biostatfinal_(2).ppt

Lectures of Stat -145
(Biostatistics)
Text book
Biostatistics
Basic Concepts and Methodology for the Health
Sciences
By
Wayne W. Daniel

Text Book : Basic Concepts and
Methodology for the Health
Sciences 2
Chapter 1
Introduction To
Biostatistics

Sciences 3
 Key words :
 Statistics , data , Biostatistics,
 Variable ,Population ,Sample

Text Book : Basic Concepts
and Methodology for the Health
4
Introduction
Some Basic concepts
Statistics is a field of study concerned
with
1- collection, organization, summarization
and analysis of data.
2- drawing of inferences about a body of
data when only a part of the data is
observed.
Statisticians try to interpret and
communicate the results to others.

5
* Biostatistics:
The tools of statistics are employed in
many fields:
business, education, psychology,
agriculture, economics, … etc.
When the data analyzed are derived from
the biological science and medicine,
we use the term biostatistics to
distinguish this particular application of
statistical tools and concepts.

6
Data:
• The raw material of Statistics is data.
• We may define data as figures. Figures
result from the process of counting or
from taking a measurement.
• For example:
• - When a hospital administrator counts
the number of patients (counting).
• - When a nurse weighs a patient
(measurement)

7
We search for suitable data to serve as
the raw material for our investigation.
Such data are available from one or more
of the following sources:
1- Routinely kept records.
For example:
- Hospital medical records contain
immense amounts of information on
patients.
- Hospital accounting records contain a
wealth of data on the facility’s business
- activities.
* Sources of Data:

8
2- External sources.
The data needed to answer a question may
already exist in the form of
published reports, commercially available
data banks, or the research literature,
i.e. someone else has already asked the
same question.

9
3- Surveys:
The source may be a survey, if the data
needed is about answering certain
questions.
For example:
If the administrator of a clinic wishes to
obtain information regarding the mode of
transportation used by patients to visit
the clinic,
then a survey may be conducted among
patients to obtain this information.

10
4- Experiments.
Frequently the data needed to answer
a question are available only as the
result of an experiment.
For example:
If a nurse wishes to know which of several
strategies is best for maximizing patient
compliance,
she might conduct an experiment in which the
different strategies of motivating compliance
are tried with different patients.

11
* A variable:
It is a characteristic that takes on
different values in different persons,
places, or things.
For example:
- heart rate,
- the heights of adult males,
- the weights of preschool children,
- the ages of patients seen in a dental
clinic.

12
Quantitative Variables
It can be measured
in the usual sense.
For example:
- the heights of
adult males,
- the weights of
preschool children,
- the ages of
patients seen in a
- dental clinic.
Qualitative Variables
Many characteristics are
not capable of being
measured. Some of them
can be ordered or
ranked.
For example:
- classification of people into
socio-economic groups,
- social classes based on
income, education, etc.
Types of variables
Quantitative Qualitative

13
A discrete variable
is characterized by
gaps or interruptions
in the values that it
can assume.
For example:
- The number of daily
admissions to a
general hospital,
- The number of
decayed, missing or
filled teeth per child
- in an
- elementary
- school.
A continuous variable
can assume any value within a
specified relevant interval of
values assumed by the variable.
For example:
- Height,
- weight,
- skull circumference.
No matter how close together the
observed heights of two people,
we can find another person
whose height falls somewhere
in between.
Types of quantitative variables
Discrete Continuous

14
* A population:
It is the largest collection of values of a
random variable for which we have an
interest at a particular time.
For example:
The weights of all the children enrolled in
a certain elementary school.
Populations may be finite or infinite.

15
* A sample:
It is a part of a population.
For example:
The weights of only a fraction of
these children.

Types of Data
Data
Constant Variable
16

These are observations that remain the same
from person to person, from time to time, or
from place to place.
Examples;
1- number of eyes, fingers, ears… etc.
2- number of minutes in an hour
3- the speed of light
4- no. of centimeters in an inch
CONSTANT DATA
17

VARIABLE DATA 1
These are observations, which vary from one
person to another or from one group of members
to others and are classified as following:
 Statistically:
1. Quantitative variable data
2. Qualitative variable data
 Epidemiologically:
1. Dependant (outcome variable)
2. Independent (study variables)
 Clinically:
- Measured (BP, Lab. parameters, etc.)
- Counted (Pulse rate, resp. rate, etc.)
- Observed (Jaundice, pallor, wound infection)
- Subjective (headache, colic, etc.)
18

VARIABLE DATA 2
Statistically, variable could be:
- Quantitative variable:
a- Continuous quantitative
b- Discrete quantitative
- Qualitative variable:
a- Nominal qualitative
b- Ordinal qualitative
19

VARIABLE DATA 3
1- Quantitative variable:
These may be continuous or discrete.
a- Continuous quantitative variable:
Which are obtained by measurement and its value could
be integer or fractionated value.
Examples: Weight, height, Hgb, age, volume of urine.
b-Discrete quantitative variable:
Which are obtained by enumeration and its value is
always integer value.
Examples: Pulse, family size, number of live births.
20

Continuous Variable
0 3
2
1
-2 -1
-
3
0 1 2 3
Discrete Variable
Continuous &
Discrete Variables
21

2- Qualitative variable:
Which are expressed in quality and cannot be
enumerated or measured but can be categorized only.
They can be ordinal or nominal.
a- Nominal qualitative: can not be put in order, and is
further subdivided into dichotomous (e.g. sex,
male/female and Yes/No variables) and
multichotomous (e.g. blood groups, A, B, AB, O).
b- Ordinal qualitative: can be put in order. e.g. degree of
success, level of education, stage of disease.
VARIABLE DATA 4
22

Epidemiologically, variable could be:
Dependent Variable:
Usually the health outcome(s) that you are studying.
Independent Variables:
Risk factors, casual factors, experimental treatment,
and other relevant factors. They also termed
“predictors”.
e.g. Cancer lung is the dependent variable while
smoking is independent variable.
VARIABLE DATA 5
23

Section (2.4) :
Descriptive Statistics
Measures of Central
Tendency
Page 38 - 41

Methodology for the Health Sciences 25
key words:
Descriptive Statistic, measure of
central tendency ,statistic, parameter,
mean (μ) ,median, mode.

