Unit - 6

Statistics and finite differences

6.1 (A) Statistics

Fitting of straight line y = a + bx, Parabola y = a + bx + cand Exponential curves by method of least squares

Method of Least Squares

Let     (1)

Be the straight line to be fitted to the given data points

Let be the theoretical value for

Then,

For S to be minimum

On simplification equation (2) and (3) becomes

The equation (3) and (4) are known as Normal equations.

On solving ( 3) and (4) we get the values of a and b

(b)To fit the parabola

The normal equations are

On solving three normal equations we get the values of a,b and c.

Example. Find the best values of a and b so that y = a + bx fits the data given in the table

Solution.

y = a + bx

x	y	Xy
0	1.0	0	0
1	2.9	2.0	1
2	4.8	9.6	4
3	6.7	20.1	9
4	8.6	13.4	16
x = 10	y ,= 24.0	xy = 67.0

Normal equations, y= na+ bx  (2)

On putting the values of

On solving (4) and (5) we get,

On substituting the values of a and b in (1) we get

Example. By the method of least squares, find the straight line that best fits the following data:

Solution. Let the equation of the straight line best fit be y = a + bx.     (1)

x	y	x y
1	14	14	1
2	27	54	4
3	40	120	9
4	55	220	16
5	68	340	25
x=15	y=204	xy=748

Normal equations are

On putting the values of x, y, xy and in (2) and (3) we have

On solving equations (4) and (5) we get

On substituting the values of (a) and (b) in (1) we get,

Example: Find the straight line that best fits of the following data by using method of least square.

Sol.

Suppose the straight line

y = a + bx…….. (1)

Fits the best-

Then-

x	y	Xy
1	14	14	1
2	27	54	4
3	40	120	9
4	55	220	16
5	68	340	25
Sum = 15	204	748	55

Normal equations are-

Put the values from the table, we get two normal equations-

On solving the above equations, we get-

So that the best fit line will be- (on putting the values of a and b in equation (1))

Example. Find the least squares approximation of second degree for the discrete data

Solution. Let the equation of second degree polynomial be

x	y	Xy
-2	15	-30	4	60	-8	16
-1	1	-1	1	1	-1	1
0	1	0	0	0	0	0
1	3	3	1	3	1	1
2	19	38	4	76	8	16
x=0	y=39	xy=10

Normal equations are

On putting the values of x, y, xy,

have

On solving (5),(6),(7), we get,

The required polynomial of second degree is

Second degree parabolas and more general curves

Change of scale

If the data is of equal interval in large numbers then we change the scale as

Example. Fit a second degree parabola to the following data by least square method:

Solution. Taking

Taking

The equation is transformed to

x		y		Uv
1929	-4	352	-5	20	16	-80	-64	256
1930	-3	360	-1	3	9	-9	-27	81
1931	-2	357	0	0	4	0	-8	16
1932	-1	358	1	-1	1	1	-1	1
1933	0	360	3	0	0	0	0	0
1934	1	361	4	4	1	4	1	1
1935	2	361	4	8	4	16	8	16
1936	3	360	3	9	9	27	27	81
1937	4	350	2	8	16	32	64	256
Total	u=0		y=11	uv=51

Normal equations are

On solving these equations we get

Example. Fit a second degree parabola to the following data.

Solution. We shift the origin to (2.5, 0) antique 0.5 as the new unit. This amounts to changing the variable x to X, by the relation X = 2x – 5.

Let the parabola of fit be y = a + bX The values of X etc. Are calculated as below:

x	X	y	Xy
1.0	-3	1.1	-3.3	9	9.9	-27	81
1.5	-2	1.3	-2.6	4	5.2	-5	16
2.0	-1	1.6	-1.6	1	1.6	-1	1
2.5	0	2.0	0.0	0	0.0	0	0
3.0	1	2.7	2.7	1	2.7	1	1
3.5	2	3.4	6.8	4	13.6	8	16
4.0	3	4.1	12.3	9	36.9	27	81
Total	0	16.2	14.3	28	69.9	0	196

The normal equations are

7a + 28c =16.2; 28b =14.3; 28a +196c=69.9

Solving these as simultaneous equations we get

Replacing X bye 2x – 5 in the above equation we get

Which simplifies to y = This is the required parabola of the best fit

Example: Find the least squares approximation of second degree for the discrete data

Solution. Let the equation of second degree polynomial be

x	y	Xy
-2	15	-30	4	60	-8	16
-1	1	-1	1	1	-1	1
0	1	0	0	0	0	0
1	3	3	1	3	1	1
2	19	38	4	76	8	16
x=0	y=39	xy=10

Normal equations are

On putting the values of x, y, xy,

have

On solving (5),(6),(7), we get,

The required polynomial of second degree is

Example: Fit a second degree parabola to the following data.

