in a lacuna.
Osteocytes maintain the hardness of the matrix by continuous deposition of
calcium salts.
d- Osteoclasts:
They are bone resorptive cells.
They originate from blood monocytes.
They are found on the surfaces of bones in small depressions termed Howship's
lacunae.

Bone marrow
It resides in the space enclosed by the cortical bone termed the marrow space or the
medullary canal.
It is supported by a delicate connective tissue framework that includes marrow cells
and blood vessels.
They are three types of marrow:
a- Red marrow:
It corresponds to hemopoietic tissue.
b- Yellow marrow:
It appears microscopically as fat tissue.
c- Gray or White marrow:
It is deficient in hemopoietic elements and it is often fibrotic.
Note: It is always a pathological tissue in a nongrowing adult bone.

The blood supply of bone marrow is through:
a- Haversian canals which are spaces in the bone of the cortex that course
parallel to the long axis of the bone and then they branch and communicate with
other similar canals.
Each canal contains one or two blood vessels, lymphatics and some nerve fibers.
b- Volkmann's canals which are spaces within the cortex that run perpendicular
to the long axis of the cortex to connect adjacent Haversian canals.
They also contain blood vessels.
Note:
Each artery has its paired vein and perhaps free nerve endings.
Drainage of the veins proceeds from the cortex outward to the periosteal veins.

Microscopic organization of bone tissue
Microscopic examination reveals two types of bone tissues which are:
1- Lamellar bone.
2- Woven bone.
Both varieties may be mineralized or unmineralized.
termed osteoid.

1- Lamellar bone
It is produced slowly and it is highly organized.
It forms the adult skeleton.
Anything other than lamellar bone in the adult skeleton is abnormal.
It is defined by three characteristics which are:
1- Type I collagen fibers have a parallel arrangement.
2- There are few osteocytes in the matrix.
3- Uniform osteocytes in lacunae parallel to the long axis of the collagen fibers.

Types of lamellar bone:
There are four types of lamellar bone which are:
1- Circumferential bone:
It forms the outer periosteal and the inner endosteal lamellar envelopes of the cortex.
2- Concentric lamellar bone:
It is arranged around the Haversian canals.
Haversian system Concentric lamellar bone + Haversian artery and vein inside
haversian canal.
Constitute osteon which compose the haversian system.

3- Interstitial lamellar bone:
It represents the remnants of either circumferential or concentric lamellar bone,
which have been remodeled and are wedged between the osteons.
4- Trabecular lamellar bone:
It forms the coarse cancellous bone of the medullary cavity.
It exhibits the plates of lamellar bone perforated by marrow spaces.
With the exception of trabecular bone, lamellar bone is found in the cortex.

2- Woven bone
It is defined by:
1- Type I collagen fibers have an irregular arrangement, hence the term woven.
2- There are numerous osteocytes in the matrix.
3- There is variation in the size and shape of the osteocytes.
It is more rapidly deposited than lamellar bone.
It is haphazardly arranged and of low tensile strength.
Collagen fibers of
woven bone

Sites:
1- In the areas surrounding tumors and infections.
2- As a part of the healing fracture.
3- In the developing fetus.
Note:
The presence of woven bone in the adult skeleton always represents a pathological
condition and indicates that reactive tissue has been produced in response to some
stress in the bone.

Osteomyelitis
It is the inflammation of bone and bone marrow.
But it is really an inflammation of the soft parts of bone which are the contents of the
medullary cavity and the Haversian canals and the periosteum.
The inflammation of bone may be:
Osteitis:
It is the inflammation of bone cortex.
ex. It occurs in the case of non-specific chronic osteomyelitis.
Chronic Focal Diffuse
Suppurative Sclerosing Sclerosing
Osteomyelitis Osteomyelitis Osteomyelitis

Periosteitis:
It is the inflammation of periosteum.
ex. Garre's Osteomyelitis.
The inflammation of the alveolar bone may be occur as Periodontitis.

Osteomyelitis
Etiology:
The inflammatory lesions in bone are caused mainly by bacteria, rarely by fungi.
The most common bacterial causative agents are Staphylococcus aureus, Escherichia
coli, Pseudomonas and Klebsiella. They can cause the infection.

Infection
Direct infection Indirect infection
ex. ex.
1- By penetrating wounds. 1- Direct spread from an adjacent
2- By penetrating fractures. infection.
ex. Open fracture of jaw. ex. Periapical infections as:
3- By penetrating surgery. a. Abscess e.g. infection of the jaw
4- By penetrating trauma. from a dental abscess.
5- By penetrating gunshot injuries. b. Infected granuloma.
c. Infected cyst.
2- Spread of infection following
extraction of an infected tooth without
antibiotic coverage.
3- Hematogenous spread of bacteria
from a distant focus or sepsis.
4- Invasion of bone from adjacent septic
arthritis or soft tissue abscesses.

Osteomyelitis may be acute to start with and may become chronic or it is a chronic
inflammation like tuberculosis from its inception.
Classification:
Osteomyelitis
Acute Osteomyelitis Chronic Osteomyelitis
Specific Chronic Nonspecific Chronic
Osteomyelitis Osteomyelitis
Usually due to nonspecific mixed infections

Acute Osteomyelitis
Definition:
It is a type of osteomyelitis.
It is a boil in a bone.
The calcified portion takes no active part in the process, but it suffers secondarily
from the loss of blood supply and a greater or less portion may die.
Acute Suppurative Osteomyelitis:
It is a serious form of diffusely spreading acute inflammation of the bone that
often causes extensive tissue necrosis.

Acute Osteomyelitis
Histopathology:
The metaphyseal area is susceptible to the acute osteomyelitis
Because of the unique vascular supply in this region.
Normally, arterioles enter the calcified portion of the growth plate from a
loop.
And then drain into the medullary cavity without establishing a capillary
bed.
This loop system permits slowing and sludging of blood flow.
Thereby allowing bacteria time to penetrate the walls of the blood
vessels and to establish an infective focus within the marrow.
This initiates an acute inflammatory response with exudation of protein-
rich fluid and neutrophil polymorphs.

If the organism is virulent and continues to proliferate, it increases pressure on the
adjacent thin-walled vessels because they lie in a closed space which is the marrow
cavity of bone.
This pressure further compromises the vascular supply in this region and
produces bone necrosis.
By allowing further bacterial proliferation, the necrotic areas coalesce into an
avascular zone.
Pus and bacteria extend into the endosteal vascular channels that supply the cortex
and spread throughout the Volkmann and Haversian canals of the cortex.

A sinus tract that extends from the cloaca to the skin may become epithelialized by
epidermis that grows into the sinus tract, so the sinus tract invariably remains
open, continually draining pus, necrotic bone and bacteria.
Eventually, pus forms underneath the periosteum, shearing off the perforating arteries
of the periosteum and further devitalizing the cortex.
The pus flows between the periosteum and the cortex, isolating more bone from its
blood supply and may even invade the joint.
Eventually, the pus penetrates the periosteum and the skin forming a draining sinus.

Periosteal new bone formation and reactive bone formation in the marrow tend to
wall off the infection.
At the same time, the osteoclastic activity resorbs the bone.
If the infection is virulent, this attempt to contain it is overwhelmed and the infection
races through the bone with virtually no bone formation but rather extensive bone
necrosis.
More commonly, pluripotential cells modulate into osteoblasts in an attempt to wall
off the infection.

Several lesions may develop. These lesions are:
1- Cloaca.
2- Sequestrum.
3- Brodie abscess.
4- Involucrum.

1- Cloaca:
It is the hole formed in the bone during the formation of a draining sinus.
2- Sequestrum:
It is a fragment of necrotic bone that is embedded in the pus.
It is separated from the living bone by the action of the osteoclasts.

4- Involucrum:
It is a lesion in which the periosteal new bone formation forms a sheath around the
necrotic sequestrum because some of the cells of the osteogenic layer usually
survive and when the acuteness of the infection is past these osteoclasts lay down
new bone over the sequestrum in the form of a new case or involucrum.
3- Brodie abscess:
It consists of reactive bone from the periosteum and the endosteum, which
surrounds and contains the infection.

1- Throbbing or intense local pain which is the primary feature of this
inflammatory process with tenderness over the accepted region.
2- Pyrexia (high fever).
3- Some tissue swelling.
4- Rapid pulse.
5- ESR (Erythrocyte Sedimentation Rate) is almost elevated.
6- Leucocytosis (white blood count shows an increase in neutrophils).
7- Paresthesia of the lower lip with the mandible may occasionally occur.
8- Malaise.
9- Chills. Specific for indirect acute inflammation.
10- Painful lymphadenopathy.
Clinical picture:

Radiographic diagnosis:
1- As the initial lesions are confined to the soft part of the bone, there are no
characteristic x-ray changes in the earlier stages of the disease in the first one or
two weeks.
2- So x-ray may be normal until the bone resorption takes place surrounded by a
zone of sclerosis, or the medullary cavity may show increased density (diffuse
radiolucency).
3- When infection extends through the cortical bone to the periosteal layer, soft
tissue swelling and the periosteal elevation can be detected radiologically.