The Statistic and The Parameter
• A Statistic:
It is a descriptive measure computed from the
data of a sample.
• A Parameter:
It is a a descriptive measure computed from
the data of a population.
Since it is difficult to measure a parameter from the
population, a sample is drawn of size n, whose
values are  1 ,  2 , …,  n. From this data, we
measure the statistic.

Measures of Central Tendency
A measure of central tendency is a measure which
indicates where the middle of the data is.
The three most commonly used measures of central
tendency are:
The Mean, the Median, and the Mode.
The Mean:
It is the average of the data.

The Population Mean:
 = which is usually unknown, then we use the
sample mean to estimate or approximate it.
The Sample Mean:
=
Example:
Here is a random sample of size 10 of ages, where
 1 = 42,  2 = 28,  3 = 28,  4 = 61,  5 = 31,
 6 = 23,  7 = 50,  8 = 34,  9 = 32,  10 = 37.
= (42 + 28 + … + 37) / 10 = 36.6
x
1
N
i
i
N
X


x
1
n
i
i
n
x



Properties of the Mean:
• Uniqueness. For a given set of data there is
one and only one mean.
• Simplicity. It is easy to understand and to
compute.
• Affected by extreme values. Since all
values enter into the computation.
Example: Assume the values are 115, 110, 119, 117, 121 and
126. The mean = 118.
But assume that the values are 75, 75, 80, 80 and 280. The
mean = 118, a value that is not representative of the set of
data as a whole.

The Median:
When ordering the data, it is the observation that divide the
set of observations into two equal parts such that half of
the data are before it and the other are after it.
* If n is odd, the median will be the middle of observations. It
will be the (n+1)/2 th ordered observation.
When n = 11, then the median is the 6th observation.
* If n is even, there are two middle observations. The median
will be the mean of these two middle observations. It will
be the (n+1)/2 th ordered observation.
When n = 12, then the median is the 6.5th observation, which
is an observation halfway between the 6th and 7th ordered
observation.

Example:
For the same random sample, the ordered
observations will be as:
23, 28, 28, 31, 32, 34, 37, 42, 50, 61.
Since n = 10, then the median is the 5.5th
observation, i.e. = (32+34)/2 = 33.
Properties of the Median:
• Uniqueness. For a given set of data there is
one and only one median.
• Simplicity. It is easy to calculate.
• It is not affected by extreme values as
is the mean.

The Mode:
It is the value which occurs most frequently.
If all values are different there is no mode.
Sometimes, there are more than one mode.
Example:
For the same random sample, the value 28 is
repeated two times, so it is the mode.
Properties of the Mode:
• Sometimes, it is not unique.
• It may be used for describing qualitative
data.

Section (2.5) :
Descriptive Statistics
Measures of Dispersion
Page 43 - 46

key words:
Descriptive Statistic, measure of
dispersion , range ,variance, coefficient of
variation.

2.5. Descriptive Statistics –
Measures of Dispersion:
• A measure of dispersion conveys information
regarding the amount of variability present in a set of
data.
• Note:
1. If all the values are the same
→ There is no dispersion .
2. If all the values are different
→ There is a dispersion:
3.If the values close to each other
→The amount of Dispersion small.
b) If the values are widely scattered
→ The Dispersion is greater.

Ex. Figure 2.5.1 –Page 43
• ** Measures of Dispersion are :
1.Range (R).
2. Variance.
3. Standard deviation.
4.Coefficient of variation (C.V).

1.The Range (R):
• Range =Largest value- Smallest value =
• Note:
• Range concern only onto two values
• Example 2.5.1 Page 40:
• Refer to Ex 2.4.2.Page 37
• Data:
• 43,66,61,64,65,38,59,57,57,50.
• Find Range?
• Range=66-38=28
S
L x
x 

2.The Variance:
• It measure dispersion relative to the scatter of the values
a bout there mean.
a) Sample Variance ( ) :
• ,where is sample mean
• Example 2.5.2 Page 40:
• Refer to Ex 2.4.2.Page 37
• Find Sample Variance of ages , = 56
• Solution:
• S2= [(43-56) 2 +(66-56) 2+…..+(50-56) 2 ]/ 10
• = 900/10 = 90
x
2
S
1
)
(
1
2
2





n
x
x
S
n
i
i
x

• b)Population Variance ( ) :
• where , is Population mean
3.The Standard Deviation:
• is the square root of variance=
a) Sample Standard Deviation = S =
b) Population Standard Deviation = σ =
2

N
x
N
i
i



 1
2
2
)
( 

Varince
2
S
2


4.The Coefficient of Variation
(C.V):
• Is a measure used to compare the
dispersion in two sets of data which is
independent of the unit of the
measurement .
• where S: Sample standard
deviation.
• : Sample mean.
)
100
(
.
X
S
V
C 
X

Example 2.5.3 Page 46:
• Suppose two samples of human males yield the
following data:
Sampe1 Sample2
Age 25-year-olds 11year-olds
Mean weight 145 pound 80 pound
Standard deviation 10 pound 10 pound

• We wish to know which is more variable.
• Solution:
• c.v (Sample1)= (10/145)*100%= 6.9%
• c.v (Sample2)= (10/80)*100%= 12.5%
• Then age of 11-years old(sample2) is more
variation

Chapter 4:
Probabilistic features of
certain data Distributions
Pages 93- 111

Text Book : Basic Concepts and Methodology for the
Health Sciences
44
Key words
Probability distribution , random variable ,
Bernolli distribution, Binomail distribution,
Poisson distribution

Health Sciences
45
The Random Variable (X):
When the values of a variable (height,
weight, or age) can’t be predicted in
advance, the variable is called a random
variable.
An example is the adult height.
When a child is born, we can’t predict
exactly his or her height at maturity.

Health Sciences
46
4.2 Probability Distributions for
Discrete Random Variables
Definition:
The probability distribution of a
discrete random variable is a table,
graph, formula, or other device used
to specify all possible values of a
discrete random variable along with
their respective probabilities.

Health Sciences
47
The Cumulative Probability
Distribution of X, F(x):
It shows the probability that the
variable X is less than or equal to a
certain value, P(X  x).

