UNIT 5

Statistics

5.1 Fitting of straight line

Curve Fitting of Type y=axb Algorithm

In this article we are going to develop an algorithm for fitting curve of type

y = axb using least square regression method.

Procedure for fitting y = axb

We have,

Example 1:

Fit a least square line for the following data. Also find the trend values and show that ∑(Y–)=0 ∑(Y–)=0.

Solution:

The equation of least square line Y=a +bX 

Normal equation for ‘a’ ∑Y=na+b                    25=5a+15b —- (1)

Normal equation for ‘b’ ∑XY = a∑X+b∑X2     88=15a+55b —-(2)

Eliminate a a from equation (1) and (2), multiply equation (2) by 3 and subtract from equation (2).

Eliminate  a  from equation (1) and (2), multiply equation (2) by (3) and subtract from equation (2). Thus we get the values of  a and b 

Here    a=1.1  and b=1.3 , the equation of least square line becomes

 Y=1.1+1.3X

example 2:

using least square method to fit a straight line of the following data

Solution:

First we calculate for the given data

Now we calculate

Calculate the slope

  m = = -131/118.4

calculate the y-intercept

use the formula to calculate the y-intercept

b =

   = 7-(-1.1*6.4)

The required line equation is

Y= -1.1x+14.0

5.2 Second degree parabola

Second-order curve

A plane curve whose rectangular Cartesian coordinates fulfil an algebraic calculation of the second degree:

Equation need not define a real geometrical form, but to reserve generality in such situations one says that it describes an imaginary second-order curve. Depending on the values of the coefficients of the equation  it can be changed by parallel movement and rotation of the coordinate system through some angle to one of the 9 canonical forms given under, to each of which there agrees a definite class of curves. Viz.,

non-degenerate curves:

, ellipses (cf. Ellipse);

, hyperbolas (cf. Hyperbola);

, parabolas (cf. Parabola);

, imaginary ellipses;

degenerate curves:

, pairs of imaginary intersecting lines;

, pairs of real intersecting lines;

, pairs of real parallel lines;

, pairs of imaginary parallel lines;

,a pair of coincident real lines.

A second-order curve that has a unique centre of symmetry (the centre of the second-order curve) is named a central curve. The coordinates of the centre of a second-order curve are determined by the explanation of the system

A second-order curve without a centre of symmetry or with an indefinite centre is named a non-central curve.

Example 1:

Solution:

∑Y i =Na+b∑X i +c∑

∑X i Y i =a∑X i +b∑+c∑

 ∑ Y i =a∑ +b∑+c∑

The required parabola is of the form y= ax2+bx+c

∴74=9a+b(0)+60c∴9a+60c=74…(i)

51=a(0)+60b+0c  ∴60b=51   ∴b=5160 =0.85411=60a+0b+708 c∴60a+708c=411…(ii) 

Solving (i) and (ii) simultaneously, we get
a=10.004 , c=-0.267

The Equation of parabola is therefore,

y=10.004+0.85X−0.267X 2

 =10.004+0.85(x−5)−0.267(x−5) 2 

=10.004+0.85x−4.25−0.267(x 2 −10x+25)

=10.004+0.85x−4.25−0.267x 2 +2.67x−6.675

∴  y = −0.921+3.52x−0.267x 2 

Example 2:

Find the least square approximation of  degree two to the data

Solution:

the normal equations are:

Here,

n = 5,

by substituting all the above values in normal equations we get,

30 = 5a+10b+30c

120=10a+30b+100c

434=30a+100b+354c

By solving the above equations we get

 a = -4, b=2,c=1.

Therefore the required polynomial is

Y= -4x+2x+x2 and errors =0

5.3 Exponential curve by the method of Least Squares

Example 1:

Determine the constants a and b by the method of least square such that

Solution:

The given relation is

Taking logarithms on both sides we get,

log y = log a+ bx…..(1)

let,

log y = Y

x = X

log a = A

b = B

now we have,

….(2)

….(3)

Now we need to find

The normal equations to fit the straight line is

Y = logey

Y= ln(y)

17.025 = 5A +30B…..(4)

122.150 = 30A+220B….(5)

By solving 4 and 5 we get

30A +180B = 102.15…(4)

30A+220B = 122.150…(5)

We get a =   0.405   ,b = 0.5

A =log a

 a = 1.499

since we have  X=x and Y=y

log y=Y,

And we  know y= aebx

Y = (1.499)e0.5x is the required exponential curve.

