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BUSINESS STATISTICS (PART-25)
UNIT- VIII
APPROXIMATIONS AND INTERVAL ESTIMATION
1. INTRODUCTION
Hello viewers, in my last three lectures we have discussed
about the Binomial, Poisson & the Normal distributions. These
are the basic distributions which we make use of, in many
applications. The first two distributions namely the Binomial &
Poisson are the discrete distributions whereas; the Normal
distribution is a continuous distribution. We have also
discussed the applications of these distributions in these
lectures. Now here the question arises:
Is there any relationship between these three distributions?
And can we also estimate the unknown parameters of these
distributions?
The answer of these questions is YES. Therefore, in this lecture
we shall first discuss at length the relationship between the
Binomial, Poisson & Normal distributions. Then we shall discuss
interval estimation of population mean of a normal and a nonnormal population.
2. POISSON APPROXIMATION
DISTRIBUTION
TO
BINOMIAL
Now here we first discuss the Poisson approximation to
Binomial distribution. We know that the Binomial distribution
function is given by:
f ( x)  ncx p x q nx
where, x  0,1,2,..., n
0  p, q  1, q  1  p
Whereas the Poisson distribution function is given by:
e   x
g ( x) 
x!
x  0,1,...
And λ, which is the parameter of the distribution, is a positive
constant. Now, if in case of the Binomial distribution
 the n is very large i.e. 𝑛 → ∞ and,
 If p is very small .Mathematically ,speaking that, p→ 0,
 such that np = λ, a constant or rather it is a finite constant
then
f(x)  g(x).
That is,
Binomial distribution function  the Poisson distribution function
Now let us look at this example which makes this approximate
relation more clear.
We know that, if the distribution of a random variable X is
Binomial with parameters n and p then
f ( x)  n c x p x q n  x
Then for n=10 & p =0.2
f (0)  10C0 (0.2) 0 (0.8)10  0.1074
That is, this gives me the probability that x takes value 0. Now
here if I take   np
then it is equal to 2. Now using the
Poisson distribution function, if we calculate the same
probability, i.e., probability of x takes value 0 then this comes
out to be 0.1353.
Now, in this table we consider the probabilities which are
calculated using the Binomial & Poisson distribution functions.
In the first column we take different values of ‘n’ in Binomial
distribution function. And in the next column we take the value
of ‘p’. And the third column gives me the value of ‘np’, which
we know is λ here. And the fourth column gives me the P(x=0)
and the last column gives the P(x=5). Now, in the first case
when n=10, I take the value p=0.2. We have just seen that for
this the P(x=0) is 0.1704. And when we use Binomial
distribution function for (x=5) then this probability comes out
to be 0.0881.
Now we increase the value of ‘n’ from 10 to 20 and reduce the
value of ‘p’ from 0.2 to 0.1 such that this product remains the
same i.e. 2 (np=2). For this P(x=0) is 0.1216 and P(x=5) is
0.0898. This way when we increase the value of ‘n’ from 20 to
50 (say) and reduce the value of ‘p’ now from 0.1 to 0.04 such
that the product remains the same i.e. 2. For this the P(x=0) is
0.1299 and P(x=5) is 0.0902. Now, when we increase the value
of ‘n’ from 50 to 100 and the value of ‘p’ reduce from 0.4 to 0.2
such that the product remains the same i.e. 2, then p(x=0) is
0.1326 and P(x=5) is 0.0902.
All these probabilities we have calculated by using the Binomial
distribution function. Now we calculate these probabilities
using the Poisson distribution function, where we take λ=2 as a
product of n and p. We remember here that the λ we keep
constant. That is, the value of λ is always 2 in this example. So
here when we use Poisson distribution function the p(x=0) is
0.1353 and the p(x=5) is 0.0902. So we can see that these two
values are very close to Binomial distribution probabilities for
n=100 & p=0.02. Now there is another justification for this
approximation. We know that the mean and variance of
Poisson distribution with parameter λ, is equal to λ. That is, the
mean and variance of Poisson distribution are equal. Whereas
the mean of Binomial distribution is np and the variance is npq.
