1
00:00:20,029 --> 00:00:27,039
So, we will continue estimation last class
what we have seen that we have seen the confidence
2
00:00:27,039 --> 00:00:28,399
interval for…
3
00:00:28,399 --> 00:00:39,100
mu 1 minus mu 2, where mu 1 is the mean population
1 mu 2 is a mean population 2. And we have
4
00:00:39,100 --> 00:00:51,379
used this formula that x 1 bar minus x 2 bar
minus z alpha by 2, then our square root of
5
00:00:51,379 --> 00:01:00,100
sigma 1 square by n 1 plus sigma 2 square
by into less then equal to mu 1 minus mu 2
6
00:01:00,100 --> 00:01:09,140
less than equal to x 1 bar minus x 2 bar plus
z alpha by 2 square root of sigma 1 square
7
00:01:09,140 --> 00:01:20,060
by n 1 plus sigma 2 square by n 2, this is
a case and we have seen one example also.
8
00:01:20,060 --> 00:01:27,070
What I have that method a and method b that
these are two type to teaching methods and
9
00:01:27,070 --> 00:01:34,520
this is a formulation only thing that z 0.025
you write, it this one this is not 0.05 this
10
00:01:34,520 --> 00:01:39,170
will be written as 0.025.
11
00:01:39,170 --> 00:01:47,789
So, using this formula like this one, what
are the prerequisite. So, the conditions,
12
00:01:47,789 --> 00:01:57,619
conditions are many first a fall sigma 1 and
sigma 2 must be known and will be collecting
13
00:01:57,619 --> 00:02:12,480
in n 1 and n 2 in to this step sample size,
and it is for normal populations and but there
14
00:02:12,480 --> 00:02:24,540
is difficulty arises here knowing that sigma
1, sigma 2. Another thing it may also happen
15
00:02:24,540 --> 00:02:31,599
that that sigma 1 equal to sigma 2 maybe that
maybe the case sigma 1 equal to sigma 2 equal
16
00:02:31,599 --> 00:02:42,050
to sigma, what does it mean? The population
rinses are same equal if population variance
17
00:02:42,050 --> 00:02:54,090
are same, then we will not use this equation
we will go for because of this special condition,
18
00:02:54,090 --> 00:03:05,489
we will go for another type of estimation,
why? We will use first we find out that what
19
00:03:05,489 --> 00:03:12,629
is the estimate of sigma.
The reason is as sigma 1 equal to sigma 2
20
00:03:12,629 --> 00:03:19,000
equal to sigma we have n 1 n 2, two set sample
are available from the different population.
21
00:03:19,000 --> 00:03:26,390
So, we will instead of using one that means
I you can go for a sample 1 and calculate
22
00:03:26,390 --> 00:03:32,620
sigma estimate also you can that is basically
sigma 1 estimate equal to sigma estimate because
23
00:03:32,620 --> 00:03:38,470
they are same, sigma 2 estimate equal to sigma
estimate, but in that case what will happen
24
00:03:38,470 --> 00:03:42,629
that find out the sigma 1, sigma 2 will be
differentially different because they are
25
00:03:42,629 --> 00:03:48,689
coming from a two population. So, n sample
only that’s why what happen you say that
26
00:03:48,689 --> 00:03:59,709
we will calculate a spooled is that basically
that two sample will be spooled gather held
27
00:03:59,709 --> 00:04:08,230
then you will be calculating the estimate
of that estimate of sigma cap, which is basically
28
00:04:08,230 --> 00:04:19,130
spooled. Now, what is this s spooled, s spooled
your s spooled will be like this…
29
00:04:19,130 --> 00:04:34,099
s square spooled you write this will be suppose
the sample 1 of the size 1 minus 1 is 1 square
30
00:04:34,099 --> 00:04:54,370
plus population 2 k is n 2 minus 1 s 2 square
divided by n 1 plus n 2 minus 2. Getting me?
