Question

NCERT Solutions Class 11, Economics, Statistics for Economics, Chapter- 6, Correlation

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In these NCERT Solutions for Class 11 Economics, we have discussed all types of NCERT intext questions and exercise questions.

Concepts covered in Class 11 Economics, Statistics for Economics, Chapter- 6 Correlation, are-

• Introduction
• What does Correlation measure?
• Types of relationship
• Types of Correlation
• Scatter Diagram
• Techniques for measuring correlation
• Karl Pearson’s Coefficient of Correlation
• Spearman’s rank correlation
• Calculation of Rank Correlation Coefficient

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Answer 1

NCERT Solutions Class 11, Economics, Statistics for Economics, Chapter- 6, Correlation

1. The unit of correlation coefficient between height in feet and weight in kgs is:

(i) kg/feet
(ii) percentage
(iii) non-existent

Solution:

(iii) non-existent

There is non-existence of correlation between the height in feet and weight in kilograms as both the figures have different measures. It is a situation of no relation. Hence, the unit of correlation is zero.

2. The range of simple correlation coefficient is

(i) 0 to infinity
(ii) minus one to plus one
(iii) minus infinity to infinity

Solution:

(ii) minus one to plus one

The value of correlation coefficient is between –1 and +1.

However, if the value of correlation is not inside this range, it indicates an error of calculation.

3. If r_xy is positive the relation between X and Y is of the type:

(i) When Y increases X increases
(ii) When Y decreases X increases
(iii) When Y increases X does not change

Solution:

(i) When Y increases X increases

When the variables, say X and Y, share a positive correlation which means both X and Y increase simultaneously then the value of rxy is positive.

4. If r_xy = 0 the variable X and Y are:

(i) linearly related
(ii) not linearly related
(iii) independent

Solution:

(ii) not linearly related

If R_xy = 0, the two variables X and Y are not correatled and there is no linear relation between them. It doesn’t mean that X and Y are completely independent, they might have other types of relationships. 

5. Of the following three measures which can measure any type of relationship :

(i) Karl Pearson’s coefficient of correlation
(ii) Spearman’s rank correlation
(iii) Scatter diagram

Solution:

(iii) Scatter diagram

The scatter diagram is not just confined to linear relations. It is a useful technique for visually examining any form of relationship without calculating any mineral value.

6. If precisely measured data are available the simple correlation coefficient is

(i) more accurate than rank correlation coefficient
(ii) less accurate than rank correlation coefficient
(iii) as accurate as the rank correlation coefficient

Solution:

(ii) less accurate than rank correlation coefficient

Generally, all the properties of Karl Pearson’s coefficient of correlation are similar to that of the rank correlation. However, it is slightly less accurate because in rank correlation, ranks are used instead of the full set of observations.

7. Why is r preferred to covariance as a measure of association?

Solution:

Correlation coefficient r is preferred to covariance as a measure of variance because:

(i) The correlation coefficient is independent of scale.

(ii) The value of correlation coefficient (r) lies between –1 and 1.

i.e., –1 ≤ r ≤ 1.

8. Can r lie outside the – 1 and 1 range depending on the type of data?

Solution:

No, the value of r cannot lie outside the range of –1 to 1. If r = –1, there is a perfect negative correlation. If r = 1, there is a perfect positive correlation between the two variables. If the value of r is not within this range, there must be some mistake or error in the calculation.

9. Does correlation imply causation?

Solution:

No, correlation does not imply causation. Correlation measures covariation and not causation. The correlation between two variables does not signify that one variable causes the other. Hence correlation does not measure the cause and effect relationship between them.

10. When is rank correlation more precise than simple correlation coefficient?

Solution:

Rank correlation method is more percise than simple correlation coefficient due to the following reasons:

1. When there is a reason to suspect the variables. Height and weight of people cannot be measured precisely but the people can be easily ranked in terms of height and weight.
2. In case of qualitiative data, it is difficult to quantify qualities such as fairness, honesty, etc. Hence Ranking would be a better alternative for the quantifications of qualities.

