Samacheer Kalvi Class 11 Business Maths Solutions Chapter 9 Correlation and Regression Analysis Ex 9.2

Question 1. From the data given below:

Find
(a) The two regression equations
(b) The coefficient of correlation between marks in Economics and Statistics
(c) The most likely marks in Statistics when the marks in Economics is 30.
Answer:
First, we need to calculate the mean of X (Marks in Economics) and Y (Marks in Statistics) and other sums. Let's create a table to help with the calculations.

We have \(N = 10\). From the table, we get the following sums:
\( \sum X = 320, \sum Y = 380, \sum x^2 = 140, \sum y^2 = 398, \sum xy = -93 \).
Now, let's find the means:
\( \overline{X} = \frac{\sum X}{N} = \frac{320}{10} = 32 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{380}{10} = 38 \)

(a) The two regression equations:
Regression equation of X on Y:
First, calculate the regression coefficient \(b_{xy}\):
\( b_{xy} = \frac{\sum xy}{\sum y^2} \)
\( b_{xy} = \frac{-93}{398} = -0.2336 \)
We can round this to \( -0.234 \).
Now, use the formula for the regression line of X on Y:
\( X - \overline{X} = b_{xy}(Y - \overline{Y}) \)
\( X - 32 = -0.234(Y - 38) \)
\( X = -0.234Y + (-0.234 \times -38) + 32 \)
\( X = -0.234Y + 8.892 + 32 \)
\( X = -0.234Y + 40.892 \)

Regression equation of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{\sum xy}{\sum x^2} \)
\( b_{yx} = \frac{-93}{140} = -0.6642 \)
We can round this to \( -0.664 \).
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 38 = -0.664(X - 32) \)
\( Y = -0.664X + (-0.664 \times -32) + 38 \)
\( Y = -0.664X + 21.248 + 38 \)
\( Y = -0.664X + 59.248 \)

(b) The coefficient of correlation \(r\):
The correlation coefficient \(r\) can be found using the formula \( r = \pm \sqrt{b_{xy} \times b_{yx}} \). We use the negative sign because both regression coefficients (\(b_{xy}\) and \(b_{yx}\)) are negative.
\( r = - \sqrt{(-0.234)(-0.664)} \)
\( r = - \sqrt{0.155376} \)
\( r = -0.394 \)

(c) The most likely marks in Statistics when marks in Economics (X) is 30:
To find the most likely marks in Statistics (Y) when marks in Economics (X) is 30, we use the regression equation of Y on X:
\( Y = -0.664X + 59.248 \)
Substitute \(X = 30\):
\( Y = -0.664(30) + 59.248 \)
\( Y = -19.92 + 59.248 \)
\( Y = 39.328 \).
In simple words: First, we find the average marks for both subjects. Then, we use special formulas to create two lines that show how the marks are related. We also find a number called 'r' that tells us how strongly they are connected. Finally, we use one of the lines to guess the Statistics mark if the Economics mark is 30.

🎯 Exam Tip: Always check the sign of the correlation coefficient \(r\). It must have the same sign as both regression coefficients \(b_{xy}\) and \(b_{yx}\). If one is positive and the other negative, or if the calculated \(r\) has a different sign, there is a calculation error.

Question 2. The heights (in cm.) of a group of fathers and sons are given below.

Find the lines of regression and estimate the height of son when the height of the father is 164 cm.
Answer:
Let X be the height of fathers and Y be the height of sons.
We need to calculate the mean of X and Y, and other sums. Let's create a table for calculations:

We have \( N = 10 \). From the table, we get the following sums:
\( \sum X = 1686, \sum Y = 1690, \sum dx = 6, \sum dy = 0, \sum dx^2 = 410, \sum dy^2 = 446, \sum dxdy = 248 \).
Now, let's find the means using the original sums:
\( \overline{X} = \frac{\sum X}{N} = \frac{1686}{10} = 168.6 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{1690}{10} = 169 \)

The two regression equations:

