Mathematical Models
I. Deterministic Models
1) Statistical Models
Answer the
following question
What methods do
we use to estimate the parameters of the underlying relationship?
1)
Dependent variable (y)
2)
Independent variable (x)
Equation of a
straight Line =
y =
bo +
b1 x
bo = y-intercept
b1 = slope
(no error in reading y – for a given value of x we can
predict y exactly using the above equation)
Simple to use –
unrealistic
1)
Plot a scatter diagram
A model that
allows for the possibility that the observations do not all lie on a straight
line is
y =
bo + b1 x + e
where e is a
random error
the average value of e for a given value of x =
0
Thus, the
expected value of (y) (average value) for a fixed value of x is the line
E(y) = bo + b1 x
The observed
values of y deviate above or below the line by a random error e
To estimate the
values of the constants bo and b1
Let y` denote
the predicted value of y for a given value of x
Then the error
of prediction (residual) is y-y` (the difference between the actual value
of y and what we predict it to be)
The least
squares method chooses the prediction line
y` = b`o + b`1 x
that minimizes the sum of the squared errors of prediction
å (y-y`)2
for all sample points
Thus, the
method of least squares consists of finding those estimates b`o and b`1 that minimize å (y-y`)2
The estimates:
b`1 =
Sxy / Sxx and b`o = Y - b`1 X
Where,
Sxx
= å (x – X )2 = å x2 – (åx)2/n
Sxy = å (x – X
) ( y – Y ) = å xy – (
( åx ) ( å y ) )
/n
Example
|
x |
y |
|
4.2 |
2.8 |
|
3.8 |
2.5 |
|
4.8 |
3.1 |
|
3.4 |
2.1 |
|
4.5 |
2.9 |
|
4.6 |
2.8 |
|
4.3 |
2.6 |
|
3.7 |
2.4 |
|
3.9 |
2.5 |
Calculate
Sxx and Sxy
|
X |
x2 |
y |
y2 |
xy |
|
4.2 |
17.64 |
2.8 |
7.84 |
11.76 |
|
3.8 |
14.44 |
2.5 |
6.25 |
9.50 |
|
4.8 |
23.04 |
3.1 |
9.61 |
14.88 |
|
3.4 |
11.56 |
2.1 |
4.41 |
7.14 |
|
4.5 |
20.25 |
2.9 |
8.41 |
13.05 |
|
4.6 |
21.16 |
2.8 |
7.84 |
12.88 |
|
4.3 |
18.49 |
2.6 |
6.76 |
11.18 |
|
3.7 |
13.69 |
2.4 |
5.76 |
8.88 |
|
3.9 |
15.21 |
2.5 |
6.25 |
9.75 |
|
37.2 |
155.48 |
23.7 |
63.13 |
99.02 |
A measure of
the strength of the relationship between two variables x and y
r = Sxy /
sqr ( SxxSyy)
where
Syy
= å y2 - (åy)2 / n
Or
r = b`1 sqr ( Sxx / Syy )
Properties of r
1)
r lies between –1 and
+1. r > 0 indicates a positive linear
relationship and r < 0 a negative linear relationship between x and y. r = 0
indicates no linear relationship between x and y
2)
r2 gives the proportion of the total
variability of the y values that can be accounted for by the independent
variable x
r2 = å( y` - Y )2 / Syy = S2xy
/ SxxSyy
1
- r2 = å ( y - y` )2
/ Syy
Example
|
Y |
X |
|
1.4 |
1 |
|
2.3 |
2 |
|
3.1 |
3 |
|
4.2 |
4 |
|
5.1 |
5 |
|
5.8 |
6 |
|
6.8 |
7 |
|
7.6 |
8 |
|
8.7 |
9 |
|
9.5 |
10 |
Calculate Syy , Sxx , and Sxy
Calculate r
1)
scatter plot
2)
plot of the residuals ( y – y`) versus y` may give
an indication of the need for higher order terms in the model
Linear = no
apparent pattern
Higher order
model = apparent pattern
The expected
value of bo = mbo and the expected value of
b1 = m b1
The standard
error of bo = s bo = s e sqr ( å x2/nSxx
)
The standard
error of b1 = s b1 = s e / sqr ( Sxx )
Estimates of
s b1 and s bo
S2e = å(y- Y)2
/ n-2
Se = √ (S2e )
a) 100( 1- a )% confidence intervals
bo ± ta/2 Se sqr ( å x2 / nSxx
)
b1 ± ta/2 Se / sqr ( Sxx )
Note
: d.f. = n –2
b)
Statistical tests
Ho: bo = 0
Ha : bo
> 0
b0
< 0
bo ¹ 0
T.S.:
t = bo` / Se
sqr ( å x2 / nSxx )
R.R.: for a given value of a and d.f. = n –2
1.
Reject Ho
if t > ta
2.
Reject Ho
if t < - ta
3.
Reject Ho
if | t | > ta/2
Ho: b1 = 0
Ha: b1 > 0
b1 < 0
b1 ¹ 0
T.S. : t = b1` / (
Se / sqr( Sxx) )
R.R.: for a given value of a and d.f. = n – 2
Same as for bo