INTRODUCTION TO BIO STATISTICS

Introduction
• Key words :
– Statistics , data , Biostatistics,
– Variable ,Population ,Sample
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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.

* 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.

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)

Sources of
data
Records Surveys Experiments
Comprehensive Sample

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:

2- 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.

3- 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.

* 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.

Types of variables
Quantitative Qualitative
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 quantitative
variables
Discrete Continuous
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.

Interval
Types of variables &
scale of measurement
Quantitative variables
Numerical
Qualitative variables
Categorical
Ratio
Nominal
Ordinal
13Mekele University: Biostatistics

Nominal
unordered categories
numbers used to represent categories
averages are meaningless; look at
frequency/proportion in each category
dichotomous e.g. gender: male = 1, female = 0
polytomous e.g. blood type: O = 1, A = 2, B = 3,
AB = 4

Ordinal
ordered categories
numbers used to represent categories
order matters; magnitude does not
differences between categories are
meaningless
Example:- severity of injury:
fatal = 1,
severe = 2,
moderate = 3,
minor = 4

Interval
The differences between observational units is
equal
The zero point is arbitrary and does not infer the
absence of the property being measured
Examples:
Degrees Fahrenheit
The difference between 30 and 40 is the same as that
between 70 and 80 degrees. But 80 is not twice as hot as 40.
years:
The difference between 1993-1994 is the same as 1995-
1996, but year 0 was not the beginning of time.

Ratio
The most detailed and objectively
interpretable of the measurement scales.
Interval scale with an absolute zero-it has a
true zero point (absence of property being
measured) as well as equal intervals
E.g. Height, weight, money, age, time, speed,
class size, the Kelvin scale of temperature

Cont…
Independent variables
Precede dependent variables in time
Are often manipulated by the researcher
The treatment or intervention that is used in a
study
Dependent variables
What is measured as an outcome in a study
Values depend on the independent variable

* 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:
• headache patients in a chiropractic office;
automobile crash victims in an emergency room
• In research, it is not practical to include all members
of a population
• Thus, a sample (a subset of a population) is taken
• Populations may be finite or infinite.

A Sample
it is a part of a population
e.g. the fraction of these patients
Random sample
Subjects are selected from a population so that each
individual has an equal chance of being selected
Random samples are representative of the source
population
Non-random samples are not representative
May be biased regarding age, severity of the
condition, socioeconomic status etc

Types of statistical methods
Descriptive statistics
Describe the data by summarizing them
Inferential statistics
Techniques, by which inferences are drawn for the
population parameters from the sample statistics
OR
sample statistics observed are inferred to the
corresponding population parameters

Cont…
Parameter
Summary data from a population
Statistic
Summary data from a sample

Examples of Scales of
Measurements
• Low income
ordinal
• CD4 count
ratio
• Year of birth
interval
• IQ scores
interval
• Severe injury
ordinal
• Raw score on a
statistics exam
interval
• Room temperature in
Kelvin
ratio
• Nationality of MU
students
nominal

Strategies for understanding
the meanings of Data

• Key words
Frequency table, bar chart ,range
width of interval , mid-interval
Histogram , Polygon

Before performing any analyses, you must
first get to know your data
Descriptive statistics are used to
summarize data in the form of tables,
graphs and numerical measures
The summary technique used depends on
the data type under consideration

Presentation techniques for
qualitative/categorical data
statistics
Frequency
Relative frequency
Cumulative frequency
Figure/chart
Pie chart
Bar chart

Frequency Distribution
for Discrete Random Variables
Example:
Suppose that we take a sample of size 16 from children in
a primary school and get the following data about the
number of their decayed teeth,
3,5,2,4,0,1,3,5,2,3,2,3,3,2,4,1
To construct a frequency table:
1- Order the values from the smallest to the largest.
0,1,1,2,2,2,2,3,3,3,3,3,4,4,5,5
2- Count how many numbers are the same.

Relative
Frequency
FrequencyNo. of decayed
teeth
0.0625
0.125
0.25
0.3125
0.125
0.125
1
2
4
5
2
2
0
1
2
3
4
5
116Total

Representing the simple frequency table
using the bar chart
Number of decayed teeth
5.004.003.002.001.00.00
Frequency
6
5
4
3
2
1
0
22
5
4
2
1
We can represent
the above simple
frequency table
using the bar
chart.
Ordinal or nominal
data
Height of each bar
is the frequency of
that category

CONTI …

Cont…
0
10
20
30
40
50
%
Single Married Divorced Widowed
Marital status
Male
Female

Cont…
instead of “stacks” rising up from
horizontal (bar chart), we could plot
instead the shares of a pie
Recalling that a circle has 360 degrees
50% means 180 degrees
25% means 90 degrees

Cont…

Frequency Distribution
for Continuous Random Variables
For large samples, we can’t use the simple frequency table to
represent the data.
We need to divide the data into groups or intervals or
classes.
So, we need to determine:
1- The number of intervals (k).
Too few intervals are not good because information will be
lost.
Too many intervals are not helpful to summarize the data.
A commonly followed rule is that 6 ≤ k ≤ 15,
or the following formula may be used,
k = 1 + 3.322 (log n)

2- The range (R).
It is the difference between the largest and the
smallest observation in the data set.
3- The Width of the interval (w).
Class intervals generally should be of the same
width. Thus, if we want k intervals, then w is
chosen such that
w ≥ R / k.

