undefined | unit 2 correlation analysis

Unit – II

Correlation Analysis

Q1) Explain Scattered diagram.

A1)

Scatter diagram method 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”.

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

Scatter diagram-1

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

Scatter diagram-2

c) High Degree of +Ve Correlation (r= + High): all the points plotted close to the straight line rising from left to right.

Scatter diagram-3

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

Scatter diagram-4

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

Scatter diagram-5

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

Scatter diagram-6

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

Scatter diagram-7

Q2) Compute Pearsons coefficient of correlation between advertisement cost and sales as per the data given below:

Advertisement cost	39	65	62	90	82	75	25	98	36	78
sales	47	53	58	86	62	68	60	91	51	84

A2)

X	Y	X - X	(X - X)2	Y – Y	(Y - Y)2
39	47	-26	676	-19	361	494
65	53	0	0	-13	169	0
62	58	-3	9	-8	64	24
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

Q3) Find Spearman's rank correlation coefficient between X and Y for this set of data:

X	13	20	22	18	19	11	10	15
Y	17	19	23	16	20	10	11	18

A3)

X	Y	Rank X	Rank Y	d	d2
13	17	3	4	-1	1
20	19	7	6	1	1
22	23	8	8	0	0
18	16	5	3	2	2
19	20	6	7	-1	1
11	10	2	1	1	1
10	11	1	2	-1	1
15	18	4	5	-1	1
					8

R =

R = 1 – 6*8/8(82 – 1) = 1 – 48 = 0.90

504

Q4) Calculation of equal ranks or tie ranks.

Find Spearman's rank correlation coefficient:

Commerce	15	20	28	12	40	60	20	80
Science	40	30	50	30	20	10	30	60

A4)

C	S	Rank C	Rank S	d	d2
15	40	2	6	-4	16
20	30	3.5	4	-0.5	0.25
28	50	5	7	-2	4
12	30	1	4	-3	9
40	20	6	2	4	16
60	10	7	1	6	36
20	30	3.5	4	-0.5	0.25
80	60	8	8	0	0
					81.5

R = 1 – (6*81.5)/8(82 – 1) = 0.02

Q5) Difference between correlation and regression.

A5)

Correlation	Regression
‘Correlation’ as the name says it determines the interconnection or a co-relationship between the variables.	‘Regression’ explains how an independent variable is numerically associated with the dependent variable.
In Correlation, both the independent and dependent values have no difference.	However, in Regression, both the dependent and independent variable are different.
The primary objective of Correlation is, to find out a quantitative/numerical value expressing the association between the values.	When it comes to regression, its primary intent is, to reckon the values of a haphazard variable based on the values of the fixed variable.
Correlation stipulates the degree to which both of the variables can move together.	However, regression specifies the effect of the change in the unit, in the known variable(p) on the evaluated variable (q).
Correlation helps to constitute the connection between the two variables.	Regression helps in estimating a variable’s value based on another given value.

Q6) How to find a linear regression equation.

Subject	X	Y
1	43	99
2	21	65
3	25	79
4	42	75
5	57	87
6	59	81

A6)

Subject	X	Y	Xy	X2	Y2
1	43	99	4257	1849	9801
2	21	65	1365	441	4225
3	25	79	1975	625	6241
4	42	75	3150	1764	5625
5	57	87	4959	3249	7569
6	59	81	4779	3481	6521
Total	247	486	20485	11409	40022

To find a and b, use the following equation

Find a:

((486 × 11,409) – ((247 × 20,485)) / 6 (11,409) – 247*247)

484979 / 7445

=65.14

Find b:

(6(20,485) – (247 × 486)) / (6 (11409) – 247*247)

(122,910 – 120,042) / 68,454 – 2472

2,868 / 7,445

= .385225

y’ = a + bx

y’ = 65.14 + .385225x

Q7) Find the two regression equation of X on Y and Y on X from the following data:

X: 10 12 16 11 15 14 20 22

Y: 15 18 23 14 20 17 25 28

A7)

Here N = Number of elements in either series X or series Y = 8

Now we will proceed to compute regression equations using normal equations.