Chronic Osteomyelitis:
It is the type of osteomyelitis that may occur after the acute phase or it may even
develop without having any preceding acute phase.
Etiology:
The causative agent is usually a mixed infection.
They are most commonly Streptococci and Staphylococci.
Histopathology:
The factors that maintain chronicity are:
1- Bone cavity which surrounded by dense sclerosis.
2- Sequestrum which acts as an irritant and harbours bacteria.
3- Bacteria are imprisoned in the fibrous tissue where they remain dormant
and may be activated at any time.
4- Sinuses which lead to the skin surface favouring secondary infection.

Clinical feature:
1- History of acute osteomyelitis may be given.
2- The commonest presentatic feature (lesion) is the sinus which is discharging pus
and sometimes small pieces of sequestrum but is less frequently seen in the jaw.
3- Pain and swelling of the jaw at the affected bone.
4- Atrophy of the surrounding tissues may be found.
The mandible especially the molar area is more frequently affected than the
maxilla.

Radiographically:
It appears primarily as a radiolucent lesion that may show focal zones of
opacification.
The radiolucent pattern is often described a moth-eaten because of its mottled
radiographic appearance.
Lesions may be very extensive and margins are often indistinct.

Non-specific chronic osteomyelitis
Suppurative Sclerosing Non- Suppurative Sclerosing
Focal Diffuse Garre's type

1- Suppurative Sclerosing Osteomyelitis:
a- Focal type:
Sometimes called Focal Sclerosing.
It is osteopetrosis when associated with good picture of normal teeth.
It is characterized by:
Condensing osteitis (focal bony reaction) usually occurs to a low grade
inflammatory stimulus of the periapical tissues.
e.g. At the apex of a tooth with long standing pulpitis.

Clinical feature:
1- Affect young individuals (below 20 years of age).
2- The associated tooth is very often grossly carious non-vital mandibular first molar.
3- Bony lesion is mostly asymptomatic.
4- On rare occasions there is little pain.
Radiographic picture:
A sharply defined, well circumscribed and radiopaque area is observed in the jaw
bone just below the root apex of the affected tooth.
The lamina dura around the root is intact.

b- Diffuse type:
Low grade infection or chronic and wide spread periodontal
disease (periodontitis) is important in etiology and progression of
diffuse sclerosing osteon, which appears to provide a portal of
entry for bacteria (Carious non-vital teeth are less frequently
implicated).
It shows both sclerotic and osteoclastic activity.
reversal lines

Clinical features:
1- It shows chronic course with acute exacerbations of pain and swelling
(usually asymptomatic lesion).
2- Occasional drainage or fistula formation may occur.
3- The mandible is more commonly affected than the maxilla.
1- Diffuse process typically affecting a large part of the jaw.
2- Ill-defined lesion.
3- In early stages: lucent zones may appear in association with sclerotic masses.
4- In advanced stages: sclerosis dominates the radiographic picture.
5- Periosteal thickening may also be seen.

2- Non-suppurative Sclerosing Osteomyelitis (Garre's type):
It is characterized by:
1- A prominent periosteal inflammatory reaction (proliferative periosteitis).
2- Subperiosteal reactive new bone deposition.
3- Focal gross thickening of the involved bone.
4- There is neither sequestration nor sinus formation.
It is most often from a periapical abscess of lower molar and due to
infection associated with tooth extraction or partially erupted molars.

Clinical features:
1- Affect the posterior of the mandible, usually unilateral.
2- Asymptomatic bony hard swelling with normally appearance overlaying skin and
mucosa.
3- On occasion, slight tenderness may be noted.
X-ray shows marked thickening and increased density of the outer cortex of the jaw
(Duplication of the cortex).
Partial obliteration of the marrow spaces (lesion appears centrally as a mottled i.e.
predominantly lucent lesion).
Presence of periapical radiolucency in relation to a grossly carious tooth.

Specific chronic osteomyelitis
Tuberculous Syphilitic Actinomycotic

1- Tuberculous Osteomyelitis:
Definition:
Tuberculosis of bone is chronic osteomyelitis occurring in early life.
Main effect:
It displays an excess of bone destruction than bone formation yet with
tendency toward limitation of spread and spontaneous healing.

Osteitis Periosteitis:
It tends to be osteitis since it begins in spongy bone
It is commonest in vertebrae, the small bones of hands and feet and the end of long
bones including both metaphysis and epiphysis.
Clinical pictures:
It appears as epithelioid granulomas (Tubercles) with central
caseous necrosis and langhans (multi-nucleated cells) with chronic
unremarkable pain.
Appear as small areas of translucency.

2- Syphilitic Osteomyelitis:
Congenital Acquired
Main effect:
It produce prominent reactive bone formation.
Osteitis Periosteitis:
Congenital syphilis tends to be periosteitis particularly of long bone.

Clinical picture:
1- Local swelling with pain and warmth.
2- Drainage of pus through skin.
3- Bone tenderness.
It appears as radiolucent lesions.

3- Actinomycotic Osteomyelitis:
It is unfairly common because of its varied presentation.
They are gram positive rods which are strict or facultative anaerobic and it is
more common in mandible than maxilla.
Morphologically:
They are filamentous and branching in nature.
A. israelli, A. bovis, A. naeslundii, A. viscous and A. odontolyticus are members of
the family Actinomycetaceae.
Except for A. bovis all the species are normal inhabitants of human oral cavity.

Precipitating factors leading to disease in the cervical facial region are:
1- Carious teeth.
2- Dental manipulation.
3- Maxillofacial trauma.
4- Deep wound.
Its pathogenesis is related to its ability to act as an intracellular parasite
and thus resist phagocytosis as well as its tendency to spread without
respect for tissue plains or anatomic barriers.

The clinical findings include:
Presence of sulphur granules seen as basophilic masses with granular center
and radiating protrusions as well as distinctive and beaded actinomyces.
Clinical picture:
1- Local mucosal trauma.
2- Soft tissue swelling.
3- Palpable mass, sometimes painful or recurrent that may be associated with drainin
sinus tract with the presence of sulphur colonies.

Treatment:
Four weeks of high dose intravenous Penicillin followed by three to six
months course of oral Penicillin + Hydrogen peroxide.
It is not conformed but there is a small punched out radiolucent areas
with irregular and ill-defined margins.

What is Statistics?
 It is the science and practice of developing knowledge through
the use of quantitative empirical data.
 It is based on statistical theory which is a branch of applied
mathematics.
 Statistical theory uses probability theory to model:
– Randomness
– Uncertainty
 Statistics may be considered a branch of decision theory
 Statistical practice includes
– Planning of observations
– Summarizing observations
– Interpreting observations
 Statistical practice allows for
– Variability
– uncertainty

Applied Statistics
 Biostatistics
 Business statistics
 Economic statistics
 Engineering statistics
 Statistical physics
 Demography
 Psychological statistics
 Social statistics
 Reliability statistics

Application Examples of
Biostatistics (Statistics in Clinical
Medicine)
 To determine the accuracy of clinical
measurements
 To compare measurement techniques
 To assess diagnostic tests
 To determine normal values
 To estimate prognosis
 To monitor patients
 To evaluate bed use
 To calculate perinatal mortality rates

Statistics and Medical
Research
 Statistics becomes most intimately involved in
medical research
 In order to read the results of the enormous
amount of research that pours into the medical
journals, all doctors should have some
understanding of the ways in which
– Studies are designed
– Data are collected
– Data are analyzed and interpreted

What Statistics actually do
Statistics do one of only 2 things:
1- they describe a set of data
2- they provide a basis for drawing generalizations
about a large group when only a small portion of the
larger group has been observed (measured)
 Category 1 is descriptive statistics
 Category 2 is inferential statistics
5:30 AM

Descriptive Statistics
Given the IQ scores for all 8th grade boys at a certain
High School.
Descriptive statistics allow me to identify…
the typical IQ
the most frequently occurring IQ
the midpoint of the range of IQ scores
But I cannot interpret the scores to have any meaning or
applications regarding other 8th grade boys in the
same town or in other locations, or to ages and
genders. This is what inferential statistics all about.