11. Does zero correlation mean independence?

Solution:

No, zero correlation does not mean independence. If there is zero correlation, it shows that X and Y are not correlated and there is no linear relationship between the two.

12. Can simple correlation coefficient measure any type of relationship?

Solution:

No, the simple correlation coefficient cannot measure any type of relationship. It can measure only the direction and magnitude of linear relationship between the two variables.

Answer 2

13. Collect the price of five vegetables from your local market every day for a week. Calculate their correlation coefficients. Interpret the result.

Day	Potato (kg)	Cabbage (per kg)	Onion (per kg)	Tomato (per kg)	Peas (per kg)
1	18	30	30	32	20
2	20	28	32	30	22
3	18	35	36	30	22
4	22	32	35	30	20
5	20	32	32	35	20
6	20	32	30	32	22
7	22	35	30	35	21

Solution:

Potato (per kg) (X)	Cabbage (per kg) (Y)	\(\bar x =20(dx = x-\bar x)\)	dx2	\(\bar y = 32 (dy = y-\bar y)\)	dy2	dxdy
20	30	-2	4	-2	4	4
18	30	0	0	-4	16	0
20	28	0	0	0	0	0
18	35	-2	4	3	9	-6
22	32	2	4	0	0	0
20	32	0	0	0	0	0
22	35	2	4	3	9	6
		Σdx = 0	Σdx2 = 16	Σdy = 0	Σdy2 = 38	Σdxdy = 4

 \( r=\frac{N \Sigma d x d y-\Sigma d x \times \Sigma d y}{\sqrt{N \Sigma d x^2-(\Sigma d x)^2} \times \sqrt{N \Sigma d y^2-(\Sigma d y)^2}} \)

\( =\frac{7 \times(4)-4 \times 0}{\sqrt{7 \times 4-(0)^2} \times \sqrt{7 \times 38-(0)^2}}\)

\(=\frac{28-0}{\sqrt{112-0} \times \sqrt{266-0}} \)

\(=\frac{28}{\sqrt{112} \times \sqrt{266}}\)

\( =\frac{28}{10.58 \times 16.30}\)

\(=\frac{28}{172.4}\)

\(=0.162\)

Likewise, we can calculate correlation coefficient between different pairs of vegetables.

14. Measure the height of your classmates. Ask them the height of their benchmate. Calculate the correlation coefficient of these two variables. Interpret the result.

Solution:

\(r=\frac{N \Sigma d x d y-\Sigma d x \times \Sigma d y}{\sqrt{N \Sigma d x^2-(\Sigma d x)^2} \times \sqrt{N \Sigma d y^2-(\Sigma d y)^2}}\)

\(=\frac{7 \times(561)-5 \times(-26)}{\sqrt{7 \times 833-(5)^2} \times \sqrt{7 \times 578-(-26)^2}} \)

\( =\frac{392+130}{\sqrt{5831-(25)} \times \sqrt{40460-(676)}} \)

\( =\frac{4057}{\sqrt{5806)} \times \sqrt{3370}}\)

\( =\frac{4057}{76.19 \times 58.05}\)

\( =\frac{4057}{4422.82}\)

\(=0.91\)

15. List some variables where accurate measurement is difficult.

Solution:

Accurate measurement can be difficult in case of:

1. Qualitative variables such as beauty, intelligence, honesty, etc.
2. Subjective variables such as poverty, development, etc. as they are interpreted differently by different people.

16. Interpret the values of r as 1, –1 and 0.

Solution:

1. r as 1 indicates perfect positive relationship between two variables.
2. r as –1 indicates, that there is perfect negative relationship between two variables.
3. r as 0 indicates that there is a lack of correlation between two variables.

17. Why does rank correlation coefficient differ from Pearson's correlation coefficient?

Solution:

Following are the reasons due to which rank correlation coefficient differs from Pearson's correlation coefficient:

1. Rank correlation coefficient is used to measure the linear relationship between the qualitative variables whereas Pearson's correlation coefficient measures the linear relationship between the quantitative variables.
2. Rank correlation coefficient is generally lower or equal to Pearson's coefficient.
3. The rank correlation coefficient is more accurate and reliable, if extreme values are given in the data.