31
00:04:54,370 --> 00:05:06,740
What you are doing here, what is this? n1
minus 1 s 1 square that is the total variability
32
00:05:06,740 --> 00:05:13,840
in the sample from population one. Second
one is total variable in the sample from the
33
00:05:13,840 --> 00:05:21,030
population 2 point to mu and this n 1 minus
n 2 minus 2, this is a degree of available.
34
00:05:21,030 --> 00:05:27,690
The reason is the calculated x 1 bar 1 degree
is the last x 2 bar is calculated under degree
35
00:05:27,690 --> 00:05:33,460
is lost, this is a resultant this is what
we are saying that s spooled are variance
36
00:05:33,460 --> 00:05:43,639
combined variance. Now, this one we are saying
that this is a estimate of sigma square, if
37
00:05:43,639 --> 00:05:53,360
this is a case and if we consider that things
are coming from normal population and sample
38
00:05:53,360 --> 00:06:05,580
size is large in that case, what will be your
interval x 1 bar minus x 2 bar minus z alpha
39
00:06:05,580 --> 00:06:11,690
by 2. Then the variance what you have written
in the earlier case what we have written,
40
00:06:11,690 --> 00:06:21,819
we have written like this that sigma 1 square
by n 1 plus sigma 2 square by n 2 then square
41
00:06:21,819 --> 00:06:27,380
root.
Now, what is happening here your sigma 1 equal
42
00:06:27,380 --> 00:06:36,990
to sigma 2 so I can write this one like this
sigma 1 sigma square by n 1 plus sigma square
43
00:06:36,990 --> 00:06:46,610
by n 2 this is nothing but sigma will come
out and 1 by n 1 plus 1 by n 2. Now, you see
44
00:06:46,610 --> 00:06:55,810
that s spooled is the estimate of the sigma.
So, that mean I can write s p if I write s
45
00:06:55,810 --> 00:07:04,910
p equal to s spooled then s p square root
of 1 by n 1 plus 1 by n 2.
46
00:07:04,910 --> 00:07:18,520
So, you will be writing s p that this portion
is now, I am writing again sigma square root
47
00:07:18,520 --> 00:07:33,860
of 1 by n 1 plus 1 by n 2 equal to s p 1 by
n 1 plus 1 by n 2, why s p? S p is a estimate
48
00:07:33,860 --> 00:07:46,599
of sigma, then you result now your earlier
that derivation like this x 1 bar minus x
49
00:07:46,599 --> 00:07:53,509
2 bar minus z alpha by 2 sigma n square by
n 1 plus less than equal to this less than
50
00:07:53,509 --> 00:08:03,069
this. Now, this quantity will be replaced
now, it will be replaced by s p into this.
51
00:08:03,069 --> 00:08:13,039
So, I will write it here now s p 1 by n 1
plus n by n 2 square root this less than equal
52
00:08:13,039 --> 00:08:30,590
to mu 1 minus mu 2 less than equal to x 1 bar minus x 2 bar, correct? Then minus you
53
00:08:30,590 --> 00:08:40,229
write not minus plus I think you can write
the place of the portion plus this one, this
54
00:08:40,229 --> 00:08:54,360
will come here plus this. And most of the
cases will be sigma 1 knowing sigma 1 sigma
55
00:08:54,360 --> 00:09:04,309
2 is not possible population variance. And
then what will happen you will be using this
56
00:09:04,309 --> 00:09:12,059
s 1 and s 2, but if s 1 and s 2 and the formulation
like this s 1 square by n 1 plus s 1 square
57
00:09:12,059 --> 00:09:20,089
by n 2 that variable variability formula if
you use, then the sample size is larger that
58
00:09:20,089 --> 00:09:26,400
is better. But if the sample size is small there is doubt
59
00:09:26,400 --> 00:09:32,720
in this formulation. So, our one of the assumption
will be that sigma 1 square equal to sigma
60
00:09:32,720 --> 00:09:39,630
2 square sigma square. That means population
variance standard deviations are equal that
61
00:09:39,630 --> 00:09:49,089
assumptions we will do, and because of these
assumption that will be using a spool. So,
62
00:09:49,089 --> 00:09:56,140
this is a equation you find out the most of
the time this equation is used z alpha by
63
00:09:56,140 --> 00:10:05,220
2 s p square root of n 1 plus 1 by n 2. And
it is also cost to many that will go for the
64
00:10:05,220 --> 00:10:16,670
equal sample size so n 1 plus equal to n 1 equal to n 2 equal to n also, if you use n
65
00:10:16,670 --> 00:10:24,569
1 equal n 2 then what will happen ultimately,
this is n minus 1, this 1 is also n minus
66
00:10:24,569 --> 00:10:34,900
1 then it is 2 n minus 2 minus n minus, what
will happen if n 1 equal n 1 equal to n 2
67
00:10:34,900 --> 00:10:36,329
equal n.