Regression equation of X on Y:
First, calculate the regression coefficient \(b_{xy}\):
\( b_{xy} = \frac{N \sum dxdy - (\sum dx)(\sum dy)}{N \sum dy^2 - (\sum dy)^2} \)
\( b_{xy} = \frac{10(248) - (6)(0)}{10(446) - (0)^2} \)
\( b_{xy} = \frac{2480 - 0}{4460 - 0} \)
\( b_{xy} = \frac{2480}{4460} = 0.556 \)
Now, use the formula for the regression line of X on Y:
\( X - \overline{X} = b_{xy}(Y - \overline{Y}) \)
\( X - 168.6 = 0.556(Y - 169) \)
\( X = 0.556Y - (0.556 \times 169) + 168.6 \)
\( X = 0.556Y - 93.964 + 168.6 \)
\( X = 0.556Y + 74.636 \)

Regression equation of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{N \sum dxdy - (\sum dx)(\sum dy)}{N \sum dx^2 - (\sum dx)^2} \)
\( b_{yx} = \frac{10(248) - (6)(0)}{10(410) - (6)^2} \)
\( b_{yx} = \frac{2480 - 0}{4100 - 36} \)
\( b_{yx} = \frac{2480}{4064} = 0.610 \)
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 169 = 0.610(X - 168.6) \)
\( Y = 0.610X - (0.610 \times 168.6) + 169 \)
\( Y = 0.610X - 102.846 + 169 \)
\( Y = 0.610X + 66.154 \)

Estimate the height of the son (Y) when the height of the father (X) is 164 cm:
We use the regression equation of Y on X, which is \( Y = 0.610X + 66.154 \).
Substitute \( X = 164 \):
Son's height \( = 0.610 \times 164 + 66.154 \)
\( = 100.04 + 66.154 \)
\( = 169.19 \text{ cm} \)
In simple words: We first calculate the average heights for fathers and sons. Then, we find numbers called regression coefficients, which show how changes in father's height affect son's height and vice versa. Using these, we create two equations that predict one height from the other. Finally, we use the equation for son's height to guess how tall a son would be if his father is 164 cm tall.

🎯 Exam Tip: Remember to use the correct regression equation for estimation. If you need to estimate Y given X, use Y on X. If you need to estimate X given Y, use X on Y.

Question 3. The following data give the height in inches (X) and the weight in lb. (Y) of a random sample of 10 students from a large group of students of age 17 years:

Estimate weight of the student of a height 69 inches.
Answer:
Let X be the height and Y be the weight. We need to estimate Y when X is given, so we will find the regression equation of Y on X. First, calculate the means and other sums for the formula. Let's make a table for this:

We have \( N = 10 \). From the table, we get the following sums:
\( \sum X = 652, \sum Y = 1169, \sum dx = 2, \sum dy = -1, \sum dx^2 = 78, \sum dy^2 = 717, \sum dxdy = 182 \).
Now, let's find the means:
\( \overline{X} = \frac{\sum X}{N} = \frac{652}{10} = 65.2 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{1169}{10} = 116.9 \)

Regression equation of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{N \sum dxdy - (\sum dx)(\sum dy)}{N \sum dx^2 - (\sum dx)^2} \)
\( b_{yx} = \frac{10(182) - (2)(-1)}{10(78) - (2)^2} \)
\( b_{yx} = \frac{1820 - (-2)}{780 - 4} \)
\( b_{yx} = \frac{1822}{776} = 2.3479 \)
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 116.9 = 2.3479(X - 65.2) \)
\( Y = 2.3479X - (2.3479 \times 65.2) + 116.9 \)
\( Y = 2.3479X - 153.08308 + 116.9 \)
\( Y = 2.3479X - 36.18308 \)
(Rounding to two decimal places, \( Y = 2.35X - 36.18 \))

Estimate the weight (Y) when the height (X) is 69 inches:
Using the regression equation \( Y = 2.3479X - 36.18308 \):
Weight \( Y = 2.3479(69) - 36.18308 \)
\( = 161.9951 - 36.18308 \)
\( = 125.81202 \)
\( = 125.81 \text{ lb} \)
In simple words: We find the average height and weight, then use a formula to figure out how height affects weight. This gives us an equation that predicts weight from height. We then use this equation to guess a student's weight if their height is 69 inches.

