WEBVTT
Kind: captions
Language: en
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.
Welcome students to the MOOC's lecture series
00:00:21.699 --> 00:00:31.579
on statistical inference . In the last lecture
, we have talked about point estimation of
00:00:31.579 --> 00:00:40.600
the parameters of the underlying distribution;
from which a sample x 1, x 2, x n is taken
00:00:40.600 --> 00:01:07.080
. In particular, , we have discussed 2 methods
method of moments .
00:01:07.080 --> 00:01:24.670
And method of maximum likelihood .
In method of moments what we have done ? We
00:01:24.670 --> 00:01:32.030
have calculated sample raw moments from the
sample x 1, x 2, x n and we have equated that
00:01:32.030 --> 00:01:41.659
with the theoretical moments as given by the
distribution under consideration. And by equating
00:01:41.659 --> 00:01:50.229
them we have formed equations and we obtained
the estimate for different parameters by solving
00:01:50.229 --> 00:01:57.250
those equations . On the other hand, the philosophy
of maximum likelihood is slightly different
00:01:57.250 --> 00:02:05.520
.
Here we try to differentiate the likelihood
00:02:05.520 --> 00:02:13.220
function of x 1, x 2, x n under the parameter
theta by taking it is derivative or partial
00:02:13.220 --> 00:02:24.209
derivative with respect to theta, and equating
it to 0 so that by solving for theta we get
00:02:24.209 --> 00:02:34.930
a function of x 1, x 2, x n which maximizes
the likelihood function. Or in other words,
00:02:34.930 --> 00:02:42.940
which maximizes the probability of obtaining
the sample x 1, x 2, x n and there theta for
00:02:42.940 --> 00:02:51.610
which that is maximized is considered to be
the maximum likelihood estimated.
00:02:51.610 --> 00:03:04.630
However, , the problem of point estimation
is that
00:03:04.630 --> 00:03:23.260
it is on the basis of sample x 1, x 2, x N
. And based on that we are trying to estimate
00:03:23.260 --> 00:03:35.870
the value of the parameter theta . As I have
already shown you an example suppose a coin
00:03:35.870 --> 00:04:00.220
is tossed N times and the outcomes are say
1 0 0 1 0 0 0 1 0 0 ; that is there are 3
00:04:00.220 --> 00:04:12.239
heads and remaining tails .
Based on that , if we try to estimate p then
00:04:12.239 --> 00:04:21.970
p hat comes out to be say 0.3 .
But suppose we make another experiment of
00:04:21.970 --> 00:04:29.810
the same coin, there may be 4 heads or there
may be 5 heads. And accordingly the p hat
00:04:29.810 --> 00:04:39.009
will change according to the number of samples
.
00:04:39.009 --> 00:05:12.250
Hence , it is prudent to obtain and interval
of the form say a b ;
00:05:12.250 --> 00:05:35.210
such that probability theta belongs to a b
is very high , say 95 percent or 99 percent
00:05:35.210 --> 00:05:45.319
. What is the advantage ? The advantage that
we are very confident that based on the sample
00:05:45.319 --> 00:05:54.300
we can see that; theta is going to belong
to this interval with a very high probability
00:05:54.300 --> 00:06:00.409
. That assurance we cannot give with respect
to point estimation .
00:06:00.409 --> 00:06:14.009
Hence comes the idea of
00:06:14.009 --> 00:06:33.760
interval estimation .
So, what we do here ? We want to obtain statistics
00:06:33.760 --> 00:07:02.219
T 1 and T 2 based on the sample
x 1, x 2, x n such that
00:07:02.219 --> 00:07:15.490
probability theta less than equal to T 1 is
equal to say alpha 1 , and probability theta
00:07:15.490 --> 00:07:23.830
greater than equal to T 2 is equal to alpha
2 .
00:07:23.830 --> 00:07:43.509
So, pictorially suppose, this is the range
of theta ; which we have already denoted as
00:07:43.509 --> 00:07:53.589
capital theta . Depending upon the distribution
and the parameter , the theta will change
00:07:53.589 --> 00:08:08.120
, but suppose we have T 1 here .
Such that probability theta belongs to this
00:08:08.120 --> 00:08:16.199
interval, or theta less than equal to T 1
is equal to alpha 1. And suppose this is the
00:08:16.199 --> 00:08:27.759
T 2 such that probability theta greater than
equal to T 2 is alpha 2 .