68
00:10:36,329 --> 00:10:49,449
What will happen then you s p square equal
to or s p spooled equal to n minus 1, s 1
69
00:10:49,449 --> 00:11:01,709
square plus n minus 1 is 2 square by n minus
n plus 2. So, this will become n minus 1 s
70
00:11:01,709 --> 00:11:11,709
1 square plus s 2 square by 2 into n minus
1. So, resultant will become s 1 square plus
71
00:11:11,709 --> 00:11:28,360
s 2 square by 2, and in the same equation
you use. And when another condition is suppose
72
00:11:28,360 --> 00:11:38,100
if I go to the I think next one is not there
what will happen if n 1 less than an n into
73
00:11:38,100 --> 00:11:52,579
and n 1 into are both small, we will using
t distribution, what will be the degree of
74
00:11:52,579 --> 00:12:04,689
the degree distribution when you use s spool
minus 2 getting that is the that is a.
75
00:12:04,689 --> 00:12:09,610
Ultimately you are calculating this an you
have computed you have this much degree of
76
00:12:09,610 --> 00:12:37,040
freedom available. So, you have to go for
n 1 plus n 2 minus 2 degree of freedom, then
77
00:12:37,040 --> 00:12:45,819
I ask when you use this formulation are this
other formulation first one that is this formulation
78
00:12:45,819 --> 00:12:56,140
z, this formulation form the normal population
it is coming sigma 1 sigma 2 is known, irrespective
79
00:12:56,140 --> 00:13:02,419
of this sample size sigma 1 sigma 2 is not
known sample size is large, you have assumed
80
00:13:02,419 --> 00:13:11,040
this distribution. If sample size is small
sigma 1 sigma 2 is not known.
81
00:13:11,040 --> 00:13:17,530
Then you have to first find out that the assumption
is sigma 1 equal to sigma 2 that answer is
82
00:13:17,530 --> 00:13:26,319
found out, that assumption to be checked if
that assumption is true, then you have to
83
00:13:26,319 --> 00:13:38,949
calculate s p square. And then you formulate
like this provocation and go for t distribution
84
00:13:38,949 --> 00:13:47,100
getting me. If how do you know that sigma
1 square sigma 1 equal to sigma 2 or sigma
85
00:13:47,100 --> 00:13:56,709
2 square 1 square equal to sigma 2 square,
they are equality of variance.
86
00:13:56,709 --> 00:14:00,900
How do you know that is the equal that is
to be tested because if you want to use this
87
00:14:00,900 --> 00:14:08,370
formulation, we require to test also that
the two population variances are equal. If
88
00:14:08,370 --> 00:14:16,949
two population are variances are not equal
then you cannot go for this. Suppose population
89
00:14:16,949 --> 00:14:29,600
variances are not equal sample size are different
you cannot and also population is not equal
90
00:14:29,600 --> 00:14:41,490
sample size are small also, and then you cannot
use z distribution. And another issue will
91
00:14:41,490 --> 00:14:47,250
taken place that because of this differentiation,
so ultimately there are some other type of
92
00:14:47,250 --> 00:14:50,980
derivation is that we are not considering
here.