Example:
Assume that the number of observations
equal 100, then
k = 1+3.322(log 100)
= 1 + 3.3222 (2) = 7.6  8.
Assume that the smallest value = 5 and the largest one of
the data = 61, then
R = 61 – 5 = 56 and
w = 56 / 8 = 7.
To make the summarization more
comprehensible, the class width may be 5
or 10 or the multiples of 10.

Example 2.3.1
• We wish to know how many class interval to have in the
frequency distribution of the data in Table 1.4.1 of ages of 189
subjects who Participated in a study on smoking cessation
Solution :
• Since the number of observations
equal 189, then
• k = 1+3.322(log 189)
• = 1 + 3.3222 (2.276)  9,
• R = 82 – 30 = 52 and
• w = 52 / 9 = 5.778
 It is better to let w = 10, then the intervals will be in the
form:

FrequencyClass interval
1130 – 39
4640 – 49
7050 – 59
4560 – 69
1670 – 79
180 – 89
189Total
Sum of frequency
=sample size=n

The Cumulative Frequency:
It can be computed by adding successive frequencies.
The Cumulative Relative Frequency:
It can be computed by adding successive relative
frequencies.
The Mid-interval:
It can be computed by adding the lower bound of the
interval plus the upper bound of it and then divide
over 2.

For the above example, the following table represents the cumulative
frequency, the relative frequency, the cumulative relative frequency and the
mid-interval.
Cumulative
Relative
Frequency
Relative
Frequency
R.f
Cumulative
Frequency
Frequency
Freq (f)
Mid –
interval
Class
interval
0.05820.0582111134.530 – 39
-0.2434574644.540 – 49
0.6720-127-54.550 – 59
0.91010.2381-45-60 – 69
0.99480.08471881674.570 – 79
10.0053189184.580 – 89
1189Total
R.f= freq/n

Example :
• From the above frequency table, complete the table then
answer the following questions:
1-The number of objects with age less than 50 years ?
2-The number of objects with age between 40-69 years ?
3-Relative frequency of objects with age between 70-79 years ?
4-Relative frequency of objects with age more than 69 years ?
5-The percentage of objects with age between 40-49 years ?
6- The percentage of objects with age less than 60 years ?
7-The Range (R) ?
8- Number of intervals (K)?
9- The width of the interval ( W) ?

Representing the grouped frequency table using the
histogram
To draw the histogram, the true classes limits should be used. They can be
computed by subtracting 0.5 from the lower limit and adding 0.5 to the upper
limit for each interval.
FrequencyTrue class limits
1129.5 – <39.5
4639.5 – < 49.5
7049.5 – < 59.5
4559.5 – < 69.5
1669.5 – < 79.5
179.5 – < 89.5
189Total
0
10
20
30
40
50
60
70
80
34.5 44.5 54.5 64.5 74.5 84.5

Representing the grouped frequency table using
the Polygon
0
10
20
30
40
50
60
70
80
34.5 44.5 54.5 64.5 74.5 84.5

Histogram
•continuous data divided into categories
•graphical representation of frequency
distribution
•height of each bar is the frequency of that
category
•assess skewness and modality of the data
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CONTI…
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frequency polygon
- is an alternative to the histogram
whereas in a histogram
- X-axis shows intervals of values
- Y-axis shows bars of frequencies
. in a frequency polygon
X-axis shows midpoints of intervals of
values
Y-axis shows dot instead of bars
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CONTI…
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Box plots
- discrete or continuous data
- displays the 25th, 50th and 75th percentiles of
the data also known as the first, second and third
quartiles respectively
- whiskers extend to adjacent values which are not
outliers
- outliers indicated as circles
- box shows the interquartile range of the data
can be used to assess skewness
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CONT…
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Two-way scatter plots
• used to assess the relationship between
two discrete or continuous measures
nature of the relationship described as
positive, negative or no relationship
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Line graph
• two continuous measures each x value
has only one corresponding y value useful
for looking at patterns over time can be
used to compare 2 or more groups
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CONTI…
0
0.5
1
1.5
2
2.5
3
3.5
0 0.5 1 1.5 2 2.5 3
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Line Graph
0
10
20
30
40
50
60
1960 1970 1980 1990 2000
Year
MMR/1000
Year MMR
1960 50
1970 45
1980 26
1990 15
2000 12
Figure (1): Maternal mortality rate of (country), 1960-2000

GRAPHS & CHARTS – LINE GRAPH
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Chapter-3
Measures of Central
Tendency

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/2)th+((n/2)+1)th]/2 ordered observation.
When n = 12, then the median 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.

