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Hello and welcome to this lecture today. We are starting a new module.
This module is on probability and statistics. In last couple of modules, we have seen, we
have discussed about different theories related to probability. Basicallythose, whatever we
have discussed, is associated with the population. Now, when we talk about the statistics, this
statistics is related to the to few samples taken from this population.More precisely
speaking, it is some of this random sampling. So, the the sample that we take that should
be the random sample for from the population. So, now, as we are using the word the population
sample, so, at the starting of this lecture, we should know what does this actually means.You
know that when we are talking about some of this parameters for those distribution, that
we have seen in the last lecture in this last module, that there are certain parameters
are there which, is associated for some standard probability distribution, that we have seen.
So, basically we have to estimate those parameters. Those parameters when we estimate, we should
estimate, we should have some sample data.From there, we should estimate those those parameters.Now
again, the samples means, when we are taking the random samples from the population, now
once we change the sample; obviously, that that estimation of those parameters may also
change.Obviously, it will change because two samples will never be exactly identical.So,
those those estimations also we will have some some distribution. So, those are basically
called the sampling distribution. So, basically our first lecture that we are going to take
in this module is is on this sampling distributions and the parameter estimation.
So, in this parameter estimation that what we will do we will do is based on this, some
of this sampling distribution.You know that, whenever we talk about the mean or standard
deviation or this type of parameters and which is estimated from the samples, those are also
follow some distribution. So, we will see what that distribution refers to and then
we can go for its estimation.Now, in the in the estimation theory also, we can do this
estimation in terms of for two different types.One is that, that it is the point estimation.The
point estimation means just one value. I can estimate from whatever information that is
available to me through the random samples.Or, what I can do is that, I can look for some
of this interval of this of this estimation. So that, this is the interval, this is the
lower limit,and this is the upper limit of this particular parameter. So, that it will
be more logical to say or in the in the statistical sense, that instead of giving a single value
as the estimate, sometimes it is preferable to give some of this interval.So, the all
these things we will discuss in this lecture.
So, our outline of today’s lecture is that, first we will discuss about the population
and sample, then the random sampling and the point estimation.As I was mentioning, thus
the single value, we will just pick up and we will just estimate from the sample.Then,
properties of this point estimation and there are two methods for this estimation.One is
that called Method of moments and other one is Method of maximum likelihood.Then, the
interval estimation of this mean, and theninterval estimation of the variance, and then estimation
of the proportion. So, all these things will go, basically this
outlines should not be a should not be a dead end here. So, it should continue and we should
have this hypothesis testing also.Because, these are all very well related to whatever
we are going to discuss. So, may be one after another, we will take all these topics. So,
to discuss about our the statistics, which is related to each and every point, whatever
we will be discussing. We will see that what and where these things are useful to handle
in some of the civil engineering problems. So, that, we will see. So to start with, we
will start that about the concept of this population and sample.
So, the population is the complete set of all values representing a particular random
process. So, when we talk about this, all values which represents a particular random
random process; that means, this all is very very I should say, that is including of, whatever
the possible range is there, that should include everything.
So, what I can give the example is that, the stream flow in a certain stream over the infinite
timeline. So, if we are collecting the time series of the of the stream flow value, so,
there should not be a finite time over which we are looking the data. So, it should be
on this infinite timeline of whatever has happened and what is going to happen, everything.
So, that is what is our our population is.When we talk about a sample, that sample is any
subset of the entire population. Its corresponding example for this stream flow that we have
used here is, the stream flow in the stream over the last 30 years. So, if I just take
over the last 30 years, then that particulardata set that is available to us is the sample
of this population. Now, you can see that whatever the probability
theory that we have discussed earlier, is basically related to its itsentire entire
possible values.That is, all possible values of that of that random process, but in reality,
we never get, we never have this entire or whatever whateverall possible values, that
is possible that we may not have. So, we have to rely on some some sample and
somehow we have to relate whatever we can estimate from the sample.We have to relate
it to to its population. So, this is basically the role of this statistics and with this,
we will just see how we can estimate the parameters how we can estimate the required information
from the samples.Then we will infer something about its population.Because as I told that,
having the entire population is not possible in for almost all the cases.