93
00:14:50,980 --> 00:14:58,059
So, essentially we are considering things
sampling from the normal population either
94
00:14:58,059 --> 00:15:05,780
the variance component are known. And either
sample in variances are not known that is
95
00:15:05,780 --> 00:15:11,860
a large sample and when variances are not
known also small sample, we are assuming that
96
00:15:11,860 --> 00:15:18,040
the variances are equal and then we are using
a spool then we are going for a t distribution
97
00:15:18,040 --> 00:15:25,510
under this case. If they are equal and some
sample size is large then you can go for z
98
00:15:25,510 --> 00:15:34,579
distribution no problem. So, j and t distribution
for the equality of means of the two population
99
00:15:34,579 --> 00:15:42,160
all these things mostly we are used, but where
are we ended now we ended that if we want
100
00:15:42,160 --> 00:15:47,199
to use this, we require to know that the population
variance are equivalent.
101
00:15:47,199 --> 00:15:58,720
So, can we not find out the taste by which
we can do that this is a this case.
102
00:15:58,720 --> 00:16:06,600
Can we not find out the other one before going
to the other one, I think you can simply find
103
00:16:06,600 --> 00:16:08,179
out that how we can calculate…
104
00:16:08,179 --> 00:16:11,870
The construction interval always the we have
discussed now confidence interval all the
105
00:16:11,870 --> 00:16:27,699
things we discussed, now confidence interval
for the ratio of two population variances.
106
00:16:27,699 --> 00:16:36,500
How do you test the two population variances
are equal in any static any test. So, eleven
107
00:16:36,500 --> 00:16:50,130
test is there in annova, annova analysis of anova that is this what is annova in annova
108
00:16:50,130 --> 00:16:56,120
what will happen will find out that one of
the important test is there eleven test, we
109
00:16:56,120 --> 00:17:04,900
want to do that the population. The level
variances is equal generally go for mean test,
110
00:17:04,900 --> 00:17:10,150
eleven test is one that we will see later
on if time permits.
111
00:17:10,150 --> 00:17:20,449
Now, confidence interval for the ratio of
two population variances, how do you go around
112
00:17:20,449 --> 00:17:30,009
it, what is that say usual of that our, our
total when your structure what is usual states
113
00:17:30,160 --> 00:17:38,900
we follow, we will follow the steps like this,
I have two population both population either
114
00:17:38,910 --> 00:17:44,720
here you are interested to test the ratio
the sigma 1 square by…
115
00:17:44,720 --> 00:17:55,360
That you want to get something like this sigma
2 square you want to get something like this.
116
00:17:55,360 --> 00:18:11,860
Something like this what will be the l and
u and you are interested to know this is the
117
00:18:11,860 --> 00:18:27,720
case. Now, if you want to get this interval
you must know that, what is statistic you
118
00:18:27,720 --> 00:18:34,720
will generate, and that based on that statistic
you will go for sampling distribution. So,
119
00:18:34,720 --> 00:18:41,310
the few things are important you must know
the appropriate statistic, then distribution
120
00:18:41,310 --> 00:18:54,250
of sampling distribution of statistic.
And then you basically choose the alpha value
121
00:18:54,250 --> 00:19:05,440
appropriate alpha that what do you want then
you construct the interval? These are these
122
00:19:05,440 --> 00:19:12,980
are the steps and every where you are doing
like this only, if it is mean the statistic
123
00:19:12,980 --> 00:19:21,750
is mean minus expected value or mean by variance
it will be z or t then if it is variance you
124
00:19:21,750 --> 00:19:27,500
are using n minus 1, s square by sigma square
that is statistic square distribution.
125
00:19:27,500 --> 00:19:33,950
Now, if it is mean difference again you are
finding out the x 1 minus x 2 bar expected
126
00:19:33,950 --> 00:19:39,810
value by its standard deviation, that also
follow z distribution depending on the conditions.
127
00:19:39,810 --> 00:19:48,750
Now, we are talking about ratio of two and
I think you all know because we have discussed
128
00:19:48,750 --> 00:19:59,450
this that n minus s square by sigma square
follows distribution with n minus 1 degree
129
00:19:59,450 --> 00:20:09,860
of freedom yes or no?