Quintiles
Quintiles
• Dividing the distribution of ordered values into
equal-sized parts
– Quartiles: 4 equal parts
– Deciles: 10 equal parts
– Percentiles: 100 equal parts
First 25% Second 25% Third 25% Fourth 25%
Q1 Q2 Q3
Q1:first quartile
Q2:second quartile = median
Q3:third quartile

Example:
Given the following data set (age of patients):-
18,59,24,42,21,23,24,32 find the third quartile
Solution:
sort the data from lowest to highest
18 21 23 24 24 32 42 59
3rd quartile = {3/4 (n+1)}th observation = (6.75)th
observation
= 32 + (42-32)x .75 = 39.5

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

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 is small.
b) If the values are widely scattered
→ The Dispersion is greater.

Example
• ** 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
– Highly sensitive to outliers
– Data: 43,66,61,64,65,38,59,57,57,50.
• Find Range? Range=66-38=28
• Inter-quartile range
– 3rd quartile – 1st quartile (75th – 25th percentile)
– Robust to outliers
– Middle 50% of observations
SL xx 

2.The Variance:
• It measure dispersion relative to the scatter of the values a
bout their mean.
a) Sample Variance ( ) :
• ,where is sample mean
• 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
xx
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



STANDARD DEVIATION SD
7 7
7 7 7
7
7 8
7 7 7
6
3 2
7 8 13
9
Mean = 7
SD=0
Mean = 7
SD=0.63
Mean = 7
SD=4.04

Standard deviation
Caution must be exercised when using standard
deviation as a comparative index of dispersion
Weights of newborn
elephants (kg)
929 853
878 939
895 972
937 841
801 826
Weights of newborn
mice (kg)
0.72 0.42
0.63 0.31
0.59 0.38
0.79 0.96
1.06 0.89
n=10 =887.1
sd = 56.50
X n=10 = 0.68
sd = 0.255
X
Incorrect to say that elephants show greater variation for birth-
weights than mice because of higher standard deviation77Mekele University: Biostatistics

4.The Coefficient of Variation (C.V):
• Is a measure use 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
VC 
X

Coefficient of variance
Coefficient of variance expresses standard deviation
relative to its mean
X
s
cv 
Weights of newborn
elephants (kg)
929 853
878 939
895 972
937 841
801 826
Weights of newborn
mice (kg)
0.72 0.42
0.63 0.31
0.59 0.38
0.79 0.96
1.06 0.89
n=10, = 887.1
s = 56.50 cv = 0.0637
X n=10, = 0.68
s = 0.255 cv = 0.375
X
Mice show
greater birth-
weight
variation

Example:
• 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

When to use coefficient of variance
• When comparison groups have very different means
(CV is suitable as it expresses the standard deviation
relative to its corresponding mean)
• When different units of measurement are involved,
e.g. group 1 unit is mm, and group 2 unit is gm (CV is
suitable for comparison as it is unit free)
• In such cases, sd should not be used for comparison

Chapter-4
Elementary Probability and
probability distribution

• Key words:
• Probability, objective Probability, subjective
probability, equally likely Mutually exclusive,
multiplicative rule , Conditional Probability,
independent events

Introduction
• The concept of probability is frequently encountered in
everyday communication. For example, a physician may
say that a patient has a 50-50 chance of surviving a certain
operation.
Another physician may say that she is 95 percent certain
that a patient has a particular disease.
• Most people express probabilities in terms of percentages.
• But, it is more convenient to express probabilities as
fractions. Thus, we may measure the probability of the
occurrence of some event by a number between 0 and 1.
• The more likely the event, the closer the number is to one.
An event that can't occur has a probability of zero, and an
event that is certain to occur has a probability of one.

Two views of Probability objective and
subjective:
• *** Objective Probability
• ** Classical and Relative
• Some definitions:
1.Equally likely outcomes:
Are the outcomes that have the same chance of
occurring.
2.Mutually exclusive:
Two events are said to be mutually exclusive if they
cannot occur simultaneously such that A B =Φ .


