UNIT III
INFERENTIAL STATISTICS
Introdution
Inferential statistics is defined as a branch of stastistics that is used
To make inferences about the charecters of population based on sample data.the
aim is to go beyond the data and make inferences about population parameter.
For example the result obtained from the analysis of the income of 1000
randomly selected citizens in a particular city in U.S suggests that the average
Monthly income of a citizen is estimated to be 600 dollors.
In this example we are trying to represent the entire population of
the city by a sample of 1000 citizen.
Descriptive statistics. (L1)
Descriptive statistics is the method of data that is presently occurring with in all
the subjects.
Example: Tables,graphs,Averages
Descriptive statistics uses data to provide description of population either
through numerical calculation or graph or table taken from the population.
inferencial statistics makes inferences and predictions about a populat ion
based on sample data.
Population. (L1)
A statistical population is the set of all possible measurements on data
Corresponding to the entire collection of units for which an inference is to be
made.
Sample. (L1)
A sample is a part of the statistical population (i.e) it is a subset which is
collected to draw an inference about the population
Methods of sampling. (L1)
(1) Simple random sampling
(2)Stratified random sampling.
(3)Systamatic sampling.
simple random sampling:(L6)
Simple random sampling is a method of selecting n units,out of N units such
that every one
unit samples has an equal chance of being chosen.This type of
sampling simply refers to random sampling.
A simple random sampling is drawn unit by unit.The units of the
population are numbered from 1to N.Aseris of random numbers or anyother
method can be used to select a sample of size n.This process of drawing units
gives an equal chance to all numbers not previously drawn. It is easy to verify that
all
samples have an equal chance.
Stratified random sampling:(L6)
In stratified sampling the population of N units is first divided into sub
population of
These sub population are non –
overlapping and together they comprise the whole of population so that
The sub population are called strata.To get the full benefit
from stratification,the values of the must known. When the strata have been
determined ,a sample is drawn from each ,the drawings being made
independently in different strata.the sample sizes with in the strata are denoted
by
respectively.If a single random sampling is taken in each
stratum,the whole procedure is known as stratified random sampling.
Stratification may produce a gain in the presence in the
estimation of the charecteristics of the whole population. It may be possible to
divide a heterogeneous population into sub population into sub population.each
of which is internally homogeneous. If each stratum is homogeneous ,the
measurements vary little from one unit to other.then a precise estimate of any
stratum mean can be obtained. These estimates of stratum means can be
combined to get a precise estimate of population.
Systematic random sampling.(L6)
This method of sampling slightly differ from simle random
sampling.Suppose N units of the population are numbered from 1 to N in some
order,to select a sample of n units ,we take a unit at random from the first k units
and every k th unit there after.
For instance ,if k is 12 and the first unit drawn is 15 ,the
subsequent units are numbers 27,42,57,72.and so on.The selection pf the first
unit determines the whole sample.
The advantages of this method are
1. It is easier to draw a sample and it is easy to execute without mistakes.
2. It is likely to be more precise than simple random sampling.
In effect it stratifies the population into n strata,which consist of the
first k units,the second k units and so on.We might expect the systematic
sample to be as the corresponding stratified random sample wiyh one unit
per stratum.The difference is that in systematic sample the unit occurs at
the same relative position in the stratum,where as in the stratified random
sample ,the position of the stratum is spread more evenly over the
population and this fact has made systematic sampling corresponding ly
more precise than stratified random sampling.
Random variable.(L1)
Let E be an experiment and S a sample space associated with
the experiment.A function X assigning to each element s€S, a real number X is
called a random variable.
Types of random variable
There are two types of random variable they are
(i).Discrete Random variable,
(ii).Continious random variable
Discrete random variable.(L1)
A random variable is said to be discrete if it assumes only a finite or countably
infinite values of X.
Example.
Suppose a random experiment consists of throwing 3 coins to record
the number of heads,then
X is the discrete random variable
Continuous random variable. (L1)
A random variable is said to be continuous if it can assume all points on the real
line.
Probability density function. (L1)
Discrete random variable:
Let ,
be possible values of a discrete random variable then P(x) is
called the p.d.f of the discrete variable X if
(i)
) ≥ 0 for i=1,2,3…
(ii)
=1
Continious random variable:
A function f is said to beb the p.d.f of a continuous random variable
X if the following conditions are satisfied .
(i)f(x)
(ii)
Mathematical expectation. (L1)
Let X be a discrete random variable with possible values ,
… ..
And P(X) is the p.d.f then the expected value of X denoted by E(x)=
.
If X is a continuous random variable with p.d.f. f
then
=
.
Variance. (L1)
Let X be a random variable.then the variance of X is denoted by
defined by
)
1. For the following data compute
(i)
,
,(iii) E
(iv) Var
-3
-2
-1 0 1
2
3
0.05 0.10 0.3 0 0.3 0.15 0.10
Solution:
E
=
=(-0.15-0.2-0.3+0.3+0.3+0.3)
=0.25
=
=2.95
=2
=2(0.25)
Var
=
) =11.55
2. When a die is thrown,X denotes the no.of turns
up,Find
(L1)
Solution
X
1 2 3 4 5 6
=
=
=
=
=
=
=15.167-12.25 =2.917
3. The monthly demand for watches known to have the following
Demand
1
2
3
4
5
6
7
8
probability 0.08 0.12 0.19 0.24 0.16 0.10 0.07 0.04
Find expected demand and variance. (L1)
Solution:
Mean = E
=
=
=4.06
=
=
is
(L3)
4. The c.d.f of a random variable x is F(X)=1-(
probability function of
Solution:
P.d.f =
=
=
=
=
=
-.
,
=
Let u=
=
=0+2
=2
=2
III) variance=E(
=
=
=
-
=0+3
=3
=3+
=3
=6
=6
-
=6
5. For triangular distribution
F(x)=
=6
,Find the mean and variance. (L1)
Solution:
E(x)=
=
=
-
find the
=
=
=3-2
=1
E(
=
=
=
=
=
=
-
=
=
=
.
.