Now, let us see that it is our population
130
00:20:09,860 --> 00:20:24,010
one that sigma 1 is the variance, sigma 1
square is the variance you collected simple
131
00:20:24,010 --> 00:20:38,450
n 1 and you have computed also s 1 square,
then your statics here is n 1 minus 1, s 1
132
00:20:38,450 --> 00:20:49,230
square by sigma 1 square will follow chi square
distribution with n 1 minus 1 degree of freedom,
133
00:20:49,230 --> 00:21:04,430
yes or no? You go for population 2. That the
variation is sigma 2 square you collected
134
00:21:04,430 --> 00:21:14,250
n 2 simple, size is basically n 2 then you
have calculated variance simple variance.
135
00:21:14,250 --> 00:21:25,980
Can I not say that n 2 minus 1 is 2 square
by sigma 2 square, this is chi square distributed
136
00:21:25,980 --> 00:21:38,250
with n 2 minus 1 degrees of freedom, we can
say, if we can say this we also can say that…
137
00:21:38,250 --> 00:21:53,160
F equal to n 1 minus 1 s 1 square by sigma
1 square divided by n 2 minus 1 s 2 square
138
00:21:53,160 --> 00:22:07,800
by sigma 2 square. We can clear like this,
but what we say the earlier that that if these
139
00:22:07,800 --> 00:22:15,650
quantity is divided, what do you say the earlier
for a F distribution, F distribution we say
140
00:22:15,650 --> 00:22:25,660
F is the ratio of 2 chi square variable, chi
square. Suppose nu 1 degrees of freedom divided
141
00:22:25,660 --> 00:22:36,180
by nu 1 followed by the denominator case chi
square in nu two degrees of freedom divided
142
00:22:36,180 --> 00:22:46,700
by nu two.
If this is the case then we will cleared like
143
00:22:46,700 --> 00:22:51,340
this what is your chi square variable here
chi square instead of write these I am writing
144
00:22:51,340 --> 00:23:00,790
now chi square, you please write see that
new 1 is n 1 minus 1 divided by n 1 minus
145
00:23:00,790 --> 00:23:10,770
1 then chi square n 2 minus 1 divided by n
2 minus 1. Can you find out what will be these
146
00:23:10,770 --> 00:23:26,290
from these? We say this one is chi square
n minus 1. So, these divided by n 1 minus
147
00:23:26,290 --> 00:23:37,060
1 means, this is s 1 square by sigma 1 square.
Now, but denominator will be s 2 square by
148
00:23:37,060 --> 00:23:44,270
sigma 2 square, so this will this definitely
follow chi square F distribution. That is
149
00:23:44,270 --> 00:23:52,460
the F what to will be the numerated degrees
of freedom follow from here what you have
150
00:23:52,460 --> 00:24:00,330
divided here n 1 minus 1.
So, n 1 minus 1 now what is the degree of
151
00:24:00,330 --> 00:24:13,610
n that this one n 2 minus 1, so you have tell
like the n 2 minus 1 so this is the distribution,
152
00:24:13,610 --> 00:24:25,420
getting me? So you know the distribution,
you very well know s 1 square by sigma 1 square
153
00:24:25,420 --> 00:24:34,090
divided s 2 square by sigma 2 square follow
F distribution with n 1 minus 1 and n 2 minus
154
00:24:34,090 --> 00:24:45,980
1, that numerator and denominator degrees
of freedom. Can you now develop? You cannot
155
00:24:45,980 --> 00:24:53,809
develop you know that so what is our what
is our statistic your statistics is s 1 square
156
00:24:53,809 --> 00:25:00,050
by sigma 1 square s 2 square by sigma 2 square
that is a divided by this, this follows F
157
00:25:00,050 --> 00:25:00,460
distribution.
158
00:25:00,460 --> 00:25:13,930
So, here F distribution is like this F you
are saying that this one follows F distribution.