00:08:27.759 --> 00:08:40.039
Then probability T 1 less than equal to theta
less than equal to T 2 or say let them be
00:08:40.039 --> 00:08:46.480
strict inequality, it does not matter because
it is a continuous case and this probability
00:08:46.480 --> 00:09:02.890
is therefore, going to be 1 minus alpha 1
minus alpha 2 . If alpha 1 plus alpha 2 is
00:09:02.890 --> 00:09:33.821
equal to alpha , then this interval
has a probability say 1 minus alpha , that
00:09:33.821 --> 00:10:13.130
theta will lie in this interval .
Question is how to obtain this interval and
00:10:13.130 --> 00:10:34.340
the confidence of these intervals . By confidence
we mean, the probability that the parameter
00:10:34.340 --> 00:10:55.790
will remain within that interval T 1 and T
2 . Typically, we consider 95 percent and
00:10:55.790 --> 00:11:17.250
99 percent confidence intervals .
Consider 95 percent . Suppose this is the
00:11:17.250 --> 00:11:33.480
range of theta , we want T 2 such that in
this there is 2.5 percent chance that the
00:11:33.480 --> 00:11:43.350
variable will occur on this side of T 2 , and
say T 1 such that on this side also there
00:11:43.350 --> 00:11:54.610
is 2.5 percent chance, that the theta will
occur in this region. So, the 95 percent confidence
00:11:54.610 --> 00:12:08.980
interval for theta is this .
If it is 99 percent, then we look at a T 2
00:12:08.980 --> 00:12:19.490
such that the occurrence here is only 0.5
percent and the occurrence probability of
00:12:19.490 --> 00:12:29.410
here is 0.5 percent. Therefore, that theta
will occurred in this interval has the chance
00:12:29.410 --> 00:12:40.420
99 percent. So, these are that confidence
intervals that we try to obtain for the parameter
00:12:40.420 --> 00:12:46.320
theta or with the help of the sample x 1,
x 2, x n .
00:12:46.320 --> 00:13:09.310
To obtain this limits
we have different tables ,
00:13:09.310 --> 00:13:29.930
for this class
I shall show the normal table and chichi square
00:13:29.930 --> 00:13:51.700
table . So, let us first understand how to
see the normal table .
00:13:51.700 --> 00:14:04.320
We know the shape of the normal curve is somewhat
like this .
00:14:04.320 --> 00:14:17.120
Consider this is 0 . Let us just look at normal
0 1 , because for any other normal with mu
00:14:17.120 --> 00:14:28.390
and sigma square, we can convert it to the
standard normal distribution . We know that
00:14:28.390 --> 00:14:42.800
probability z which is standard normal less
than equal to 0 is equal to 0.5 . So, for
00:14:42.800 --> 00:15:02.920
95 percent we will look at the value say call
it z 0.975.
00:15:02.920 --> 00:15:32.839
Because the area below this is equal to 97.5
percent ; that is this area is 2.5 percent
00:15:32.839 --> 00:15:52.280
. And since z is symmetric , this value such
that below this is 2.5 percent is going to
00:15:52.280 --> 00:16:12.070
be minus of this . Therefore, z of 0.025 is
equal to minus of z of 0.975 .
00:16:12.070 --> 00:16:24.890
So, there are tables where these values are
tabulated . First let me show you the normal
00:16:24.890 --> 00:16:31.370
table .
So, if you look at that table, it is showing
00:16:31.370 --> 00:16:44.480
standard normal distribution table values
represent area to the left of the z score
00:16:44.480 --> 00:16:53.280
as you can see that .
So, in this table there are all the values
00:16:53.280 --> 00:17:00.730
tabulated . Let us see the value corresponding
to 0.975 .
00:17:00.730 --> 00:17:17.130
Consider this value , this is 0.975. So, we
have to obtain the x value for the value on
00:17:17.130 --> 00:17:24.639
the real line such that probability a standard
normal variate less than equal to this is
00:17:24.639 --> 00:17:36.370
0.975. How to obtain this ? We first look
at the left side of this table and we can
00:17:36.370 --> 00:17:43.950
see the value given is 1.96, 1.9 as you can
see .
00:17:43.950 --> 00:18:01.009
So, that is 1.9 , and for the second decimal
place we go above and we can find 6.
