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Making the t-test using tables of the critical values of t for various levels of confidence.
But first - some review
es
DDt 21
The test statistic
We have the means of two samples
1D 2D&
es is the standard error,But its computed differently from the
single sample standard error which is just
N
sse
ˆ s is the unbiased estimate of
the standard deviation
Note:
How do we compute the critical value of t - the test
statistic?
es
DDt 21
In this case 21
11
nnss pe wher
e
2
)1()1(
21
222
2112
nn
snsnsp
sp is the pooled estimate of the standard deviation derived as
follows
N
sse
ˆ
Sim
ilar fo
rm
Going through this computation for the samples of strike at locations A and B yields t ~ 5.2.
Evaluating the statistical significance of differences in strike at locations A and B using the t-test
The value of our test statistic, t = 5.2
Degrees of freedom = n1+n2-2 = 38
Closest value in the table is 40
= 0.1 %
=0.001 as a one-tailed probability
= 1 chance in 1000 that these two means are actually equal or were drawn from the same parent population.
PsiPlot returns a two-tailed probability for a t 5.2. That probability is 0.000007155 (about one chance in 140,000). Note that this is twice the value returned by Excel.
The book works through the differentiation of y = x2, so let’s try y =x4.
4)( dxxdyy
multiplying that out -- you get ... 432234 )()(4)(64 dxdxxdxxdxxxdyy
432234 )()(4)(64 dxdxxdxxdxxxdyy
Remember this idea of dy and dx is that the differential changes are infinitesimal - very small.
So if dx is 0.0001 (that’s 1x10-4) then (dx)2 = 0.00000001 (or 1x10-8) (dx)3 = 1x10-12 and (dx)4 = 1x10-16.
So even though dx is infinitesimally small, (dx)2 is orders of magnitude smaller
432234 )()(4)(64 dxdxxdxxdxxxdyy
so that we can just ignore all those terms with (dx)n where n is greater than 1.
dxxxdyy 34 4
Our equation gets simple fast
Also, since y =x4, we have dxxydyy 34
dxxdy 34
and then -
34xdx
dy
Divide both sides of this equation by dx to get
dxxdy 34
This is just another illustration of what you already know as the power rule,
1 nnaxdx
dyis
Just as a footnote, remember that the constant factors in an expression carry through the differentiation.This is obvious when we consider the derivative -
baxy 2
which - in general for
naxy
bdxxadyy 2)(
bdxxdxxadyy )2( 22
axdxbaxdyy 2)( 2 axdxydyy 2
)2( xadx
dy
Examining the effects of differential increments in y and x we get the
following
Don’t let negative exponents fool you. If n is -1, for example, we still have
1 nnaxdx
dy
2 axdx
dy
or just
)()()( xgxfxy Given the function -
what is dx
dy?
dx
dg
dx
df
dx
dy
We just differentiate f and g individually and take their sum, so that
Take the simple example )()( 42 baxcxy
- what is
dx
dy?
What are the individual derivatives of )( 2 cx )( 4 bax and ?
)( 2 cxf let
then - dx
cxd
dx
df )( 2
We just apply the power rule and obtain
xdx
df2
We know from the forgoing note that the c disappears.
We use the power rule again to evaluate the second term, letting
g = (ax4+b)34ax
dx
dg
Thus - 342 axx
dx
dy
)()( 42 baxcxdx
d
dx
dy
Differences are treated just like sums
so that
is just
342 axxdx
dy
Recall how to handle derivatives of functions like
)()()( xgxfxy
?
or
)(
)()(
xg
xfxy
fgy
Removing explicit reference to the independent variable x, we have
))(( dggdffdyy Going back to first principles, we have
Evaluating this yields dfdgfdggdffgdyy
Since df x dg is very small and since y=fg, the above becomes -
fdggdfdy
Which is a general statement of the rule used to evaluate the derivative of a product of functions
The quotient rule is just a variant of the product rule, which is used to differentiate functions like
g
fy
2gdx
dgfdxdfg
g
f
dx
d
The quotient rule states that
And in most texts the proof of this relationship is a rather tedious one.
The quotient rule is easily demonstrated however, by rewriting the quotient as a product and applying the product rule. Consider
1 fgg
fy
fhy
We could let h=g-1 and then rewrite y as
Its derivative using the product rule is just
dx
dhf
dx
dfh
dx
dy
dh = -g-2dg and substitution yields
2gdx
dgf
gdx
df
dx
dy
2gdx
dgf
gdx
df
g
g
dx
dy
Multiply the first term in the sum by g/g (i.e. 1) to get >
Which reduces to
2gdx
dgfdxdfg
dx
dy
i.e. the quotient rule
•The derivative of an exponential functions
xey
xedx
dy
Given >
In general for axey xedx
axd
dx
dy )( axae
xay If express a as en so that nxxn eey
then nxnx needx
d
dx
dy
)ln()ln( aen n Note
nxnx needx
d
dx
dy
Since nxx ea and
)ln(an
xaadx
dy . )ln(
in general
a can be thought of as a general base. It could be 10 or 2, etc.
•The derivative of logarithmic functions
Given >
)ln(xy
xdx
dy 1
We’ll talk more about these special cases after we talk about the chain
rule.
Differentiating functions of functions -
Given a function
22 )1( xy we consider
)()1( 2 xhx write 2hy compute
hhdh
d
dh
dy22
Then compute
xxdx
d
dx
dh212 an
d
take the product of the two, yielding dx
dh
dh
dy
dx
dy.
xxdx
dh
dh
dy
dx
dy2).1(2. 2
)1(4 2 xx
22 )1( xy
We can also think of the application of the chain rule especially when powers are involved as working form the outside to inside of a function
22 )1( xyWhere
xxdx
dy2.)1(2 12
Derivative of the quantity squared viewed from the outside.
Again use power rule to differentiate the inside term(s)
Using a trig function such as )2sin( axy
let axh 2
then dx
dh
dh
dy
dx
dy.
Which reduces to aaxdx
dy2).2cos( or just
)2cos(2 axadx
dy
In general if
))...))))((...(((( xqihgfy
then
dx
dq
di
dh
dh
dg
dg
df
df
dy
dx
dy........
axey ( ) axdy d axe
dx dx axae
Returning to those exponential and natural log cases - we already implemented the chain rule when differentiating
h in this case would be ax and, from the chain rule,
dx
dh
dh
dy
dx
dy. become
s dx
dh
dh
de
dx
dy h
. or
dx
dhe
dx
dy h. and finally
axaedx
dy
since
axh and
adx
dh
For functions like 2axey
we follow the same procedure.
Let 2axh and then
From the chain rule we have dx
dh
dh
dy
dx
dy.
axdx
dh2
hh eedh
d
dh
dy
22. axaxe
dx
dh
dh
dy
dx
dyhence
Thus for that porosity depth relationship we were working with
- /
0)( zez
?
)( /0
dz
ed
dz
zd z
/0 ze
For logarithmic functions like )ln( 2xy
We combine two rules, the special rule for natural logs and the chain rule.
Let 2xhdx
dh
dh
dy
dx
dy.Chain
rule
Log
rule xdx
xd 1)(ln
then
2
1
xdh
dy an
dx
dx
dh2
soxx
x
dx
xd 22)ln(2
2
For next Tuesday answer question 8.8 in Waltham (see page 148).
xexi . )( 2
)sin(.3 )( 2 yii
)tan(.xx.cos(x) )( 2 xziii 24 17)ln(.3 )( Biv
Find the derivatives of