• The universal Set (S): The set all possible outcomes.
• The empty set Φ : Contain no elements.
• The event ,E : is a set of outcomes in S which has a
certain characteristic.
• Classical Probability : If an event can occur in N
mutually exclusive and equally likely ways, and if m
of these possess a triat, E, the probability of the
occurrence of event E is equal to m/ N .
• For Example: in the rolling of the die , each of the six
sides is equally likely to be observed . So, the
probability that a 4 will be observed is equal to 1/6.

• Relative Frequency Probability:
• Def: If some posses is repeated a large number of
times, n, and if some resulting event E occurs m
times , the relative frequency of occurrence of E ,
m/n will be approximately equal to probability of E .
P(E) = m/n .
• *** Subjective Probability :
• Probability measures the confidence that a particular
individual has in the truth of a particular proposition.
• For Example : the probability that a cure for cancer
will be discovered within the next 10 years.

Elementary Properties of Probability:
• Given some process (or experiment ) with n
mutually exclusive events E1, E2, E3,…………,
En, then
1. P(Ei ) ≥ 0, i= 1,2,3,……n
2. P(E1 )+ P(E2) +……+P(En )=1
3. P(Ei +EJ )= P(Ei )+ P(EJ ), Ei ,EJ are mutually
exclusive

Rules of Probability
1-Addition Rule
P(A U B)= P(A) + P(B) – P (A∩B )
2- If A and B are mutually exclusive (disjoint) ,then
P (A∩B ) = 0
Then , addition rule is
P(A U B)= P(A) + P(B) .
3- Complementary Rule
P(A' )= 1 – P(A)
where, A' = complement event

Example
TotalLater >18
(L)
Early = 18
(E)
Family history of
Mood Disorders
633528Negative(A)
573819Bipolar
Disorder(B)
854441Unipolar (C)
1136053Unipolar and
Bipolar(D)
318177141Total

**Answer the following questions:
Suppose we pick a person at random from this sample.
1-The probability that this person will be 18-years old or younger?
2-The probability that this person has family history of mood orders
Unipolar(C)?
3-The probability that this person has no family history of mood
orders Unipolar( )?
4-The probability that this person is 18-years old or younger or has
no family history of mood orders Negative (A)?
5-The probability that this person is more than18-years old and has
family history of mood orders Unipolar and Bipolar(D)?
C

Solution:
1. P(E)=141/318
2. P(C)=41/318
3. P( )= 1-P(C)=1-41/318
4. P(E U A)=P(E)+P(A)-P(E n A)
= (141/318) + (63/318) - 28/318
=141/318
5. P(L n D) = 60/318
C

Conditional Probability:
P(AB) is the probability of A assuming that B has
happened.
• P(AB)= , P(B)≠ 0
• P(BA)= , P(A)≠ 0
)(
)(
BP
BAP 
)(
)(
AP
BAP 

Example
From previous example , answer
• suppose we pick a person at random and find he is 18
years or younger (E),what is the probability that this
person will be one who has no family history of mood
disorders (A)?
• Solution:
• P(A/E)=28/141, P(E)=141/318, P(AnE)=(28/318)

exercise
• suppose we pick a person at random and
find he has family history of mood (D) what
is the probability that this person will be 18
years or younger (E)?

Multiplicative Rule:
• P(A∩B)= P(AB)P(B)
• P(A∩B)= P(BA)P(A)
Where,
• P(A): marginal probability of A.
• P(B): marginal probability of B.
• P(BA):The conditional probability.

Independent Events:
• If A has no effect on B, we said that A,B are
independent events.
• Then,
1- P(A∩B)= P(B)P(A)
2- P(AB)=P(A)
3- P(BA)=P(B)

Example
• In a certain high school class consisting of 60 girls
and 40 boys, it is observed that 24 girls and 16 boys
wear eyeglasses . If a student is picked at random
from this class ,the probability that the student
wears eyeglasses , P(E), is 40/100 or 0.4 .
• What is the probability that a student picked at
random wears eyeglasses given that the student is a
boy?
• What is the probability of the joint occurrence of the
events of wearing eye glasses and being a boy?

Example
• Suppose that of 1200 admission to a general
hospital during a certain period of time,750 are
private admissions. If we designate these as a set A,
then compute P(A) , P( ).A

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.

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.

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).

Example :
F(x)=
P(X≤ x)
P(X=x)frequencyNumber of
Programs
0.20880.2088621
0.36700.1582472
0.49830.1313393
0.62960.1313394
0.82490.1953585
0.94950.1246376
0.96300.013547
1.00000.0370118
1.0000297Total

• 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 ( ) 1P X x  
( ) 1P X x 

4.3 The Binomial Distribution:
• 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.

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

• 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 Mekele University: Biostatistics 108
( ) ( ) , 0,1,2,....,X n X
n
f x P X x p q x n
x

 
     
 
n
x
 
  
 
!
!( )!
n n
x n xx
 
    
! ( 1)( 2)....(1)x x x x  

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

Example
• 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?