So, when we take as p→0 in Binomial distribution, then
automatically q→1. And when we do this then the mean and
variances of Binomial distribution will be same as like of
Poisson distribution.
3. NORMAL APPROXIMATION
DISTRIBUTION
TO
BINOMIAL
Earlier we discuss the Poisson approximation to Binomial
distribution. Now, we are discussing Normal approximation to
Binomial distribution. Now here is the example of electric fuses
produced by a factory. Here we consider a sample of 1000 fuses
and this sample size we denote by ‘n’. In this example there are
27 defective fusses and 973 non- defective fusses. Also the
P(observing a defective when a single fuse is tested)=0.02
That is p=0.02 in Binomial distribution. Now the question is:
What is P(x ≥ 27)=???
Here x is the number of defective fuses. So,
P(x  27)  1000C2 7 (0.02) 2 7 (0.098)9 7 3  ......  (0.02)1 0 0 0
We can see that the calculation of this probability is becomes
very tedious when we use the Binomial distribution.
This can be simplified if we use the Normal approximation or
the Normal probability distribution function to calculate this
probability. So, in Normal distribution when we take µ=np then
the value of µ is 20. And the value of σ= √npq, i.e., 4.43. Now
we make a transformation on x variable. And this
transformation is given by z. That is
When we substitute the values of µ & σ, I get the value of z as
1.47. Now P (x ≥ 27) = P (z ≥ 1.47). Now we are in a position to
make use of the Normal table to calculate this probability. And
using the normal table the P (z ≥ 1.47) = 0.0708.
Now we consider another example of Binomial distribution;
where n=15 and the value of p is 0.04. So if we use Normal
approximation the value of µ = np = 6 and the value of σ = √npq
=1.897. Now here in this table calculate the P(x=7) using the
Binomial distribution function and as well as the Normal
distribution function. These values are given in these two
columns. So, in the first case the P(x=7) = 0.177, whereas using
the Normal distribution function this probability is 0.1826. Now
in the next case it gives the P(5 ≤ x<9) this probability is 0.698
when we use Binomial distribution function and this is 0.6918
when we use the Normal distribution function. The last row of
this table gives the P(x<7). This probability is 0.62 when we use
Binomial distribution function and the same is 0.6026 when we
use Normal approximation.
So, we can see that these probabilities are very close to each
other and we can see that how fair the Normal approximation
works in case of the Binomial distribution when ‘n’ is large and
‘p’ is neither close to 1 nor close to 0. So in general the
Binomial distribution can be approximated to normal random
variable, as
1
1
( x  )  np
( x  )  np
2
2
b(x; n, p)  P(
Z
)
npq
npq
If n is large and p is not close to 0 or 1. Then Normal
approximation to Binomial distribution is satisfactory. Also if p
is close to ½, the approximation is rather close, even for
n as low as 15. We are saying that if n is large and p is not
close to 0 or 1 then this approximation is satisfactory.
But even if n is small and p=1/2 then this approximation
still works very well. So the
Working Rule: If both np and nq are greater than 5,
Normal approximation to Binomial distribution is
satisfactory.
Now we shall discuss the Normal approximation to
Poisson distribution. As we know that the Poisson
distribution function is given by
e   x
f ( x) 
x!
where,
x  0,1,...
And λ is a positive constant. Now this Poisson distribution can
be approximated with Normal distribution for large value of λ.
That is, as λ→∞ the
Poisson distribution → Normal distribution.
Now having discussed the relationship between Binomial,
Poisson and Normal distribution we come to the point of
estimation of the parameters of these distributions. And
especially today we shall discuss about the estimation of
population mean µ. Now in order to estimate the population
parameters we select a random sample from the given
population and by taking the observations on the selected
sample, we calculate certain statistics to estimate the
parameters of the distribution.
4. INTERVAL ESTIMATION
Now we shall discuss the estimation of population mean µ.