00:18:01.009 --> 00:18:15.559
Therefore we get 1.96 . In a similar way,
suppose we want 99 percent confidence interval.
00:18:15.559 --> 00:18:23.760
Therefore, what we will be looking at ? We
will be looking at a value such that below
00:18:23.760 --> 00:18:36.850
that the probability of occurrence is going
to be 0.995. So, again we look at the table
00:18:36.850 --> 00:18:50.419
, and we find that
00:18:50.419 --> 00:19:03.100
this is the value which is 0.99492 and if
we go further we get 0.99506 .
00:19:03.100 --> 00:19:16.610
Therefore, 0.995 we would expect is somewhere
at the middle . Now if I go along the row
00:19:16.610 --> 00:19:37.299
, very slowly we can see that the value corresponding
to this is 2.5 . As I am dragging the my pen
00:19:37.299 --> 00:19:49.220
slowly you can see that the values are coming
out to be 0.99492 and 0.99506 .
00:19:49.220 --> 00:20:05.740
So, if we go above
we can find the values are here along this
00:20:05.740 --> 00:20:20.080
column are 0.7 and 0.8 . So, we can say say
2.58 is the value such that a standard normal
00:20:20.080 --> 00:20:30.119
distribution will have a value less than 2.58
has the chance 0.995, what has the probability
00:20:30.119 --> 00:20:40.210
0.995 or it is 99.5 percent chance that the
standard normal value will obtain a will have
00:20:40.210 --> 00:20:53.179
a value between below or a standard normal
variate will have a value below 2.58. Therefore,
00:20:53.179 --> 00:20:59.169
we need to remember at least these 2 values
.
00:20:59.169 --> 00:21:30.119
So, you remember this table, for standard
normal , I write it both sided
00:21:30.119 --> 00:21:46.080
for one percent the value is 2.58 as we have
just observed. Above 2.58 the chance is only
00:21:46.080 --> 00:21:59.159
0.5 percent. And below minus 2.58 the chance
is only 0.5 percent so, together we have one
00:21:59.159 --> 00:22:16.320
percent.
Or the 99 percent confidence interval is minus
00:22:16.320 --> 00:22:33.289
2.58 to 2.58 . Similarly, when we are looking
at say 5 percent here then this value we have
00:22:33.289 --> 00:22:58.389
just seen is 1.96 or 95 percent confidence
interval is minus 1.96 to 1.96 .
00:22:58.389 --> 00:23:07.089
So, these 2 values you need to remember. Because
typically these are the values that we use
00:23:07.089 --> 00:23:13.039
. Of course, there are many other values as
you have seen in the table it is full with
00:23:13.039 --> 00:23:20.279
values. So, for normal for different level
of confidence we can actually obtain the value
00:23:20.279 --> 00:23:31.600
from the table, but for this class we restrict
ourselves altitude of 2 cases ; both sided
00:23:31.600 --> 00:23:48.710
interval and 95 percent confidence and 99
percent confidence . How to use this table
00:23:48.710 --> 00:23:57.700
?
So, consider this following problem .
00:23:57.700 --> 00:24:36.340
Suppose a coin is tossed 4,000 times and number
of heads obtained is 2,400 . Question is,
00:24:36.340 --> 00:24:55.200
can we consider
the coin to be unbiased .
00:24:55.200 --> 00:25:10.159
And second question , what will be the interval
00:25:10.159 --> 00:25:45.700
between which the number of heads should lie
so that we can consider the coin to be unbiased
00:25:45.700 --> 00:26:08.279
with 95 percent confidence .
So, let me explain the problem again .
00:26:08.279 --> 00:26:25.389
So, there is a coin we tossed it 4,000 times
. If it is unbiased ,
00:26:25.389 --> 00:26:44.910
expected number of heads is equal to 2,000
and expected number of tail is also 2,000
00:26:44.910 --> 00:26:54.169
. But it does not actually mean that in the
result we will get exactly 2,000 heads and
00:26:54.169 --> 00:26:59.759
exactly 2,000 tells, that will not happen
in generally.