Sciences 48
Example 4.2.1 page 94:
F(x)=
P(X≤ x)
P(X=x)
frequenc
y
Number of
Programs
0.2088
0.2088
62
1
0.3670
0.1582
47
2
0.4983
0.1313
39
3
0.6296
0.1313
39
4
0.8249
0.1953
58
5
0.9495
0.1246
37
6
0.9630
0.0135
4
7
1.0000
0.0370
11
8
1.0000
297
Total

Health Sciences
49
See figure 4.2.1 page 96
See figure 4.2.2 page 97
Properties of probability distribution
of discrete random variable.
1.
2.
3. P(a  X  b) = P(X  b) – P(X  a-1)
4. P(X < b) = P(X  b-1)
0 ( ) 1
P X x
  
( ) 1
P X x
 


Health Sciences
50
Example 4.2.2 page 96: (use table
in example 4.2.1)
What is the probability that a randomly
selected family will be one who used
three assistance programs?
in example 4.2.1)
selected family used either one or two
programs?

Health Sciences
51
Example 4.2.4 page 98: (use table in
example 4.2.1)
What is the probability that a family picked
at random will be one who used two or
fewer assistance programs?
example 4.2.1)
selected family will be one who used fewer
than four programs?
example 4.2.1)
selected family used five or more
programs?

Health Sciences
52
in example 4.2.1)
selected family is one who used
between three and five programs,
inclusive?

Health Sciences
53
4.3 The Binomial Distribution:
The binomial distribution is one of the most
widely encountered probability distributions
in applied statistics. It is derived from a
process known as a Bernoulli trial.
Bernoulli trial is :
When a random process or experiment
called a trial can result in only one of two
mutually exclusive outcomes, such as dead
or alive, sick or well, the trial is called a
Bernoulli trial.

Health Sciences
54
The Bernoulli Process
A sequence of Bernoulli trials forms a Bernoulli
process under the following conditions
1- Each trial results in one of two possible,
mutually exclusive, outcomes. One of the
possible outcomes is denoted (arbitrarily) as a
success, and the other is denoted a failure.
2- The probability of a success, denoted by p,
remains constant from trial to trial. The
probability of a failure, 1-p, is denoted by q.
3- The trials are independent, that is the outcome
of any particular trial is not affected by the
outcome of any other trial

Health Sciences
55
The probability distribution of the binomial
random variable X, the number of
successes in n independent trials is:
Where is the number of combinations
of n distinct objects taken x of them at a
time.
* Note: 0! =1
( ) ( ) , 0,1,2,....,
X n X
n
f x P X x p q x n
x

 
   
 
 
 
n
x
 
 
 
 
!
!( )!
n n
x n x
x
 

 
  
 
! ( 1)( 2)....(1)
x x x x
  

Health Sciences
56
Properties of the binomial
distribution
1.
2.
3.The parameters of the binomial
distribution are n and p
4.
5.
( ) 0
f x 
( ) 1
f x 

( )
E X np
  
2
var( ) (1 )
X np p
   

Health Sciences
57
Example 4.3.1 page 100
If we examine all birth records from the North
Carolina State Center for Health statistics for
year 2001, we find that 85.8 percent of the
pregnancies had delivery in week 37 or later
(full- term birth).
If we randomly selected five birth records from
this population what is the probability that
exactly three of the records will be for full-term
births?
Exercise: example 4.3.2 page 104

Health Sciences
58
Suppose it is known that in a certain
population 10 percent of the population is
color blind. If a random sample of 25
people is drawn from this population, find
the probability that
a) Five or fewer will be color blind.
b) Six or more will be color blind
c) Between six and nine inclusive will be color
blind.
d) Two, three, or four will be color blind.
Exercise: example 4.3.4 page 106

Health Sciences
59
4.4 The Poisson Distribution
If the random variable X is the number of
occurrences of some random event in a certain
period of time or space (or some volume of
matter).
The probability distribution of X is given by:
f (x) =P(X=x) = ,x = 0,1,…..
The symbol e is the constant equal to 2.7183.
(Lambda) is called the parameter of the
distribution and is the average number of
occurrences of the random event in the interval
(or volume)
!
x
x
e  



Health Sciences
60
Properties of the Poisson
distribution
1.
2.
3.
4.
( ) 0
f x 
( ) 1
f x 

( )
E X
 
 
2
var( )
X
 
 

Health Sciences
61
In a study of a drug -induced anaphylaxis
among patients taking rocuronium bromide
as part of their anesthesia, Laake and
Rottingen found that the occurrence of
anaphylaxis followed a Poisson model with
=12 incidents per year in Norway .Find
1- The probability that in the next year,
among patients receiving rocuronium,
exactly three will experience anaphylaxis?


Health Sciences
62
2- The probability that less than two patients
receiving rocuronium, in the next year will
experience anaphylaxis?
3- The probability that more than two patients
receiving rocuronium, in the next year will
experience anaphylaxis?
4- The expected value of patients receiving
rocuronium, in the next year who will
experience anaphylaxis.
5- The variance of patients receiving
experience anaphylaxis
6- The standard deviation of patients receiving
experience anaphylaxis

Health Sciences
63
Example 4.4.2 page 111: Refer to
example 4.4.1
1-What is the probability that at least three
patients in the next year will experience
anaphylaxis if rocuronium is administered
with anesthesia?
2-What is the probability that exactly one
patient in the next year will experience
with anesthesia?
3-What is the probability that none of the
patients in the next year will experience
with anesthesia?

Health Sciences
64
4-What is the probability that at most
two patients in the next year will
experience anaphylaxis if rocuronium
is administered with anesthesia?
Exercises: examples 4.4.3, 4.4.4
and 4.4.5 pages111-113
Exercises: Questions 4.3.4 ,4.3.5,
4.3.7 ,4.4.1,4.4.5

4.5 Continuous
Probability Distribution
Pages 114 – 127

66
• Key words:
Continuous random variable, normal
distribution , standard normal
distribution , T-distribution

67
• Now consider distributions of
continuous random variables.

68
1- Area under the curve = 1.
2- P(X = a) = 0 , where a is a constant.
3- Area between two points a , b =
P(a<x<b) .
Properties of continuous
probability Distributions:

69
4.6 The normal distribution:
• It is one of the most important probability
distributions in statistics.
• The normal density is given by
• , - ∞ < x < ∞, - ∞ < µ < ∞, σ > 0
• π, e : constants
• µ: population mean.
• σ : Population standard deviation.
2
2
2
)
(
2
1
)
( 






x
e
x
f

70
Characteristics of the normal
distribution: Page 111
• The following are some important
characteristics of the normal distribution:
1- It is symmetrical about its mean, µ.
2- The mean, the median, and the mode are all
equal.
3- The total area under the curve above the
x-axis is one.
4-The normal distribution is completely
determined by the parameters µ and σ.

