Today, we will start a structural equation
modelling part 1; that is measurement model. SEM, that is structural equation modelling,
measurement modelling, measurement model. Now, let me see the content of today’s presentation.
That we will start with conceptual model, then the assumptions of the model, then how
to estimate the model parameters and model adequacy test, followed by a case study. You
see in last class, I have explained that measure structural equation modelling has two components,
one is measurement component, and another one is structural component.
Measurement component is essentially a confirmative factor analysis, and structural part or we
can say the structural model is equivalent to your path model or path analysis. Both
the model those measurement as well as structural path, there are three important steps, one
is model identification, then parameter estimation and model adequacy test. This is true for
structural part, also in this lecture we will consider the measurement part which is confirmatory
factor analysis. So, let us start with a conceptual model first. You see here. In last class I have shown you similar diagram
and you see there are three factors Xi 1, Xi 2 and Xi 3 and each of the factors are
manifested by a different variables starting from X 1, X 2, X 3 for Xi 1, X 4 and X 5 for
Xi 2 and X 6, X 7, X 8 for Xi 3. So, in confirmatory factor analysis the basis is that that there
are hidden constructs which are this or latern construct. Other you can say hidden variable
that Xi 1, Xi 2, Xi 3 which are the causes of some manifest variable like X 1, X 2, X
3 to X 8 by putting this arrogate. For example, from Xi 1 to X 1, Xi 1 to X 2 and Xi 1 to
x 3 we are restricting the model here in such a sense that we know that Xi 1 is manifested
by X 1, X 2 and X 3. This manifest variable. Similarly, Xi 2 is manifested by X 4 and X
5. Similarly, Xi 3 is manifested by X 6, X 7 and X 8, another issue here that this hidden
construct or latren construct, latern variable they co vary in the sense that if Xi 1 change
there may be change of Xi 2. There may be change of Xi 3 where the correlation components
is there. So, these type of, this is a typical structure of confirmatory factor model and
other issues here, apart from this correlation between the construct which is denoted by
phi. This come phi 2 1, this is phi 3 1, this curvature
line, this curvature line is basically phi 3 2. Now, we are saying X 1 is caused by Xi
1 and as a result a causal linkage is given, Xi 1 to X 1 and parameter which basically
depicting the relationship between Xi 1 and X 1 is lambda 1 1. Earlier, I also told you
this lambda 1 1, this 1 1 this suffix comes that X 1 from X 1 this point is taken, Xi
1 1 is taken. So, it is not possible that the variability
of X 1 will be fully explained by Xi 1. So, here is possibility of some other variables
or hidden causes which may affect X 1 or we can say the errors part, noise part, all those
things are considered by delta 1. So, in the same manner you have to explain X 1, X 2,
X 3 and up to X 8. What I said verbally, this is depicted in equation form. We are saying
X 1 is represented by lambda 1 or Xi 1 plus delta 1. If you think of regression line point
of view you will be getting this equation. So, in the same manner there are as there
are 8 X variables, so you are getting 8 linear equation and collective if you write in matrix
form then that will be X 8 cross 1 equal to that capital lambda, which is a matrix of
the dimension 8 cross 3. And which you can see here 8 cross 3 dimensions here, because
Xi 1, Xi 2, Xi 3 that 3 and 8 X variables, this 8 plus this is Xi plus delta this one.
So, this is the, this is the matrix form equation for this particular example. Now, you can go for a general equation from
there that means what I means to say here the general equation will be X p cross 1 where
we are saying. There are P number of manifest variables and
which can be represented by this manner, the lambda which is a matrix of matrix relating
P manifest variable to m factors. And this m factors are denoted by like this m cross
1 and definitely for every manifest variable there will be an error. So, this is the equation
for confirmatory factor model, if you see recall that factor analysis you have found
out there also similar relationship, but there are there is difference in the structure of
the model then the model assumption in the covariance structure.
So, what are the assumption? Here we assume that the expected value of delta equal to
0, that mean the noise variable, the error terms that mean is 0 and covariance of delta,
this one is your theta. Or you can write theta delta, also m times theta delta and it is
symmetric. So, if I say like this will be P cross P.