Example
• 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.

Properties of continuous probability Distributions:
*continuous variable is one that can assume any
value within a specified interval of values
assumed by the variable.
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) .

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
exf

Characteristics of the normal distribution
• 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.

5- The normal distribution
depends on the two
parameters  and .
 determines the
location of
the curve.
But,  determines
the scale of the curve, i.e.
the degree of flatness or
peaked ness of the curve.
1 2 3
1 < 2 < 3

1
2
3
1 < 2 < 3

Note that :
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

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
ezf




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.

“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

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

INTRODUCTION TO BIO STATISTICS

Example :
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

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
36






x
z

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

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
4.53

X
3.1
4.55 

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

X
3.1
4.55.4 
3.1
4.53.7 

• Key words:
• Point estimate, interval estimate, estimator,
Confident level ,α , Confident interval for mean μ,
Confident interval for two means,
Confident interval for population proportion P,
Confident interval for two proportions

• 6.1 Introduction:
• Statistical inference is the procedure by which we reach to
a conclusion about a population on the basis of the
information contained in a sample drawn from that
population.
• Suppose that:
• an administrator of a large hospital is interested in the
mean age of patients admitted to his hospital during a
given year.
1. It will be too expensive to go through the records of all
patients admitted during that particular year.
2. He consequently elects to examine a sample of the records
from which he can compute an estimate of the mean age
of patients admitted to his that year.

• To any parameter, we can compute two types of estimate: a
point estimate and an interval estimate.
• A point estimate is a single numerical value used to estimate
the corresponding population parameter.
• An interval estimate consists of two numerical values defining
a range of values that, with a specified degree of confidence,
we feel includes the parameter being estimated.
• The Estimate and The Estimator:
• The estimate is a single computed value, but the estimator is
the rule that tell us how to compute this value, or estimate.
• For example,
• is an estimator of the population mean,. The single
numerical value that results from evaluating this
formula is called an estimate of the parameter .
n
x
x i
i


Confidence Interval for a Population
Mean: (C.I)
Suppose researchers wish to estimate the mean of
some normally distributed population.
• They draw a random sample of size n from the
population and compute , which they use as a point
estimate of .
• Because random sampling involves chance, then
can‟t be expected to be equal to .
• The value of may be greater than or less than .
• It would be much more meaningful to estimate  by
an interval.
x
x

The 1- percent confidence interval (C.I.)
for :
• We want to find two values L and U between which  lies with
high probability, i.e.
P( L ≤  ≤ U ) = 1-

For example:
• When,
•  = 0.01,
then 1-  =
•  = 0.05,
then 1-  =
•  = 0.05,
then 1-  =

We have the following cases
a) When the population is normal
1) When the variance is known and the sample size is large or small,
the C.I. has the form:
P( - Z (1- /2) /n <  < + Z (1- /2) /n) = 1- 
2) When variance is unknown, and the sample size is small, the C.I. has
the form:
P( - t (1- /2),n-1 s/n <  < + t (1- /2),n-1 s/n) = 1- 
x x
xx

b) When the population is not normal and n
large (n>30)
1) When the variance is known the C.I. has the
form:
P( - Z (1- /2) /n <  < + Z (1- /2) /n) = 1- 
2) When variance is unknown, the C.I. has the
form:
P( - Z (1- /2) s/n <  < + Z (1- /2) s/n) = 1- 
x x
x x

Example:
• Suppose a researcher , interested in obtaining an estimate
of the average level of some enzyme in a certain human
population, takes a sample of 10 individuals, determines
the level of the enzyme in each, and computes a sample
mean of approximately
Suppose further it is known that the variable of
interest is approximately normally distributed with a
variance of 45. We wish to estimate . (=0.05)
22x

Solution:
• 1- =0.95→ =0.05→ /2=0.025,
• variance = σ2 = 45 → σ= 45,n=10
• 95%confidence interval for  is given by:
P( - Z (1- /2) /n <  < + Z (1- /2) /n) = 1- 
• Z (1- /2) = Z 0.975 = 1.96 (refer to table D)
• Z 0.975(/n) =1.96 ( 45 / 10)=4.1578
22 ± 1.96 ( 45 / 10) →
• (22-4.1578, 22+4.1578) → (17.84, 26.16)
• Exercise example 6.2.2 page 169
22x
x
x

Example
The activity values of a certain enzyme measured in normal
gastric tissue of 35 patients with gastric carcinoma has a
mean of 0.718 and a standard deviation of 0.511.We want
to construct a 90 % confidence interval for the population
mean.
• Solution:
• Note that the population is not normal,
• n=35 (n>30) n is large and  is unknown ,s=0.511
• 1- =0.90→ =0.1
• → /2=0.05→ 1-/2=0.95,