71
5- The normal distribution
depends on the two
parameters  and .
 determines the
location of
the curve.
(As seen in figure 4.6.3) ,
But,  determines
the scale of the curve, i.e.
the degree of flatness or
peaked ness of the curve.
(as seen in figure 4.6.4)
1 2 3
1 < 2 < 3

1
2
3
1 < 2 < 3

72
Note that : (As seen in Figure
4.6.2)
1. P( µ- σ < x < µ+ σ) = 0.68
2. P( µ- 2σ< x < µ+ 2σ)= 0.95
3. P( µ-3σ < x < µ+ 3σ) = 0.997

73
The Standard normal
distribution:
• Is a special case of normal distribution
with mean equal 0 and a standard deviation
of 1.
• The equation for the standard normal
distribution is written as
• , - ∞ < z < ∞
2
2
2
1
)
(
z
e
z
f




74
Characteristics of the
standard normal distribution
1- It is symmetrical about 0.
2- The total area under the curve
above the x-axis is one.
3- We can use table (D) to find the
probabilities and areas.

75
“How to use tables of Z”
Note that
The cumulative probabilities P(Z  z) are given in
tables for -3.49 < z < 3.49. Thus,
P (-3.49 < Z < 3.49)  1.
For standard normal distribution,
P (Z > 0) = P (Z < 0) = 0.5
Example 4.6.1:
If Z is a standard normal distribution, then
1) P( Z < 2) = 0.9772
is the area to the left to 2
and it equals 0.9772.
2

76
Example 4.6.2:
P(-2.55 < Z < 2.55) is the area between
-2.55 and 2.55, Then it equals
P(-2.55 < Z < 2.55) =0.9946 – 0.0054
= 0.9892.
Example 4.6.2:
P(-2.74 < Z < 1.53) is the area between
-2.74 and 1.53.
P(-2.74 < Z < 1.53) =0.9370 – 0.0031
= 0.9339.
-2.74 1.53
-2.55 2.55
0

77
Example 4.6.3:
P(Z > 2.71) is the area to the right to 2.71.
So,
P(Z > 2.71) =1 – 0.9966 = 0.0034.
Example :
P(Z = 0.84) is the area at z = 2.71.
So,
P(Z = 0.84) =1 – 0.9966 = 0.0034
0.84
2.71

78
How to transform normal
distribution (X) to standard
normal distribution (Z)?
• This is done by the following formula:
• Example:
• If X is normal with µ = 3, σ = 2. Find the
value of standard normal Z, If X= 6?
• Answer:




x
z
5
.
1
2
3
6







x
z

79
4.7 Normal Distribution Applications
The normal distribution can be used to model the distribution of
many variables that are of interest. This allow us to answer
probability questions about these random variables.
Example 4.7.1:
The ‘Uptime ’is a custom-made light weight battery-operated
activity monitor that records the amount of time an individual
spend the upright position. In a study of children ages 8 to 15
years. The researchers found that the amount of time children
spend in the upright position followed a normal distribution with
Mean of 5.4 hours and standard deviation of 1.3.Find

80
If a child selected at random ,then
1-The probability that the child spend less than 3
hours in the upright position 24-hour period
P( X < 3) = P( < ) = P(Z < -1.85) = 0.0322
-------------------------------------------------------------------------
2-The probability that the child spend more than 5
P( X > 5) = P( > ) = P(Z > -0.31)
= 1- P(Z < - 0.31) = 1- 0.3520= 0.648
-----------------------------------------------------------------------
3-The probability that the child spend exactly 6.2
P( X = 6.2) = 0



X
3
.
1
4
.
5
3 



X
3
.
1
4
.
5
5 

81
4-The probability that the child spend from 4.5 to
7.3 hours in the upright position 24-hour period
P( 4.5 < X < 7.3) = P( < < )
= P( -0.69 < Z < 1.46 ) = P(Z<1.46) – P(Z< -0.69)
= 0.9279 – 0.2451 = 0.6828
• Hw…EX. 4.7.2 – 4.7.3



X
3
.
1
4
.
5
5
.
4 
3
.
1
4
.
5
3
.
7 

82
6.3 The T Distribution:
(167-173)
1- It has mean of zero.
2- It is symmetric about the
mean.
3- It ranges from - to .
0

83
4- compared to the normal distribution,
the t distribution is less peaked in the
center and has higher tails.
5- It depends on the degrees of freedom
(n-1).
6- The t distribution approaches the
standard normal distribution as (n-1)
approaches .

84
Examples
t (7, 0.975) = 2.3646
------------------------------
t (24, 0.995) = 2.7696
--------------------------
If P (T(18) > t) = 0.975,
then t = -2.1009
-------------------------
If P (T(22) < t) = 0.99,
then t = 2.508
0.005
t (24, 0.995)
0.995
t (7, 0.975)
0.025
0.975
t
0.975
0.025
0.99
0.01
t

Chapter 7
Using sample statistics to
Test Hypotheses
about population
parameters
Pages 215-233

Methodology for the Health Sciences
86
 Key words :
 Null hypothesis H0, Alternative hypothesis HA ,
testing hypothesis , test statistic , P-value

87
Hypothesis Testing
 One type of statistical inference, estimation,
was discussed in Chapter 6 .
 The other type ,hypothesis testing ,is discussed
in this chapter.