159
00:25:13,930 --> 00:25:19,170
What do we want to create? We want to create
the interval you create like this earlier
160
00:25:19,170 --> 00:25:31,570
also you have seen this, this is my u and
let this is my l. So, this one is if I consider
161
00:25:31,570 --> 00:25:39,309
alpha then this side only this will be alpha
by 2 this side wall it will be alpha by 2,
162
00:25:39,309 --> 00:25:47,990
but the right hand side probability if I consider
the integrity here, this total probability
163
00:25:47,990 --> 00:25:58,430
here this is 1 minus alpha by 2. So, 1 minus
alpha by 2 will be this probability then corresponding
164
00:25:58,430 --> 00:26:04,850
f values you will be getting, what will be
the F value here?
165
00:26:04,850 --> 00:26:14,970
F is our n 1 minus 1 n 2 minus 1 that is the
degrees of freedom numerator and denominator
166
00:26:14,970 --> 00:26:22,820
degrees of freedom, but this corresponding
1 minus alpha that probability value that
167
00:26:22,820 --> 00:26:28,320
probability value for that probability value,
what is the value of this. Then that means
168
00:26:28,320 --> 00:26:47,980
l equal to this what is your u f n 1 minus
1 n 2 minus 1, this is alpha by 2. Now, again
169
00:26:47,980 --> 00:27:00,290
you know that F equal to s 1 square by sigma
1 square s 2 square by sigma 2 square.
170
00:27:00,290 --> 00:27:11,650
Also we say that probability that this quantity
less than equal to F less than equal to this,
171
00:27:11,650 --> 00:27:24,550
this equal to 1 minus alpha, it is now instead
of s what we required to prove it you write
172
00:27:24,550 --> 00:27:38,320
down this one s 1 square by sigma 1 square
divided by whole divided by s 2 square by
173
00:27:38,320 --> 00:27:50,890
sigma 2 square. So, I want what you want ultimately
from here, you want a expression where in
174
00:27:50,890 --> 00:27:55,910
the middle portion there will be sigma 1 square
by sigma 1 square by sigma 2 square, then
175
00:27:55,910 --> 00:28:07,470
shake the left and right hand side. Can you
recap now, you take 2 minutes take 2 minutes
176
00:28:07,470 --> 00:28:13,820
and do it.
Essentially why I am putting so much of effort
177
00:28:13,820 --> 00:28:21,400
here because you see the all the cases, procedure
remains same you have to know the statistics
178
00:28:21,400 --> 00:28:28,980
after the distribution, lower bound and upper
bound that based on alpha value and probability
179
00:28:28,980 --> 00:28:33,970
of this will be this then your interval is
same. Now, not necessarily that this quantity
180
00:28:33,970 --> 00:28:38,420
suppose some other quantity you are deriving
you are able to approve, find out the proof
181
00:28:38,420 --> 00:28:45,540
in statistics as well as its distribution.
You can create the interval getting me it
182
00:28:45,540 --> 00:28:58,120
is very simple. So, what we have done then
we found out that…
183
00:28:58,120 --> 00:29:08,290
F n 1 minus 1 into minus 1, 1 minus alpha
by 2 less than equal to s 1 square by sigma
184
00:29:08,290 --> 00:29:18,400
1 square s 2 square by sigma 2 square less
than equal to F n 1 minus 1 n 2 minus 1 alpha
185
00:29:18,400 --> 00:29:27,640
by 2. You want sigma 1 by sigma 2 that whole
square. So, you have to what you are required
186
00:29:27,640 --> 00:29:38,059
to do now? You require first take the inverse
in the sense that a reciprocal, if you do
187
00:29:38,059 --> 00:29:43,910
the reciprocal part what will happen that
this will be greater than greater than.
188
00:29:43,910 --> 00:29:51,150
So, if I write like this ultimately what will
happen, the resultant part if you write like
189
00:29:51,150 --> 00:30:07,440
this can you not write like this
190
00:30:07,440 --> 00:30:14,120
then I am writing, it is going up sigma 1
square by sigma 2 square into what is after
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remaining here, that is s 1 square by s 2
square that is basically this one I want to
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give a inverse sign, so that is why so s 2
square is there it will go up s 1 square will
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come down here sigma 1 square, where in that
denominator its goes up sigma 2 square come
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down then this is 1 by f n 1 minus 1 n 2 minus 1 alpha by 2.