So, this you variance component of delta of diagonal in the covariance that will be equal,
that is why symmetric and delta is multivariate normal with mean 0 and covariance matrix theta
delta. Or you can write theta also another important issue here is that assumption is
that covariance between Xi, that mean manifest, sorry latern construct and the error term
related to the manifest variable, they are 0. So, this is our your assumptions related
to confirmatory factor model. Now, there as I told you there are covariance.
So, the covariance structure of your see covariance structure, first one is covariance of X, this
will be capital sigma P cross P. There will be covariance structure for the Xi latern
construct, this one will be phi which is again m cross m matrix diagonal, not diagonal this
is symmetric matrix. Then covariance of delta which we say theta P. P it is mostly assumed as diagonal matrix, assumed as diagonal not necessary always it will be diagonal,but it is assumed like this So, you have seen earlier in your exploiting factoring, also we have written X equal to
that delta Xi plus, I think we had lambda Xi plus delta, means this capital lambda,
replace delta. Now, if you find out the covariance of X then ultimately what you will be finding
out? You will be finding out something like this plus covariance of this delta. This is
theta in X and this phi was not there in excusive factor, it was I if phi equal to I then it
is orthogonal factor analysis. So, this is null set that covariance structure and the
relationship between this, okay? So, now let us stick to come to the model
identification part, what I mean to say here by model identification if you clearly look
into the model and the parameters that to be estimated as well as the information, what
is available. There must be sufficient necessity and sufficiency of the information available
to estimate the parameters of the confirmatory factor model. So, in order to do so you should
now let us find out that what are the parameters, we require to be estimated if you see this
slide you will see that we have few parameters to be estimated from in C F A. One is your lambda, capital lambda which is
p cross m matrix. So, these many parameters to be estimated, there is phi which is also
a p cross p matrix which is p cross p matrix, but being symmetric it has phi is m cross
m matrix. I am sorry phi is m cross m matrix which is symmetric matrix. So, numbers of
parameters will be m into m plus 1 by 2 to be estimated and then there is theta delta,
as I told you that theta delta. What is this theta delta? This is delta, this is theta
delta. So, theta delta or theta so there if we assume that it is diagonal then there will
be p number of parameters to be estimated. So. we require to estimate capital lambda,
we require to estimate phi, we require to estimate theta delta. Here in case of capital
lambda p cross m, this number of parameters phi m into m plus 1 by 2, this number of parameter
theta can number or parameters are there. Then in total the number of parameters to
be estimated, number of parameter to be estimated we can write p m plus m into m plus 1 by 2
plus p, okay? Now, what you require to know that if t, suppose t equal to 50. If we require
to estimate p 50 parameters and you require at least 50 simultaneous equations, getting
me? Now, what information we have in case of confirmatory factor analysis? We have only one information, that is capital
sigma which is the variance, co variance matrix of X, that is co variance matrix of X. Now,
how many unique or non-redundant elements, this is also symmetric matrix. So, it has
number of know redundant element equals to p into p plus 1 by 2. Now, see this what type
of situation will occur? We require to estimate p number of parameters, it may so happen that
number of parameters to be estimated is greater than number of independent non redundant elements
in p. That is the available information, it may so happen that t equal to p into p plus
1 by 2. It may so happen that t less than p into p plus 1 by 2.
Now, the first case, this is model cannot be identified. This is model is unidentified
or under identified, un identified this case number of parameters or the number of unknows and knows are equal. This is uniquely identified case and this case this is over identified, because we have more information available over identified case. So, at least this two are necessary if you have this case type of situation which is uniquely estimated. If you have this over estimation, that is the desirable one over
estimate is the desirable one. This condition, particularly this two, if
this two conditions are either of the two is satisfying you are saying that necessary
condition is satisfied. Necessary condition which is also known as ordered condition,
okay? But ordered condition alone is not sufficient, this is necessary condition necessity is satisfied.