The Population and the Sample
A population is any group of people, all of whom have at
least on characteristic in common.
A sample is a selected smaller subset of the population
5:30 AM
Statistic Parameter
Sample Population

Sampling
To make generalizations from a sample, it needs to be
representative of the larger population from which it is
taken.
In the ideal scientific world, the individuals for the
sample would be randomly selected.
This requires that each member of the population has an
equal chance of being selected each time a selection is
Statistic Parameter
Sample Population
Draw Generalizations
 

Basic concepts and notation
N
n
X
S


Population
Sample

Statistical Inference
N
n
X
S


Population
Sample
Descriptive
Statistics
Probability

Statistical Methods
 Plan observations to control their variability
(experiment design)
 Summarize a collection of observations
(descriptive statistics)
 Reach conclusions about what the observations
can tell us (statistical inference)

Experiment Design
 In medical research, statistical thinking is heavily
involved in the design of experiments, particularly
comparative experiments where we wish to study the
difference between the effects of two or more
treatments.
 These experiments may be carried out
– in the laboratory
– on animals
– on human volunteers
– on human patients in the hospital or community
 In the case of preventive trials, they may be carried
out on currently healthy patients

Comparison Techniques
 We could compare the results of the new treatment on new
patients with records of previous results using the old
treatment.
– This is seldom convincing, because there are many differences between
the patients who received the old treatment and the patients who are
going to receive the old one.
 We could ask people to volunteer for the new treatment and
give the standard treatment to those who do not volunteer.
– The difficulty here is that people who volunteer and people who do not
volunteer are likely to be different in many ways apart from the treatment.
 We can allocate patients to the new or standard treatment and
observe the outcome. The way in which patients are allocated
can influence the results enormously. There are two basic
approaches to patient allocation:
– Random
– Quasi-random

Random Allocation
 Assume that are 20 subjects to be allocated to two groups, which we shall
label A and B.
 Those 20 subjects are further assigned the following random number
sequence:
 (3, 4, 6, 2, 9, 7, 5, 3, 2, 6, 9, 7, 9, 3, 7, 2, 3, 3, 2, 4)
Subject Random
Number
Group Subject Random
Number
Group
1 3 A 11 9 A
2 4 B 12 7 A
3 6 B 13 9 A
4 2 B 14 3 A
5 9 A 15 7 A
6 7 A 16 2 B
7 5 A 17 3 A
8 3 A 18 3 A
9 2 B 19 2 B
10 6 B 20 4 B

Random Allocation
 The system above gives unequal numbers in
the two groups, 12 in A and 8 in B.
 Sometimes it is desired that the groups be of
equal size.
 In that case we can assign Group A to the first
ten entries only in the table labeled A, and
assign Group B to the remainder of the twenty
subjects.

Types of Variables
 Qualitative Variables
 Attributes, categories
 Examples: male/female, registered to vote/not,
ethnicity, eye color....
 Quantitative Variables
 Discrete - usually take on integer values but can
take on fractions when variable allows - counts, how
many
 Continuous - can take on any value at any point
along an interval - measurements, how much

Discrete Data
A set of data is said to be discrete if the values / observations
belonging to it are distinct and separate.
They can be counted (1,2,3,.......).
Examples:
the number of kittens in a litter;
the number of patients in a doctors surgery;
the number of flaws in one metre of cloth;
gender (male, female);
blood group (O, A, B, AB).

Continuous Data
A set of data is said to be continuous if the values / observations
belonging to it may take on any value within a finite or infinite interval.
You can count, order and measure continuous data.
Examples:
height;
weight;
temperature;
the amount of sugar in an orange;
the time required to run a mile.

Scales of Measurement
 Nominal Scale - Labels represent various levels of a
categorical variable.
 Ordinal Scale - Labels represent an order that
indicates either preference or ranking.
 Interval Scale - Numerical labels indicate order and
distance between elements. There is no absolute zero
and multiples of measures are not meaningful.
 Ratio Scale - Numerical labels indicate order and
distance between elements. There is an absolute zero
and multiples of measures are meaningful.

Diagrammatic Representation of
Data
 It is often convenient to present data pictorially.
 Information can be conveyed much more
quickly by a diagram than by a table of
numbers.
 This is particularly useful when data are being
presented to an audience.
 A diagram can also help the reader get the
salient points of a table of numbers.
 Unfortunately, unless great care is taken,
diagrams can be very misleading and should
only serve an illustrative purpose and not a

Histogram
A histogram is a way of summarizing data that are measured on
an interval scale (either discrete or continuous).
It is often used in exploratory data analysis to illustrate the major features of the
distribution of the data in a convenient form.
It divides up the range of possible values in a data set into classes or groups.
For each group, a rectangle is constructed with a base length equal to the range of
values in that specific group, and an area proportional to
the number of observations falling into that group.
This means that the rectangles might be drawn of non-uniform height.

Example
 The average daily cost to community hospitals for
patient stays during 1993 for each of the 50 U.S.
states was given in the next table.
 a) Construct a frequency distribution. State interval width
and class mark.
 b) Construct a histogram,
 c) Construct a relative frequency distribution,
 d) Construct a cumulative frequency distribution.

Example –Data List
AL $775 HI 823 MA 1,036 NM 1,046 SD 506
AK 1,136 ID 659 MI 902 NY 784 TN 859
AZ 1,091 IL 917 MN 652 NC 763 TX 1,010
AR 678 IN 898 MS 555 ND 507 UT 1,081
CA 1,221 IA 612 MO 863 OH 940 VT 676
CO 961 KS 666 MT 482 OK 797 VA 830
CT 1,058 KY 703 NE 626 OR 1,052 WA 1,143
DE 1,024 LA 875 NV 900 PA 861 WV 701
FL 960 ME 738 NH 976 RI 885 WI 744
GA 775 MD 889 NJ 829 SC 838 WY 537

Example – Data Array
CA 1,221 TX 1,010 RI 885 NY 784 KS 666
WA 1,143 NH 976 LA 875 AL 775 ID 659
AK 1,136 CO 961 MO 863 GA 775 MN 652
AZ 1,091 FL 960 PA 861 NC 763 NE 626
UT 1,081 CH 940 TN 859 WI 744 IA 612
CT 1,058 IL 917 SC 838 ME 738 MS 555
OR 1,052 MI 902 VA 830 KY 703 WY 537
NM 1,046 NV 900 NJ 829 WV 701 ND 507
MA 1,036 IN 898 HI 823 AR 678 SD 506
DE 1,024 MD 889 OK 797 VT 676 MT 482

Example – Frequency Distribution
Average daily cost Number Mark
$450 – under $550 4 $500
$550 – under $650 3 $600
$650 – under $750 9 $700
$750 – under $850 9 $800
$850 – under $950 11 $900
$950 – under $1,050 7 $1,000
$1,050 – under $1,150 6 $1,100
$1,150 – under $1,250 1 $1,200
Interval width: $100

Example – Histogram
0
2
4
6
8
10
12
500 600 700 800 900 1000 1100 1200

Example – Relative Frequency Distribution
(Polygon)
0
0.05
0.1
0.15
0.2
0.25
0 200 400 600 800 1000 1200 1400

Example – Cumulative Frequency
Distribution
Average daily cost Number Cum. Freq.
$450 – under $550 4 4
$550 – under $650 3 7
$650 – under $750 9 16
$750 – under $850 9 25
$850 – under $9 11 36
$950 – under $1,050 7 43
$1,050 – under $1,150 6 49
$1,150 – under $1,250 1 50

Example – Cumulative Frequency Distribution &
Cumulative Relative Frequency Distribution
Average daily cost Cum.Freq. Cum.Rel.Freq.
$450 – under $550 4 4/50 = .02
$550 – under $650 7 7/50 = .14
$650 – under $750 16 16/50 = .32
$750 – under $850 25 25/50 = .50
$850 – under $950 36 36/50 = .72
$950 – under $1,050 43 43/50 = .86
$1,050 – under $1,150 49 49/50 = .98
$1,150 – under $1,250 50 50/50 = 1.00

Example – Cumulative Distribution
0
5
10
15
20
25
30
35
40
45
50
0 200 400 600 800 1000 1200

Pie Charts
 The pie chart or pie diagram is the equivalent of
the histogram for qualitative data.
 It shows the relative frequency for each
category by dividing a circle into sectors, the
angles of which are proportional to the relative
frequency.
 We thus multiply each relative frequency by
360°, to give the corresponding angle in
degrees.

Pie Chart Calculations of the Distribution of Causes of
Death
Angle
(degrees)
Relative
frequency (%)
FrequencyCause of death
17849.471143559Circulatory system
7821.63062767Neoplasms
(cancers)
5415.12343886Respiratory
system
113.1529147Digestive system
62.6667736Injury and
poisoning
297.95823094Others

Major Causes of Death
Pie Chart

Bar Charts
 Histograms and pie charts depict the
distribution of a single variable.
 A bar chart or bar diagram shows the
relationship between two variables, usually one
being quantitative and the other qualitative or a
grouped quantitative variable, such as time in
years.
 The values of the first variable height are shown
by the heights of bars, one bar for each
category of the second variable.
 Bar charts can be used to represent

Annual Standardized Mortality Rate from Cancer
of Esophagus (England & Wales, 1960-1969)
Mortality rateYear
5.11960
5.01961
5.21962
5.21963
5.21964
5.41965
5.41966
5.61967
5.81968
5.91969

Bar chart showing the relationship between mortality due to
cancer of the esophagus and year

Key Terms
 Measures of
Central
Tendency,
The Center
 Mean
 µ, population; , sample
 Weighted Mean
 Median
 Mode
x

Key Terms
 Measures of
Dispersion,
The
Spread
 Range
 Mean absolute deviation
 Variance
 Standard deviation
 Inter-quartile range
 Inter-quartile deviation
 Coefficient of variation

Key Terms
 Measures of
Relative
Position
 Quantiles
 Quartiles
 Deciles
 Percentiles
 Residuals
 Standardized values

The Mean
 Mean
 Arithmetic average = (sum all values)/# of values
 Population: µ = (Sxi)/N
 Sample: = (Sxi)/nx

The Weighted Mean
 When what you have is grouped data,
compute the mean using µ = (Swixi)/Swi

The Median
 To find the median:
1. Put the data in an array.
2A. If the data set has an ODD number of numbers, the
median is the middle value.
2B. If the data set has an EVEN number of numbers, the
median is the AVERAGE of the middle two values.
(Note that the median of an even set of data values is not
necessarily a member of the set of values.)

