Unit - 5

Statistics

5.1 Fitting of curve- Least squares principle , fitting of straight line , fitting of second degree parabola, fitting of curves of the form y= , y = a , y =

Method of least square-

Suppose

y = a + bx ………. (1)

Is the straight line has to be fitted for the data points given-

Let be the theoretical value for

Now-

For the minimum value of S -

Or

Now

Or

On solving equation (1) and (2), we get-

These two equation are known as the normal equations.

Now on solving these two equations we get the values of a and b.

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 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

To fit the parabola

The normal equations are

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

Note- Change of scale-

We change the scale if the data is large and given in equal interval.

As-

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

	1929	1930	1931	1932	1933	1934	1935	1936	1937
	352	356	357	358	360	361	361	360	359

Solution: Taking x0 = 1933,  y0 = 357

Taking           u = x – x0,                 v = y – y0

                       u = x – 1933,             v = y – 357

The equation y = a+bx+cx2  is transformed to v = A + Bu + Cu2    …(2)

	u = x – 1933					u2v	u3	u4
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	359	2	8	16	32	64	256
Total

Normal equations are

On solving these equations, we get    A = 694/231,   B = 17/20,    C = - 247/924

V = 694/231 + 17/20 u – 247/924 u2V = 694/231 + 17/20 u – 247/924 u2

y – 357 = 694/231 + 17/20  (x – 1933) – 247/924 (x – 1933)2

= 694/231 + 17x/20  - 32861/20 – 247x2/924 (-3866x) – 247/924 + (1933)2

y = 694/231 – 32861/20 – 247/924 (1933)2 + 17x/20 + (247  3866)x/924  - 247 x2/924

y = 3 – 1643.05 – 998823.36 + 357 + 0.85 x + 1033.44 x – 0.267 x2

y = - 1000106.41 + 1034.29x – 0.267 x2

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 nomal equations are-

X                                 Y                         Y* = ln Y                               X2                         XY*

2                                1.8                      0.5878                                    4                          0.1756

4                                1.5                      0.4055                                  16                            1.622

6                                1.4                      0.3365                                  36                             2.019

8                                1.1                      0.0953                                  64                             0.7264

10                              1.1                     0.0953                                100                                0.953

12                              0.9                  -0.10536                                144                           -1.26432

42                                                            1.415                               364                             5.26752

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-

Key takeaways-

5.2 Coefficient of correlation by Karl Pearson‟s method and lines of regression of bivariate data

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.

Scatter diagram-

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.

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

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.

Example:

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

Calculate the of correlation.

Roll No.
Marks in Economics
Marks in

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:

x	y	X = x - 65	Y = y - 66	X2	Y2	X.Y
78 36 98 25 75 82 90 62 65 39 650	84 51 91 60 68 62 86 58 53 47 660	13 -29 33 -40 10 17 25 -3 0 -26 0	18 -15 25 -6 2 -4 20 -8 -13 -19 0	169 841 1089 1600 100 289 625 9 0 676 5398	324 225 625 36 4 16 400 64 169 361 2224	234 435 825 240 20 -68 500 24 0 494 2704

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-

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

Sol.

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 Y are uncorrelated random variables, the of correlation between and

Solution.

Let and

Then

Now

Similarly

Now

Also

(As and are not correlated, we have )

Similarly

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.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, aa2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Unit - 5

Statistics

5.1 Fitting of curve- Least squares principle , fitting of straight line , fitting of second degree parabola, fitting of curves of the form y= , y = a , y =

Method of least square-

Suppose

y = a + bx ………. (1)

Is the straight line has to be fitted for the data points given-

Let be the theoretical value for

Now-

For the minimum value of S -

Or

Now

Or

On solving equation (1) and (2), we get-

These two equation are known as the normal equations.

Now on solving these two equations we get the values of a and b.

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 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

To fit the parabola

The normal equations are

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

Note- Change of scale-

We change the scale if the data is large and given in equal interval.

As-

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

	1929	1930	1931	1932	1933	1934	1935	1936	1937
	352	356	357	358	360	361	361	360	359

Solution: Taking x0 = 1933,  y0 = 357

Taking           u = x – x0,                 v = y – y0

                       u = x – 1933,             v = y – 357

The equation y = a+bx+cx2  is transformed to v = A + Bu + Cu2    …(2)

	u = x – 1933					u2v	u3	u4
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	359	2	8	16	32	64	256
Total

Normal equations are

On solving these equations, we get    A = 694/231,   B = 17/20,    C = - 247/924

V = 694/231 + 17/20 u – 247/924 u2V = 694/231 + 17/20 u – 247/924 u2

y – 357 = 694/231 + 17/20  (x – 1933) – 247/924 (x – 1933)2

= 694/231 + 17x/20  - 32861/20 – 247x2/924 (-3866x) – 247/924 + (1933)2

y = 694/231 – 32861/20 – 247/924 (1933)2 + 17x/20 + (247  3866)x/924  - 247 x2/924

y = 3 – 1643.05 – 998823.36 + 357 + 0.85 x + 1033.44 x – 0.267 x2

y = - 1000106.41 + 1034.29x – 0.267 x2

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 nomal equations are-

X                                 Y                         Y* = ln Y                               X2                         XY*

2                                1.8                      0.5878                                    4                          0.1756

4                                1.5                      0.4055                                  16                            1.622

6                                1.4                      0.3365                                  36                             2.019

8                                1.1                      0.0953                                  64                             0.7264

10                              1.1                     0.0953                                100                                0.953

12                              0.9                  -0.10536                                144                           -1.26432

42                                                            1.415                               364                             5.26752

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-

Key takeaways-

5.2 Coefficient of correlation by Karl Pearson‟s method and lines of regression of bivariate data

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.

Scatter diagram-

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.

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

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.

Example:

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

Calculate the of correlation.

Roll No.
Marks in Economics
Marks in

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:

x	y	X = x - 65	Y = y - 66	X2	Y2	X.Y
78 36 98 25 75 82 90 62 65 39 650	84 51 91 60 68 62 86 58 53 47 660	13 -29 33 -40 10 17 25 -3 0 -26 0	18 -15 25 -6 2 -4 20 -8 -13 -19 0	169 841 1089 1600 100 289 625 9 0 676 5398	324 225 625 36 4 16 400 64 169 361 2224	234 435 825 240 20 -68 500 24 0 494 2704

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-

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

Sol.

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 Y are uncorrelated random variables, the of correlation between and

Solution.

Let and

Then

Now

Similarly

Now

Also

(As and are not correlated, we have )

Similarly

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