So in now, in sample, I was mentioning that the random samples, when should we call that
a sample is a random is a random sample. So, as it is impractical or and or uneconomical
to observe the entire population that I was mentioning. So, a sample that is a subset
is selected from the population for analysis.A sample is said to be random sample, when these
two conditions are satisfying.One is that, it is the representative of the population.
So, whatever the possible cases that we can see in the population, the sample should have
have that representation in it.Thesecond one is that, when the probability theory can be
applied to it, to infer the results that pertain to the entire population.
So, the probability theory as I was telling, it is basically relates to the population.Now,
if we can apply the probability theory to the sample to infer the results that pertains
to the population, then we can say that, whatever the sample we are taking, that is the random
sample. So, these are just the concept and with this concept, what we can do is that,
we can use whatever we have seen from the probability theory, we can apply to the to
this random samples.
Now, the random sample for finite and infinite populations. So, you know that there are certain
cases where the population can be finite, or in other cases it can be infinite, or sometimes
it can be count ably infinite.Even though I know that; obviously, there is some maximum
number is there. But we can sometimes; we can also say that that is count ably infinite.
So, in such cases, what should be the random sample?
So, first we will take the case of this finite population.An observation set x 1, x 2, x
3 up to x n is selected from a finite population of size n. So, that entire size of the population
is n is said to be a random sample, if its values are such that, each x I of n has the
same probability of being selected.Now, so as we can see that there is this, the size
of this population is finite that its maximum number is n. So, this x 1, x 2, x 3,so, we
can say that it is a random sample, if each of this item have the, so this, each of this
element of this set have the equal probability of being selected from the population.
The second one, if the population is infinite, an observation set again that x 1, x 2, x
3, x n is selected from an infinite population f x.So obviously, when we talk about this
infinite series, we generally denote it through its probability density function.Here, it
is that f x is said to be a random sample.If its values are such that, each x I of n have
the same distribution f x and the n random variables are independent.
So each observation, now the first observation,second observation,third up to n, these are also;
we can say that these are also a random variables. So, these variables should have the same distribution,
which is f x and which is same to the distribution of the population.And all these n elements
of this set are independent to each other, then we can say that this set is one random
sample.
The classical approach to estimation of this parameter. So, the classical approach of the
estimation of this of the distribution parameters is of two types.One is the point estimation
and other one is the interval estimation. So, this point estimation means a single parameter
value is estimated from the observed dataset that is from the sample. So, from the sample
a single parameter value is estimated. Now,we will we will discuss how to estimate
that parameter.That we will see so, but, here I want to trace the word single. So, when
the single parameter is estimated, that is known as the point estimation.On the other
hand, it will be the interval estimation, when a certain interval is determined from
the observed dataset.It can be said with a definite confidence level that the parameter
value will lie within that interval. So, there are now, might be many new things
here.One is that it is the interval. So, this word, the certain interval when we are predicting,
then it is the interval estimation.Now, when we say that this is the interval estimation,
then certain confidence level comes. So that, with this much confidence or with this much
confidence and this confidence are also in this in the statistical sense. So, sometimees
the general values that we use for this confidence levels are the 90 percent confidence,95 percent
confidence, 99 percent confidence. So, this confidence level can vary from the 0 to 1
basically. So, means 0 percent to 100 percent. So…So, so whenever we say that it is an
interval estimate, we this interval estimations are always associated with some confidence
level.We will see all these things, how we can how we candeclare that, this is the confidence
for this this interval. But, one thing is should be cleared here now is that, this is
this confidence level.If I increase the confidence level, so that means, as you can read from
this sentence, that it can be said with a definite confidence level, that the parameter
value will lie within that interval. Now, suppose that I take two cases; one is
that 95 percent confidence level, other one is the 99 percent confidence level.Then the,
obviously, the 99 percent confidence level is more. So that, thethat the chance of the
exact parameter value will lie within that interval.In case of the 99 percent confidence,
should be more so; obviously, the interval estimate that we are doing should be more
should be more wider in case of 99 percent confidence interval, compared to what we can
estimate in case of the 95 percent confidence interval. So, more the confidence level, wider
is the interval. So, that is what you can we we can see at least at this test.We will
discuss all these things.
Now, the random sampling and point estimation. So, first we will take that point estimation
out of these two estimates. So, as the parameters of the distribution of a population are unknown
and it is not feasible to obtain them by studying the entire population, hence, the random sample
is generally selected.The parameters of the distribution that are computed based on the
analysis of sample values are called the estimator of the parameters.