The Mode
 The mode is the most frequent value.
 While there is just one value for the mean
and one value for the median, there may be
more than one value for the mode of a data
set.
 The mode tends to be less frequently used
than the mean or the median.
0
2
4
6
8
10
12
500 600 700 800 900 1000 1100 1200

Comparing Measures of Central
Tendency
 If mean = median = mode, the shape of the distribution is
symmetric.
 If mode < median < mean or if mean > median > mode,
the shape of the distribution trails to the right,
is positively skewed.
 If mean < median < mode or if mode > median > mean,
the shape of the distribution trails to the left,
is negatively skewed.

The Range
 The range is the distance between the
smallest and the largest data value in the
set.
 Range = largest value – smallest value
 Sometimes range is reported as an
interval, anchored between the smallest
and largest data value, rather than the
actual width of that interval.

Residuals
 Residuals are the differences between
each data value in the set and the
group mean:
 for a population, xi – µ
 for a sample, xi – x

The Variance
 Variance is one of the most frequently used
measures of spread,
 for population,
 for sample,
 The right side of each equation is often used
as a computational shortcut.
2 
S(x
i
–)2
N

S(x
i
)2 – N2
N
s2
S(x
i
–x)2
n–1

S(x
i
)2–nx2
n–1

The Standard Deviation
 Since variance is given in squared units,
we often find uses for the standard
deviation, which is the square root of
variance:
 for a population,
 for a sample,
  2
s s2

Quartiles
 One of the most frequently used quantiles is the
quartile.
 Quartiles divide the values of a data set into four
subsets of equal size, each comprising 25% of the
observations.
 To find the first, second, and third quartiles:
 1. Arrange the N data values into an array.
 2. First quartile, Q1 = data value at position (N + 1)/4
 3. Second quartile, Q2 = data value at position 2(N + 1)/4
 4. Third quartile, Q3 = data value at position 3(N + 1)/4

Quartiles
0.0 1.5 3.0 4.5 6.0
0
25
50
75
100
Ln_YarnS
CumulativeFrequency
Q1 Q2 Q3

Standardized Values
 How far above or below the individual value is
compared to the population mean in units of
standard deviation
 “How far above or below” (data value – mean)
which is the residual...
 “In units of standard deviation” divided by 
Standardized individual value:
A negative z means the data value falls below the mean.
x–
 z

Probability of an Event
soccurrenceofNo.Total
occurcanAeventtimesofNo
AEventofobability
.
Pr 

Success = at least one “1”

Success = at least one “1,1”

P (Failure) = P(no “1”)
  482.
6
5
4

P (Success) = .518
P (Failure) = P(no “1,1”)
  509.
36
35
24

P (Success) = .491

To add or to multiply ?
P(outcomes add up to “10”) =?

P(outcomes add up to “10”) 36/3=
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,2 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6

Mutually Exclusive Events
Two events are mutually exclusive (or disjoint) if it is impossible
for them to occur together.
If two events are mutually exclusive, they cannot be independent
and vice versa.

Example:
A subject in a study cannot be both male and female,
A subject cannot be aged 20 and 30.
However:
A subject could be both male and 20.
A subject could be both female and 30.

Independent Events
Two events are independent if the occurrence of one of the
events gives us no information about whether or not the other
event will occur; that is, the events have no influence on each other.
If two events are independent then they cannot be mutually
exclusive (disjoint) and vice versa.

Example
Suppose that a man and a woman each have a pack of 52 playing
cards. Each draws a card from his/her pack. Find the probability
that they each draw the ace of clubs.
We define the events
A = 'the man draws the ace of clubs'
B = 'the woman draws the ace of clubs'
Clearly events A and B are independent so,
P(A and B) = P(A)•P(B) = (1/52)×(1/52) = 1/2704
That is, there is a very small chance that the man and the woman
will both draw the ace of clubs.

Conditional Probability
Suppose
You go out for lunch at the same place and time every Friday.
You are served lunch within 15 minutes with probability 0.9.
However,
given that you notice that the restaurant is exceptionally busy, the
probability of being served lunch within 15 minutes may reduce to
0.7.
This is the conditional probability of being served lunch
within 15 minutes given that the restaurant is exceptionally busy.

The usual notation for "event A occurs given that event B has
occurred" is A|B (A given B).
The symbol | is a vertical line and does not imply division.
P(A|B) denotes the probability that event A will occur given that
event B has occurred already.

A rule that can be used to determine a conditional probability
from unconditional probabilities is
P(A|B) = P(A and B) / P(B)
where,
P(A|B) = the (conditional) probability that event A will occur given
that event B has occurred already
P(A and B) = the (unconditional) probability that event A and event
B occur
P(B) = the (unconditional) probability that event B occurs

Binomial (Bernoulli)
Distribution
Bernoulli trial: a random event that can take on
only one of two possible outcomes, with the
outcomes
arbitrarily denoted as either
– “success” or “yes” or “true”
– “failure” or “no” or “false”
For example, flipping a coin is an example of a
Bernoulli trial, since the outcome is either a head
(yes) or a tail (no).

X = Number of successes in n trials
xnpxp
xnx
nxP 

 )1(
)!(!
!)(
n = 6, x = 0
n = 6, x = 1
n = 6, x = 3 x0 n
Binomial Distribution

0 10 20 30 40 50
0.00
0.05
0.10
i
b(i;50,p)
p=.25
p=.5
p=.75

7 9 11 13 15 17 19 21
0
10
20
30
C1
0 10 20 30 40 50
0.00
0.05
0.10
i
b(i;50,p)
p=.25
p=.5
p=.75
13 16 20 11 12
10 12 20 16 15
10 12 9 18 12
11 11 9 11 14
13 8 4 13 12
14 11 14 15 12
18 13 7 11 9
15 11 8 11 16
9 12 12 18 15
13 9 15 12 12

APPLICATION EXAMPLES (from
Medicine)
• The number of patients out of n that respond to treatment.
• The number of people in a community of n people that have asthma.
• The number of people in a group of n intravenous drug users who are
HIV positive.

Probability that a patient with swollen leg has a clot = 0.3
What is the distribution of patients with a clot among a total of 10 patients with
swollen leg.
No. of Patients
Probability

POISSON DISTRIBUTION
 Used to describe a number of events (usually
rare) in an interval given an average number of
events.
 Gives the number of times a particular event
occurs in a given unit interval.
 The mean number of events in each unit will
be denoted by .
 Unit intervals may be in units of area, volume,
etc.
 Most commonly, however, unit intervals are

Poisson Distribution
P(x)  xe–
x!
x = 0
x = 2
x = 8 x0
events
X = Number of occurrences of an event

Poisson Distribution
50403020100
0.2
0.1
0.0
j
p(j;lambda)
lambda=5
lambda=10
lambda=15

Applications of Poisson Distribution
 Modeling the distribution of phone calls
 The arrivals of trucks and cars at a tollbooth
 The number of accidents at an intersection
 Counting nuclear decay events
 The demand of patients for service at a
health institution

Normal Distribution
The normal distribution (also called the Gaussian
distribution) is a family of distributions recognized as
being
symmetrical
unimodal
bell-shaped.
The normal distribution is characterized by two parameters:
1. mean (µ) determines the distribution’s location. Fig. 1
shows two normal distributions with different means
2. The standard deviation (σ) of a particular normal
distribution determines its spread. Fig. 2 demonstrates two
normal distributions with different spreads:

The z-score
 Assume a population of normal distribution with statistic X.
 The z-score (i.e., standardized X-statistic) is defined by:
And, if the distribution of X was normal, or at least approximately
normal, you could then take that z-score, and refer it to a table of the
standard normal distribution to figure out the proportion of scores
higher than X, or lower than X, etc.
X
XXX
z



 




Fig. 1. Two Normal distributions with different means

Fig. 2. Two Normal distributions with different standard deviations

Normal Distribution
0 10 20
0.0
0.1
0.2
0.3
0.4
x
y
Sigma = 1
Sigma = 2
Sigma = 3