00:26:59.759 --> 00:27:10.570
So, the what you want to estimate it? That
what is the the value of p or the probability
00:27:10.570 --> 00:27:20.620
of success or the probability of getting a
head in a toss given this result ? Okay so,
00:27:20.620 --> 00:27:41.440
we solve it in the following way . Since , the
number of trials is high that is 4,000 is
00:27:41.440 --> 00:28:06.879
a large number , we can consider
normal approximation
00:28:06.879 --> 00:28:31.889
to the proportion of heads obtained .
Let p be the random variable denoting the
00:28:31.889 --> 00:28:45.909
proportion of heads . Therefore, expected
value of p is equal to half , and variance
00:28:45.909 --> 00:29:05.739
of p as we know is equal to pq by N is equal
to half into half divided by 4,000 .
00:29:05.739 --> 00:29:32.820
Therefore , p minus half upon root over half
into half upon 4,000 may be considered as
00:29:32.820 --> 00:29:49.600
normal 0 1 .
Since , we obtained 2,400 number of heads
00:29:49.600 --> 00:30:08.940
therefore, obtained proportion is equal to
2,400 divided by 4,000 is equal to 6 upon
00:30:08.940 --> 00:30:28.029
10 is equal to 3 upon 5 . Therefore, 3 upon
5 minus half divided by root over half into
00:30:28.029 --> 00:30:38.919
half into 4,000 can be assumed to be normal
0 1 .
00:30:38.919 --> 00:30:55.879
Now this denominator is
half into root over 1 upon 4,000 is equal
00:30:55.879 --> 00:31:23.450
to half into 1 by 20 into 1 upon root 10 is
equal to 1 upon 40 into 1 upon root 10 .
00:31:23.450 --> 00:31:53.139
Therefore, the obtained value of the normal
0 1 variable is 3 by 5 minus half upon 1 by
00:31:53.139 --> 00:32:12.539
40 into 1 by root 10 ; is equal to 6 minus
5 upon 10 divided by 1 upon 40 into 1 upon
00:32:12.539 --> 00:32:27.649
root 10 ; is equal to 1 upon 10 into 40 into
root 10 , which is equal to 4 into root 10
00:32:27.649 --> 00:32:49.080
and which is much greater than 3 .
Typically, almost 100 percent observation
00:32:49.080 --> 00:33:06.779
of normal 0 1 variable lies in the interval
minus 3 to plus 3 .
00:33:06.779 --> 00:33:28.049
Since , the obtained value
is much greater than 3 we can say
00:33:28.049 --> 00:33:44.960
the coin is not unbiased . If the coin were
unbiased, then it is the value obtained should
00:33:44.960 --> 00:33:54.159
lie in the interval minus 3 to plus 3 .
The second part of the question says that
00:33:54.159 --> 00:34:11.040
what should be the number of heads ? Part
b is , what should be the number of heads
00:34:11.040 --> 00:34:36.460
to consider the coin to be unbiased with 95
percent confidence .
00:34:36.460 --> 00:34:59.549
We know , number of heads is follow binomial
in p . So, here N is equal to 4,000 , p is
00:34:59.549 --> 00:35:11.170
equal to half .
Therefore expected number of heads is equal
00:35:11.170 --> 00:35:28.980
to N by 2 is equal to 2,000 , and variance
is equal to npq is equal to 4 thousand into
00:35:28.980 --> 00:35:46.540
half into half is equal to 1,000 .
Therefore, N , minus 2,000 divided by root
00:35:46.540 --> 00:36:01.000
over 1,000 has to be less than equal to 1.96
, this we have already obtained as we are
00:36:01.000 --> 00:36:24.049
looking for 95 percent confidence . Or in
other words, if N is the number of heads then
00:36:24.049 --> 00:36:46.059
modulus of N minus 2,000 divided by 10 root
10 has to be less than equal to 1.96 .
00:36:46.059 --> 00:37:04.940
Or modulus of N minus 2,000 is less than equal
to 1.96 into 10 root 10 .
00:37:04.940 --> 00:37:22.730
Or 2,000 minus 1.96 into 10 root 10 has to
be less than equal to N, less than equal to
00:37:22.730 --> 00:37:48.150
2,000 plus 1.96 into 10 root 10 . Now 1.96
into 10 root 10 is approximately 62 .
00:37:48.150 --> 00:38:10.530
Therefore, the 95 percent confidence interval
for number of heads
00:38:10.530 --> 00:38:34.380
to consider that coin to be unbiased is 2,000
minus 62 to 2,000 plus 62 is equal to 1938
00:38:34.380 --> 00:38:43.660
comma 2,062 .