Answer 3

18. Calculate the correlation coefficient between the heights of fathers in inches (X) and their sons (Y):

X	65	66	57	67	68	69	70	72
Y	67	56	65	68	72	72	69	71

Solution:

X	Y	XY	X2	Y2
65	67	4355	4225	4489
66	56	3696	4356	3136
57	65	3705	3249	4225
67	68	4556	4489	4624
68	72	4896	4624	5184
69	72	4968	4761	5184
70	69	4830	4900	4761
72	71	5112	5184	5041
ΣX = 534	ΣY = 1540	ΣXY = 36118	ΣX2 = 35788	ΣY2 = 36644

 \( r=\frac{\Sigma X Y-\frac{(\Sigma X)(\Sigma Y)}{N}}{\sqrt{\Sigma X^2-\frac{(\Sigma X)^2}{N}} \sqrt{\Sigma Y^2-\frac{(\Sigma Y)^2}{N}}}\)

\(=\frac{36118-\frac{(534) \times(540)}{8}}{\sqrt{35788-\frac{(534)^2}{8}} \sqrt{36644-\frac{(540)^2}{8}}} \)

\(r=\frac{36118-36045}{\sqrt{35788-35644.5} \sqrt{36644-36450}}\)

\(r=\frac{73}{\sqrt{143.5} \sqrt{194}}\)

\(r=\frac{73}{11.93 \times 13.93}\)

\(r=\frac{73}{166.88} \)

\( r=0.437\)

19. Calculate the correlation coefficient between X and Y and comment on their relationship:

X	– 3	– 2	– 1	1	2	3
Y	9	4	1	1	4	9

Solution:

X	Y	XY	X2	Y2
-3	9	-27	9	81
-2	4	-8	4	16
-1	1	-1	1	1
1	1	1	1	1
1	1	1	1	1
1	1	1	1	1
1	2	2	1	4
3	9	27	9	81
ΣX = 0	ΣY = 28	ΣXY = 0	ΣX2 = 28	ΣY2 = 196

\( r=\frac{\Sigma X Y-\frac{(\Sigma X)(\Sigma Y)}{N}}{\sqrt{\Sigma X^2-\frac{(\Sigma X)^2}{N}} \sqrt{\Sigma Y^2-\frac{(\Sigma Y)^2}{N}}}\) 

\(=\frac{0-\frac{(0) \times(28)}{6}}{\sqrt{28-\frac{(0)^2}{6}} \sqrt{196-\frac{(28)^2}{6}}} \)

\(=\frac{0}{\sqrt{28-0} \sqrt{196-31}}\)

\(=0\)

There is non-linear correlation between the two variables as y = x². So, in this question, there is a failure to find the correct relationship between these two variables. 

Answer 4

20. Calculate the correlation coefficient between X and Y and comment on their relationship:

X	1	3	4	5	7	8
Y	2	6	8	10	14	16

Solution:

X	Y	XY	X2	Y2
1	2	2	1	4
3	6	18	9	36
4	8	32	16	64
5	10	50	25	100
7	14	98	49	196
8	16	128	64	256
ΣX = 28	ΣY = 56	ΣXY = 328	ΣX2 = 164	ΣY2 = 656

 \(r=\frac{\Sigma X Y-\frac{(\Sigma X)(\Sigma Y)}{N}}{\sqrt{\Sigma X^2-\frac{(\Sigma X)^2}{N}} \sqrt{\Sigma Y^2-\frac{(\Sigma Y)^2}{N}}}\)

\(=\frac{328-\frac{(288 \times) \times(56)}{6}}{\sqrt{164-\frac{784}{6}} \sqrt{656-\frac{3136}{6}}} \)

\(r=\frac{328-261}{\sqrt{33} \sqrt{133}} \)

\( r=\frac{67}{5.74 \times 11.5}\)

\( r=\frac{67}{66}\)

\( r=1 .01\)

or \(r=1\)

As, the correlation coefficient is +1 between two variables. So, we can say that the variables are perfectly positively corrected.