Regression equation of X on Y: X = a + bY

The two normal equations are:

Substituting the values in above normal equations, we get

120 = 8a + 160b ..... (i)

2542 = 160a + 3372b ..... (ii)

Let us solve these equations (i) and (ii) by simultaneous equation method

Multiply equation (i) by 20 we get 2400 = 160a + 3200b

Now rewriting these equations:

2400 = 160a + 3200b

2542 = 160a + 3372b

(-) (-) (-) .

-142 = -172b

Therefore, now we have -142 = -172b, this can re-written as 172b = 142

Now, b = 142/172 = 0.8256 (rounded off)

Substituting the value of b in equation (i), we get

120 = 8a + (160 * 0.8256)

120 = 8a + 132 (rounded off)

8a = 120 - 132

8a = -12

a = -12/8

a = -1.5

Thus, we got the values of a = -1.5 and b = 0.8256

Hence the required regression equation of X on Y:

X = a + bY => X = -1.5 + 0.8256Y

Regression equation of Y on X: Y = a + bX

The two normal equations are:

∑Y = Na + b∑X

∑XY = a∑X + b∑X2

Substituting the values in above normal equations, we get

160 = 8a + 120b ..... (iii)

2542 = 120a + 1926b ..... (iv)

Let us solve these equations (iii) and (iv) by simultaneous equation method

Multiply equation (iii) by 15 we get 2400 = 120a + 1800b

Now rewriting these equations:

2400 = 120a + 1800b

2542 = 120a + 1926b

(-) (-) (-) .

-142 = -126b

Therefore, now we have -142 = -126b, this can re-written as 126b = 142

Now, b = 142/126 = 1.127 (rounded off)

Substituting the value of b in equation (iii), we get

160 = 8a + (120 * 1.127)

160 = 8a + 135.24

8a = 160 - 135.24

8a = 24.76

a = 24.76/8

a = 3.095

Thus, we got the values of a = 3.095 and b = 1.127

Hence the required regression equation of Y on X:

Y = a + bX => Y = 3.095 + 1.127X

Q8) Compute correlation coefficient from the following data

Hours of sleep (X)	Test scores (Y)
8	81
8	80
6	75
5	65
7	91
6	80

A8)

X	Y	X - X	(X - X)2	Y - Y	(Y - Y)2
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

X = 40/6 =6.7

Y = 472/6 = 78.7

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

Thus, Correlation coefficient is positively correlated

Q9) Calculate coefficient of correlation between X and Y series using Karl Pearson shortcut method

X	14	12	14	16	16	17	16	15
Y	13	11	10	15	15	9	14	17

A9)

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

X	Y	dx	dx2	dy	dy2	dxdy
14	13	-1.0	1.0	-1.0	1.0	1.0
12	11	-3.0	9.0	-3.0	9.0	9.0
14	10	-1.0	1.0	-4.0	16.0	4.0
16	15	1.0	1.0	1.0	1.0	1.0
16	15	1.0	1.0	1.0	1.0	1.0
17	9	2.0	4.0	-5.0	25.0	-10.0
16	14	1	1	0	0	0
15	17	0	0	3	9	0
120	104	0	18	-8	62	6

r = 8 *6 – (0)*(-8)

√8*18-(0)2 √8*62 – (-8)2

r = 48/√144*√432 = 0.19

Q10) Calculate coefficient of correlation between X and Y series using Karl pearson shortcut method

X	1800	1900	2000	2100	2200	2300	2400	2500	2600
F	5	5	6	9	7	8	6	8	9

A10)

Assumed mean of X and Y is 2200, 6

X	Y	dx	dx (i=100)	dx2	dy	dy2	dxdy
1800	5	-400	-4	16	-1.0	1.0	4.0
1900	5	-300	-3	9	-1.0	1.0	3.0
2000	6	-200	-2	4	0.0	0.0	0.0
2100	9	-100	-1	1	3.0	9.0	-3.0
2200	7	0	0	0	1.0	1.0	0.0
2300	8	100	1	1	2.0	4.0	2.0
2400	6	200	2	4	0	0	0.0
2500	8	300	3	9	2	4	6.0
2600	9	400	4	16	3	9	12.0

			0	60	9	29	24

Note – we can also proceed dividing x/100

r = (9)(24) – (0)(9)

√9*60-(0)2 √9*29– (9)2

r = 0.69

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