Solution

We shift the origin to (2.5, 0) antique 0.5 as the new unit. This amounts to changing the variable x to X, by the relation X = 2x – 5.

Let the parabola of fit be y = a + bXThe values of X etc. Are calculated as below:

x	X	y	Xy
1.0	-3	1.1	-3.3	9	9.9	-27	81
1.5	-2	1.3	-2.6	4	5.2	-5	16
2.0	-1	1.6	-1.6	1	1.6	-1	1
2.5	0	2.0	0.0	0	0.0	0	0
3.0	1	2.7	2.7	1	2.7	1	1
3.5	2	3.4	6.8	4	13.6	8	16
4.0	3	4.1	12.3	9	36.9	27	81
Total	0	16.2	14.3	28	69.9	0	196

The normal equations are

7a + 28c =16.2; 28b =14.3;. 28a +196c=69.9

Solving these as simultaneous equations we get

Replacing X bye 2x – 5 in the above equation we get

Which simplifies to y =

This is the required parabola of the best fit.

Example: Fit the curve by using the method of least square.

Sol.

Here-

Now put-

Then we get-

x	Y		XY
1	7.209	1.97533	1.97533	1
2	5.265	1.66108	3.32216	4
3	3.846	1.34703	4.04109	9
4	2.809	1.03283	4.13132	16
5	2.052	0.71881	3.59405	25
6	1.499	0.40480	2.4288	36
Sum = 21		7.13988	19.49275	91

Normal equations are-

Putting the values form the table, we get-

7.13988 = 6c + 21b

19.49275 = 21c + 91b

On solving, we get-

b = -0.3141 and c = 2.28933

c =

Now put these values in equations (1), we get-

Example: Estimate the chlorine residual in a swimming pool 5 hours after it has been treated with chemicals by fitting an exponential curve of the form

of the data given below-

Sol.

Taking log on the curve which is non-linear,

We get-

Put

Then-

Which is the linear equation in X,

Its normal equations are-

Here N = 6,

Thus the normal equations are-

On solving, we get

Or

A = 2.013 and B = 0.936

Hence the required least square exponential curve-

Prediction-

Chlorine content after 5 hours-

6.2 Lines of regression and Correlation

Correlation:

So far we have confined our attention to the analysis of observation on a single variable. There are, however, many phenomena where the changes in one variable are related to the changes in the other variable.  For instance, the yield of crop varies with the amount of rainfall, the price of a commodity increases with the reduction in its supply and so on. Such a simultaneous variation, i.e., when the changes in one variable are associated or followed by change in the other, is called correlation. Such a data connecting two variables is called bivariate population.

If an increase (or decrease) in the values of one variable corresponds to an increase (or decrease) in the other, the correlation is said to be positive. If the increase (or decrease) in one corresponds to the decrease (or increase) in other, the correlation is said to be negative.  If there is no relationship indicated between the variables, they are said to be independent or uncorrelated.

When two variables are related in such a way that change in the value of one variable affects the value of the other variable, then these two variables are said to be correlated and there is correlation between two variables.

Example- Height and weight of the persons of a group.

The correlation is said to be perfect correlation if two variables vary in such a way that their ratio is constant always.

Types of correlation:

According to the direction of change in variables there are two types of correlation

1. Positive Correlation

2. Negative Correlation

1. Positive Correlation:

Correlation between two variables is said to be positive if the values of the variables deviate in the same direction i.e. if the values of one variable increase (or decrease) then the values of other variable also increase (or decrease). For example:

1. Heights and weights of group of persons;

2. House hold income and expenditure;

3. Amount of rainfall and yield of crops

2. Negative Correlation:

Correlation between two variables is said to be negative if the values of variables deviate in opposite direction i.e. if the values of one variable increase (or decrease) then the values of other variable decrease (or increase). Some examples of negative correlations are correlation between

1. Volume and pressure of perfect gas;

2. Price and demand of goods;

3. Literacy and poverty in a country

Scatter diagram-

Scatter diagram is a statistical tool for determining the potentiality of correlation between dependent variable and independent variable. Scatter diagram does not tell about exact relationship between two variables but it indicates whether they are correlated or not.

To obtain a measure of relationship between the two variables, we plot their corresponding values on the graph, taking one of the variables along the x-axis and the other along the y-axis.

Correlation measures the nature and strength of relationship between two variables. Correlation lies between +1 to -1. A correlation of +1 indicates a perfect positive correlation between two variables. A zero correlation indicates that there is no relationship between the variables. A correlation of -1 indicates a perfect negative correlation.