88
Definition of a hypothesis
 It is a statement about one or more populations .
It is usually concerned with the parameters of
the population. e.g. the hospital administrator
may want to test the hypothesis that the average
length of stay of patients admitted to the
hospital is 5 days

89
Definition of Statistical hypotheses
 They are hypotheses that are stated in such a way that
they may be evaluated by appropriate statistical
techniques.
 There are two hypotheses involved in hypothesis
testing
 Null hypothesis H0: It is the hypothesis to be tested .
 Alternative hypothesis HA : It is a statement of what
we believe is true if our sample data cause us to reject
the null hypothesis

90
7.2 Testing a hypothesis about the
mean of a population:
 We have the following steps:
1.Data: determine variable, sample size (n), sample
mean( ) , population standard deviation or sample
standard deviation (s) if is unknown
2. Assumptions : We have two cases:
 Case1: Population is normally or approximately
normally distributed with known or unknown
variance (sample size n may be small or large),
 Case 2: Population is not normal with known or
unknown variance (n is large i.e. n≥30).
x

91
 3.Hypotheses:
 we have three cases
 Case I : H0: μ=μ0
HA: μ μ0
 e.g. we want to test that the population mean is
different than 50
 Case II : H0: μ = μ0
HA: μ > μ0
 e.g. we want to test that the population mean is greater
than 50
 Case III : H0: μ = μ0
HA: μ< μ0
 e.g. we want to test that the population mean is less
than 50



92
4.Test Statistic:
 Case 1: population is normal or approximately
normal
σ2
is known σ2
is unknown
( n large or small)
n large n small
 Case2: If population is not normally distributed and n is
large
 i)If σ2
is known ii) If σ2
is unknown
n
X
Z

o
-

n
s
X
Z o
- 

n
s
X
T o
- 

n
s
X
Z o
- 

n
X
Z

o
-


93
5.Decision Rule:
i) If HA: μ μ0
 Reject H 0 if Z >Z1-α/2 or Z< - Z1-α/2
(when use Z - test)
Or Reject H 0 if T >t1-α/2,n-1 or T< - t1-α/2,n-1
(when use T- test)
 __________________________
 ii) If HA: μ> μ0
 Reject H0 if Z>Z1-α (when use Z - test)
Or Reject H0 if T>t1-α,n-1 (when use T - test)


94
 iii) If HA: μ< μ0
Reject H0 if Z< - Z1-α (when use Z - test)
 Or
Reject H0 if T<- t1-α,n-1 (when use T - test)
Note:
Z1-α/2 , Z1-α , Zα are tabulated values obtained
from table D
t1-α/2 , t1-α , tα are tabulated values obtained from
table E with (n-1) degree of freedom (df)

95
 6.Decision :
 If we reject H0, we can conclude that HA is
true.
 If ,however ,we do not reject H0, we may
conclude that H0 is true.

96
An Alternative Decision Rule using the
p - value Definition
 The p-value is defined as the smallest value of
α for which the null hypothesis can be rejected.
 If the p-value is less than or equal to α ,we
reject the null hypothesis (p ≤ α)
 If the p-value is greater than α ,we do not
reject the null hypothesis (p > α)

97
Example 7.2.1 Page 223
 Researchers are interested in the mean age of a
certain population.
 A random sample of 10 individuals drawn from the
population of interest has a mean of 27.
 Assuming that the population is approximately
normally distributed with variance 20,can we
conclude that the mean is different from 30 years ?
(α=0.05) .
 If the p - value is 0.0340 how can we use it in making
a decision?

98
Solution
1-Data: variable is age, n=10, =27 ,σ2=20,α=0.05
2-Assumptions: the population is approximately
normally distributed with variance 20
3-Hypotheses:
 H0 : μ=30
 HA: μ 30
x


99
4-Test Statistic:
 Z = -2.12
5.Decision Rule
 The alternative hypothesis is
 HA: μ > 30
 Hence we reject H0 if Z >Z1-0.025/2= Z0.975
 or Z< - Z1-0.025/2= - Z0.975
 Z0.975=1.96(from table D)

100
 6.Decision:
 We reject H0 ,since -2.12 is in the rejection
region .
 We can conclude that μ is not equal to 30
 Using the p value ,we note that p-value
=0.0340< 0.05,therefore we reject H0

101
Example7.2.2 page227
 Referring to example 7.2.1.Suppose that the
researchers have asked: Can we conclude
that μ<30.
1.Data.see previous example
2. Assumptions .see previous example
3.Hypotheses:
 H0 μ =30
 HِA: μ < 30

102
4.Test Statistic :
 = = -2.12
5. Decision Rule: Reject H0 if Z< Z α, where
 Z α= -1.645. (from table D)
6. Decision: Reject H0 ,thus we can conclude that the
population mean is smaller than 30.
n
X
Z

o
-

10
20
30
27 

103
 Among 157 African-American men ,the mean
systolic blood pressure was 146 mm Hg with a
standard deviation of 27. We wish to know if
on the basis of these data, we may conclude
that the mean systolic blood pressure for a
population of African-American is greater than
140. Use α=0.01.

104
Solution
1. Data: Variable is systolic blood pressure,
n=157 , =146, s=27, α=0.01.
2. Assumption: population is not normal, σ2 is
unknown
3. Hypotheses: H0 :μ=140
HA: μ>140
4.Test Statistic:
 = = = 2.78
n
s
X
Z o
- 

157
27
140
146 
1548
.
2
6

105
5. Desicion Rule:
we reject H0 if Z>Z1-α
= Z0.99= 2.33
(from table D)
6. Desicion: We reject H0.
Hence we may conclude that the mean systolic
blood pressure for a population of African-
American is greater than 140.

106
7.3 Hypothesis Testing :The Difference
between two population mean :
1.Data: determine variable, sample size (n), sample means,
population standard deviation or samples standard deviation
(s) if is unknown for two population.
2. Assumptions : We have two cases:
 Case1: Population is normally or approximately normally
distributed with known or unknown variance (sample size
n may be small or large),
 Case 2: Population is not normal with known variances (n
is large i.e. n≥30).

107
 3.Hypotheses:
 Case I : H0: μ 1 = μ2 → μ 1 - μ2 = 0
 HA: μ 1 ≠ μ 2 → μ 1 - μ 2 ≠ 0
 e.g. we want to test that the mean for first population is
different from second population mean.
 Case II : H0: μ 1 = μ2 → μ 1 - μ2 = 0
HA: μ 1 > μ 2 → μ 1 - μ 2 > 0
 e.g. we want to test that the mean for first population is
greater than second population mean.
 Case III : H0: μ 1 = μ2 → μ 1 - μ2 = 0
HA: μ 1 < μ 2 → μ 1 - μ 2 < 0
 e.g. we want to test that the mean for first population
is greater than second population mean.