So, whatever the quantity that we try to estimate as the estimate of that parameter is known
as the estimator. So, there are two things, one is the estimation and the other one is
the estimators. So, through what through what function I can say, through what function
I am trying to estimate that parameter that is known as the estimator.
Thus parameters correspond to the population; while estimators corresponds to the sample.
So that, so this is what, when we are talking about the parameters of a distribution or
of a of a particular probability distribution, say so, that is generally corresponds to the
population.When we talk about the estimator that is correspond to the sample. So, if I
take the example of this normal distribution that we have that we have seen earlier, is
that there are two parameters now.One is the mu and the other one is the sigma square,
mu is the mean and the sigma square is the variance.
So that, mu and sigma square, these are the properties,these are the these are related,
these are associated with the population.Now, how we will estimate that mean that is mu
and how we will estimate that variance from a sample. So, those are the estimator and
those are associated with the sample. That we will see now.
So, when we say that there is a point estimator and that point estimator is, you know that
should have some properties.This properties, there are four different properties that a
point estimator should have before we can use that one.Because, you see that there could
be the many functions that we can use to estimate that particular parameter. But, so, we have
to take that particular function, which is a satisfying all these all these properties
or at least the the more maximum number of this properties should be satisfied by those
estimators. That these four properties are the unbiasedness,
consistency, efficiency and sufficiency. So, we will take one by one.The unbiasedness is
the bias of an estimator is equal to the difference between the estimator’s expected value and
its true value.For an unbiased estimator of the parameter, expected value should be equal
to the true value. Now, when we are talking about this true value;
that means, it is the value for the population that is mu. Now, whatever the estimator, from
the estimator, which is, we are getting which we should, for which we should use the sample
that is available with us.And I, as I told that this estimators also have some distribution.
So, as these estimators are also having some distributions so, we can also calculate, whatever
the properties that we knew from our earlier lectures and modules is that, so one is that
the expected value. So, if I take the expectation of that estimator itself, so, that will reach
to one one value. Now, the difference between the the this expected
value of the estimator and its true value should be obviously, the desirable thing is
that, it should be as minimum as possible. So, and that is the bias. So, the estimators
should be unbiased and this is the property that is written here.At least, when when we
see that if the n tends to infinity, means n, when the n is the number of sample that
we take. So, when the n tends to infinity, then if that expectation of the estimator
should be equal to its the populationparameter. So, if that is the case that is basically
the check for this unbiasedness.
Second is the consistency.It refers to the asymptotic property, whereby, the error in
the estimator decreases with the increase in the sample size of n.Thus as n tends to
infinity, estimated value approaches to the true value of the parameter. So, this is the
consistency.That is, as we are increasing, the increasing the number of element in the
sample, the estimators should be such that that, the value of the estimators should be
should approach to the to the true value of the parameter, that is the parameter of the
population.
Third one is the efficiency.An estimator with lesser variance is said to be more efficient
compared to that with a greater variance keeping,keeping other conditions same. So here, you can see
that suppose, we got two estimators and both the estimators can satisfy the first two properties;
one is the both are unbiased and both are what is called, that consistent. So, if both,
are both are satisfying the first two condition, then we have to we have to select the estimator,
which is having the less amount of the variance. So, this is basically related thing, which
one is more efficient more efficient estimator, the the estimator having the lesser variance
is is more efficient in such cases. Sufficiency, if a point estimator utilizes all the information
that is available from the random sample, then it is called a sufficient estimator.
Now, now we will discuss this thing through, there are two commonly used method of this
point estimator. This will apply to one of the known distribution and can easily be handled
through the, this hand calculation problem, that we will see. So, basically the two commonly
used method for this point estimation is are the point estimation of the parameter are
the method of moments and other one is method of maximum likelihood.
So, first we will take this method of moments.The method of moment is based on the fact, that
the moments of a random variable have some relationship with the parameters of the distribution.
So, whatever the sample that we are having, we will calculate what are the moments of
thatrandom variable and that should have some relationship with the parameters of the distribution.We
have discussed this in the first moment, second moment with respect to the origin.