There is another condition called rank condition, because then you will see that we have basically
talking about the matrices. So, the rank condition, rank of my matrix
is important issue here and it is little bit complicated one also. Rank conditions also
verbally you have to satisfy then the rank condition is the sufficient condition and
for example, is given that Willey in 1973 that is the reference. Now, order conditions
satisfy the necessity and rank conditions satisfied the sufficiency. Let us assume that
it is done. In the sense model e is identified, if model
is identified. The next step is how to estimate the model. So, estimation of model parameters,
okay? So, you have seen that we assume that ultimately
that X, the manifest variable is normally distributed. It is the primitive multivariate
normal p X variable p, p number of variable are there, multivariate normal with mean 0
and variance, co variance matrix sigma, capital sigma. Then for any observation multivariate
observation Xi you can write that this is the PDF can be written like this, 2 pi to
the power p by 2 sigma determinant to the power half e to the power minus half. Then
X minus mu e, e is to write that is X i minus mu e is zero here.
So, X minus m transpose, that means Xi transpose sigma inverse Xi, this is the multivariate
normal distribution per density function for a particular multivariate observation. Now,
we collect in observation i equal to 1 to n, we want to know the log first the likelihood.
So, likelihood if you see this equation you find, you see the there is only one parameter
which is sigma and mu e is 0. So, only one parameter is there, so we can
write log of likelihood of sigma, not log likelihood of sigma which can be written as
multiple equation of this, i equal to 1 to n f xi which will be multiplying i equal to
X equal to 1, X 1, X 2 up to X n. Then the resultant will be like this 1 by 2 pi to the
power n p by 2. Then determinant to the power n by 2, then e to the power minus half, then
sum i equal to 1 to n. Then Xi transpose sigma inverse Xi that is what will be the likelihood
and it is customary to take log likelihood. So, if you take log likelihood then what you
get? You get minus m p by 2 log 2 pi for this term, minus n by 2 log this for this term
minus half, i equal sum of i equal to 1 to n Xi transpose in sigma inverse Xi. So, obviously
this from our we want to estimate this sigma that parameters it will be will go for some
optimization root and there this constant term in the equation has whether you had keep
it or do not keep it this is immaterial. So, we remove this constant. So, you can write
this as minus n by 2 log of this plus half of I can write like this, minus n by 2 into
this minus. So, minus half of this Xi transpose this xi. So, let me write a phrase that log
of l, this sigma equal to minus n by 2 log determinant sigma minus half Xi transpose
sigma inverse Xi. Now, this term can be written like this minus
n by 2 log this minus, if I write n by 2 then summation i equal to 1 to n. This can be written
like this, this can be written like this 1 by n, I am teaching because I have considered
n here. So, this one can be written like this, write again, you write like this like this,
then I come to R. R form n by 2, this minus n by 2 we can write this quantity as trace
1 by trace of 1 by xi transpose xi sigma inverse this, this is possible. Now, 1 by n Xi transpose Xi, this is nothing but variance, covariance matrix of the sample provided, n is large, 1 by n
minus 1 and 1 by n become same. So, with modelling equation we can write like
this minus n by 2, then this is trace of s sigma inverse. This one is minus n by 2 log
of this plus trace of s sigma inverse. So, this is your log likelihood. Now, see what
is the condition here in our estimation here, actually this procedure is like this, you
will from the model from the model covariance of X, your lambda phi, lambda transpose plus
theta delta. From this you will get covariance X sigma, that is sigma in terms of model parameter
you collect sample, then you get the covariance matrix.
Also, from sample there will be S. S is the again p cross q that sample covariance matrix,
there what we want to do? We want to match this two, this two suppose a condition is
such that S is equal to sigma, then if I put here what I can write? Log of l of s, this
can be written like this, this log of S determinant of S plus trace of S S inverse. Now, S S inverse
is I, so you can write like this. This log of S plus sum of the diagonal elements of
the matrix I, that is p this is your equation number 1. Now, another one is we have already seen. The likelihood one, this equal to minus n
by 2 log of this plus trace of this, this is our equation 2. So, what you want? We want
to find out parameters. Now, this sigma this will be in terms of model parameter S here,
and here when we are talking about S it is basically the numerical values and here it
is in terms of model parameters like lambda phi and all those things.
So, we will create a function now that we want to minimize that we are saying F theta
which is nothing but S minus sigma theta of this nature which I am saying, not exactly
which will be of this nature. So, then we can write like this F theta equal to log of
l S minus log of l sigma, which if you write this is minus n by 2 log of determinant of
S plus p plus n by 2 log of determent of sigma plus trace of S, this trace of S sigma inverse.