Normal Distribution
X P(<X) P(Xi< <Xi+1)
-3.0 0.001350 0.004432
-2.9 0.001866 0.005953
-2.8 0.002555 0.007915
-2.7 0.003467 0.010421
-2.6 0.004661 0.013583
-2.5 0.006210 0.017528
-2.4 0.008198 0.022395
-2.3 0.010724 0.028327
-2.2 0.013903 0.035475
-2.1 0.017864 0.043984
-2.0 0.022750 0.053991
-1.9 0.028717 0.065616
-1.8 0.035930 0.078950
-1.7 0.044565 0.094049
-1.6 0.054799 0.110921
-1.5 0.066807 0.129518
-1.4 0.080757 0.149727
-1.3 0.096800 0.171369
-1.2 0.115070 0.194186
-1.1 0.135666 0.217852
-1.0 0.158655 0.241971
-0.9 0.184060 0.266085
-0.8 0.211855 0.289692
-0.7 0.241964 0.312254
-0.6 0.274253 0.333225
-0.5 0.308538 0.352065
-0.4 0.344578 0.368270
-0.3 0.382089 0.381388
-0.2 0.420740 0.391043
-0.1 0.460172 0.396953
0.0 0.500000 0.398942
X P(<X) P(Xi< <Xi+1)
0.0 0.500000 0.398942
0.1 0.539828 0.396953
0.2 0.579260 0.391043
0.3 0.617911 0.381388
0.4 0.655422 0.368270
0.5 0.691462 0.352065
0.6 0.725747 0.333225
0.7 0.758036 0.312254
0.8 0.788145 0.289692
0.9 0.815940 0.266085
1.0 0.841345 0.241971
1.1 0.864334 0.217852
1.2 0.884930 0.194186
1.3 0.903200 0.171369
1.4 0.919243 0.149727
1.5 0.933193 0.129518
1.6 0.945201 0.110921
1.7 0.955435 0.094049
1.8 0.964070 0.078950
1.9 0.971283 0.065616
2.0 0.977250 0.053991
2.1 0.982136 0.043984
2.2 0.986097 0.035475
2.3 0.989276 0.028327
2.4 0.991802 0.022395
2.5 0.993790 0.017528
2.6 0.995339 0.013583
2.7 0.996533 0.010421
2.8 0.997445 0.007915
2.9 0.998134 0.005953
3.0 0.998650 0.004432
N(0,1)

Normal Distribution
7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5
0
10
20
C1
0 10 20
0.0
0.1
0.2
0.3
0.4
x
y
Sigma = 1
Sigma = 2
Sigma = 3
7.9006 11.5151 9.9542 9.4493 8.2387
10.4707 9.4041 9.3517 10.5664 10.9079
10.0077 12.5188 9.6937 10.0757 10.1616
10.2881 9.8560 10.0014 9.8467 11.5006
10.2982 9.6023 9.7238 11.5413 8.4595
9.2372 11.0408 12.8996 9.5590 9.1041
8.9170 9.7734 7.9844 8.3484 11.3703
10.6260 10.0952 11.4019 8.9842 9.3783
9.7574 7.9312 8.1566 9.9305 9.1158
8.6436 10.4689 9.3356 10.8788 7.8790

Sampling Distributions
 Imagine drawing (with replacement) all possible
samples of size n from a population, and for each
sample, calculating a statistic -- e.g., the sample
mean.
 The frequency distribution of those sample means
would be the sampling distribution of the mean (for
samples of size n drawn from that particular
population).
 Normally, one thinks of sampling from relatively large
populations, but the concept of a sampling distribution
can be illustrated with a small population.

Example
 A population consisted of the following 5
scores: 2, 3, 4, 5, and 6.
 The population mean = 4, and the population
standard deviation (dividing by N) = 1.414.
 If we draw (with replacement) all possible
samples of 2 from this population, we would
end up with the 25 samples shown in Table.
 This distribution (histogram) of sample means is
called the sampling distribution of the mean for
samples of n=2 from the population of interest
(i.e., our population of 5 scores)

All possible samples of n=2 from a population of 5
scores
Mean of the sample means = 4.000
SD of the sample means = 1.000
(SD calculated with division by N)

The Central Limit Theorem (CLT)
 The mean of the sampling distribution of the mean =
the population mean
 The SD of the sampling distribution of the mean = the
standard error (SE) of the mean = the population
standard deviation divided by the square root of the
sample size
 Putting these statements into symbols:
XX
  { mean of the sample means = the population mean }
n
X
X

  { SE of mean = population SD over square root of n }

What the CLT tells us about the shape of the
sampling distribution
 The central limit theorem also provides us with some
very helpful information about the shape of the
sampling distribution of the mean.
 Specifically, it tells us the conditions under which the
sampling distribution of the mean is normally
distributed, or at least approximately normal, where
approximately means close enough to treat as normal
for practical purposes.
 The shape of the sampling distribution depends on
two factors:
– the shape of the population from which the sample has been
drawn, and
– The sample size.

The Shape of the Sampling Distribution
 If the population from which you sampled is itself normally
distributed, then the sampling distribution of the mean will be
normal, regardless of sample size. (Even for sample size =
1, the sampling distribution of the mean will be normal, because
it will be an exact copy of the population distribution).
 If the population distribution is reasonably symmetrical (i.e., not
too skewed, reasonably normal looking), then the sampling
distribution of the mean will be approximately normal for
samples of 30 or greater.
 If the population shape is as far from normal as possible, the
sampling distribution of the mean will still be approximately
normal for sample sizes of 300 or greater.

The z-scores for the Sampling Distribution of the
Means
 Based on what we learned from the central limit
theorem, we are now in a position to compute a
z-score as follows:
n
XX
z X
X
X
/


 



And, if the sampling distribution of X is normal, or at least
approximately normal, we may then refer this value of z to the
standard normal distribution, just as we did when we were using raw
scores.

An Example.
 Here is a (fictitious) newspaper advertisement for a
program designed to increase intelligence of school
children:

Example (Contd.)
 An expert on IQ knows that in the general population of
children, the mean IQ = 100, and the population SD = 15 (for
the WISC, at least).
 He also knows that IQ is (approximately) normally distributed in
the population.
 Equipped with this information, you can now address questions
such as:
 If the n = 25 children from Dundas are a random sample from
the general population of children,
A. What is the probability of getting a sample mean of 108 or higher?
B. What is the probability of getting a sample mean of 92 or lower?
C. How high would the sample mean have to be for you to say that the
probability of getting a mean that high (or higher) was 0.05 (or 5%)?
D. How low would the sample mean have to be for you to say that the

Solution
 If we have sampled from the general population of children, as we are
assuming, then the population from which we have sampled is at least
approximately normal.
 Therefore, the sampling distribution of the mean will be normal, regardless of
sample size.
 Therefore, we can compute a z-score, and refer it to the table of the
standard normal distribution.
 So, for part (A):
And from a table of the standard normal distribution we can see that the
probability of a z-score greater than or equal to 2.667 = 0.0038.
Translating that back to the original units, we could say that the probability of
getting a sample mean of 108 (or greater) is .0038 (assuming that the 25 children are
a random sample from the general population).
667.2
3
8
25
15
100108







n
XX
z
X
X
X
X





 For part (B), do the same, but replace 108 with 92:
667.2
3
8
25
15
10092









n
XX
z
X
X
X
X




And the probability of a sample mean less than or equal to 92 is also equal to
0.0038.
Had we asked for the probability of a sample mean that is either 108 or greater,
or 92 or less, the answer would be 0.0038 + 0.0038 = 0.0076.
Part (C) above amounts to the same thing as asking, "What sample mean
corresponds to a z-score of 1.645?", because we know that p (z≥ = 1.645) = 0.05.
We can start out with the usual z-score formula and try to determine the
corresponding value of X
Because
X
X
X
z


 we should have
935.104100
25
15
645.1  XX
zX 
So, had we obtained a sample mean of 105, we could have concluded that the
probability of a mean that high or higher was .05 (or 5%).

 For part (D), because of the symmetry of the standard
normal distribution about 0, we would use the same
method, but substituting -1.645 for 1.645.
 This would yield an answer of 100 - 4.935 = 95.065.
 So the probability of a sample mean less than or equal
to 95 is also 5%.

What hypotheses would come out as a byproduct of
the analysis of the above data?
 One has two hypotheses:
– Null hypothesis
– Alternative hypothesis
 These two hypotheses are mutually exclusive and exhaustive.
 In other words, they cannot share any outcomes in common, but
together must account for all possible outcomes.
 Informally, the null hypothesis typically states something along the
lines of, "there is no treatment effect", or "there is no difference
between the groups".
 The alternative hypothesis typically states that “there is a treatment
effect “, or that there is a difference between the groups.
 Furthermore, an alternative hypothesis may be directional or non-
directional.
 That is, it may or may not specify the direction of the difference
between the groups.

A directional alternative hypothesis
 Ho: μ ≤ 100
 H1: μ > 100
 This pair of hypotheses can be summarized as follows.
 If the alternative hypothesis is true, the sample of 25 children
we have drawn is from a population with mean IQ greater than
100.
 But if the null hypothesis is true, the sample is from a
population with mean IQ equal to or less than 100.
 Thus, we would only be in a position to reject the null
hypothesis if the sample mean is greater than 100 by a
sufficient amount.
 If the sample mean is less than 100, no matter by how much,
we would not be able to reject Ho .