You know the first moment with respect to the origin is the mean, and second moment
with respect to the origin, we generally we can take, but, we generally do not take from
the second moment onwards.We take it with respect to the mean and that gives the expression
for the spread of the distribution.All these things, we have discussed earlier and we have
also discussed that the first moment with respect to the mean is zero. So, that concept
we have discussed earlier. So, here we will use that concept to use in this method of
moments. So, if a probability distribution has m number
of parameters, then the first m moments of the distribution are equated to the first
m sample moments.The resulting m number of equations can then be solved to determine
the m number of parameters. So, we will take two examples.One is that; say for example,
that exponential distribution. So, exponential distribution is having one parameter, that
you know, the lambda, one parameter is having. So, the first moments, the first moment of
the sample that you are having should be equated to the first sample moment.Whatever the sample
that you are having, we can also calculate what is this first moment, with respect to
origin,of course, and that we can equate to to get that what is the estimate for this
lambda. We take another example.That is, say that
normal distribution or gamma distribution, both are having two parameters. So,for normal
it is mu and sigma square and for gamma you know that it is alpha and beta. So that, two
parameters are there, then the first two moments, first moment and the second moment should
be equated to the first two moments of the sample and this should be equated to those
those parameters.That is, first one should be equated to the mu in case of normal distribution,
and second one should be equated to the sigma square in case of normal distribution. So,
there are two unknowns now and there are two equations also.That we can solve, to get what
are the estimate of those parameters.
Now, for a sample size n, the sample mean and sample variance that what we use as their
estimator, is that x bar is 1 by n summation of all the all the sample that we are having.Here,
the sample size is n as it is mentioned here. So, we will sum up all the sample and divide
by n, that is, basically the arithmetic mean.And this is, we can show that this is the estimator
for the sample mean, which is satisfying all the properties that we can, that we have discussed.That
is, four requirements for the estimator. Now this this is s square, which is the sample
variance, which is 1 by n summation of x I minus x bar.Again, this x bar is that mean
estimated through this equation and this is square summing up from this I minus I equals
to 1 to n.Now, this is the point estimator for the variance.
Now, this one, it can, we can show thatif we use this one, this will not be not be unbiased.Now,
to make this one unbiased, that is, as I was telling that if the n tends to infinity, it
should reach to the actual value.Then, what we have to use?We have to use that 1 by n
minus 1. So, if we use that n minus 1, so, in many text what you will see that this estimator
for this variance is that a 1 by n minus 1. So, that minus 1 is basically to make the
estimator unbiased.And you know that, this minus 1, that we are talking about is basically
a, we can say that it is a, it is that degrees of freedom. So, one degree of freedom is lost
here, that is why we have to, we need to get that n minus 1.Why it is lost is that, we
are using this mean which is also estimated from the sample itself.
So, that is why. So, one degrees of freedom is lost here, and so, we have to write that
n minus 1.Now, somehow if you know what is the population mean and if you can replace
this x bar with respect to the population mean, that is x I minus mu, that square and
if you if we use this quantity, then there is no need to make that n minus 1, then that
1 minus n is sufficient. So, thus this x bar and s square are the point
estimates of the population mean and the population variance. So, these are the point estimate
for those things, which is obtained from the sample and the parameters of the distribution
can be determined from these. So, these are the first two,that mean the population are
are the estimate for the population mean and population variance. So if needed, other higher
order sample moments can also be obtained, to calculate all the parameters.
So that means, that we can go to the skewness.With sample estimate of the skewness, we can go
to the sample estimate of the catharsis and all.
So now, the relation between the parameter of some common distribution and the moments
are say for example, here we are taking the normal distribution first.There are two parameters
that you know, one is that mu and other one is the sigma square are equal to the mean
and variance, like this. So, expectation of x is equals to mean and the variance is equal
to sigma square. So, what you can see is that, directly we
can use whatever, we, whatever the estimate that we have done, that we can put here to
get what is this population mean and the population variance.
Now, in case of the gamma distribution, now the parameters are the alpha and beta, that
relates to the mean and variance as as follows.That you know, that expectation of a x in case
of the gamma distribution is the alpha beta and the and the variance is alpha beta square.
So, all this things we have discussed in the earlier modules.You can refer to that those
lectures.Here, we are just using that what we have seen in the earlier modules.