108
4.Test Statistic:
 Case 1: Two population is normal or approximately
normal
σ2
is known σ2
is unknown if
( n1 ,n2 large or small) ( n1 ,n2 small)
population population Variances
Variances equal not equal
where
2
2
2
1
2
1
2
1
2
1 )
(
-
)
X
-
X
(
n
n
Z







2
1
2
1
2
1
1
1
)
(
-
)
X
-
X
(
n
n
S
T
p 




2
2
2
1
2
1
2
1
2
1 )
(
-
)
X
-
X
(
n
S
n
S
T





2
)
1
(n
)
1
(n
2
1
2
2
2
2
1
1
2






n
n
S
S
Sp

109
 Case2: If population is not normally distributed
 and n1, n2 is large(n1 ≥ 0 ,n2≥ 0)
 and population variances is known,
2
2
2
1
2
1
2
1
2
1 )
(
-
)
X
-
X
(
n
n
Z








110
5.Decision Rule:
i) If HA: μ 1 ≠ μ 2 → μ 1 - μ 2 ≠ 0
(when use Z - test)
Or Reject H 0 if T >t1-α/2 ,(n1+n2 -2) or T< - t1-α/2,,(n1+n2 -2)
(when use T- test)
 __________________________
 ii) HA: μ 1 > μ 2 → μ 1 - μ 2 > 0
 Reject H0 if Z>Z1-α (when use Z - test)
Or Reject H0 if T>t1-α,(n1+n2 -2) (when use T - test)

111
 iii) If HA: μ 1 < μ 2 → μ 1 - μ 2 < 0
Reject H0 if Z< - Z1-α (when use Z - test)
 Or
Reject H0 if T<- t1-α, ,(n1+n2 -2) (when use T - test)
Note:
Z1-α/2 , Z1-α , Zα are tabulated values obtained
from table D
t1-α/2 , t1-α , tα are tabulated values obtained from
table E with (n1+n2 -2) degree of freedom (df)
6. Conclusion: reject or fail to reject H0

112
 Researchers wish to know if the data have collected provide
sufficient evidence to indicate a difference in mean serum
uric acid levels between normal individuals and individual
with Down’s syndrome. The data consist of serum uric
reading on 12 individuals with Down’s syndrome from
normal distribution with variance 1 and 15 normal individuals
from normal distribution with variance 1.5 . The mean
are and α=0.05.
Solution:
1. Data: Variable is serum uric acid levels, n1=12 , n2=15,
σ2
1=1, σ2
2=1.5 ,α=0.05.
100
/
5
.
4
1 mg
X  100
/
4
.
3
2 mg
X 

113
2. Assumption: Two population are normal, σ2
1 , σ2
2
are known
3. Hypotheses: H0: μ 1 = μ2 → μ 1 - μ2 = 0
 HA: μ 1 ≠ μ 2 → μ 1 - μ 2 ≠ 0
4.Test Statistic:
 = = 2.57
5. Desicion Rule:
Reject H 0 if Z >Z1-α/2 or Z< - Z1-α/2
Z1-α/2= Z1-0.05/2= Z0.975=1.96 (from table D)
6-Conclusion: Reject H0 since 2.57 > 1.96
Or if p-value =0.102→ reject H0 if p < α → then reject H0
2
2
2
1
2
1
2
1
2
1 )
(
-
)
X
-
X
(
n
n
Z







15
5
.
1
12
1
)
0
(
-
3.4)
-
(4.5



114
Example7.3.2 page 240
The purpose of a study by Tam, was to investigate wheelchair
Maneuvering in individuals with over-level spinal cord injury (SCI)
And healthy control (C). Subjects used a modified a wheelchair to
incorporate a rigid seat surface to facilitate the specified
experimental measurements. The data for measurements of the
left ischial tuerosity (
‫المتحرك‬ ‫الكرسي‬ ‫من‬ ‫وتأثيرها‬ ‫الفخذ‬ ‫عظام‬
) for SCI and
control C are shown below
169
150
114
88
117
122
131
124
115
131
C
143
130
119
121
130
163
180
130
150
60
SCI

115
We wish to know if we can conclude, on the
basis of the above data that the mean of
left ischial tuberosity for control C lower
than mean of left ischial tuerosity for SCI,
Assume normal populations equal
variances. α=0.05, p-value = -1.33

116
Solution:
1. Data:, nC=10 , nSCI=10, SC=21.8, SSCI=133.1 ,α=0.05.
 , (calculated from data)
2.Assumption: Two population are normal, σ2
1 , σ2
2 are
unknown but equal
3. Hypotheses: H0: μ C = μ SCI → μ C - μ SCI = 0
HA: μ C < μ SCI → μ C - μ SCI < 0
4.Test Statistic:

Where,
1
.
126

C
X 1
.
133

SCI
X
569
.
0
10
1
10
1
04
.
756
0
)
1
.
133
1
.
126
(
1
1
)
(
-
)
X
-
X
(
2
1
2
1
2
1









n
n
S
T
p


04
.
756
2
10
10
)
3
.
32
(
9
)
8
.
21
(
9
2
)
1
(n
)
1
(n 2
2
2
1
2
2
2
2
1
1
2











n
n
S
S
Sp

117
5. Decision Rule:
Reject H 0 if T< - T1-α,(n1+n2 -2)
T1-α,(n1+n2 -2) = T0.95,18 = 1.7341 (from table E)
6-Conclusion: Fail to reject H0 since -0.569 < - 1.7341
Or
Fail to reject H0 since p = -1.33 > α =0.05

118
Dernellis and Panaretou examined subjects with hypertension
and healthy control subjects .One of the variables of interest was
the aortic stiffness index. Measures of this variable were
calculated From the aortic diameter evaluated by M-mode and
blood pressure measured by a sphygmomanometer. Physics wish
to reduce aortic stiffness. In the 15 patients with hypertension
(Group 1),the mean aortic stiffness index was 19.16 with a
standard deviation of 5.29. In the30 control subjects (Group 2),the
mean aortic stiffness index was 9.53 with a standard deviation of
2.69. We wish to determine if the two populations represented by
these samples differ with respect to mean stiffness index .we wish
to know if we can conclude that in general a person with
thrombosis have on the average higher IgG levels than persons
without thrombosis at α=0.01, p-value = 0.0559