So, this one you can write minus that n by 2. Then log of determinant of sigma plus trace of S sigma inverse minus log of S determent S minus p, we want to minimize this function. So, keeping this constant n by 2 again is of no use. So, final equation will be for our estimation, is this log of determinant of sigma plus trace of S sigma inverse minus log of S minus p. This is the equation, which want to minimize, okay? So, its null issue you have to use Newton
Rapson or similar method, Newton Rapson similar method of numerical that optimization part.
So, this is what is in the nut shell, the parameter estimation in place of confirmatory
factor analysis, which is basically our measurement model, you see here. This is the theta log sigma theta trace of
this, this is the case ignoring the constant. So, ultimately minimize this one using Newton
Rapson or Gauss algorithms, okay? Now, let us see that whatever mathematics
we have described now can it be put into a case study as a real life example. I will
show you the example here which I have shown you earlier also in the in last, in the first
class of structural equation modelling. I have shown you this one, but there what I
have done actually, I have given you a glimpse of this things, just like scrawling down the
slide. Because just the glimpse of what is this not will describe in detail that what
is this measurement model and with the same case study, okay? So, the case study as you
know it is a role of personal and socio technical factors in work injuries in mines and a study
based on employee’s perceptions and you can see that the source is Paul and Maiti,
2008. It is published in the ergonomics, okay? So, let us start like this we have several
manifest variables here. There are 18 manifest variables for example, age, experience, impulsivity,
negative, affectivity, depression, risk taking, safety training, safety practice, safety equipment,
availability, maintenance, job stress. Like this we are wondering that how I am saying
these are manifest variable although, most of the things cannot be observed. So, actually
what happen for every of the variable, we ask several questions and then those questions
are summed into a particular quantity. And then that summed up values we have taken as
the value for each of the variable, for each of the observations or individuals who participated
in this study. So, in that sense it is manifested, means
observed in that sense. Otherwise, its two layer questions actual it was like this, only
one questions, then there sum to this. This manifest variables what we are saying there
sum then further level of aggression actually. Suppose, there are question 1 to let it be
150 then there are manifest variable like Xi 1, like there are 18. Then this again it
is aggregated into what I can say these are X 1, sorry these are X 1 to X 18 related to
Xi 1 to some Xi, let it be Xi 9. So, this level of aggression is done here.
So, we are taking in this level of aggregation, we are considering here this is manifest variable,
but you may start from here to here. That will be combos some we have this. So, the
same thing if you write in the confirmatory factor analysis form, it will be something
like this. You see all the this covariance structure
between this 9 gita Xi variables, it is not pictorially shown because of space concept.
Otherwise, this will be this and then what will happen? We will immediately, you can
go for the equations also for this, getting me? So, like I am giving one equation only
here if I want to know what is X 1? Then this is nothing. I can write lambda 1 1 Xi 1 plus delta 1 if
you consider this. So, here lambda 1 1 Xi 1 plus delta 1. Similarly, this one lambda
2 1 Xi 1 plus, so x 2 will be lambda 2 1 Xi 1 plus delta 2, if you consider X 3. So, X
3 is ultimately it is the single indicator manifest variable for the constant Xi 2. So,
X 3 can be written like this that 3 2 xi 2 plus delta 3.