So, fine. So, as to conclude this one is that, now you are having, basically, this you can
estimate from this sample.This also you can estimate from the sample.Now, if you equate
this one with these two parameters, that is, what we are saying, so, we are having that
two unknown shear alpha and beta and there are two equations. So, these two can be can
be solved to get what is the estimate of this alpha and beta.And here, it is straightforward
because, we are getting this mu, is the expectation which is directly should be the equal to this
x bar.Whatever you have seen and the variance, should be that sigma square, what you have
seen in this slide. Obviously, if you are using this x bar, that is, which is also estimated
from this sample, this instead of 1 by n, it should be n minus 1.
Now, the second method is that method of maximum likelihood.The method of maximum likelihood,
can be used to obtain the point estimator of the of the parameters of a distribution
directly. So, there are some shortcomings of this method
of moments, which we generally suppose that, sometimes the estimator that is using the
method of moments, what we get is that, sometimes the solutions are once it is solved, we get
that the, it is not within the within the range of these parameters. So, sometimes this
kind of things are observed. So, that is the criticism over the method of moments. So there,
this method of likelihood has found to be moremore effective.Because, here directly
the distribution we are using and we are we are developing a likelihood function and that
likelihood function is maximized to estimate thatthat particular parameter.Now, you know
that how to maximize that likelihood function.First of all, we will see what is the likelihood
function and then we will maximize that one.Now, as many parameters are there, so, we have
to maximize all those parameters. So that, we will get those many equation to have to
solve them to get those estimate. So, suppose that if a sample value of a random
variable x with density function f x and the parameter is theta here are this x 1 x 2 x
n then, the maximum likelihood method is aimed at finding that value of theta, which maximizes
the likelihood of obtaining the set of observations x 1 x 2 x n.
So, what we have to do is that the likelihood of the of obtaining a particular sample x
Iis proportional to the function value of the pdf at x i. So, so this x I means from
this x 1 x 2 x 3, we have to calculate what is that likelihood function. So, the likelihood
function for obtaining the set of this observation x 1 x 2 x n is given by these values of this
distribution at each point.That is, what is the value of that function at x 1, at x 2,
at x 3 and all and their multiplication. So, this differentiating the likelihood function
with respect to theta now and equating it to 0, so, this is basically we are we are
maximizing the likelihood function.We are finding where this likelihood function will
be maximized.We get the value, that is theta value of of that estimate, that is theta hat,
we can just set which is the maximum likelihood estimator of this parameter theta, that is
this 1 will equate to 0 and then we will get that estimate of that theta hat.
The solution can also be obtained by maximizing the logarithm of this likelihood function.
So, if we take the log also, sometimes this it will be more that. So, far as mathematical
calculation is concerned, it a may become easier that will take the log of this likelihood
function and will differentiate with respect to the parameter, can if there are m numbers
of parameters of the distribution, then the likelihood function is like this. So, there
are theta 1 theta 2 up to theta m, these are the parameters of the distribution and this
is you know, that this is a multiplication sign of this at each sample point x i.
The maximum likelihood estimators are obtained by solving the following simultaneous equations.
So, we will, we have to take the differentiation with respect to each parameter theta j, j
can vary from 1 to up to m.And we can take this; we can take this parcel derivatives
equated to 0. So, we will get m equations to solve them, to get that estimate of this
m parameters.
Now, we will take one example using both the methods that we have seen just now.Both, that
thatmethod of method of moment and method of maximum likelihood.
So, and here we have taken that example of this exponential distribution.You know this
exponential distribution here, the example is on this interarrival time of this vehicle
on a certain stretch of a highway is expressed by an exponential distribution, where this
f t is equals to 1 by lambda e power minus t by lambda.
Now, means some places or even in this lecture also, earlier we have, we might have used
some other form, that is, lambda e power minus lambda t. So, it does not matter. So, there
just parameter is taken that 1 by lambda here. So, here also. So, the form is not changing,
only thing the parameter represents in a different way. So, if we use that, the the other form
also that is lambda e power minus lambda t, then also we can we can get the same result
that we see now. So, and obviously, here this t is greater
than equal to 0.Now, there are some samples has been taken, that isthe time between the
successive arrival of the vehicle was observed as 2.2 seconds, 4 seconds, 7.3 seconds, 1.1
seconds, 6.2 second, 3.4 seconds and 8.1 seconds. So, this is a sample that we have collected.Now,
determine the mean, inter arrival time that is lambda by, so, or I, if I do not even want
to mention this; what is this, I just want to estimate, what is the parameter lambda
for this distribution by two methods.One is the method of moments and other one is the
method of maximum likelihood.