119
Solution:
1. Data:, n1=53 , n2=54, S1= 44.89, S2= 34.85 α=0.01.
2.Assumption: Two population are not normal, σ2
1 , σ2
2
are unknown and sample size large
3. Hypotheses: H0: μ 1 = μ 2 → μ 1 - μ 2 = 0
HA: μ 1 > μ 2 → μ 1 - μ 2 > 0
4.Test Statistic:

ِ
standard
deviation
Sample
Size
Mean LgG level
Group
44.89
53
59.01
Thrombosis
34.85
54
46.61
No
Thrombosis
59
.
1
54
85
.
34
53
89
.
44
0
)
61
.
46
01
.
59
(
)
(
-
)
X
-
X
(
2
2
2
2
2
1
2
1
2
1
2
1








n
S
n
S
Z



120
5. Decision Rule:
Reject H 0 if Z > Z1-α
Z1-α = Z0.99 = 2.33 (from table D)
6-Conclusion: Fail to reject H0 since 1.59 > 2.33
Or
Fail to reject H0 since p = 0.0559 > α =0.01

121
7.5 Hypothesis Testing A single
population proportion:
 Testing hypothesis about population proportion (P) is carried out
in much the same way as for mean when condition is necessary for
using normal curve are met
1.Data: sample size (n), sample proportion( ) , P0
2. Assumptions :normal distribution ,
p̂
n
a
p 

sample
in the
element
of
no.
Total
istic
charachtar
some
with
sample
in the
element
of
no.
ˆ

122
 3.Hypotheses:
 Case I : H0: P = P0
HA: P ≠ P0
 Case II : H0: P = P0
HA: P > P0
 Case III : H0: P = P0
HA: P < P0
4.Test Statistic:
Where H0 is true ,is distributed approximately as the standard
normal
n
q
p
p
p
Z
0
0
0
ˆ 


123
5.Decision Rule:
i) If HA: P ≠ P0
 _______________________
 ii) If HA: P> P0
 Reject H0 if Z>Z1-α
 _____________________________
 iii) If HA: P< P0
Reject H0 if Z< - Z1-α
Note: Z1-α/2 , Z1-α , Zα are tabulated values obtained from
table D

124
2. Assumptions : is approximately normaly distributed
3.Hypotheses:
 H0: P = 0.063
HA: P > 0.063
 4.Test Statistic :
5.Decision Rule: Reject H0 if Z>Z1-α
Where Z1-α = Z1-0.05 =Z0.95= 1.645
21
.
1
301
)
0.937
(
063
.
0
063
.
0
08
.
0
ˆ
0
0
0





n
q
p
p
p
Z
p̂

125
6. Conclusion: Fail to reject H0
Since
Z =1.21 > Z1-α=1.645
Or ,
If P-value = 0.1131,
fail to reject H0 → P > α

126
Wagen collected data on a sample of 301 Hispanic women
Living in Texas .One variable of interest was the percentage
of subjects with impaired fasting glucose (IFG). In the
study,24 women were classified in the (IFG) stage .The article
cites population estimates for (IFG) among Hispanic women
in Texas as 6.3 percent .Is there sufficient evidence to
indicate that the population Hispanic women in Texas has a
prevalence of IFG higher than 6.3 percent ,let α=0.05
Solution:
1.Data: n = 301, p0 = 6.3/100=0.063 ,a=24,
q0 =1- p0 = 1- 0.063 =0.937, α=0.05
08
.
0
301
24
ˆ 


n
a
p

127
7.6 Hypothesis Testing :The
Difference between two
population proportion:
 Testing hypothesis about two population proportion (P1,, P2 ) is
carried out in much the same way as for difference between two
means when condition is necessary for using normal curve are
met
1.Data: sample size (n1 ‫و‬n2), sample proportions( ),
Characteristic in two samples (x1 , x2),
2- Assumption : Two populations are independent .
2
1
ˆ
,
ˆ P
P
2
1
2
1
n
n
x
x
p




128
 3.Hypotheses:
 Case I : H0: P1 = P2 → P1 - P2 = 0
HA: P1 ≠ P2 → P1 - P2 ≠ 0
 Case II : H0: P1 = P2 → P1 - P2 = 0
HA: P1 > P2 → P1 - P2 > 0
 Case III : H0: P1 = P2 → P1 - P2 = 0
HA: P1 < P2 → P1 - P2 < 0
4.Test Statistic:
Where H0 is true ,is distributed approximately as the standard
normal
2
1
2
1
2
1
)
1
(
)
1
(
)
(
)
ˆ
ˆ
(
n
p
p
n
p
p
p
p
p
p
Z








129
5.Decision Rule:
i) If HA: P1 ≠ P2
 _______________________
 ii) If HA: P1 > P2
 Reject H0 if Z >Z1-α
 _____________________________
 iii) If HA: P1 < P2
 Reject H0 if Z< - Z1-α
Note: Z1-α/2 , Z1-α , Zα are tabulated values obtained from
table D

130
Noonan is a genetic condition that can affect the heart growth,
blood clotting and mental and physical development. Noonan examined
the stature of men and women with Noonan. The study contained 29
Male and 44 female adults. One of the cut-off values used to assess
stature was the third percentile of adult height .Eleven of the males fell
below the third percentile of adult male height ,while 24 of the female
fell below the third percentile of female adult height .Does this study
provide sufficient evidence for us to conclude that among subjects with
Noonan ,females are more likely than males to fall below the respective
of adult height? Let α=0.05
Solution:
1.Data: n M = 29, n F = 44 , x M= 11 , x F= 24, α=0.05
479
.
0
44
29
24
11







F
M
F
M
n
n
x
x
p 545
.
0
44
24
ˆ
,
379
.
0
29
11
ˆ 





F
F
F
M
m
M
n
x
p
n
x
p

131
2- Assumption : Two populations are independent .
3.Hypotheses:
 Case II : H0: PF = PM → PF - PM = 0
HA: PF > PM → PF - PM > 0
 4.Test Statistic:
5.Decision Rule:
Reject H0 if Z >Z1-α , Where Z1-α = Z1-0.05 =Z0.95= 1.645
6. Conclusion: Fail to reject H0
Since Z =1.39 > Z1-α=1.645
Or , If P-value = 0.0823 → fail to reject H0 → P > α
39
.
1
29
)
521
.
0
)(
479
.
0
(
44
)
521
.
0
)(
479
.
0
(
0
)
379
.
0
545
.
0
(
)
1
(
)
1
(
)
(
)
ˆ
ˆ
(
2
1
2
1
2
1












n
p
p
n
p
p
p
p
p
p
Z

Chapter 9
Statistical Inference and The
Relationship between two variables
Prepared By : Dr. Shuhrat Khan