 How much greater than 100 must the sample mean be for us to
be comfortable in rejecting the null hypothesis?
 The answer that most disciplines use by convention the
following: The difference between and µ must be large
enough that the probability this large difference occurred by
chance (given a true null hypothesis) is 5% or less.
 The observed sample mean for this example was 108. As we
saw earlier, this corresponds to a z-score of 2.667, and p (z≥ =
2.667) 0.0038. (The value of 0.0038 is as a matter of fact that
the probability that the big difference between and µ came
out as a matter of chance.
 Therefore, we could reject Ho , and we would act as if the
sample was drawn from a population in which mean IQ is
greater than 100.
X
X

A non-directional alternative hypothesis
 Ho: μ = 100
 H1: μ ≠ 100
 In this case, the null hypothesis states that the 25 children are a random
sample from a population with mean IQ = 100, and the alternative
hypothesis says they are not ― but it does not specify the direction of the
difference from 100.
 In the directional test, we needed to have > 100 by a sufficient amount, in
order to reject Ho.
 But in this case, with a non-directional alternative hypothesis, we may reject
Ho if < 100 or if > 100 , provided the difference is large enough.
 For this example, the sample mean = 108. This represents a difference of +8
from the population mean (under a true null hypothesis).
 Because we are interested in both tails of the distribution, we must figure out
the probability of a difference of +8 or greater, or a change of -8 or greater.
 In other words, p ( ≥ 108) + p ( < 92) = .0038 + .0038 = .0076.
X
X X
X X

Single sample t-test (when σ is not
known)
 In many real-world cases of hypothesis testing, one does not
know the standard deviation of the population.
 In such cases, it must be estimated using the sample standard
deviation. That is, s (calculated with division by n-1) is used to
estimate σ.
 Other than that, the calculations are as we saw for the z-test for
a single sample ― but the test statistic is called t, not z.
X
X
n
s
X
t

1 degrees of freedom (df = n – 1). Here we have
n
s
sX
 and
 
11
1
2






n
SS
n
XX
s X
n
i
i
N.B. There are n-1 degrees of freedom whenever you calculate a
sample variance (or standard deviation).

 To calculate the p-value for a single sample z-test, we used the
standard normal distribution.
 For a single sample t-test, we must use a t-distribution with n-1
degrees of freedom.
 As this implies, there is a whole family of t-distributions, with
degrees of freedom ranging from 1 to infinity ∞.
 All t-distributions are symmetrical about 0, like the standard
normal.
 In fact, the t-distribution with df = ∞ is identical to the standard
normal distribution.
 t-distributions with df < ∞ have lower peaks and thicker tails
than the standard normal distribution.

Probability density functions of: the standard normal distribution (the highest peak with the thinnest
tails); the t-distribution with df =10 (intermediate peak and tails); and the t-distribution with df=2
(the lowest peak and thickest tails). The dotted lines are at -1.96 and +1.96, the critical values of z
for a
two-tailed test with alpha = .05. For all t-distributions with df < ∞ , the proportion of area beyond -
1.96
and +1.96 is greater than .05. The lower the degrees of freedom, the thicker the tails, and the
greater the

Area beyond critical values of t = ±1.96 in various t-distributions.
The t-distribution with df = ∞ is identical to the standard normal distribution.

Example of single-sample t-test
 A researcher believes that in recent years women have been
getting taller.
 She knows that 10 years ago the average height of young adult
women living in her city was 63 inches.
 The standard deviation is unknown.
 She randomly samples eight young adult women currently
residing in her city and measures their heights.
 The following data are obtained: [64, 66, 68, 60, 62, 65, 66, 63.]
 The null hypothesis is that these 8 women are a random
sample from a population in which the mean height is 63
inches.
 The non-directional alternative states that the women are a
random sample from a population in which the mean is not 63
inches.

Solution
 The sample mean is 64.25.
 Because the population standard deviation is not known, we must estimate it
using the sample standard deviation.
        5495.2
7
25.646325.646625.6464
1
222
1
2






 
n
XX
s
n
i
i
We can now use the sample standard deviation to estimate the standard error of the
mean:
901.0
8
5495.2
meanofSEEstimated 
n
s
sX
And finally:
387.1
901.0
6325.64





X
X
s
X
t


 This value of t can be referred to a t-distribution
with df = n-1 = 7.
 Doing so, it is found that the conditional
probability of obtaining a t-statistic with absolute
value equal to or greater than 1.387 is equal to
0.208.
 Therefore, assuming that alpha had been set at
the usual 0.05 level, the researcher cannot
reject the null hypothesis.

Paired (or related samples) t-test
 Suppose you have either 2 scores for each person
 (e.g., before and after), or when you have matched pairs of
scores (e.g., husband and wife pairs, or twin pairs).
 The paired t-test may be used in this case, given that its
assumptions are met adequately.
 Quite simply, the paired t-test is just a single-sample t-test
performed on the difference scores.
 That is, for each matched pair, compute a difference score.
 Whether you subtract Score (1) from Score (2) or vice versa
does not matter, so long as you do it the same way for each
pair.
 Then perform a single-sample t-test on those differences.
 The null hypothesis for this test is that the difference scores are
a random sample from a population in which the mean
difference has some value which you specify.

 For example, suppose you found some old research
which reported that on average, husbands were 5
inches taller than their wives.
 If you wished to test the null hypothesis that the
difference is still 5 inches today (despite the overall
increase in height), your null hypothesis would state
that your sample of difference scores (from
husband/wife pairs) is a random sample from a
population in which the mean difference = 5 inches.
 In the equations for the paired t-test, is often
replaced with , which stands for the mean difference.
X
D

where
D = the (sample) mean of the difference scores
D = the mean difference in the population, given a true Ho (often D = 0, but not always)
Ds = sample standard deviation of the difference score (dividing by n-1)
n = number of matched pairs (number of individuals = 2n)
D
s = SE of the mean difference
df = n - 1
n
s
D
s
D
t
D
D
D
D  




Example of paired t-test
 A political candidate wishes to determine if
endorsing increased social spending is likely to
affect her standing in the polls.
 She has access to data on the popularity of
several other candidates who have endorsed
increases spending.
 The data was available both before and after
the candidates announced their positions on the
issue [see Table].

Data for paired t-test example
Popularity
Ratings
Candidate
DifferenceAfterBefore
143421
445412
656503
254524
765585
-329326
746397
648428
-147489
6534710

Solution
 Examining the last column we find out that:
D = 3.5
If the null hypothesis is true, we shall assume that D = 0
D
s = 3.5668and
Since n = 10, it turns out that SE = 1.1279; and thus t = 3.103
Now df = 9
The null hypothesis for this test states that the mean difference in the population is
zero; that in other words, endorsing increased social spending has no effect on
popularity ratings in the population from which we have sampled.
If that is true, the probability of seeing a difference of 3.5 points or more is 0.0127
(the p-value).
Therefore, the politician would likely reject the null hypothesis, and would endorse
increased social spending, once again since there is only a probability of 0.0127
justifying that his rejection came out by chance.
As a matter of fact, a two-sided p-value for significance is given by p = 0.0127
(using SPSS or MATLAB).

Unpaired (or independent samples) t-test
 Another common form of the t-test may be used if you have 2 independent
samples (or groups).
 The formula for this version of the test is given by:
   
21
2121
XX
s
XX
t




is the difference between the means of two (independent) samples, or the
difference between group means. is the difference between the
corresponding population means, assuming that Ho is true.
 21 XX 
 21  







21
2 11
21
nn
ss pooledXX
groupswithin
groups
21
212
df2
var
within
pooled
SS
nn
SSSS
estimateiancepooleds 




 
2
1
1
1


n
i
i XXSS
 
2
1
2
2


n
i
i XXSS
n1 = sample size for Group 1
n2 = sample size for Group 2
df = n1 + n2 - 2

Example of unpaired t-test
 A nurse was hired by a governmental ecology agency to
investigate the impact of a lead smelter on the level of lead in
the blood of children living near the smelter.
 Ten children were chosen at random from those living near the
smelter.
 A comparison group of 7 children was randomly selected from
those living in an area relatively free from possible lead
pollution.
 Blood samples were taken from the children, and lead levels
determined.
 Given the tabulated results (scores are in micrograms of lead
per 100 milliliters of blood) and using α2−tailed = 0.01 , what do
you conclude?

Lead Levels
Children Living in Unpolluted AreaChildren Living near Smelter
918
1316
821
1514
1717
1219
1122
24
15
18

Solution
 The null hypothesis for this example is that the 2 groups of children are 2
random samples from populations with the same mean levels of lead
concentration in the blood. Thus, = 0. Now
n1 = 10, n2 = 7, and
= 18.4 and = 12.1429. Thus, = 6.2571
 21  
2X1X 21 XX 
SS1 = 90.4 and SS2 = 60.8571
Further
df = 10 +7 – 2 = 15
So, SS1 + SS2 ≈ 90.4 + 60.9 = 151.3
0867.10
15
3.1512
pooleds
and
565.1
7
1
10
1
0867.10
21






XX
s
998.3
565.1
2571.6
t
and a two-tailed significance is given by p = 0.0012 < α2-tailed
The null hypothesis of two equal means is likely to be rejected in that case.