So, first the method of moment.When we have said that, we should take the moment with
respect to the origin and that we should equate with this mean. So, if we take this moment,
you know that this is a first moment with respect to mean that we have discussed in
the earlier classes, is that that t multiplied by that distribution.Taking this integration
over the entire support of this distribution and here this exponential distribution having
the support 0 to infinity. So, we will take this integration and if we solve this one,
we can see that this mu is equals to lambda here. So, this is mu. So, the lambda is equals
to mu, which is, we can obtained from this sample of this x bar, which is the estimator
for this mean. So, this 1 by 7 summation of all this t I, so, it is the arithmetic mean.
So, 6.04 second is the estimate for this lambda. Now, if we use the other form of this exponential
distribution that is that is lambda e power minus lambda t then, you can see that this
mu will become that 1 by lambda. So, there the lambda will be equals to 1 by x bar and
this will be 1 by 6.04 second. So, it depends on what way parameters is used here.
And, the other one that is use of this maximum likelihood maximum likelihood function.That
is, so, assuming the random sampling, the likelihood function of the observed value
is that is t 1, t 2, t 3 up to t 7. So, all this sample will take and this will get what
is the likelihood function first. So, 1 by lambda e power. So, e power minus
t I by lambda. So, at all this observation, we are getting this value and we are multiplying
them with each other. So, this lambda power minus 7 exponential of this value, we will
get and the estimator can now be obtained by differentiating the likelihood function
l with respect to the parameter that is lambda.
So, if we do that that, if we do this derivative, we take with respect to the lambda, then we
will get this form.If we just equate it to the 0, then after solving this form, you will
get that again the lambda is equals to 6.04 second.
So, for this one, this example, that we have shown for both the method, whatever the parameter
that we have that we have estimated are same for the lambda is equals to 6.04 second for
both the methods; that is method of moment and method of maximum likelihood. But sometimes,
in some problems, in some distribution this two estimate may not be same.
So, next we will take that interval estimation. So, as I was telling that in this point estimation,
we generally get a single value.That is, what we have seen in in the previous example also,
in both the methods of method of moments or method of maximum likelihood, we get the single
value.That is, the lambda value that you have seen that 6.04 second. So, only a single value
that you have seen in the point estimate. Now, the the interval estimate, we generally
look for a interval, in which, with some confidence that actual value of the parameter should
lie. So, this is what this interval estimation. So, in case of point estimate, the chances
are very low that the true value of the parameter will exactly coincide with the estimated value.
So, as the sample is finite, always there will be some error. So, hence sometimes, it
is desirable or it is useful to specify an interval within which the parameter is expected
to lie. The interval is associated with certain confidence
level that is, it can be stated with certain degree of confidence that the parameter will
lie within that interval. So, this is what we have, we are discussing few minutes ago,
that always this estimate whenever we say that this is the this is the interval, so,
that interval must be associated with some of some confidence level, in statistical sense.
Well. So, the confidence interval of the mean with known variance. So, so whatever the estimators
that we have seen for this mean, and as we are taking it from the sample, so, that will
also have some sampling distribution; it should have.And if we somehow know what is the variance
and known variance means, we know the population variance.If we know that, one then, how we
can get that confidence interval for the mean. So, for a large sample n, so, large sample
is again, this is a subjective word. So, generally we can, we have seen that, if n is greater
than equal to 30, then you can say that in case of mean only, so, not for all in general
case, in case of mean, if the sample size is greater than 30, then we can say that it
is a large sample. So, in such case, if x bar is calculated sample mean and the sigma
square is the known variance of the population. So, sigma square I know which is exactly which
is actual value of this population.How we know, that is the second issue, but, we know
this sample population we know the variance of the population.Only thing we are interested
to know, what is the interval for this sample mean.