REGRESSION
CORRELATION
ANALYSIS OF VARIANCE
•
Regression, Correlation and Analysis of
Covariance are all statistical techniques that
use the idea that one variable say, may be
related to one or more variables through an
equation. Here we consider the relationship of
two variables only in a linear form, which is
called linear regression and linear correlation;
or simple regression and correlation. The
relationships between more than two
variables, called multiple regression and
correlation will be considered later.
•
Simple regression uses the relationship
between the two variables to obtain
information about one variable by knowing
the values of the other. The equation showing
this type of relationship is called simple linear
regression equation. The related method of
correlation is used to measure how strong the
relationship is between the two variables is.
133
EQUATION OF REGRESSION

Line of Regression
•
Simple Linear Regression:
•
Suppose that we are interested in a variable Y, but we want to
know about its relationship to another variable X or we want
to use X to predict (or estimate) the value of Y that might be
obtained without actually measuring it, provided the
relationship between the two can be expressed by a line.’ X’ is
usually called the independent variable and ‘Y’ is called the
dependent variable.
•
•
We assume that the values of variable X are either fixed or
random. By fixed, we mean that the values are chosen by
researcher--- either an experimental unit (patient) is given this
value of X (such as the dosage of drug or a unit (patient) is
chosen which is known to have this value of X.
•
By random, we mean that units (patients) are chosen at
random from all the possible units,, and both variables X and
Y are measured.
•
We also assume that for each value of x of X, there is a whole
range or population of possible Y values and that the mean of
the Y population at X = x, denoted by µy/x , is a linear
function of x. That is,
•
•
µy/x = α +βx
DEPENDENT VARIABLE
INDEPENDENT VARIABLE
TWO RANDOM VARIABLE
OR
BIVARIATE
RANDOM
VARIABLE

ESTIMATION
•
Estimate α and β.
•
Predict the value of Y at a
given value x of X.
•
Make tests to draw
conclusions about the model
and its usefulness.
•
We estimate the parameters α
and β by ‘a’ and ‘b’
respectively by using sample
regression line:
•
Ŷ = a+ bx
•
Where we calculate
•
We select a sample of
n observations (xi,yi)
from the population,
WITH
the goals

B =
ESTIMATION AND CALCULATION OF CONSTANTS , ‘’a’’ AND ‘’b’’

EXAMPLE
•
investigators at a sports health centre are
interested in the relationship between oxygen
consumption and exercise time in athletes
recovering from injury. Appropriate mechanics
for exercising and measuring oxygen
consumption are set up, and the results are
presented below:
–
x variable

exercise
time
(min)
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
y variable
oxygen consumption
620
630
800
840
840
870
1010
940
950
1130

calculations
•
o
r

Pearson’s Correlation Coefficient
• With the aid of Pearson’s correlation coefficient (r),
we can determine the strength and the direction of
the relationship between X and Y variables,
• both of which have been measured and they must
be quantitative.
• For example, we might be interested in examining
the association between height and weight for the
following sample of eight children:

Height and weights of 8 children
Child Height(inches)X Weight(pounds)Y
A 49 81
B 50 88
C 53 87
D 55 99
E 60 91
F 55 89
G 60 95
H 50 90
Average ( = 54 inches) ( = 90 pounds)

Scatter plot for 8 babies
height weight
49 81
50 88
53 83
55 99
60 91
55 89
60 95
50 90
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70
1‫متسلسلة‬

Table : The Strength of a Correlation
•
• Value of r (positive or negative) Meaning
• _______________________________________________________
•
• 0.00 to 0.19 A very weak correlation
• 0.20 to 0.39 A weak correlation
• 0.40 to 0.69 A modest correlation
• 0.70 to 0.89 A strong correlation
• 0.90 to 1.00 A very strong correlation
• _______________________________________________________
_

FORMULA FOR CORRELATION
COEFFECIENT ( r )
• With Pearson’s r,
• means that we add the products of the deviations to see if the positive
products or negative products are more abundant and sizable. Positive
products indicate cases in which the variables go in the same direction (that is,
both taller or heavier than average or both shorter and lighter than average);
• negative products indicate cases in which the variables go in opposite
directions (that is, taller but lighter than average or shorter but heavier than
average).
•

Computational Formula for Pearsons’s Correlation Coefficient r
•
Where SP (sum of the product), SSx (Sum of
the squares for x) and SSy (sum of the squares
for y) can be computed as follows:

Child
X
Y
X2
Y2
XY
A 12
12 144
144
144
B
10
8
100
64
80
C
6
12
36
144
72
D
16
11
256
121
176
E
8
10
64 100
80
F
9
8
81
64
72
G
12
16
144
256
192
H
11
15
121
225
165
∑ 84 92 946 1118 981

Table 2 : Chest circumference and Birth
Weight of 10 babies
• X(cm) y(kg) x2 y2 xy
• ___________________________________________________
• 22.4 2.00 501.76 4.00 44.8
• 27.5 2.25 756.25 5.06 61.88
• 28.5 2.10 812.25 4.41 59.85
• 28.5 2.35 812.25 5.52 66.98
• 29.4 2.45 864.36 6.00 72.03
• 29.4 2.50 864.36 6.25 73.5
• 30.5 2.80 930.25 7.84 85.4
• 32.0 2.80 1024.0 7.84 89.6
• 31.4 2.55 985.96 6.50 80.07
• 32.5 3.00 1056.25 9.00 97.5
• TOTAL
• 292.1 24.8 8607.69 62.42 731.61

Checking for significance
• There appears to be a strong between chest circumference and birth
weight in babies.
• We need to check that such a correlation is unlikely to have arisen by
in a sample of ten babies.
• Tables are available that gives the significant values of this correlation
ratio at two probability levels.
• First we need to work out degrees of freedom. They are the number
of pair of observations less two, that is (n – 2)= 8.
• Looking at the table we find that our calculated value of 0.86 exceeds
the tabulated value at 8 df of 0.765 at p= 0.01. Our correlation is
therefore statistically highly significant.

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