Example 2:

Fit the curve of the form y= aebx for the following data

Solution:

The given relation is

Taking logarithms on both sides we get,

log y = log a+ bx logee…..(1)

the required normal equations are,

….(2)

….(3)

We have n=3

The normal equations become

3A +6b = 7.8750

6A + 20 b = 18.4456

By solving the above two equations we get

A = 1.361  and b = 0.3415

Since A =logea a = e1.361 = 6.9317

The curve of the fit is

Thus,the required  equation is,

5.4 Coefficient of linear correlation

Whenever two variables x and y are so related that an increase in the one is accompanied by an increase or decrease in the other, then the variables are said to be correlated.

For example, the yield of crop varies with the amount of rainfall.

If an increase in one variable corresponds to an increase in the other, the correlation is said to be positive. If increase in one corresponds to the decrease in the other the correlation is said to be negative. If there is no relationship between the two variables, they are said to be independent.

Perfect Correlation: If two variables vary in such a way that their ratio is always constant, then the correlation is said to be perfect.

KARL PEARSON’S COEFFICIENT OF CORRELATION:

rbetween two variables x and y is defined by the relation

Where,X = x –, Y = y –

i.e. X, Y are the deviations measured from their respective means,

Example:Ten students got the following percentage of marks in Economics and Statistics

Calculate the of correlation.

Solution:. Let the marks of two subjects be denoted by and respectively.

Then the mean for marks and the mean ofy marks

and are deviations ofx’s and ’s from their respective means, then the data may be arranged in the following form:

5.5 Regression lines

If the scatter diagram indicates some relationship between two variables and , then the dots of the scatter diagram will be concentrated round a curve. This curve is called the curve ofregression. Regression analysis is the method used for estimating the unknown values of one variable corresponding to the known value of another variable.

LINE OF REGRSSION

When the curve is a straight line, it is called a line of regression. A line of regression is the straight line which gives the best fit in the least square sense to the given frequency.

Example:Find the correlation betweenx and , when the lines ofregression are: and

Solution. Let the line of regression ofx on be

Then, the line ofregressionofy on is

and

which is not possible. So our choice of regression line is incorrect.

The regression line ofx on is

And, the regression line ofy on is

And

Hence the correlation coefficient between and is

Example: The following regression equations were obtainedfrom a correlation table:

Find the value of the correlation coefficient,

(b) the mean and

(c) the mean of

Solution.

(a)  From (1),

(b) From (2),

From (3) and (4)

Coefficient of correlation

(b) (1) and (2) pass through the point .

   (5)

   (6)

On solving (5) and (6), we get

5.6 Rank correlation

Solution. Let be the ranks of individuals corresponding to two characteristics.

Assuming nor two individuals are equal in either classification, each individual takes the values 1, 2, 3, and hence their arithmetic means are, each

Let , , , be the values of variable and , , those of

Then

where and y are deviations from the mean.

Clearly, and

SPEARMAN’S RANK CORRELATION COEFFICIENT:

where denotes rank coefficient of correlation and refers to the difference ofranks between paired items in two series.

Example:Compute Spearman’s rank correlation coefficient r for the following data:

Solution:

Example:IfXand Yare uncorrelated random variables, the of correlation between and

Solution.

Let and

Then

Now

Similarly

Now

Also

(As and are not correlated, we have )

Similarly

Reference Books:

1. A text book of Applied Mathematics Volume I and II by J.N. Wartikar and P.N. Wartikar

2. Higher Engineering Mathematics by Dr. B. S. Grewal

3. Advanced Engineering Mathematics by H. K. Dass

4. Advanced Engineering Mathematics by Erwins Kreyszig