ML Aggarwal Class 12 Maths Solutions Section A Chapter 12 Correlation And Regression

12.1 Introduction

Up to now, we have focused on one set of observations or measurements — for example, heights of students in a class, exam marks of students, weekly wages of workers, and so on. Now we move into the study of joint distributions — two or more sets of observations or measurements collected on the same sample — and the relationships that exist between them, such as the link between children's heights and weights, the connection between income and expenditure, and the relationship between demand and supply.

If (x, y) are pairs of measurements, we say that the two measurements have a statistical relationship if knowing the value of one member x in any pair helps us estimate the second member y more accurately.

Simple Correlation refers to the degree and direction of similarity in how two variables change together across corresponding pairs of observations.

The link between two series, or Correlation, has the following key aspects:
(i) checking if a relationship is present, and measuring its strength or size.
(ii) determining the direction of the link (positive or negative); for example, spending rises as income goes up; demand for a product falls as its cost increases.
(iii) determining whether the relationship is meaningful or just due to chance.
(iv) finding a cause-and-effect link, if one exists.

12.2 Scatter Diagrams

When pairs (x, y) of joint observations from N samples are shown as points on the X and Y axes of a plane, we get a scatter plot, or scatter diagram. These plots give a clear visual sense of how two variables relate to each other.

For example, consider a scatter diagram for the observations (1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1).

From this scatter diagram, we can see that there is a perfect negative relationship between X and Y. As X increases, Y decreases in a perfectly linear manner.

12.3 Covariance of X and Y

When we plot scatter diagrams with the arithmetic means \( (\bar{x}, \bar{y}) \) as the origin, we see patterns that help us understand the relationship:

In the first case (independent variables), the products \( (x - \bar{x})(y - \bar{y}) \) turn out to be positive in the first and third quadrants, and negative in the second and fourth quadrants, resulting in \( \Sigma(x - \bar{x})(y - \bar{y}) = 0 \).

In the second case (strong positive relationship), most pairs lie in the first and third quadrants, so \( \Sigma(x - \bar{x})(y - \bar{y}) \) produces a high positive value. Similarly, in the third case (moderate positive relationship), \( \Sigma(x - \bar{x})(y - \bar{y}) \) gives a moderate positive result. In the fourth case (moderate negative relationship), more pairs sit in the second and fourth quadrants, so \( \Sigma(x - \bar{x})(y - \bar{y}) \) yields a moderate negative value. This analysis leads us to define covariance:

Definition: The mean of the product of deviation scores \( (x_i - \bar{x}) \) and \( (y_i - \bar{y}) \) is called the covariance of X and Y, that is,
\[ \text{Cov}(X, Y) \text{ or } C_{XY} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{N} \text{ ...(1)} \]

It is straightforward to show that
\[ \text{Cov}(X, Y) = \frac{1}{N}\left[\Sigma xy - \frac{1}{N}\Sigma x \Sigma y\right] \text{ ...(2)} \]

When x and y are small numbers, it is simpler to compute Cov(X, Y) using formula (2). When \( x - \bar{x} \) and \( y - \bar{y} \) are small whole numbers, formula (1) is preferred. Otherwise, we can pick assumed means A and B, use \( u = x - A \) and \( v = y - B \), then
\[ \text{Cov}(X, Y) = \frac{1}{N}\left[\Sigma uv - \frac{1}{N}\Sigma u \Sigma v\right] \text{ ...(3)} \]

Illustrative Examples

Example 1. Find Cov(X, Y) for the following data:
X: 3, 4, 5, 6, 7
Y: 8, 7, 6, 5, 4
Solution: Here N = 5. We build the following table:

Here \( \Sigma x = 25 \), \( \Sigma y = 30 \), \( \Sigma xy = 140 \)
\[ \therefore \text{Cov}(X, Y) = \frac{1}{5}\left[140 - \frac{1}{5}(25)(30)\right] = \frac{1}{5}[140 - 150] = -2 \]
Therefore, we see a negative relationship between X and Y.