Sample Size, Precision, and
Power A study that is insufficiently precise or lacks the power to reject
a false null hypothesis is a waste of time and money.
 A study that collects too much data is also wasteful.
 Therefore, before collecting data, it is essential to determine the
sample size requirements of a study.

Sample Size Calculation
 Before calculating the sample size requirements of a
study you must address some questions:
 The theme of learning from the study
 mean
 mean difference
 proportion
 proportion (risk) ratio
 odds ratio
 Slope
• The estimation methodology
• with a given precision
 with a given power
• The type of sample
• A single group
• Two or more independent groups
• Matched pairs

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 227
Parametric Statistical Inference
Instructor: Ron S. Kenett
Email: ron@kpa.co.il
Course Website: www.kpa.co.il/biostat
Course textbook: MODERN INDUSTRIAL STATISTICS,
Kenett and Zacks, Duxbury Press, 1998

Null and Alternative Hypotheses
 In any experiment, there are two hypotheses that attempt to
explain the results. They are the null hypothesis and
alternative hypothesis.
 Alternative Hypothesis (H1 or HA). In experiments that entail
manipulation of an independent variable, the alternative
hypothesis states that the results of the experiment are due to
the effect of the independent variable. In a coin tossing
experiment, H1 would state that the biased coin had been
selected, and that p(Head) = 0.15.
 Null Hypothesis (H0). The null hypothesis is the complement of
the alternative hypothesis. In other words, if H1 is not true, then
H0 must be true, and vice versa. In the foregoing coin tossing
situation, H0 asserts that the fair coin was selected, and that
p(Head) = 0.50.

Null and Alternative
Hypotheses
 Thus, the decision rule to minimize the overall
p(error) can be restated as follows:
 if p(X | H0 ) > p(X | H1 ) then do not reject H0
 if p(X | H0 ) < p(X | H1 ) then reject H0
where
X = independent random variable (usually
statistic)
 According to statistical purists, it is only proper
to reject the null hypothesis or fail to reject the
null hypothesis. Acceptance of either
hypothesis is strictly forbidden.

Rejection Region
 The rejection region is a range containing
outcomes that lead to rejection of H0 .

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 231
 Step 1.
A claim is made.
 A new claim is asserted that
challenges existing thoughts
about a population
characteristic.
 Suggestion: Form the alternative
hypothesis first, since it
embodies the challenge.
The Logic of Hypothesis Testing

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 232
 Step 2.
How much
error are you
willing to
accept?
 Select the maximum acceptable
error, a. The decision maker must
elect how much error he/she is
willing to accept in making an
inference about the population. The
significance level of the test is the
maximum probability that the null
hypothesis will be rejected
incorrectly, a Type I error.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 233
 Step 3.
If the null
hypothesis were
true, what would
you expect to
see?
 Assume the null hypothesis is
true. This is a very powerful
statement. The test is always
referenced to the null hypothesis.
Form the rejection region, the
areas in which the decision maker
is willing to reject the
presumption of the null
hypothesis.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 234
 Step 4.
What did you
actually see?
 Compute the sample statistic.
The sample provides a set of data
that serves as a window to the
population. The decision maker
computes the sample statistic and
calculates how far the sample
statistic differs from the presumed
distribution that is established by
the null hypothesis.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 235
 Step 5.
Make the
decision.
 The decision is a conclusion supported by
evidence. The decision maker will:
 reject the null hypothesis if the sample
evidence is so strong, the sample statistic so
unlikely, that the decision maker is convinced H1
must be true.
 fail to reject the null hypothesis if the sample
statistic falls in the nonrejection region. In this
case, the decision maker is not concluding the
null hypothesis is true, only that there is
insufficient evidence to dispute it based on this
sample.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 236
 Step 6.
What are the
implications of
the decision for
future actions?
 State what the decision means in
terms of the research program.
The decision maker must draw out the
implications of the decision. Is there
some action triggered, some change
implied? What recommendations
might be extended for future attempts
to test similar hypotheses?

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 237
 Type I Error:
 Saying you reject H0 when it really is true.
 Rejecting a true H0.
 Type II Error:
 Saying you do not reject H0 when it really is
false.
 Failing to reject a false H0.
Two Types of Errors

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 238
What are acceptable error levels?
 Decision makers frequently use a 5%
significance level.
 Use a = 0.05.
 An a-error means that we will decide to adjust the
machine when it does not need adjustment.
 This means, in the case of the robot welder, if the
machine is running properly, there is only a 0.05
probability of our making the mistake of concluding
that the robot requires adjustment when it really
does not.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 239
Three Types of Tests
 Nondirectional, two-tail test:
 H1: pop parameter n.e. value
 Directional, right-tail test:
 H1: pop parameter > value
 Directional, left-tail test:
 H1: pop parameter < value
Always put hypotheses in terms of population
parameters and have
H0: pop parameter = value

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 240
Two tailed test
a/2 a/21a
–z +z
Do Not
Reject H 0
00
Reject HReject H
H0: pop parameter = value
H1: pop parameter n.e. value

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 241
Right tailed test
H0: pop parameter  value
H1: pop parameter > value
a1a
+z
Do Not Reject H 00 Reject H

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 242
Left tailed test
H0: pop parameter  value
H1: pop parameter < value
a 1a
–z
Do Not Reject H 0Reject H 0

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 243
H1
Ho
Ho H1
OK
OK
Type I
Error
Type II
Error

Example
 Suppose you have two coins in your pocket.
– One is a fair coin—i.e., p(Head) = p(Tail) = 0.5.
– The other coin is biased toward Tails: p(Head) = .15, p(Tail)
= .85.

 Then place the two coins on a table, and choose one
of them.
 You take the selected coin, and flip it 11 times, noting
each time whether it showed Heads or Tails.
 Let X = the number of Heads observed in 11 flips.
 Let Hypothesis A (Null Hypothesis) be that you
selected and flipped the fair coin.
 Let Hypothesis B (Alternate Hypothesis) be that you
selected and flipped the biased coin.
 Under what circumstances would you decide that
Hypothesis A is true?
 Under what circumstances would you decide that
Hypothesis B is true?

 A good way to start is to think about what kinds
of outcomes you would expect for each
hypothesis.
 For example, if hypothesis A is true (i.e., the
coin is fair), you expect the number of Heads to
be somewhere in the middle of the 0-11 range.
 But if hypothesis B is true (i.e., the coin is
biased towards tails), you probably expect the
number of Heads to be quite small.
 Note as well that a very large number of Heads
is improbable in either case, but is even less
probable if the coin is biased towards tails.

Then what is the decision?
 IF the number of heads is LOW THEN decide that
the coin is biased towards TAILS (Hypothesis B)
 ELSE decide that the coin is fair (Hypothesis A)
 But an obvious problem now is, how low is low?
 The answer is really quite simple.
 The key is to recognize that the variable X (the
number of Heads) has a binomial distribution.
 Furthermore, if Hypothesis A is true, X will have a
binomial
 distribution with N = 11, p = .5, and q = .5.
 But if hypothesis B is true, then X will have a binomial

Two Binomial Distributions with N = 11 and X = # of
Heads

 We are now in a position to compare conditional probabilities
for particular experimental outcomes.
 For example, if we actually did carry out the coin tossing
experiment and obtained 3 Heads (X=3), we would know that
the probability of getting exactly 3 Heads is lower if Hypothesis
A is true (.0806) than it is if Hypothesis B is true (.1517).
 Therefore, we might decide that Hypothesis B is true if the
outcome was X = 3 Heads.
 But what if we had obtained 4 Heads (X=4) rather than 3?
 In this case the probability of exactly 4 Heads is higher if
Hypothesis A is true (.1611) than it is if hypothesis B is true
(.0536).
 So in this case, we would probably decide that Hypothesis A is
true (i.e., the coin is fair).

And there is always a chance of an
error!
 Note that even if the coin is biased towards tails, it is
possible for the number of Heads to be very large; and
if the coin is fair, it is possible to observe very few
Heads.
 No matter which hypothesis we choose, therefore,
there is always the possibility of making an error.
 However, the use of the decision rule described here
will minimize the overall probability of error.
 In the present example, this rule would lead us to
decide that the coin is biased if the number of Heads
was 3 or less; but for any other outcome, we would
conclude that the coin is fair.

Decision rule to minimize the overall probability of error
Rejection region

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 252
What Test to Apply?
Ask the following questions:
 Are the data the result of a measurement (a
continuous variable) or a count (a discrete
variable)?
 Is  known?
 What shape is the distribution of the
population parameter?
 What is the sample size?