Now there are two types of estimation:
Point
estimation
Interval
estimation
Now, if on the basis of a random sample of cars we say that the
average of a car is 20 km/liter. Then we say that estimated
value of µ =20. So this is Point estimation of µ. But if we make a
statement on the basis of the sample observations that the
average of a car is between 18 and 22 km/liter. Then it provides
me an interval estimate of µ. That is, the µ lies between 18 &
22. Now in order to find the interval estimate of µ, if x1, x2,. . .,
xn the random sample is selected from the normal population
then, the
P(1.96 
x
 1.96)  0.95......(1)
/ n
Now, for any value of n if sample is selected from a normal
population then this relation is true, that we know. But even if
sample is not selected from the Normal population but n is
large, then using the approximate relationship between
Binomial and Normal distribution the relation (1) is true. Now
the relation (1), which is the
P(1.96 
gives me
x
 1.96)  0.95......(1)
/ n
P( x  1.96  / n    x  1.96  / n )  0.95
and this gives me 95% confidence interval estimate of µ as:
( x  1.96 / n , x  1.96 / n )
So this gives me the lower and upper limits of the confidence
interval. Now in general
P( z / 2 
x
 n
 z / 2 )  1  
this is clear from this graph.
1
/2
 z / 2
/2
1.96
z / 2
And this in turn gives me the ( 1   ) X 100% confidence interval
estimate of µ as:
( x  z / 2

n
, x  z / 2

n
)
5. INTERVAL ESTIMATION : AN EXAMPLE
Now here we consider an example to see how a 90%
Confidence Interval can be constructed for µ.
Suppose we wish to estimate (µ) the average daily yield of a
chemical manufactured in a chemical plant. In order to find an
estimate of µ, a random sample of 50 days is selected. The daily
yield recorded for these n=50 days. It gives (say) x  871tons
and the value of σ is 21 tons. Then the 90% Confidence Interval
estimate of µ is given by ( x  1.645  / n ) . And that in turns
gives me the interval estimate of µ as the interval (866.11,
875.89). That is, the lower limit of this Confidence Interval is
866.11 and the upper limit of this Confidence Interval estimate
is 875.89. Therefore, estimated average daily yield of µ is
between 866.11 and 875.89 tons.
The 90% confidence interval implies that in repeated sampling,
if we calculate confidence interval for each sample, then 90% of
the confidence intervals would enclose µ.
Here, (1   )  0.90 or in terms of percentage it is
(1   ) 100  90% is called confidence coefficient. Now here
in this table we give the confidence intervals for the µ i.e.
different values of (1- α). So, the first column of this table gives
the values of the confidence coefficient. The second column
gives me the lower confidence limit and the third column gives
the upper confidence limit. so, when
(1- α) = 0.90 then the LCL= ( x  1.645  / n ) and
UCL= ( x  1.645  / n ) .
If I come to the last row of this table then (1- α) = 0.99 then the
confidence limits will be ( x  2.58  / n ) and ( x  2.58  / n ) .
These values 1.645 or 1.96 or 2.58 all these values comes from
the normal table. And these values depend: what is my
confidence coefficient? That depends the value of α or (1-α).
And we know that the α is the level of significance and (1-α) we
are calling the confidence coefficient. So, In general
( x  z / 2  / n )
gives an interval estimate of µ with confidence coefficient (1-α).
So, using this formula one can construct the different
confidence intervals by taking different values of (1-α) not
necessarily 0. 90 or 0.95 or 0.99.
6. SUMMARY
In today’s lecture we have discussed two important aspects of
the three basic distributions namely the Binomial, Poisson &
Normal. First we have discussed the relationship between these
three distributions. That is,
 when Poisson distribution can be taken as an approximation
to Binomial distribution, or
 when the Normal distribution can be taken as an
approximation of the Binomial distribution, or
 when we can take Normal distribution as an approximation
to Poisson distribution.
These approximate relationships help us to calculate the
probabilities in a more simplified manner. Then in the second
part of this lecture we have discussed the interval estimation
of population parameter especially, the population mean µ.
This estimation of µ we have discussed in both the cases
when the sample is selected from the normal population and
even when the sample is not selected from the normal
population. So, with this lecture we have come to an end of
our discussions on these three basic distributions of statistics
namely: the Binomial, Poisson & the Normal.
Thank You!