Definition-

“Correlation analysis deals with the association between two or more variables.” —Simpson and Kafka

“Correlation is an analysis of the co-variation between two variables.” —A.M. Tuttle

Methods of computing coefficient of correlation

1. Scatter diagram method-

It is the simplest method to study correlation between two variables. The correlations of two variables are plotted in the graph in the form of dots thereby obtaining as many points as the number of observations. The degree of correlation is ascertained by looking at the scattered points over the charts.

The more the points plotted are scattered over the chart, the lesser is the degree of correlation between the variables. The more the points plotted are closer to the line, the higher is the degree of correlation. The degree of correlation is denoted by “r”.

• Perfect positive correlation (r = +1) – All the points plotted on the straight line rising from left to right

• Perfect negative correlation (r=-1) – all the points plotted on the straight line  falling from left to right

• High Degree of +Ve Corr elation (r= + High): all the points plotted close to the straight line rising  from left to right

• High Degree of –Ve Correlation (r= – High) - all the points plotted close to the straight line falling from left to right.

• Low degree of +Ve Correlation (r= + Low): all the points are highly scattered to the straight line rising  from left to right

• Low Degree of –Ve Correlation (r= - Low): all the points are highly scattered to the straight line falling from left to right

• No Correlation (r= 0) – all the points are scattered over the graph and do not show any pattern

2.     Karl Pearson’s coefficient of correlation:

Coefficient of correlation measures the intensity or degree of linear relationship between two variables. It was given by British Biometrician Karl Pearson (1867-1936).

Karl Pearson’s Coefficient of Correlation is widely used mathematical method is used to calculate the degree and direction of the relationship between linear related variables. The coefficient of correlation is denoted by “r”.

If X and Y are two random variables then correlation coefficient between X and Y is denoted by r and defined as-

Karl Pearson’s coefficient of correlation-

Here- and

Note-

1. Correlation coefficient always lies between -1 and +1.

2. Correlation coefficient is independent of change of origin and scale.

3. If the two variables are independent then correlation coefficient between them is zero.

Example: Find the correlation coefficient between Age and weight of the following data-

Sol.

x	Y					( ) )
30	56	-10	100	-4	16	40
44	55	4	16	-5	25	-20
45	60	5	25	0	0	0
43	64	3	9	4	16	12
34	62	-6	36	2	4	-12
44	63	4	16	3	9	12
Sum= 240	360	0	202	0	70	32

Karl Pearson’s coefficient of correlation-

Here the correlation coefficient is 0.27.which is the positive correlation (weak positive correlation), this indicates that the as age increases, the weight also increase.

Short-cut method to calculate correlation coefficient-

Here,

Example: Find the correlation coefficient between the values X and Y of the dataset given below by using short-cut method-

Sol.

X	Y
10	90	-20	400	20	400	-400
20	85	-10	100	15	225	-150
30	80	0	0	10	100	0
40	60	10	100	-10	100	-100
50	45	20	400	-25	625	-500
Sum = 150	360	0	1000	10	1450	-1150

Short-cut method to calculate correlation coefficient-

Example

Psychological tests of intelligence and of engineering ability were applied to 10 students. Here is a record of ungrouped data showing intelligence ratio (I.R) and engineering ratio (E.R). Calculate the co-efficient of correlation.

Solution:

We construct the following table:

Student	Intelligence ratio	Engineering ratio
A B C D E F G H I J	100                      6     104                     5     102                     3     101                     2     100                     1       99                     0       98                    -1       96                    -3       93                    -6       92                    -7	101                 3   103                 5   100                 2    98                  0    95                 -3    96                 -2   104                 6    92                 -6    97                 -1    94                 -4	36 25 9 4 1 0 1 9 36 49	9 25 4 0 9 4 36 36 1 16	18 25 6 0 -3 0 -6 18 6 28
Total	990                    0	980                 0	170	140	92

From this table, mean of i.e., and mean of , i.e.,

Substituting these values in the formula (1), we have

Other examples:

Example-1-Compute Pearsons coefficient of correlation between advertisement cost and sales as per the data given below:

Solution

X	Y
39	47	-26	676	-19	361	494
65	53	0	0	-13	169	0
62	58	-3	9	-8	64	24s
90	86	25	625	20	400	500
82	62	17	289	-4	16	-68
75	68	10	100	2	4	20
25	60	-40	1600	-6	36	240
98	91	33	1089	25	625	825
36	51	-29	841	-15	225	435
78	84	13	169	18	324	234
650	660		5398		2224	2704

r = (2704)/√5398 √2224 = (2704)/(73.2*47.15) = 0.78

Thus Correlation coefficient is positively correlated

Example 2

Compute correlation coefficient from the following data

X	Y
8	81	1.3	1.8	2.3	5.4	3.1
8	80	1.3	1.8	1.3	1.8	1.8
6	75	-0.7	0.4	-3.7	13.4	2.4
5	65	-1.7	2.8	-13.7	186.8	22.8
7	91	0.3	0.1	12.3	152.1	4.1
6	80	-0.7	0.4	1.3	1.8	-0.9
40	472		7		361	33

= 40/6 =6.7

= 472/6 = 78.7

r = (33)/√7 √361 = (33)/(2.64*19) = 0.66

Thus Correlation coefficient is positively correlated

Example 3

Calculate coefficient of correlation between X and Y series using Karl pearson shortcut method

Solution

Let assumed mean for X = 15, assumed mean for Y = 14

r = 48/√144*√432 = 0.19

Regression-

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 of regression. Regression analysis is the method used for estimating the unknown values of one variable corresponding to the known value of another variable.

Or in other words, Regression is the measure of average relationship between independent and dependent variable

Regression can be used for two or more than two variables.

There are two types of variables in regression analysis.

1. Independent variable

2. Dependent variable

The variable which is used for prediction is called independent variable.

It is known as predictor or regressor.

The variable whose value is predicted by independent variable is called dependent variable or regressed or explained variable.

The scatter diagram shows relationship between independent and dependent variable, then the scatter diagram will be more or less concentrated round a curve, which is called the curve of regression.

When we find the curve as a straight line then it is known as line of regression and the regression is called linear regression.

Note- regression line is the best fit line which expresses the average relation between variables.

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.

Equation of the line of regression-

Let

y = a + bx  …………..  (1)

Is the equation of the line of y on x.

Let be the estimated value of for the given value of .

So that, According to the principle of least squares, we have the determined ‘a’ and ‘b’ so that the sum of squares of deviations of observed values of y from expected values of y,

That means-

Or

   …….. (2)

Is minimum.

Form the concept of maxima and minima, we partially differentiate U with respect to ‘a’ and ‘b’ and equate to zero.

Which means

And

These equations (3) and (4) are known as normal equation for straight line.

Now divide equation (3) by n, we get-

This indicates that the regression line of y on x passes through the point
.

We know that-

The variance of variable x can be expressed as-

Dividing equation (4) by n, we get-

From the equation (6), (7) and (8)-

Multiply (5) by, we get-

Subtracting equation (10) from equation (9), we get-

Since ‘b’ is the slope of the line of regression y on x and the line of regression passes through the point (), so that the equation of the line of regression of y on x is-

This is known as regression line of y on x.

Note-

are the coefficients of regression.

2.

Example: Two variables X and Y are given in the dataset below, find the two lines of regression.

Sol.

The two lines of regression can be expressed as-

And

x	y			Xy
65	66	4225	4356	4290
66	68	4356	4624	4488
67	65	4489	4225	4355
67	69	4489	4761	4623
68	74	4624	5476	5032
69	73	4761	5329	5037
70	72	4900	5184	5040
71	70	5041	4900	4970
Sum = 543	557	36885	38855	37835

Now-

And

Standard deviation of x-

Similarly-

Correlation coefficient-

Put these values in regression line equation, we get

Regression line y on x-

Regression line x on y-

Regression line can also be find by the following method-

Example: Find the regression line of y on x for the given dataset.

Sol.

Let y = a + bx is the line of regression of y on x, where ‘a’ and ‘b’ are given as-

We will make the following table-

x	y	Xy
4.3	12.6	54.18	18.49
4.5	12.1	54.45	20.25
5.9	11.6	68.44	34.81
5.6	11.8	66.08	31.36
6.1	11.4	69.54	37.21
5.2	11.8	61.36	27.04
3.8	13.2	50.16	14.44
2.1	14.1	29.61	4.41
Sum = 37.5	98.6	453.82	188.01

Using the above equations we get-

On solving these both equations, we get-

a = 15.49 and b = -0.675

So that the regression line is –

y = 15.49 – 0.675x 

Note – Standard error of predictions can be find by the formula given below-

Difference between regression and correlation-

1. Correlation is the linear relationship between two variables while regression is the average relationship between two or more variables.

2. There are only limited applications of correlation as it gives the strength of linear relationship while the regression is to predict the value of the dependent varibale for the given values of independent variables.

3. Correlation does not consider dependent and independent variables while regression consider one dependent variable and other indpendent variables.