Then 90% confident interval for  is given by :
P( - Z (1- /2) s/n <  < + Z (1- /2) s/n) = 1- 
• Z (1- /2) = Z0.95 = 1.645 (refer to table D)
• Z 0.95(s/n) =1.645 (0.511/ 35)=0.1421
0.718 ± 1.645 (0.511) / 35→
(0.718-0.1421, 0.718+0.1421) →
(0.576,0.860).
• Exercise example 6.2.3 page 164:
xx

Example6.3.1 Page 174:
• Suppose a researcher , studied the effectiveness of early
weight bearing and ankle therapies following acute repair
of a ruptured Achilles tendon. One of the variables they
measured following treatment the muscle strength. In 19
subjects, the mean of the strength was 250.8 with
standard deviation of 130.9
we assume that the sample was taken from is
approximately normally distributed population. Calculate
95% confident interval for the mean of the strength ?

Solution:
• 1- =0.95→ =0.05→ /2=0.025,
• Standard deviation= S = 130.9 ,n=19
• 95%confidence interval for  is given by:
P( - t (1- /2),n-1 s/n <  < + t (1- /2),n-1 s/n) = 1- 
• t (1- /2),n-1 = t 0.975,18 = 2.1009 (refer to table E)
• t 0.975,18(s/n) =2.1009 (130.9 / 19)=63.1
• 250.8 ± 2.1009 (130.9 / 19) →
• (250.8- 63.1 , 22+63.1) → (187.7, 313.9)
• Exercise 6.2.1 ,6.2.2
• 6.3.2 page 171
8.250x
x
x

6.3 Confidence Interval for the difference
between two Population Means: (C.I)
If we draw two samples from two independent population
and we want to get the confident interval for the
difference between two population means , then we have
the following cases :
a) When the population is normal
1) When the variance is known and the sample sizes is large
or small, the C.I. has the form:
2
2
2
1
2
1
2
1
2121
2
2
2
1
2
1
2
1
21 )()(
nn
Zxx
nn
Zxx



 


2) When variances are unknown but equal, and the sample size is
small, the C.I. has the form:
2
)1()1(
11
)(
11
)(
21
2
22
2
112
21
)2(,
2
1
2121
21
)2(,
2
1
21
2121





nn
SnSn
S
where
nn
Stxx
nn
Stxx
p
p
nn
p
nn
 

Example 6.4.1 P174:
The researcher team interested in the difference between serum uric
and acid level in a patient with and without Down‟s syndrome .In a
large hospital for the treatment of the mentally retarded, a sample of
12 individual with Down‟s Syndrome yielded a mean of
mg/100 ml. In a general hospital a sample of 15 normal individual of
the same age and sex were found to have a mean value of
If it is reasonable to assume that the two population of values are
normally distributed with variances equal to 1 and 1.5,find the 95%
C.I for μ1 - μ2
Solution:
1- =0.95→ =0.05→ /2=0.025 → Z (1- /2) = Z0.975 = 1.96
• 1.1 1.96(0.4282) = 1.1 0.84 = ( 0.26 , 1.94 )
5.41 x
4.32 x
2
2
2
1
2
1
2
1
21 )(
nn
Zxx

 
 15
5.1
12
1
96.1)4.35.4( 

Example 6.4.1 P178:
The purpose of the study was to determine the effectiveness of an
integrated outpatient dual-diagnosis treatment program for
mentally ill subject. The authors were addressing the problem of substance abuse
issues among people with sever mental disorder. A retrospective chart review was
carried out on 50 patient ,the recherché was interested in the number of inpatient
treatment days for physics disorder during a year following the end of the program.
Among 18 patient with schizophrenia, The mean number of treatment days was 4.7
with standard deviation of 9.3. For 10 subject with bipolar disorder, the mean
number of treatment days was 8.8 with standard deviation of 11.5. We wish to
construct 99% C.I for the difference between the means of the populations
Represented by the two samples

Solution :
• 1-α =0.99 → α = 0.01 → α/2 =0.005 → 1- α/2 = 0.995
• n2 – 2 = 18 + 10 -2 = 26+n1
• t (1- /2),(n1+n2-2) = t0.995,26 = 2.7787, then 99% C.I for μ1 – μ2
• where
•
• then
(4.7-8.8) 2.7787 √102.33 √(1/18)+(1/10)
- 4.1 11.086 =( - 15.186 , 6.986)
Exercises: 6.4.2 , 6.4.6, 6.4.7, 6.4.8 Page 180
21
)2(,
2
1
21
11
)(
21 nn
Stxx p
nn