Example 2. Compute Cov(X, Y) for the following pairs of observations:
(15, 44), (20, 43), (25, 45), (30, 37), (40, 34), (50, 37)
Solution: Here
\[ \bar{x} = \text{mean of X values} = \frac{15 + 20 + 25 + 30 + 40 + 50}{6} = \frac{180}{6} = 30 \]
\[ \bar{y} = \frac{44 + 43 + 45 + 37 + 34 + 37}{6} = \frac{240}{6} = 40 \]
We build the following table:

So \[ \text{Cov}(X, Y) = \frac{1}{N}\Sigma(x - \bar{x})(y - \bar{y}) = \frac{1}{6}(-235) = -39.17 \]

Example 3. Calculate the covariance for the following bivariate data:
X: 11, 12, 13, 14, 15, 17, 18, 19, 20, 21
Y: 14, 8, 12, 21, 19, 19, 23, 22, 17, 25
Solution: Assume mean of X is A = 16, and for Y is B = 19. We build this table:

\[ \text{Cov}(X, Y) = \frac{1}{N}\left[\Sigma uv - \frac{1}{N}\Sigma u \Sigma v\right] = \frac{1}{10}\left[125 - \frac{1}{10}(0)(-10)\right] = 12.5 \]

Exercise 12.1

Question 1. Find the covariance of the data given below:
(1, 5), (2, 7), (3, 9), (4, 11), (5, 10), (6, 9), (7, 8), (8, 7), (9, 6), (10, 5)
Answer: - 1.35
In simple words: To find covariance, calculate the mean of each variable, find how far each point is from its mean, multiply those differences together, and then take the average of all those products. A negative result means as one variable goes up, the other tends to go down.

Question 2. Calculate the covariance of the following bivariate data:
X: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Y: 78, 72, 66, 60, 54, 48, 42, 36, 30, 24, 18, 12
Answer: - 71.5
In simple words: This data shows a strong negative link - as X increases steadily, Y decreases steadily. The negative covariance value confirms this opposite movement pattern.

Question 3. Calculate the covariance of observations (3, 5), (6, 7), (9, 9), (12, 11), (15, 13), (18, 15), (21, 17), (24, 19) using assumed means A = 13 and B = 12.
Answer: 31.5
In simple words: Using assumed means makes the arithmetic simpler. We subtract the assumed means from each value, multiply the results, and average them. The positive covariance shows that X and Y move together in the same direction.

Question 4. Calculate covariance for the following data:
X: 1, 2, 3, 4, 6, 7, 8, 9
Y: 16, 9, 4, 1, 1, 4, 9, 16
Answer: 0
In simple words: A covariance of zero means the two variables have no linear connection - they don't move together in any consistent direction. However, they may have other types of relationships, like a curved one.

Question 5. Prove that though covariance is independent of the choice of origin, it depends upon the scale. If u = ax + b, v = cy + d, show that cov(u, v) = a.c. cov(x, y).
Answer: When we shift the origin (add or subtract a constant - that is b and d), covariance remains the same because deviations from the mean don't change. However, when we change the scale by multiplying by constants a and c, the covariance gets multiplied by a times c. This shows that covariance depends on the units we measure with, not on where we start counting from.
In simple words: Moving the starting point doesn't affect covariance, but changing the size of the units does. If you measure in centimeters instead of meters, the covariance value changes.

12.4 Karl Pearson's Coefficient of Correlation

Although covariance is independent of the choice of origin, it relies on the measurement scale. To make it independent of both origin and scale, we use Karl Pearson's coefficient of correlation (also called Product Moment Correlation):
\[ r \text{ or } \rho(x, y) = \frac{\text{Cov}(x, y)}{\sqrt{\text{Var } x} \sqrt{\text{Var } y}} = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} \]

It is simple to show that when \( u = ax + b \) and \( v = cy + d \), then
\[ \rho(u, v) = \frac{\text{Cov}(u, v)}{\sigma_u \sigma_v} = \frac{\Sigma(ax - ax)(cy - cy)}{N\sqrt{a^2(x - \bar{x})^2}\sqrt{c^2(y - \bar{y})^2}} \]
\[ = \frac{ac \text{Cov}(x, y)}{|ab| \sigma_x \sigma_y} = \pm\frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} = \pm \rho(x, y) \]
Therefore, the coefficient of correlation is independent of both choice of origin and scale. Note that \( -1 \leq r \leq 1 \) (the complete proof is beyond our scope).