Or in
00:38:43.660 --> 00:38:53.980
other words , if we toss the coin 4,000 times
and the number of heads is obtained in this
00:38:53.980 --> 00:39:05.690
range between these 2 values, then we are
95 percent confident that this coin is unbiased
00:39:05.690 --> 00:39:07.020
.
Let us look at another problem .
00:39:07.020 --> 00:39:35.750
Consider normal mu sigma square and we want
to find
00:39:35.750 --> 00:40:00.780
confidence interval
for sigma square . Let us consider the case
00:40:00.780 --> 00:40:21.640
that mu is known .
In that case , sigma x i minus mu whole square
00:40:21.640 --> 00:40:32.700
i is equal to 1 to N divided by sigma square.
We know that , it is a chi square with n degrees
00:40:32.700 --> 00:40:47.050
of freedom where x 1 x 2 x n is the sample
obtained .
00:40:47.050 --> 00:41:00.400
In general, , the chi square distribution
will have a shape like this .
00:41:00.400 --> 00:41:23.669
So, it is different from
normal 0 1 as chi square is in 0 to infinity
00:41:23.669 --> 00:41:32.660
. It is a positive random variable therefore;
it ranges over 0 to infinity. And secondly,
00:41:32.660 --> 00:41:54.651
it is not symmetric around mean . Also the
shape of the chi square changes with that
00:41:54.651 --> 00:42:05.119
degrees of freedom as degrees of freedom increases
our chi-square may take a shape like this.
00:42:05.119 --> 00:42:23.039
Therefore
for different degrees of freedom ,
00:42:23.039 --> 00:42:49.290
we need to obtain the cut off values for getting
the desired .
00:42:49.290 --> 00:43:05.200
Interval therefore, we will be looking at
a value here such that this probability is
00:43:05.200 --> 00:43:15.859
alpha by 2 will be of looking at a value here,
such that this probability is also alpha by
00:43:15.859 --> 00:43:33.059
2. So, that this is the 100 minus alpha percent
confidence interval
00:43:33.059 --> 00:43:44.040
for chi square . So, let me show the chi square
table .
00:43:44.040 --> 00:43:55.380
So, if you look at you can see that it is
a chi square distribution table . The values
00:43:55.380 --> 00:44:04.849
are tabulated for different for different
degrees of freedom and , the values are given
00:44:04.849 --> 00:44:14.740
for this point such that the shaded area is
equal to alpha.
00:44:14.740 --> 00:44:21.589
That means, the probability the chi square
distribution is greater than this value , that
00:44:21.589 --> 00:44:32.100
probability is alpha. And therefore, when
we are looking at a 95 percent confidence
00:44:32.100 --> 00:44:41.500
interval, we look at this value such that
the above this the probability is only 2.5
00:44:41.500 --> 00:44:52.680
percent. Also we look at a value here such
that, above that the probability is 97.5 percent
00:44:52.680 --> 00:45:07.460
or in decimal form , this value will give
corresponding to chi square with 0.975 .
00:45:07.460 --> 00:45:27.829
0.975 and for this, we are looking at the
value for chi square with 0.025 .
00:45:27.829 --> 00:45:34.020
How do we get the values ?
If we look at that table, we will find that
00:45:34.020 --> 00:45:43.579
the values are given for different degrees
of freedom; 1, 2, 3, 4 and here you can see
00:45:43.579 --> 00:45:59.369
the values tabulated are for 0.975 , and if
I go further to the right , we can see the
00:45:59.369 --> 00:46:10.609
values are tabulated for point 0 to 5.
In fact, it is given for 0.005, 0.010, like
00:46:10.609 --> 00:46:19.980
that for a fixed set of alpha the values have
been computed . In particular, let us look
00:46:19.980 --> 00:46:26.740
at chi square with 100 degrees of freedom
.
00:46:26.740 --> 00:46:41.020
Look at the value it is 100 degrees of freedom
, and corresponding to 0.975, this value is
00:46:41.020 --> 00:46:52.800
74.22 ; that means, a chi square 100 degrees
of freedom will take a value greater than
00:46:52.800 --> 00:47:08.590
equal to 74.222 with probability 9.975 or
97.5 cases out of 100 .