33.102
21018
)5.119()3.917(
2
)1()1( 22
21
2
22
2
112







xx
nn
SnSn
Sp

6.5 Confidence Interval for a Population
proportion (P):
A sample is drawn from the population of interest ,then
compute the sample proportion such as
This sample proportion is used as the point estimator of the
population proportion . A confident interval is obtained by
the following formula
Pˆ
n
a
p 
samplein theelementofno.Total
isticcharachtarsomewithsamplein theelementofno.
ˆ
n
PP
ZP
)ˆ1(ˆ
ˆ
2
1





Example 6.5.1
The Pew internet life project reported in 2003 that 18%
of internet users have used the internet to search for
information regarding experimental treatments or
medicine . The sample consist of 1220 adult internet
users, and information was collected from telephone
interview. We wish to construct 98% C.I for the
proportion of internet users who have search for
information about experimental treatments or medicine

Solution :
1-α =0.98 → α = 0.02 → α/2 =0.01 → 1- α/2 = 0.99
Z 1- α/2 = Z 0.99 =2.33 , n=1220,
The 98% C. I is
0.18 0.0256 = ( 0.1544 , 0.2056 )
Exercises: 6.5.1 , 6.5.3 Page 187
18.0
100
18
ˆ p
1220
)18.01(18.0
33.218.0
)ˆ1(ˆ
ˆ
2
1




 n
PP
ZP 

Confidence Interval for the difference
between two Population proportions :
Two samples is drawn from two independent population
of interest ,then compute the sample proportion for each
sample for the characteristic of interest. An unbiased
point estimator for the difference between two population
proportions
A 100(1-α)% confident interval for P1 - P2 is given by
21
ˆˆ PP 
2
22
1
11
2
1
21
)ˆ1(ˆ)ˆ1(ˆ
)ˆˆ(
n
PP
n
PP
ZPP







Example
Connor investigated gender differences in proactive and
reactive aggression in a sample of 323 adults (68 female
and 255 males ). In the sample ,31 of the female and 53
of the males were using internet in the internet café. We
wish to construct 99 % confident interval for the
difference between the proportions of adults go to
internet café in the two sampled population .

Solution :
1-α =0.99 → α = 0.01 → α/2 =0.005 → 1- α/2 = 0.995
Z 1- α/2 = Z 0.995 =2.58 , nF=68, nM=255,
The 99% C. I is
0.2481 2.58(0.0655) = ( 0.07914 , 0.4171 )
2078.0
255
53
ˆ,4559.0
68
31
ˆ 
M
M
M
F
F
F n
a
p
n
a
p
M
MM
F
FF
MF
n
PP
n
PP
ZPP
)ˆ1(ˆ)ˆ1(ˆ
)ˆˆ(
2
1






255
)2078.01(2078.0
68
)4559.01(4559.0
58.2)2078.04559.0(





• Key words :
• Null hypothesis H0, Alternative hypothesis HA , testing
hypothesis , test statistic , P-value

Hypothesis Testing
• One type of statistical inference, estimation,
was discussed previously.
• The other type ,hypothesis testing ,is
discussed in this session.

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

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

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

• 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


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-


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)


• 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
t1-α/2 , t1-α , tα are tabulated values obtained from
table E with (n-1) degree of freedom (df)

• 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.

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)

Example
• 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?

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


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 )

• 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

Example
• Referring to previous example.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
• HA: μ < 30

4.Test Statistic :
• = = -2.12
5. Decision Rule: Reject H0 if Z< Z α, where
• Z α= -1.645. (from table)
6. Decision: Reject H0 ,thus we can conclude that the
population mean is smaller than 30.
n
X
Z

o-

10
20
3027 

Example
• 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.

Solution
1. Data: Variable is systolic blood pressure,
n=157 , x=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
140146 
1548.2
6

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.

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).

• 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.

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
2121 )(-)X-X(
nn
Z





21
2121
11
)(-)X-X(
nn
S
T
p 



2
2
2
1
2
1
2121 )(-)X-X(
n
S
n
S
T




2
)1(n)1(n
21
2
22
2
112



nn
SS
Sp

• 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
2121 )(-)X-X(
nn
Z






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)

• 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

Example
• 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.
100/5.41 mgX  100/4.32 mgX 

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
2121 )(-)X-X(
nn
Z





15
5.1
12
1
)0(-3.4)-(4.5



Example
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 for SCI and control C are shown below
16915011488117122131124115131C
14313011912113016318013015060SCI