When \( x - \bar{x} \) and \( y - \bar{y} \) are small whole numbers, we use
\[ r = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\sqrt{\Sigma(x - \bar{x})^2}\sqrt{\Sigma(y - \bar{y})^2}} \text{ ...(1)} \]

When x and y are small numbers, we use
\[ r = \frac{\Sigma xy - \frac{1}{N}\Sigma x \Sigma y}{\sqrt{\Sigma x^2 - \frac{1}{N}(\Sigma x)^2}\sqrt{\Sigma y^2 - \frac{1}{N}(\Sigma y)^2}} \text{ ...(2)} \]

Otherwise, we use assumed means A and B with \( u = x - A \) and \( v = y - B \):
\[ r = \frac{\Sigma uv - \frac{1}{N}\Sigma u \Sigma v}{\sqrt{\Sigma u^2 - \frac{1}{N}(\Sigma u)^2}\sqrt{\Sigma v^2 - \frac{1}{N}(\Sigma v)^2}} \text{ ...(3)} \]

Some remarks regarding coefficient of correlation

1. The square of r, denoted r², is called the coefficient of determination. Clearly, \( 0 \leq r^2 \leq 1 \). The variation between X and Y is indicated by r², not r alone. For example, if r = 0.9, there is a strong positive link between X and Y, but r² = (0.9)² = 0.81, which means only 81 percent of the change in Y can be explained by changes in X.

2. Correlation is considered to be of high degree if \( \frac{3}{4} \leq |r| \leq 1 \), of moderate degree if \( \frac{1}{4} \leq |r| < \frac{3}{4} \), and of low degree if \( 0 \leq |r| < \frac{1}{4} \).

3. When X and Y are independent variables, cov(X, Y) = 0 and the coefficient of correlation r = 0. Conversely, if r = 0, then X and Y have no linear link. However, Y might still have a non-linear connection with X. For instance, with observations ( - 4, 16), ( - 3, 9), ( - 2, 4), ( - 1, 1), (1, 1), (2, 4), (3, 9), (4, 16), we find that r = 0. However, we can also see that Y = X². Therefore, even though r = 0, we can still correctly estimate Y if we know X.

4. Researchers and advertisers often misuse the correlation coefficient. It may or may not show a cause-and-effect link. For example, in any school, there is typically a strong positive link between children's shoe sizes and spelling ability. Does this mean that larger feet lead to better brains or that learning to spell improves foot growth? More likely, a third factor - the age of the children - affects both factors.

Example 1. Find Karl Pearson's coefficient of correlation between X and Y for the following data:
X: 5, 4, 3, 2, 1
Y: 4, 2, 10, 8, 6
Solution: Here N = 5, and X and Y are small numbers. So we use formula (2). We build this table:

\[ r = \frac{\Sigma xy - \frac{1}{N}\Sigma x \Sigma y}{\sqrt{\Sigma x^2 - \frac{1}{N}(\Sigma x)^2}\sqrt{\Sigma y^2 - \frac{1}{N}(\Sigma y)^2}} = \frac{80 - \frac{1}{5}(15)(30)}{\sqrt{55 - \frac{1}{5}(15)^2}\sqrt{220 - \frac{1}{5}(30)^2}} \]
\[ = \frac{80 - 90}{\sqrt{55 - 45}\sqrt{220 - 180}} = \frac{-10}{\sqrt{10}\sqrt{40}} = \frac{-10}{20} = -0.5 \]

Example 2. Calculate coefficient of correlation from the following data:
X: 12, 13, 14, 15, 16, 17, 18
Y: 14, 17, 18, 19, 20, 24, 28
Solution: Here \( \bar{X} = \frac{\Sigma X}{N} = \frac{105}{7} = 15 \), \( \bar{Y} = \frac{\Sigma Y}{N} = \frac{140}{7} = 20 \). Also, \( X - \bar{X} \) and \( Y - \bar{Y} \) are small numbers, so we use formula (1). We build this table:

\[ r = \frac{\Sigma(X - \bar{X})(Y - \bar{Y})}{\sqrt{\Sigma(X - \bar{X})^2}\sqrt{\Sigma(Y - \bar{Y})^2}} = \frac{58}{\sqrt{28}\sqrt{130}} = \frac{58}{\sqrt{3640}} = 0.961 \]

Example 3. Find Karl Pearson's coefficient of correlation between x and y for the following data:
x: 16, 18, 21, 20, 22, 26, 27, 15
y: 22, 25, 24, 26, 25, 30, 33, 18
Solution: Assume mean A = 20 for the x-variate and B = 25 for y-variate, and we use formula (3). We build this table:

Hence, \[ \rho(X, Y) = \frac{\Sigma uv - \frac{1}{N}\Sigma u \Sigma v}{\sqrt{\Sigma u^2 - \frac{1}{N}(\Sigma u)^2}\sqrt{\Sigma v^2 - \frac{1}{N}(\Sigma v)^2}} = \frac{152 - \frac{1}{8}(5)(-1)}{\sqrt{135 - \frac{1}{8}(5)^2}\sqrt{221 - \frac{1}{8}(-1)^2}} \]
\[ = \frac{152.625}{\sqrt{135 - 3.125}\sqrt{221 - 0.125}} = \frac{152.625}{\sqrt{131.875}\sqrt{220.875}} = 0.894 \]

Example 4. Find the correlation coefficient between the heights of husbands and wives based on the following data (given in inches) and interpret the result.
Solution: We use assumed means A = 70 and B = 66, and we use formula (3).

\[ r = \frac{\Sigma uv - \frac{1}{N}\Sigma u \Sigma v}{\sqrt{\Sigma u^2 - \frac{1}{N}(\Sigma u)^2}\sqrt{\Sigma v^2 - \frac{1}{N}(\Sigma v)^2}} = \frac{140 - \frac{1}{15}(0)(5)}{\sqrt{194 - \frac{1}{15}(0)^2}\sqrt{127 - \frac{1}{15}(5)^2}} \]
\[ = \frac{140}{\sqrt{194}\sqrt{127 - 1.67}} = 0.89 \]
This represents a strong positive correlation. Tall men tend to marry tall women, and short men tend to marry short women - a phenomenon known as assortive mating.

Example 5. The following table shows the percentage of cats killed while falling from various storeys from skyscrapers in New York. Calculate correlation coefficient and draw scatter diagram. Comment on the results.
Fallen from number of storeys: 1, 2, 3, 4, 5, 6, 7, 8, 9
Percentage killed: 3, 9, 12, 15, 18, 15, 12, 9, 3
Solution: We use assumed means A = 5 and B = 12. We build this table:

\[ r = \frac{\Sigma uv - \frac{1}{N}\Sigma u \Sigma v}{\sqrt{\Sigma u^2 - \frac{1}{N}(\Sigma u)^2}\sqrt{\Sigma v^2 - \frac{1}{N}(\Sigma v)^2}} = \frac{0 - \frac{1}{9}(0)(-12)}{\sqrt{60 - \frac{1}{9}(0)^2}\sqrt{234 - \frac{1}{9}(-12)^2}} = 0 \]
since both \( \Sigma uv = 0 \) and \( \Sigma u = 0 \).

Although r = 0 indicates no linear link between X and Y, the scatter diagram clearly demonstrates a non-linear connection. This shows that up to the 5th storey, the percentage of cats being killed increases, but after that it starts falling, so that only 3 percent are killed when falling from the 9th storey. Cats have remarkably flexible bodies - when they drop from higher storeys, they get sufficient time to stretch their bodies like parachutes, which protects them from being killed - even from breaking their legs.

Example 6. A student while calculating correlation coefficient between two variables x and y for 25 pairs of observations obtained the following results:
\( \Sigma x = 125 \), \( \Sigma x^2 = 650 \), \( \Sigma y = 100 \), \( \Sigma y^2 = 460 \), \( \Sigma xy = 508 \).
On rechecking, it was found that he had wrongly copied two pairs as (6, 14) and (8, 6) whereas the correct values were (8, 12) and (6, 8). Calculate the correct correlation coefficient between x and y.
Solution:
Correct \( \Sigma x = 125 - 6 - 8 + 8 + 6 = 125 \)
Correct \( \Sigma y = 100 - 14 - 6 + 12 + 8 = 100 \)
Correct \( \Sigma x^2 = 650 - 6^2 - 8^2 + 8^2 + 6^2 = 650 \)
Correct \( \Sigma y^2 = 460 - 14^2 - 6^2 + 12^2 + 8^2 = 460 - 196 - 36 + 144 + 64 = 436 \)
Correct \( \Sigma xy = 508 - (6)(14) - (8)(6) + (8)(12) + (6)(8) = 508 - 84 - 48 + 96 + 48 = 520 \).