Then, it is, it can be shown that this x bar is is a normal distribution with mean equals
to mu, which is the population mean and the variance of this x bar, that is, this sample
mean, variance of the x bar is sigma square by n or the standard deviation of this x bar
is sigma by square root n. So, this what I want to repeat once again,
that this sigma square is the variance of the population for the random variable x.Now,
what we are talking about is this x bar.This x bar is again another random variable, which
is normally distributed having the same mean of the population, which is mu and its standard
deviation is sigma by square root n. Now, you know from the, from our earlier lecture,
that this if we just take this quantity now, that this, what is the random variable minus
its mean divided by its standard deviation is a is a is a standard normal distribution.
So, that is why this x bar minus mu by sigma square root n is a standard normal variate.
So, now for once we know that this is the quantity and this follow a standard normal
distribution.Now, we can calculate whatever the confidence interval that we are looking
for. So, the confidence interval of the mean mu is given by is this, that is, x bar.Basically,
we are just equating it with two side of this standard normal distribution.
Now, if you if you see this one here.So, if this is your standard normal distribution,
then basically that that quantity, that is x bar minus mu by sigma by square root n,
that should, so, this is your, this is the standard normal distribution. So, this the
distribution that I have drawn is a standard normal distribution. So, this should lie between
these two values here, should in such a way that this area should be your, that whatever
the confidence limit that you wish to specify. Now, so, so this is the confidence level of
that estimated that it should have.Now, if I just say that this this confidence level
is, say at some some level, say that 95 percent confidence level. So, whatever is remaining
here is your 0.025 and whatever is remaining here is again 0.025.
So, we have to find out these two values.Suppose that, if I just put that one, z alpha by 2
and and this is say that z alpha by 2; obviously, this will be the negative side, this is 0
for the standard 1. So, this one, so, this quantity should lie between this z alpha by
2 and this z alpha by 2, where this alpha by 2 are basically the cumulative probability
up to that point. So...So, this x bar minus mu divided by sigma
by square root n should lie between this two, z alpha by 2 and this minus z alpha by, sorry,
z alpha by 2. So, z is the is the variate, is the is the reduced variate for this standard
normal distribution.And, this alpha by 2 is this part where we are. So, the confidence,
if I just want to relate to this confidence level is that, it will be that 1 minus alpha
multiplied by 100 confidence level. So, this is the confidence level that we can say. So,
that if this area, this white area is the alpha by 2, then this 95, that is here in
case of 0.25 here, can be relate to this one. So, this is that confidence interval that
we are, that is the confidence level that we are talking about for this, once we can
put this limits as this z alpha by 2.Now, if I just do some arithmetic change, then
it will come like this; that is, mu should be equals to here that x bar plus z alpha
by 2 sigma by square root n.Here it will be, x bar minus z alpha by 2 sigma by square root
n. So, you remember, that even this alpha by
2 when we are talking about. So, so this alpha by 2, it is automatically, when it is coming
to this negative side, it have this negative value. So, do not confuse that this will have
this have the negative value and this again another negative sign is here. So, this will
not be the plus. So, this is a single value that we are using here, which should havethat
this particular quantity.From the symmetry, both the values will be same numerically,
only this one will be positive and this one will be negative.
So, this is what is mention here, that is x bar minus z alpha by 2 sigma by square root
n minus mu less than x bar plus z alpha by 2 sigma by square root n. So, you know that
from the continuous distribution less than and less than equals to are same.
Where, this 1 minus alpha into 100 percent, this is the quantity, which is the degree
of confidence.And here, this minus plus z alpha by 2 is the value of the standard normal
variate, at the cumulative probability level alpha by 2 and 1 minus alpha by 2.
So, when you are taking this minus sign minus z alpha by 2, this is is the probability level
at this alpha by 2. So, if it is 95, if you put this alpha equals to 0.95 then; obviously,
1 minus, so, sorry, this is full is equals to 0.95, then alpha will become 0.05 and this
alpha by 2 will become 0.025. Now, at this 0.025, this z alpha by 2 will
be, say if it is 0.95, you know that this value will be 1.96. So, this will be x bar
minus 1.96 multiplied by sigma by square root n and here also, plus 1.96 sigma by square
root n this will come.
Now, as we are discussing that if the sample is more, if the sample is large, if it is
more than 30, now in other case, if the sample size is small, say if it is less than 30 and
if the x bar is the calculated sample mean and this, then this s square is this calculated
sample variance, then the random variable x bar minus mu divided by s square root of
n, follow a t distribution with n minus 1 degrees of freedom.