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

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CX 1.133SCIX
569.0
10
1
10
1
04.756
0)1.1331.126(
11
)(-)X-X(
21
2121







nn
S
T
p

04.756
21010
)3.32(9)8.21(9
2
)1(n)1(n 22
21
2
22
2
112







nn
SS
Sp

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

Example
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

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 deviationSample
Size
Mean LgG levelGroup
44.895359.01Thrombosis
34.855446.61No
Thrombosis
59.1
54
85.34
53
89.44
0)61.4601.59()(-)X-X(
22
2
2
2
1
2
1
2121







n
S
n
S
Z


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

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 ,
pˆ
n
a
p 
samplein theelementofno.Total
isticcharachtarsomewithsamplein theelementofno.
ˆ

• 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
qp
pp
Z
00
0
ˆ 


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

2. Assumptions : is approximately normally 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
063.008.0ˆ
00
0





n
qp
pp
Z
pˆ

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

Example
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

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 n2), sample proportions( ),
Characteristic in two samples (x1 , x2),
2- Assumption : Two populations are independent .
21
ˆ,ˆ PP
21
21
nn
xx
p




• 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
21
2121
)1()1(
)()ˆˆ(
n
pp
n
pp
pppp
Z






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

Example
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
4429
2411







FM
FM
nn
xx
p 545.0
44
24
ˆ,379.0
29
11
ˆ 
F
F
F
M
m
M
n
x
p
n
x
p

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)379.0545.0(
)1()1(
)()ˆˆ(
21
2121









n
pp
n
pp
pppp
Z

An Introduction to the Chi-Square
Distribution

TESTS OF INDEPENDENCE
• To test whether two criteria of classification are
independent . For example socioeconomic status and
area of residence of people in a city are independent.
• We divide our sample according to status, low, medium
and high incomes etc. and the same samples is
categorized according to urban, rural or suburban and
slums etc.
• Put the first criterion in columns equal in number to
classification of 1st criteria ( Socioeconomic status) and
the 2nd in rows, where the no. of rows equal to the no.
of categories of 2nd criteria (areas of cities).

The Contingency Table
• Table Two-Way Classification of sample
First Criterion of Classification →
Second
Criterion ↓
1 2 3 ….. c Total
1
2
3
.
.
r
N11
N21
N31
.
.
Nr1
N12
N22
N32
.
.
Nr2
N13
N 23
N33
.
.
Nr3
……
……
…...
……
N1c
N2c
N3c
.
.
N rc
N1.
N2.
N3.
.
.
Nr.
Total N.1 N.2 N.3 …… N.c N

Observed versus Expected Frequencies
• Oi j : The frequencies in ith row and jth column given in any
contingency table are called observed frequencies that result
form the cross classification according to the two
classifications.
• ei j :Expected frequencies on the assumption of independence
of two criterion are calculated by multiplying the marginal
totals of any cell and then dividing by total frequency
• Formula:
N
NN
e
ji
ij
)(( 


Chi-square Test
• After the calculations of expected frequency,
Prepare a table for expected frequencies and use Chi-square
Where summation is for all values of r xc = k cells.
• D.F.: the degrees of freedom for using the table are (r-1)(c-1)
for α level of significance
• Note that the test is always one-sided.




k
i
e
eo
i
ii
1
2
]
)(
[
2


Example
The researcher are interested to determine that preconception
use of folic acid and race are independent. The data is:
Observed Frequencies Table Expected frequencies Table
Use of
Folic
Acid total
Yes
No
White
Black
Other
260
15
7
299
41
14
559
56
21
Total 282 354 636
Yes no Total
White
Black
Other
s
(282)(559)/636
=247.86
(282)(56)/636
=24.83
(282)((21)
=9.31
(354)(559)/63
6
=311.14
(354)(559)
=
31.17
21x354/636
=11.69
559
56
21

Calculations and Testing
091.969.11/.....
14.311/86.247/
)69.1114(
)14.311299()86.247260(
2
222




• Data: See the given table
• Assumption: Simple random sample
• Hypothesis: H0: race and use of folic acid are independent
HA: the two variables are not independent. Let α = 0.05
• The test statistic is Chi Square given earlier
• Distribution when H0 is true chi-square is valid with (r-1)(c-1) = (3-
1)(2-1)= 2 d.f.
• Decision Rule: Reject H0 if value of is greater than
= 5.991
• Calculations:

2

2
)1)(1(,  cr

Conclusion
• Statistical decision. We reject H0 since 9.08960> 5.991
• Conclusion: we conclude that H0 is false, and that there is a
relationship between race and preconception use of folic
acid.
• P value. Since 7.378< 9.08960< 9.210, 0.01<p <0.025
• We also reject the hypothesis at 0.025 level of significance but
do not reject it at 0.01 level.

INTRODUCTION TO BIO STATISTICS

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INTRODUCTION TO BIO STATISTICS