\[ r = \frac{\Sigma xy - \frac{1}{n}\Sigma x \Sigma y}{\sqrt{\Sigma x^2 - \frac{1}{n}(\Sigma x)^2}\sqrt{\Sigma y^2 - \frac{1}{n}(\Sigma y)^2}} = \frac{520 - \frac{1}{25}(125)(100)}{\sqrt{650 - \frac{1}{25}(125)^2}\sqrt{436 - \frac{1}{25}(100)^2}} \]
\[ = \frac{520 - 500}{\sqrt{650 - 625}\sqrt{436 - 400}} = \frac{20}{\sqrt{25}\sqrt{36}} = \frac{20}{5 \times 6} = \frac{20}{30} = 0.667 \]

Exercise 12.2

Question 1. Find ρ(x, y) if cov(x, y) = - 16.5, var(x) = 2.25 and var(y) = 144.
Answer: r = - 0.92
In simple words: Use the formula r = cov(x, y) divided by the square root of var(x) times var(y). This converts covariance into a standardized measure that always lies between - 1 and 1.

Question 2. If n = 10, Σ x = 26, Σ y = - 27, Σ x² = 226, Σ y² = 267, Σ xy = 7, find correlation coefficient.
Answer: r = 0.44
In simple words: When you have the raw sums, use formula (2) which uses these totals directly without needing the deviations from the mean. This method is faster for computational work.

Question 3. For the observations (1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12), (7, 14), (8, 16), (9, 18), (10, 20), calculate cov(X, Y) and ρ(X, Y). Also make scatter diagram. Interpret the result.
Answer: Cov(X, Y) = 16.5 and r = 1, which shows perfect positive linear relationship. This means every rise in X matches exactly with an equal proportional rise in Y - the points lie perfectly on a straight line going upward from left to right.
In simple words: When r = 1, the two variables have a perfect positive link. If you know X, you can predict Y with total accuracy using a simple formula like Y = 2X.

Question 4. Calculate the coefficient of correlation between X and Y from the following data using Karl Pearson's method:
X: 1, 2, 3, 4, 5
Y: 2, 5, 3, 8, 7
Answer: 0.806
In simple words: This shows a strong positive correlation. Although the numbers don't follow a perfect pattern, there is a clear tendency for Y to rise as X rises, though with some variation and scatter.

Question 5. Compute Karl Pearson's Coefficient of Correlation between sales and expenditures of a firm for six months.
Sales (in lakh of Rs.): 18, 20, 27, 20, 21, 29
Expenditure (in lakh of Rs.): 23, 27, 28, 28, 29, 30
Answer: 0.68 (approximately)
In simple words: A correlation of about 0.68 means there is a moderate to fairly strong positive link between sales and spending. As sales go up, expenditure also tends to climb, but the relationship is not perfectly tight.

Question 6. The coefficient of correlation between two variables X and Y is 0.64. Their covariance is 16. The variance of X is 9. Find the standard deviation of Y-series.
Answer: 8.33
In simple words: Rearrange the correlation formula to solve for the standard deviation of Y. Since r = cov(X, Y) divided by (SD of X times SD of Y), you can find the missing standard deviation.

Question 7. Find the covariance and the coefficient of correlation between x and y when n = 10, Σ x = 60, Σ y = 60, Σ x² = 400, Σ y² = 580 and Σ xy = 305.
Answer: - 5.5, - 0.5863
In simple words: The negative covariance and negative correlation both show that as x increases, y tends to decrease. The moderate strength of this negative link (around - 0.59) means the relationship exists but with some scatter.

Question 8. Calculate correlation coefficient from the following results:
N = 10, ΣX = 100, ΣY = 150, Σ(X - 10)² = 180, Σ(Y - 15)² = 215, Σ(X - 10) (Y - 15) = 60
Answer: 0.305
In simple words: When deviations from specific values (not necessarily the means) are given, use formula (1) adapted for these deviations. The low to moderate correlation of 0.305 shows a weak positive link.

Question 9. Coefficient of correlation between X and Y for 50 observations is 0.3, mean of X is 10 and that of Y is 6, S.D. of X is 3 and that of Y is 2. Later it was discovered that one pair of values (10, 6) was inaccurate and hence weeded out. Calculate the correlation between remaining 49 pairs of values.
Answer: 0.3 (no effect)
In simple words: The removed pair (10, 6) happened to exactly match the means of both variables. Removing a point at this location doesn't change the overall correlation of the remaining 49 observations.

