# NCERT Solutions for Class 11 Economics Chapter 6 - Correlation

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**10. The unit of correlation coefficient between height in feet and weight in kgs is :****(i) kg/feet****(ii) percentage****(iii) non-existent**

**Ans.** (iii) non-existent

**Explanation:**

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.

**11. The range of simple correlation coefficient is****(i) 0 to infinity****(ii) minus one to plus one****(iii) minus infinity to infinity**

**Ans.** (ii) minus one to plus one

**Explanation:**

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

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

**12. 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**

**Ans.** (i) When Y increases X increases

**Explanation:**

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.

**13. If r _{xy} = 0 the variable X and Y are :**

**(i) linearly related**

**(ii) not linearly related**

**(iii) independent**

**Ans.** (ii) not linearly related

**Explanation:**

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.

**14. 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**

**Ans.** (iii) Scatter diagram

**Explanation:**

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.

**15. 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**

**Ans.** (ii) less accurate than rank correlation coefficient

**Explanation:**

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.

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

**Ans.** 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.

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

**Ans.** 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.

**18. Does correlation imply causation?**

**Ans.** 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.

**19. When is rank correlation more precise than simple correlation coefficient?**

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

- 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.
- 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.

**20. Does zero correlation mean independence?**

**Ans.** 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.

**21. Can simple correlation coefficient measure any type of relationship?**

**Ans.** 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.

**22. 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 |

**Ans.**

Potato (per kg) (X) | Cabbage (per kg) (Y) | x̄=20(dx=x- x̄) | dX^{2} |
ȳ=32(dy=Y- ȳ) | dY^{2} |
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 | Σdx^{2} = 16 |
Σdy = 0 | Σdy^{2} = 38 |
Σdxdy = 4 |

$$r =\frac{N\Sigma dxdy-\Sigma dx×\Sigma dy}{\sqrt{N\Sigma dx^2-(\Sigma dx)^2}×\sqrt{N\Sigma dy^2-(\Sigma dy)^2}}\\ =\frac{7×(4)-4×0}{\sqrt{7×4-(0)^2}×\sqrt{7×38-(0)^2}}\\=\frac{28-0}{\sqrt{112-0}×\sqrt{266-0}}\\=\frac{28}{\sqrt{112}×\sqrt{266}}\\=\frac{28}{10.58×16.30}\\=\frac{28}{172.4}=0.162\\\text{Likewise, we can calculate correlation coefficient between different pairs of vegetables.}$$

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

**Ans .**

$$r=\frac{N\Sigma dxdy-\Sigma dx×\Sigma dy}{\sqrt{N\Sigma dx^2-(\Sigma dx)^2}×\sqrt{N\Sigma dy^2-(\Sigma dy)^2}}\\=\frac{7×(561)-5×(-26)}{\sqrt{7×833-(5)^2}×\sqrt{7×578-(-26)^2}}\\=\frac{3927+130} {\sqrt{5831-(25)}×\sqrt{40460-(676)}}\\=\frac{4057}{\sqrt{5806)}×\sqrt{3370}}\\=\frac{4057}{76.19×58.05}\\=\frac{4057}{442282}=0.91.$$

**24. List some variables where accurate measurement is difficult.**

**Ans.** Accurate measurement can be difficult in case of:

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

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

**Ans. **

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

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

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

- 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.
- Rank correlation coefficient is generally lower or equal to Pearson's coefficient.
- The rank correlation coefficient is more accurate and reliable, if extreme values are given in the data.

**27. 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 |

**Ans.**

X | Y | XY | X^{2} |
Y^{2} |

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 | ΣX^{2} = 35788 |
ΣY^{2} = 36644 |

$$r=\frac{\Sigma XY-\frac{(\Sigma X)(\Sigma Y)}{N}}{\sqrt{\Sigma X^2-\frac{(\Sigma X)^2}{N}}{\sqrt{\Sigma Y^2-\frac{(\Sigma Y)^2}{N}}}}\\=\frac{36118-\frac{(534)×(540)}{8}}{\sqrt{35788-\frac{(534)^2}{8}}{\sqrt{36644-\frac{(540)^2}{8}}}}\\r=\frac{36118-36045}{\sqrt{35788-35644.5}\sqrt{36644-36450}}\\r=\frac{73}{\sqrt{143.5}\sqrt{194}}\\r=\frac{73}{11.93×13.93}\\=\frac{73}{166.88}\\r=0.437$$

**28. 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 |

**Ans. **

X | Y | XY | X^{2} |
Y^{2} |

-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 | ΣX^{2} = 28 |
ΣY^{2} = 196 |

$$r=\frac{\Sigma XY-\frac{(\Sigma X)(\Sigma Y)}{N}}{\sqrt{\Sigma X^2-\frac{(\Sigma X)^2}{N}}{\sqrt{\Sigma Y^2-\frac{(\Sigma Y)^2}{N}}}}\\=\frac{0-\frac{(0)×(28)}{6}}{\sqrt{28-\frac{(0)^2}{6}}{\sqrt{196-\frac{(28)^2}{6}}}}\\r=\frac{0}{\sqrt{28-0}\sqrt{196-31}}\\r=0$$

There is non-linear correlation between the two variables as y = x^{2}. So, in this question, there is a failure to find the correct relationship between these two variables.

**29. 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 |

**Ans.**

X | Y | XY | X^{2} |
Y^{2} |

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 | ΣX^{2} = 164 |
ΣY^{2} = 656 |

$$r=\frac{\Sigma XY-\frac{(\Sigma X)(\Sigma Y)}{N}}{\sqrt{\Sigma X^2-\frac{(\Sigma X)^2}{N}}{\sqrt{\Sigma Y^2-\frac{(\Sigma Y)^2}{N}}}}\\=\frac{328-\frac{(28)×(56)}{6}}{\sqrt{164-\frac{784}{6}}{\sqrt{656-\frac{3136}{6}}}}\\r=\frac{328-261}{\sqrt{33}\sqrt{133}}\\r=\frac{67}{5.14×11.5}=\frac{67}{66.01 \space or\space 66}\\r=\frac{67}{66}\\r = 1·01 \\Or\\r = 1$$

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