00:47:08.590 --> 00:47:28.380
And if we go further , we can see that for
0 to 5 the values are 129.561 , okay . So,
00:47:28.380 --> 00:47:44.460
let us consider one problem .
00:47:44.460 --> 00:48:15.589
Suppose x 1, x 2 x 100 is a sample from normal
1 comma sigma square ; such that sigma x i
00:48:15.589 --> 00:48:41.039
minus 1 whole square is equal to 200 . Find
95 percent confidence interval
00:48:41.039 --> 00:49:00.680
for sigma square .
Since x 1, x 2, x 100 are from normal with
00:49:00.680 --> 00:49:11.290
1 comma sigma square ; therefore, sigma x
i minus 1 whole square upon sigma square is
00:49:11.290 --> 00:49:37.130
chi square with 100 degrees of freedom . We
have just seen the 95 percent confidence interval
00:49:37.130 --> 00:50:05.760
is 74.22 to 129.56 . Therefore, probability
74.22 less than equal to sigma x i minus 1
00:50:05.760 --> 00:50:17.059
whole square upon sigma square , less than
equal to 129.56 is equal to 95 percent .
00:50:17.059 --> 00:50:32.730
Or probability 1 upon 129.56 less than equal
to sigma square upon sigma x i minus 1 whole
00:50:32.730 --> 00:50:43.660
square less than equal to 1 upon 74.22 is
equal to 95 percent, or probability sigma
00:50:43.660 --> 00:50:53.839
x i minus 1 whole square, upon 129.56 less
that equal to sigma square, less than equal
00:50:53.839 --> 00:51:04.380
to sigma x i minus 1 whole square upon 74.22
is equal to 95 percent.
00:51:04.380 --> 00:51:17.660
Since this is given to be 200 or probability
200 upon 129.56 less than equal to sigma square
00:51:17.660 --> 00:51:33.710
less than equal to 200 upon 74.22 is equal
to 95 percent . Or probability 1.54 less than
00:51:33.710 --> 00:51:49.589
equal to sigma square less than equal to 2.69
is equal to 95 percent . Therefore, 95 percent
00:51:49.589 --> 00:52:06.779
confidence interval for
sigma square is equal to 1.54 comma 2.69 . So,
00:52:06.779 --> 00:52:12.530
the problems that we solved , they are all
both sided we have taken .
00:52:12.530 --> 00:52:30.230
If we want to take one sided interval , say
consider normal 0 1 and we want one sided
00:52:30.230 --> 00:52:46.690
interval ;
that is , if z follows normal 0 1 , we are
00:52:46.690 --> 00:53:02.490
looking at z alpha such that probability z
less than equal to z alpha is equal to say
00:53:02.490 --> 00:53:10.250
95 percent .
Then it is one sided interval ; or in other
00:53:10.250 --> 00:53:19.500
words if it is a normal curve , we are looking
at z alpha such that this area is 95 percent
00:53:19.500 --> 00:53:27.589
. That is, this area is 5 percent, these are
called one sided intervals . And for normal
00:53:27.589 --> 00:53:46.740
0 1 at 95 percent, these value is 1.64 .
If we are looking at 99 percent confidence
00:53:46.740 --> 00:54:05.680
interval
and both sided therefore we are looking at
00:54:05.680 --> 00:54:18.710
this is only half percent . And this is 0.5
percent; that means, only one percent are
00:54:18.710 --> 00:54:27.130
beyond this range ; therefore, this is 99
percent then for standard normal these value
00:54:27.130 --> 00:54:38.880
is 2.58 . You should remember these values
for solving problems with respect to interval
00:54:38.880 --> 00:54:53.019
estimation . Okay students , with that I conclude
my lectures on theory of estimation . We have
00:54:53.019 --> 00:54:59.279
learnt the properties of estimators.
We have learnt how to estimate the value of
00:54:59.279 --> 00:55:06.839
a parameter from a sample, when you are looking
at point estimation. Also today, we have seen
00:55:06.839 --> 00:55:17.280
how to look at an interval estimation for
a parameter given a sample x 1 x 2 x n . In
00:55:17.280 --> 00:55:29.440
the next class , I shall start the last chapter
of statistical inference namely , testing
00:55:29.440 --> 00:55:41.539
of hypothesis .
Thank you .