Key takeaways-

1. Karl Pearson’s coefficient of correlation-

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

3.     Short-cut method to calculate correlation coefficient-

4.     Spearman’s rank correlation-

5.     The variable which is used for prediction is called independent variable. It is known as predictor or regressor.

6.     regression line is the best fit line which expresses the average relation between variables.

7.     regression line of y on x.

6.3 Rank correlation

A group of n individuals may be arranged in order to merit with respect to some characteristic. The same group would give different orders for different characteristics. Considering the order corresponding to two characteristics A and B for that group of individuals.

Let be the ranks of the ith individuals in A and B respectively. Assuming that no two individuals are bracketed equal in either case, each of the variables taking the values 1,2,3,…,n, we have

If X,Y be the deviation of x, y from their means, then

Similarly

Now let so that

Hence the correlation coefficient between these variables is

This is called the rank correlation coefficient (Spearman’s rank corr) and is denoted by.

Spearman’s rank correlation-

When the ranks are given instead of the scores, then we use Spearman’s rank correlation to find out the correlation between the variables.

Spearman’s rank correlation coefficient can be defined as-

Example:

	1	6	5	10	3	2	4	9	7	8
	6	4	9	8	1	2	3	10	5	7

Solution:

If   , then

Hence

nearly.

Example

Three judges A, B, C, give the following ranks. Find which pair of judges has common approach

Solution: Here

	Ranks by
1 6 5 10 3 2 4 9 7 8	3 5 8 4 7 10 2 1 6 9	6 4 9 8 1 2 3 10 5 7	-2 1 -3 6 -4 -8 2 8 1 -1	-3 1 -1 -4 6 8 -1 -9 1 2	5 -2 4 -2 -2 0 -1 1 -2 -1	4 1 9 36 16 64 4 64 1 1	9 1 1 16 36 64 1 81 1 4	25 4 16 4 4 0 1 1 4 1
Total			0	0	0	200	214	60

Since is maximum, the pair of judges A and C have the nearest common approach.

Example: Compute the Spearman’s rank correlation coefficient of the dataset given below-

Solution

Person	Rank in test-1	Rank in test-2	d =
A	9	1	8	64
B	10	2	8	64
C	6	3	3	9
D	5	4	1	1
E	7	5	2	4
F	2	6	-4	16
G	4	7	-3	9
H	8	8	0	0
I	1	9	-8	64
J	3	10	-7	49
Sum				280

Example: If X and 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

Example 1 –

Solution

Here, highest value is taken as 1

R = 1 – (6*8)/5(52 – 1) = 0.60

Example 2 -

Calculate Spearman rank-order correlation

Solution

Rank by taking the highest value or the lowest value as 1.

Here, highest value is taken as 1

R = 0.67

Therefore, this indicates a strong positive relationship between the ranks individuals obtained in the math and English exam.

Example 3 –

Find Spearman's rank correlation coefficient between X and Y for this set of data:

Solution

R =

Example 4 – Calculation of equal ranks or tie ranks

Find Spearman's rank correlation coefficient:

Solution

R = 1 – (6*81.5)/8(82 – 1) = 0.02

Example 5 –

Solution

R = 1 – (6*24)/10(102 – 1) = 0.85

The correlation between X and Y is positive and very high.

6.4 (B) Finite Differences:

Operator E and , Factorial notations

Definition:

Interpolation is a technique of  estimating the value of  a  function for any  intermediate value of the  independent  variable while  the process  of  computing the  value of the function outside the given range is called   extrapolation.

Let be a function of x.

The table given below gives corresponding values of y for  different values of x.

X				….
y= f(x)				….

The process of finding the values of y corresponding to any value of x which lies between   is called interpolation.

If the given function is a polynomial it is polynomial interpolation and given function is known as interpolating polynomial.

Note- The process of computing the value of the function outside the given range is called extrapolation.

Thus interpolation is the “art of reading between the lines of a table.”

Conditions for Interpolation

1)     The function must be   a polynomial of independent variable.

2)     The function   should be either increasing or decreasing function.

3)     The value of   the function should be increase or decrease uniformly.

Finite Difference

Let be a function of x.The table given below gives corresponding values of y for  different values of x.

X				….
y= f(x)				….

There are three types of differences are useful-

a)     Forward Difference: 

The  are called differences of y, denoted by 

The  symbol   is  called the forward  difference operator.  Consider the forward  difference  table  below:

Where

And third forward difference  so  on.

b)    Backward Difference:

The difference are called first  backward difference and  is denoted by  Consider the backward  difference  table  below:

Where

And third backward differences so on.

Example: Construct a backward difference table for y = log x, given-