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Now, I am writing just in other way, so what
we are doing we are again going to the standard
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format the way we write sigma 1 square by
sigma 2 square, I am writing like this so
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what will happen this side this side portion
will come here and s 1 square if I take out
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of this to this side as well as this side,
then what will happen here s 1 square by s
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2 square divided by f n 1 minus 1 n 2 minus
1 1 minus alpha by 2.
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And this side will be s 1 square by s 2 square
by f n 1 minus 1, n 2 minus 1 alpha by 2.
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That is what is now see sigma 1 and sigma
2, which one is greater so there are again
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they are change combination changes are there
and if you go to statistical any basics statistical
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book, you may find out some other type of
derivation, presentation not derivation same
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way you derive.
You depending on that that possibility they
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are because we will not go into that there
because it is not required for all of us we
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have to proceed to this are all prerequisites
for multi variant statistical modeling. So,
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00:32:21,840 --> 00:32:26,510
we have to finish as early as possible I am
planning to finish by another lecture, the
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00:32:26,510 --> 00:32:32,500
prerequisite so that will straightly go to
the start the multi variant one that what
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00:32:32,500 --> 00:32:40,320
is the ultimate aim for, when you are looking
for if this is the case.
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So, can you not find out some suppose this
is the one of the example say two medicine
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that is treatment of asthma, there are two
different type medicine may be you are taking
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some medicine or you are using inhaler. Or
let it be that two different type of medicinal
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00:33:05,650 --> 00:33:11,929
on only you are consuming, when they have
the manufacturers, manufacturer is claiming
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00:33:11,929 --> 00:33:19,480
that may be medicine both medicine is working
as usual or not, but they are saying that
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00:33:19,480 --> 00:33:28,830
they are competitive, competitive in the sense.
Then when we talk of comparative then one
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00:33:28,830 --> 00:33:37,299
of the issue is mean value, you have to find
out the response variable we are saying that
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00:33:37,299 --> 00:33:44,820
response variable is that once I take the
medicine, how long it will take the up may
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00:33:44,820 --> 00:33:51,270
to get relief from the asthmatic problem?
That is we are saying that average is relief
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time. So, if I take medicine a and this are all
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00:33:55,450 --> 00:33:59,460
emergency medicine in the sense that when
the asthma is problem will occur that time
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you are taking for relief purpose, let it
be like this then if you take medicine a then
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00:34:04,620 --> 00:34:12,169
what we have found out that average relief
time is 2 hours. If I take medicine b average
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00:34:12,169 --> 00:34:24,210
relief time is 3 hours and we have checked
with 20 patient for medicine a and here 25
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00:34:24,210 --> 00:34:33,320
patient using medicine b, you got this average
time and also you got the sample standard
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00:34:33,320 --> 00:34:41,210
deviation both are two and two from that 20
percent from that 25 percent from medicine
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00:34:41,210 --> 00:34:47,179
a and medicine b.
Now, you may be interested to see that using
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this information you may be interested to
the confidence interval for the variability
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00:34:52,330 --> 00:35:02,220
part. It may so happen why variability is
important? It may so happen that the same
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medicine for the same medicine one patient
that let Mr. x is getting cure relief by
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1 hour, for Mr. y it may be take 4 hours
so that variability because as we have already
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00:35:20,660 --> 00:35:27,280
discussed variability is very, very important
issue. So, I we want to using this we want
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to test that what is the that interval, it
will vary. So, you can very easily calculate
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then you can may not calculate definitely
you will be able to calculate because s 1
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00:35:40,619 --> 00:35:46,480
square and s 2 square are what are those things
s 1 in this example what is our s 1 square
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00:35:46,480 --> 00:35:49,540
4 that is s 1.