So, in the same manner as there are X 18 so you will be able to write X 18, come to this
one, X 18 is again a single indicator for Xi 9. So, this is your lambda 18 9 xi 9 plus
delta 18. So, you can write in matrix form, when you write in matrix form you will be
getting a equation in matrix form, equation we said X equal to lambda Xi plus delta. This
equation you can find out here, here we have 18 cross 1. This will also be 18 cross 1,
this is our 9 cross 1. So, this will be 18 cross 9, so this type of equation you can
find out, okay? Now, let us see the model identification for
this case. You just see that lambda part that how many lambdas are there. You count 1, 2,3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18. So, 18 lambdas, so we have written
lambda 18 because others are 0. For example, lambda 3 1 if you give one linkage here lambda
3 1 that is 0, because it is a confirmatory. We know what are the manifest variable coming
out of the hidden constructs phi. There are how many X 18. So, how many Xi 9 xi. So, m
into m cross m into m plus 1 by 2 m into m plus 1 by 2. So, 9 into 10 by 2, that will
be 45. So, you have lambda that is your 18, then
your phi related variables will be 45 phi, related parameters will be 45 theta delta,
again 18 delta 1 to delta 18. So, our t is 18 plus 45 plus 18 that is 81. Now, what is
the unique elements? There we have 18 manifest variable cross 18. So, 18 into 19 by 2 this
will be 171. Now, t equal to 81 it is much less than 171, the model is over identified,
it is a good case so and necessity condition you satisfied and sufficiency we have not
tested here that says the software they test all those things. Now, let us see that data part. The data part
is actually random independent sampling. First we have taken accident group of workers followed
by with frequency matching, non accident group of workers where all together 300 observations. So, immediately as I told you that how many
independent non redundant element in your sigma matrix. This is the case, this is from
sample, these are all co relation matrix. Now, question is what we want? We want basically
to minimize this function and in this matrix is basically for S, that is sample co variance
matrix, but actually we have taken correlation. Now, question comes whether co variance or
correlation it all depends on the purpose of the study. In our purpose of the study
we are more interested in the pattern of the relationship, then the original strength of
relationship between latent variable and your manifest variable. We are more interested
in the pattern of the relationships and not the original value. So, when you are interested
in the pattern of the relationship R is a variable, R matrix should be used, that is
correlation matrix should be used. Now, through definitely this is F theta, this
is the function which is to be minimized and we have used this software, in this case LISREL
linear structural relations. So, this software we use and ultimately this
is what is the all the parameters which is estimated. You can see if you go back lambda
1 1, lambda 2 1, lambda 3 2, lambda 4 3 like this and you see that lambda 3 2. What is
the value of lambda 3 2? Here, lambda 3 2 is 1 because this one indicator with one constant,
we assume that this is the manifest variable itself is the construct, understood?
So, that will be the what is the output of this measurement model. See ultimately we
are talking about long back I think this one, these things when we have clubbed into this
factors, these are all the factors or constructs latent constructs. This Xi 1 to Xi 9 these
are not arbitrary Xi 1 to Xi 9, they have some meaning. Actually X 1, X 2 if you see
that age and experience then demographics, this is the this is Xi 1 impulsivity negativity
all four are clubbed to the var and Xi 2 value. Actually the negative personality is given
here. I think this Xi 3, X 1, X 2, X 3. X 3 is work
injuries. It is kept as it is what X 3, Xi 3 is the negative personality, then Xi 4 is
your safety environment. Again Xi 5 is job stress, Xi 6 is social support, Xi 7 is job
dissatisfaction, Xi 8 is work hazards, Xi 9 is safe work behaviour. So, these are all
latent variables in this sense now we also want to know the co relation matrix between
latent variables which is the output of this measurement model. You see that demographic work injuries negative
personality, these are the latent constants and this is your correlation matrix in then
there are little star is there. This star indicates point 0.05 probability level of
significance. I think all are significant here except this value 0.04, 0.01 some other
value, but essentially we are interested from measurement model to know that what is the
correlation matrix of the latent constructs or factors what we are going to evaluate or
estimate, getting me? That is what we have done and we have done this with the help of
this, this lambda values and the Xi and the error term you are getting this values. And
this will be this is a value, you want this is very, very important one because this will
be used in structural model as input to structural model in structural equation modelling, I
told you. That structural equation modelling two parts
SEM has two parts, measurement model and structural model. The output of this will be input to
this, fine? That I would consider. The structural measurement model would I consider
measurement model as I said or not. If the model is not fit, if it is not adequate enough
then the correlation matrix between the constructs generated they are not good. Also, we have
doubt about those correlation values, we cannot abruptly accept this one. Now, in model adequacy
test, in last class also I told you that fit index there are three types of fit index,
absolute fit index, relative fit index and parsimonious fit index. Under absolute fit
index chi square, chi square degree of freedom. So, absolute fit index, relative and parsimonious
we are discussing about and absolute fit index there are many indices like chi square, chi
square by degree of freedom, goodness of fit index, root mean square, residual root mean
square. That is RMR RMSEA standard error approximation,then relative fit index. These are the standard indices available in any literature related to structural equation modelling and most
of the, why most, I think almost all the indices are based on chi square value. So, we will
discuss little of this. For example, absolute fit index what it does? It answers this question is the residual or
unexplained variance remaining after model fitting appreciable. So, we do something like
this. There will be two that hypothesis null and alternate hypothesis. Null hypothesis
is we are saying that sigma equal to sigma theta, that actually that what you have estimated,
that is correct and alternatively you are saying no, they are not correct. So, then
this we will define one quantity called chi square, which is n minus 1 into F theta, that
F theta you have seen that the minimization function, so that value you have after estimation.