 Sample Size Requirements for Estimating a Mean or Mean
Difference
where
n = sample size
s = standard deviation
d = margin of error (at 95% confidence level)
 For paired samples
2
2
4
d
s
n 
2
2
4
d
s
n d

where
sd= standard deviation of the DELTA variable

 For independent samples
2
2
4
d
s
n
p

where
sp= pooled estimate of standard deviation

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 256
Test of µ,  Known, Population
Normally Distributed
 Test Statistic:
 where
 is the sample statistic.
 µ0 is the value identified in the null hypothesis.
  is known.
 n is the sample size.
n
x
z  0
–

x

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 257
Test of µ,  Known, Population Not
 If n > 30, Test Statistic:
 If n < 30, use a distribution-free test.
n
x
z  0
–


9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 258
Test of µ,  Unknown, Population
 Test Statistic:
 where
 is the sample statistic.
 µ0 is the value identified in the null hypothesis.
  is unknown.
 n is the sample size
 degrees of freedom on t are n – 1.
x
x–
n
st 0

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 259
Test of µ,  Unknown, Population Not
 If n > 30, Test Statistic:
 If n < 30, use a distribution-free test.
t
x –
0
s
n

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 260
Observed Significance Levels
 A p-Value is:
 the exact level of significance of the test statistic.
 the smallest value a can be and still allow us to reject the null
hypothesis.
 the amount of area left in the tail beyond the test statistic for a
one-tailed hypothesis test or
 twice the amount of area left in the tail beyond the test
statistic for a two-tailed test.
 the probability of getting a test statistic from another sample
that is at least as far from the hypothesized mean as this
sample statistic is.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 261
Several Samples
 Independent Samples:
 Testing a company’s
claim that its peanut
butter contains less fat
than that produced by a
competitor.
 Dependent Samples:
 Testing the relative fuel
efficiency of 10 trucks
that run the same route
twice, once with the
current air filter
installed and once with
the new filter.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 262
Test of (µ1 – µ2), 1 = 2, Populations
Normal
 Test Statistic
 where degrees of freedom on t = n1 + n2 – 2
2–
21
2
2
)1–
2
(2
1
)1–
1
(
2where
2
1
1
12
]
2
–
1
[–]
2
–
1
[
nn
snsn
ps
nnps
xx
t





!!
!
!
!
!
!
!!
!
!
!
!
!


9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 263
The mean of population 1 is equal to the mean of
population 2
(1) Both distributions are normal
2 1 = 2
Hypothesis
Assumption
Test Statistic t -distribution with df = n1+ n2-2
        2/11/1/1 21
2
22
2
1121
21



nnsnsnnn
XX
t
H0: pop1 = pop2
H1: pop1 ≠ pop2
Example:
Comparing Two populations

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 264
-5 0 5
0.0
0.1
0.2
0.3
0.4
0.5
t
t(x;nu)
nu=5
nu=50
t -distribution with df = n1+ n2-2
        2/11/1/1 21
2
22
2
1121
21



nnsnsnnn
XX
t
Rejection
Region
Rejection
Region
Example:
Comparing Two populations

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 265
Test of (µ1 – µ2), 1 n.e. 2,
Populations Normal, large n
 Test Statistic
 with s1
2 and s2
2 as estimates for 1
2 and 2
2
z 
[x
1
–x
2
]–[
1
–
2
]
0
s
1
2
n
1

s
2
2
n
2

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 266
Test of Dependent Samples
(µ1 – µ2) = µd
 Test Statistic
 where d = (x1 – x2)
= Sd/n, the average difference
n = the number of pairs of observations
sd = the standard deviation of d
df = n – 1
n
d
s
dt 
d

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 267
Test of Equal Variances
 Pooled-variances t-test assumes the two
population variances are equal.
 The F-test can be used to test that assumption.
 The F-distribution is the sampling distribution
of s1
2/s2
2 that would result if two samples were
repeatedly drawn from a single normally
distributed population.

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 268
Test of 1
2 = 2
2
 If 1
2 = 2
2 , then 1
2/2
2 = 1. So the hypotheses can be
worded either way.
 Test Statistic: (whichever is larger)
 The critical value of the F will be F(a/2, n1, n2)
 where a = the specified level of significance
n1 = (n – 1), where n is the size of the sample
with the larger variance
n2 = (n – 1), where n is the size of the sample
with the smaller variance
2
1
2
2or
2
2
2
1
s
s
s
s
F 

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 269
Confidence Interval for (µ1 – µ2)
 The (1 – a)% confidence interval for the
difference in two means:
 Equal variances, populations normal
 Unequal variances, large samples
!
!
!
!
!
!
!
!
!
!
!
!
!
!
‫׳‬±
2
1
1
12
2
)
2
–
1
(
nn
p
stxx a
2
2
2
1
2
1
2
)
2
–
1
(
n
s
n
s
zxx ‫׳‬± a

9/8/2017 (c) 2001, Ron S. Kenett, Ph.D. 270
The mean of population 1 is equal to the mean of
population 2
(1) Both distributions are normal
2 1 = 2
Hypothesis
Assumption
Test Statistic
The standard deviation of population 1 is equal to
the standard deviation of population 2
Both distributions are normal
The proportion of error in population 1 is equal to
the proportion of errors in population 2
n1p1 and n2p2 > 5 (approximation by normal
distribution)
F distribution with
df2 = n1-1 and df2 = n2-12
2
2
1
s
s
F 
t distribution with df = n1+ n2-2
        2/11/1/1 21
2
22
2
1121
21



nnsnsnnn
XX
t
Z - Normal distribution
  21
2211
/1/11
//
nnpp
nXnX
Z
avgavg 


21
21
nn
XX
pavg



Summary

A Case Study
Consider the development of a placement machine that picks
components from a tray and positions them on printed circuit
boards.
The customer requirements involve precision in the x-y position.
The developers of the system collected data from 26 boards,
with 16 components on each.
For each board the deviations in x and y, from the required
nominal values, were recorded, producing 416 values for x_dev
and y_dev.

A Case Study
-0.003 -0.002 -0.001 0.000 0.001 0.002 0.003 0.004 0.005
-0.003
-0.002
-0.001
0.000
0.001
0.002
0.003
x_dev
y_dev
Figure 1: Scatter plot of y deviations versus x deviations

A Case Study
1
2
3
-0.003 -0.002 -0.001 0.000 0.001 0.002 0.003 0.004 0.005
-0.003
-0.002
-0.001
0.000
0.001
0.002
0.003
x_dev
y_dev
Figure 2: Scatter plot of y deviations versus x deviations with coding variable

 
YX
YXYX
r
 


X Y
2 2
3 1
1 2
4 3
3 5
5 4
3.00 2.83
1.41 1.47
XY
4
3
2
12
15
20
9.33
Y
XX
Y
X6
Y6
Mean
StDev
r = (9.33 - 3.00*2.83) /
(1.41*1.47) = 0.41
The Correlation Coefficient

Coefficient of Correlation
 A measure of the
 Direction of the linear relationship between
x and y.
 If x and y are directly related, r > 0.
 If x and y are inversely related, r < 0.
 Strength of the linear relationship between
x and y.
 The larger the absolute value of r, the more the
value of y depends in a linear way on the value of x.

2.0
4.0
6.0
8.0
10.0
2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0
Y
X
Predicted Y
X Y
7.6 5.6
4.2 3.9
10.2 6.7
13.8 7.9
13.6 7.7
14.5 8.4
3.2 3.2
7.9 5.9
13.4 7.1
4.7 4.9
5.9 4.4
3.5 3.7
3.4 3.2
5.0 3.1
5.6 4.5
Y = 2.07 + 0.424*X +
e
R2 = 94%
Confidence Limits
The Linear Regression Model

Simple Linear Regression Model
 Probabilistic Model: yi = b0 + b1xi + ei
where yi = a value of the dependent variable, y
xi = a value of the independent variable, x
b0 = the y-intercept of the regression line
b1 = the slope of the regression line
ei = random error, the residual
 Deterministic Model:
= b0 + b1xi where
and is the predicted value of y in contrast to the actual
value of y.
ˆy
i b
0
b
0
, b
1
b
1
ˆy
i

Determining the Least Squares Regression Line
 Least Squares Regression Line:
 Slope
 y-intercept
ˆy  b0
 b1
x1
b
1

( x
i
y
i
) – n x y
( x
i
2) – n x2
b
0
 y – b
1
x



Coefficient of Determination (R2)
 A measure of the
 Strength of the linear relationship
between x and y.
 The larger the value of R2, the more the value
of y depends in a linear way on the value of x.
 Amount of variation in y that is related to
variation in x.
 Ratio of variation in y that is explained
by the regression model divided by the
total variation in y.

د حاتم البيطار استشاري وجراح الفم والاسنان 01005684344 اتصل للحجز بالعيادة Dr.Hatem EL-Bitar

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د حاتم البيطار استشاري وجراح الفم والاسنان 01005684344 اتصل للحجز بالعيادة Dr.Hatem EL-Bitar