Here you see, this is that the variance, we do not know that is the unknown variance.
So, this one also we have to calculate from the from the sample itself. So, this also,
again when you take that this x bar minus this population mean divided by this sample
variance divided by square root n. So, this one instead of following this standard
normal distribution, it will follow the t distribution with n minus 1 degrees of freedom.
So, this t distributions and all we have discussed earlier.Only thing is that, if the sample
size is small, this quantity will follow the t distribution.If the sample size is more
and thisthis variance is known, then this will follow a normal distribution.
So, here in this case, this this confidence interval will be x bar minus t alpha by 2
sigma by square root n and x bar plus t alpha by 2 sigma by square root n.So, this is a
t distribution with degrees of freedom n minus 1.And, you can see even that from standard
text book about this t distribution, when this n n goes beyond 30, then the value of
this t alpha by 2 and the z alpha by 2 are essentially same.
So,where this again, the 1 minus alpha into 100 percent is the degree of confidence and
minus t alpha by 2, is the value of the standard t distribution variate at cumulative probability
again, that alpha by 2 and this 1 minus alpha by 2. So, basically the difference between
the t and the and the standard normal distribution is at this lower end level, where you will
get a a wider estimate of the interval because,the when the sample size is less, as there are
uncertainty is more. Though it is assumed that the sample is drawn
from a normal population, the expression applies roughly for non normal population also. So,
basically when the population is normal distributions, these are very well acceptable method. But,
even though it is non normal also, this method we can apply.
So, we will take one example of this, whatever we have seen, is that 30 concrete cubes prepared
under certain condition.The sample mean of these cubes is found to be 24 kilo Newton
per meter cube, if the standard deviation is known to be 4 kilo Newton per meter cube,
determine the 99 percent and 95 percent confidence interval of the mean strength of the concrete
cube. So, this 4 kilo Newton per meter cube is known;
it is from the population.And this one, when we say that this one is obtained from this
sample.
So, if we want to solve this one, then you know, that we first of all, we will get that
what should be the quintile value for this z alpha by 2 and this is for this 99% confidence
interval from the standard normal table.You can see that it is 2.575, which is z alpha
by 2 value.So, this sigma by square root n multiplied by z alpha by 2 is equals to 1.88,
whatever the data is available.
So, the 99 percent confidence interval will be the mean minus that quantity 1.88 and mean
plus 1.88. So, the confidence interval is 22.12 and 25.88 kilo Newton per meter square.
Sorry. To determine the 95 percent confidence interval,
again we have to find out the z alpha by 2 and this is your 1.96, earlier it was 2.575.
So, it is now 1.96.
If we calculate this one, it will become 1.43 and the 95 percent confidence interval of
the mean strength of this concrete will be 24 minus 1.43 and 24 plus 1.43. So 22.57 and
25.43 kilo Newton per meter square.Sorry.This is or this is the density sorry sorry this
is the density. So, this it is kilo meter per meter cube.
But, what we should observe here is that there are two confidence interval.We have determined
one is the 99 percent confidence interval and other one is the 95 percent confidence
confidence interval. So, the 95 percent confidence interval is
22.57 to 25.43 whereas, the 99 percent confidence interval is 22.12 and 25.88. So, you can see
that this 99 percent confidence interval is wider, because it is more likely that the
larger interval will contain the mean value than the smaller one. So, hence the 99 percent
confidence interval is larger, when the 95 percent confidence interval whereas compare
to this 95 percent confidence interval. So, this example, that we have used, it is
we we we have taken that a variance is known. So, once we we have decided the variance is
known, we have used the standard normal distribution. But, in the other case, we have seen that
sometimes the sample size is less and we have to we have to use, in that case, we have to
use that, what is the t distribution.From the t distribution interval, we have to use
that one. So, maybe we will take up the same example,
but, in that time we will just declare that whatever the distribution, whatever the standard
deviation that we got, it is not from the population, but, from the sample. So, that
example we will basically start from the next lecture and after that we will take what should
be the, that estimation for the other parameters like the variance proportion and all.And,
we will also relate, we will also see about this test of hypothesis which is obviously,
essential when we are comparing the mean of from the two different samples or the variance
or proportion from. So, when it is related to two different samples, then we have to
go for those testing. So, we will start from this point in the next lecture.Thank you.