So, chi square that n minus 1 F theta, this follows this chi square distribution with
nue degrees of freedom, where nue can be estimated like this, p into p plus 1 by 2 minus t, that
is number of non-redundant elements minus number of parameters to be estimated. That
is what more degrees of freedom available here and you will find out that what chi square
value you get that should be as small as possible, because if for perfect fit F theta will be
0, the n minus 1 into F theta that it will be 0. So, 0 is the ideal value, but it will
all depends on sample size, also n minus 1. Now, see that you will never get this your
theta will be 0, because you are doing the numerical way of optimization, numerical optimization
where some convergence value will be there. Now, if n is sufficiently large what will
happen? This value will become large. So, what you want to how do then justify that
whether the model is fit or not. One way is that this value should be as small as possible
and other one is you go by chi square by degree of freedom. So, it is recommended in the later lecture
that essentially the chi square distribution is such that the expected value of chi square
nu is nu, because that is the degree of freedom. Because it is a this parameter in chi square
we use the degree of freedom only. So, actually the chi square by the degree of freedom should
be 1, but is not recommended what is said that 225 is the recommended value above 35
constraints. Now, if you use G F I that is goodness of fit index which is similar to
R square in multiple regression, you can remember or recollect that R square equal to 1 minus
SSE by SST. Now, you see this formulation here, that way
we have written here that 1 minus trace of this by this. So, this is total variability
and this one is the error term. It is similar to R square and this G F I value varies from
0 to 1 and it is desirable that G F I is greater than 0.90. So, in your model when develop
a measurement model the software will give you the G F I value, if you find out that
the G F I value is 0.9 or more, that it is good. It is desirable, but if it is less than
0.9 what you will do? You will not consider the error model, it all depends on the system
for which you are developing the model. I am telling you even 0.8 also you can consider,
absolutely no problem. If you think that the dynamics and is huge the volatility is more,
many other issues you have to take into consideration. Now, another index is RMR, I think this is
something where each of the value of the S matrix and each of the corresponding value
of the estimated matrix, that sigma basically you take S then you estimate sigma and by
that process in between the parameters are also estimated. Now, the values sigma and
S values, S this values and sigma estimated, this values are here, getting me?
Now, if you take this and this what is the difference? Take this, this what is the difference?
So, we will take this, this what is the difference? These differences are squared here, you see
what we have done S J K minus sigma J K this is the estimated one square by p into p plus
1, that is the non-redundant part. 2 is given because that twice of this, this by 2 p into
p plus 1 by 2. So, this quantity should be also as low as
possible. It varies from 0 to 1 and RMR should be less than 0.05 and for RMSEA root mean
square error approximation, this is the modulus of chi square minus its degrees of freedom
by n minus 1. It is seen that 0.03 to 0.08 in this range this lies and this range also
says substantial increment then relative fit index. Now, how well does a particular model do in
explaining a set of observed data compared with a range of other possible models. Here
what you do? You creates nested model, several models and then you compare one model with
other and then you say which model is better. And based on this you create a index and that
index talks about your model adequacy or otherwise we can say improvement in terms of model adequacy.
Here, most of mostly we will consider null model, that is the worst fit model which is
known as null model, where we think that the covariance matrix is diagonal. Only diagonal
means variance part is there of diagonal elements as 0.
Now, if you say that X chi square for the null model is chi square 0 and chi square
for the proposed model is chi square mu, where mu is the degrees of freedom. Then you are
in a position to develop or I can say quantify, this indices like NFI is chi square 0 minus
chi square nu by chi square 0 and all these indices this values lie between 0 to 1. And
it is desirable that they will be greater than 0.09, sorry 0.90 then CFI comparative
fit index, that is 1 minus chi square nu minus nu by chi square 0 minus nu 0.
