Transcript the test statistic? - West Virginia University

tom.h.wilson
[email protected]
Dept. Geology and Geography
West Virginia University
But first - some review
Making the t-test
using tables of the
critical values of t
for various levels
of confidence.
How do we compute the critical
value of t - the test statistic?
The test statistic
D1  D2
t
se
We have the means of two samples D1 & D2
se
is the standard error,
But its computed differently from the
single sample standard error which is just
sˆ
se 
N
Note:
ŝ is the unbiased estimate of
the standard deviation
D1  D2
t
se
In this case se  s p
1 1

n1 n2
sˆ
se 
N
where
sp is the pooled estimate of the
standard deviation derived as follows
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
s 2p  1
n1  n2  2
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 twotailed 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.
y  dy  ( x  dx ) 4
multiplying that out -- you get ...
y  dy  x  4 x dx  6 x (dx )  4 x(dx )  (dx )
4
3
2
2
3
4
y  dy  x 4  4 x3dx  6 x 2 (dx ) 2  4 x(dx )3  (dx ) 4
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
y  dy  x 4  4 x3dx  6 x 2 (dx ) 2  4 x(dx )3  (dx ) 4
so that we can just ignore all those
terms with (dx)n where n is greater
than 1.
Our equation gets simple fast
y  dy  x 4  4 x3dx
Also, since y =x4, we have
y  dy  y  4 x3dx
and then dy  4 x3dx
dy  4 x3dx
Divide both sides of this
equation by dx to get
dy
 4x3
dx
This is just another illustration of what
you already know as the power rule,
which - in general for y  ax n
is
dy
 naxn 1
dx
Just as a footnote, remember that
the constant factors in an expression
carry through the differentiation.
This is obvious when we consider the
derivative y  ax 2  b
Examining the effects of differential
increments in y and x we get the following
y  dy  a( x  dx )  b
2
y  dy  a( x 2  2 xdx  dx 2 )  b
y  dy  (ax  b)  2axdx
2
y  dy  y  2axdx
dy
 a(2 x)
dx
Don’t let negative exponents fool you.
If n is -1, for example, we still have
dy
 naxn 1
dx
or just
dy
 ax 2
dx
Given the function - y( x)  f ( x)  g ( x)
dy
what is
dx
?
We just differentiate f and g
individually and take their sum, so
that
dy df dg


dx dx dx
Take the simple example
y  ( x 2  c)  (ax 4  b)
dy
- what is
dx
?
What are the individual
derivatives of ( x 2  c) and (ax 4  b) ?
let
then -
f  ( x 2  c)
df d ( x 2  c)

dx
dx
We just apply the power rule and
obtain
df
 2x
dx
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)
dg
 4ax3
dx
Thus dy
 2 x  4ax3
dx
Differences are treated just like sums
so that

dy d

( x 2  c)  (ax4  b)
dx dx
is just
dy
 2 x  4ax3
dx

Recall how to handle derivatives of
functions like
y ( x)  f ( x) g ( x)
or
f ( x)
y ( x) 
?
g ( x)
Removing explicit reference to the
independent variable x, we have
y  fg
Going back to first principles, we have
y  dy  ( f  df )( g  dg)
Evaluating this yields
y  dy  fg  gdf  fdg  dfdg
Since df x dg is very small and since
y=fg, the above becomes -
dy  gdf  fdg
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
f
y
g
The quotient rule states that
df  f dg
g
d f
dx
dx
  
dx  g 
g2
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
f
y   fg 1
g
We could let h=g-1 and then rewrite y as
y  fh
Its derivative using the
product rule is just
dy
df
dh
h  f
dx
dx
dx
dh = -g-2dg and substitution yields
dy

dx
df
dx 
g
f dg
g2
dx
Multiply the first term in the sum by g/g (i.e. 1)
to get >
df
dg
f
dy g dx
dx


dx g g
g2
Which reduces to
df  f dg
g
dy
dx
dx

dx
g2
i.e. the quotient rule
Given >
•The derivative of an
exponential functions
In general for y  e
If y  a
then
ax
dy
 ex
dx
dy d (ax) x

e  ae ax
dx
dx
express a as
dy d nx
 e  nenx
dx dx
x
y  ex
en
 
so that y  e
n x
 enx
Note
n  ln( en )  ln( a )
dy d nx
 e  nenx
dx dx
Since a x  e nx and n  ln(a)
in general
dy
 ln( a) . a x
dx
a can be thought of as a general
base. It could be 10 or 2, etc.
•The derivative of
logarithmic functions
Given >
y  ln( x)
dy 1

dx x
We’ll talk more about these
special cases after we talk
about the chain rule.
Differentiating functions of functions Given a function y  ( x 2  1) 2 we consider
dy d 2
2
2
( x  1)  h( x) write y  h compute

h  2h
dh dh
 


dh d 2
Then compute

x  1  2 x and
dx dx
dy dy dh
 .
take the product of the two, yielding
dx dh dx
dy dy dh
 .  2( x 2  1).2 x
dx dh dx
 4 x( x 2  1)
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
y  ( x 2  1) 2
Where y  ( x 2  1) 2
dy
 2( x 2  1)1.2 x
dx
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
y  sin( 2ax)
let h  2ax
dy dy dh
 .
then
dx dh dx
dy
 cos(2ax).2a or just
Which reduces to
dx
dy
 2a cos(2ax)
dx
In general if
y  f ( g (h(i(...(q( x))...))))
then
dy dy df dg dh dq
 . . . .....
dx df dg dh di
dx
Returning to those exponential and natural
log cases - we already implemented the
chain rule when differentiating
ye
ax
dy d (ax) ax

e
dx
dx
 ae ax
h in this case would be ax and, from the chain rule,
h
dy dy dh
dy
de
dh
 .
becomes

.
dx dh dx
dx dh dx
dy
h dh
e .
dx
dx
dy
ax

ae
and finally
dx
dh
a
since h  ax and
dx
or
For functions like
ye
ax 2
we follow the same procedure.
dy dy dh
 .
From the chain rule we have
dx dh dx
dh
2
Let h  ax and then
 2ax
dx
dy d h
 e  eh
dh dh
dy dy dh
ax 2
 .  2axe
hence
dx dh dx
Thus for that porosity depth
relationship we were working with -
 ( z )  0 e  z / 


d ( z ) d 0e z / 

?
dz
dz

 0

e z / 
For logarithmic functions like
y  ln( x 2 )
We combine two rules, the special rule for
d (ln x) 1
natural logs and the chain rule.

dx
x
dy dy dh
Let h  x 2 then
 .
dx dh dx
dy 1
dh
 2 and
 2x
dh x
dx
d ln( x 2 ) 2 x 2
so
 2
dx
x
x
For next Tuesday answer question
8.8 in Waltham (see page 148).
Find the derivatives of
(i )   x 2 .e x
(ii) y  3 2 .sin(  )
(iii) z  x.cos(x)  x . tan( x)
2
(iv) B  3 4 . ln( )  17 2