Question 10. Calculate Karl Pearson's coefficient of correlation between the marks in English and Mathematics obtained by 10 students:
English: 20, 13, 18, 21, 11, 12, 17, 14, 19, 15
Mathematics: 17, 12, 23, 25, 14, 8, 19, 21, 22, 19
Answer: 0.748
In simple words: This strong positive correlation (about 0.75) suggests that students who score well in English also tend to score well in Mathematics. This points to an underlying link between language skills and mathematical ability.

Question 11. Find Karl Pearson's coefficient of correlation between X and Y for the following data:
X: 16, 18, 21, 20, 22, 26, 27, 15
Y: 22, 25, 24, 26, 25, 30, 33, 14
Answer: 0.894
In simple words: With a correlation near 0.9, this is a very strong positive link. The two variables move together quite closely, though not perfectly.

Question 12. From the following table, calculate the Karl Pearson's coefficient of correlation.
X: 6, 2, 10, 4, 8
Y: 9, 11, ?, 8, 7
Arithmetic means of X and Y series are 6 and 8 respectively.
Answer: - 0.919
In simple words: First, use the given means to find the missing Y value. Since the mean of Y is 8 and we have 5 values, we can calculate what the unknown entry must be. Then compute correlation normally.

Question 13. The weights of sons and fathers (in kilograms) are given below:
Weight of father: 65, 66, 67, 67, 68, 69, 70, 72
Weight of son: 67, 68, 65, 68, 72, 72, 69, 71
Find the coefficient of correlation.
Answer: r = 0.603, which shows a moderate but significant relationship between weights of fathers and sons.
In simple words: A correlation of about 0.6 suggests that sons' weights do follow their fathers' weights to some extent, but genetics and other factors create variation - heavier fathers tend to have heavier sons, but not always.

Question 14. Calculate Karl Pearson's coefficient of correlation from the following data and interpret the result:
Serial number of student: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Marks in mathematics: 15, 18, 21, 24, 27, 30, 36, 39, 42, 48
Marks in statistics: 25, 25, 27, 27, 31, 33, 35, 41, 41, 45
Answer: r = 0.98, which shows a strong positive relationship, which points to application of common skills/intelligence factor. But it does not mean that if you study only mathematics and score good marks in it, you will automatically score well in statistics. So take care!
In simple words: A very high correlation (0.98) between math and statistics marks shows they are strongly linked. However, correlation does not prove causation - high intelligence helps both subjects, but studying math alone won't automatically make you good at statistics.

Question 15. Calculate Karl Pearson's coefficient of correlation between x and y for the following data:
X: 6, 2, 4, 9, 1, 3, 5, 8
Y: 13, 8, 12, 15, 9, 10, 11, 16
[Take assumed mean for x as 5 and for y as 12.]
Answer: r = 0.946
In simple words: With a correlation of about 0.95, this is very close to a perfect positive link. The variables are strongly connected, though there is still slight scatter around the best-fit line.

Question 16. Calculate ρ(X, Y) for the following data and comment on the result.
X: - 4, - 3, - 2, - 1, 1, 2, 3, 4
Y: 16, 9, 4, 1, 1, 4, 9, 16
Answer: r = 0, which shows lack of linear relationship, though we see that Y = X².
In simple words: A correlation of zero doesn't always mean the variables are unrelated. Here Y is perfectly determined by X (Y equals X squared), but the relationship is curved, not linear. Pearson's r only measures straight-line links.

Question 17. Calculate Karl Pearson's coefficient of correlation between X and Y from the following data and comment on the result.
X: 1, 2, 3, 4, 5
Y: 7, 6, 5, 4, 3
Answer: r = - 1, which shows perfect negative correlation.
In simple words: When r = - 1, the two variables have a perfect negative link lying exactly on a straight line. As X increases by one unit, Y decreases by a constant amount every time, with no exceptions.

Question 18. A psychologist selected a random sample of 22 students. He grouped them in 11 pairs so that students in a pair have nearly equal scores in an intelligence test. In each pair one student was taught by method A and the other by method B and examined after the course. The marks obtained by them are tabulated below:

12.5 Lines of Regression

Often we need to make forecasts or estimates of future values. People make bets on cricket matches based on past performance; companies estimate (forecast) sales based on historical information, and so on. Given information pairs \( (x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n) \), we can draw a scatter plot and then imagine a smooth curve that best fits this data. This practice is known as curve fitting.