And TLI you have seen this, this also in the similar manner you see. Ultimately they take
into consideration the chi square value of the proposed model and a worspit model which
is known as null model and the comparative indices are developed and higher the index
value, it is better 0.9 or more is required. Then your parsimony fit index, it is similar
to adjust R square S A square in you regression and it basically talks about the par parameter
fit. Par parameter is estimates and AGFI here you just see that the top upper portion or
the denominator here is divided by the degrees of freedom and numerator is also divided by
the degrees of freedom. What we have done in calculating R A square, this value should
lay between 0 and 1 and AGFI greater than 0.90 is desirable. Then parsimonious non fit
index which is nu by nu zero NFI, and where this is nu is the degree of freedom proposed
model and like this, okay? Now, let us see the goodness of fit indices
for the case study here, some of the fit indices I have given there are others. So, chi square
degree of freedom of 99, chi square value is 257.24, if you divide by 99, this is almost
100. So, it will be around chi square by degree of freedom is around 2.6. So, it is good because
aim is to do 5 root mean square residual 0.06 which is little more than 0.05, CFI is 0.98
very good more than 0.90, NFI 0.97 more then 0.90, CFI is 0.99 and IFI is 0.99. So, essentially
then chi square is 257.24 chi square by degree of freedom is around 2.6, RMR is 0.06, GFI
is 0.98 like this. This model is very good, fit model. Now, see that who has basically worked in
this, who are the pioneers that Karl Joreskog and Dag Sorbom, I think in around 1978 probably
they have developed this software. First that listenery came I think in 1989 and its remarkable
development in this field and we all are tremendously benefited. What I can tell you further that for you have
to understand the structural equation modelling. I might say that Joreskog, Sorbom this LISREL
8, structural equation medalling with SIMPLIS command language. This man, this manual is
very good and you can go through and is a lot of publications by Joreskog, Soorbom.
Others is the Hayduk is one person who has written a book, this and this is a very good
book. Also in addition there are many other books available in structural equation modelling. So finally, let me just summarize to my today’s
lecture. We said that structural equation modelling has two parts, one is measurement
model measurement model, another one is structural model. So, measurement model is nothing but
confirmatory factor analysis structural model is actually path analysis. Now, we have discussed
details of CFA in terms of its identification, what I say that under identification there
will be necessary condition. There will be sufficient condition, this two
must be satisfied in necessary condition is known as order condition and this one is known
as rank condition. Other one sufficient condition is known as rank condition and in this order
condition we say the number of parameters to be estimated must be less than number of
non-redundant elements in the covariance matrix. Then this is over identification case and
it is a desirable case, then we have shown you the estimation parameter. Estimation,
now I said that parameter estimation it is basically a function you minimize, which is
basically log of determinant of sigma plus trace of S sigma inverse minus log of determinant
of S minus p. This function is minimized through Newton Rapson or similar method and then the
parameters are estimated and the actually we have sample data in terms of S or R. And
we have the population value in terms of sigma theta. We try to match this two and using
this function the better, the best match is considered and then you corresponding the
theta value, these are used. Now, theta is function of many things like
lambda, like your phi, like your theta delta. So, these are theta, means theta means so
many things are there. Any combinations that is what you are trying to estimate because
from co relation matrix to here, co relation matrix or covariance to co variance matrix,
one to one, this correspondence you are doing. This is parameter estimation, once parameters
are estimated then you can test the parameters values, this lambda using simple t test whether
it is significant or not, but apart from this the another important output from this CFA
is after parameter estimation is your co relation matrix of the latent correlation. Or covariance
latent construct which is very, very important, because this will be the input to the structural
model. Then what I have given you? I have given you the what are the model adequacy
test. So, under model adequacy we have seen that
absolute test and then your comparative test or relative test, another one is parsimonious,
absolute is similar to R square R A square, where the actual variance explain is considered
in comparative case. We compare with different model and parsimonious, it is basically fit
part parameter estimated and finally I have shown you a case study for all of you. The
case study is there, if you are interested please go through Paul P and J Maiti, the
synergic role of socio technical and personal characteristics in mines, published in ergonomics
in 2008. Thank you very much.