🎯 Exam Tip: Pay close attention to whether the question asks for Y on X or X on Y regression. Choosing the wrong equation will lead to an incorrect answer.

Question 4. Obtain the two regression lines from the following data N = 20, ΣΧ = 80, ΣΥ = 40, ΣΧ² = 1680, ΣΥ² = 320 and ΣΧΥ = 480.
Answer:
Given data:
\( N = 20, \sum X = 80, \sum Y = 40, \sum X^2 = 1680, \sum Y^2 = 320, \sum XY = 480 \).
First, calculate the means:
\( \overline{X} = \frac{\sum X}{N} = \frac{80}{20} = 4 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{40}{20} = 2 \)

Regression line of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum X^2 - (\sum X)^2} \)
\( b_{yx} = \frac{20(480) - (80)(40)}{20(1680) - (80)^2} \)
\( b_{yx} = \frac{9600 - 3200}{33600 - 6400} \)
\( b_{yx} = \frac{6400}{27200} = 0.23529 \)
Rounding to two decimal places, \( b_{yx} \approx 0.24 \).
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 2 = 0.24(X - 4) \)
\( Y = 0.24X - (0.24 \times 4) + 2 \)
\( Y = 0.24X - 0.96 + 2 \)
\( Y = 0.24X + 1.04 \)

Regression line of X on Y:
First, calculate the regression coefficient \(b_{xy}\):
\( b_{xy} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum Y^2 - (\sum Y)^2} \)
\( b_{xy} = \frac{20(480) - (80)(40)}{20(320) - (40)^2} \)
\( b_{xy} = \frac{9600 - 3200}{6400 - 1600} \)
\( b_{xy} = \frac{6400}{4800} = 1.3333 \)
Rounding to two decimal places, \( b_{xy} \approx 1.33 \).
Now, use the formula for the regression line of X on Y:
\( X - \overline{X} = b_{xy}(Y - \overline{Y}) \)
\( X - 4 = 1.33(Y - 2) \)
\( X = 1.33Y - (1.33 \times 2) + 4 \)
\( X = 1.33Y - 2.66 + 4 \)
\( X = 1.33Y + 1.34 \)
In simple words: We are given summary numbers for two variables. We use these numbers to find their average values. Then, we calculate two special coefficients which help us write two equations. Each equation shows how one variable changes when the other one changes.

🎯 Exam Tip: When using summary data, ensure you correctly substitute the sums for \(N\), \( \sum X \), \( \sum Y \), \( \sum X^2 \), \( \sum Y^2 \), and \( \sum XY \) into the regression coefficient formulas.

Question 5. Given the following data, what will be the possible yield when the rainfall is 29"

The coefficient of correlation between rainfall and production is 0.8.
Answer:
Let X be Rainfall and Y be Production.
Given data:
Mean of X (\( \overline{X} \)) = 25
Standard Deviation of X (\( \sigma_x \)) = 3
Mean of Y (\( \overline{Y} \)) = 40
Standard Deviation of Y (\( \sigma_y \)) = 6
Coefficient of correlation (r) = 0.8
We need to estimate production (Y) when rainfall (X) is 29". So, we will find the regression equation of Y on X.

First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = r \times \frac{\sigma_y}{\sigma_x} \)
\( b_{yx} = 0.8 \times \frac{6}{3} \)
\( b_{yx} = 0.8 \times 2 = 1.6 \)

Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 40 = 1.6(X - 25) \)
\( Y = 1.6X - (1.6 \times 25) + 40 \)
\( Y = 1.6X - 40 + 40 \)
\( Y = 1.6X \)

To find the yield (Y) when the rainfall (X) is 29":
Substitute \( X = 29 \) into the regression equation:
\( Y = 1.6 \times 29 \)
\( Y = 46.4 \text{ units/acre} \)
In simple words: We are given the averages and spread of rainfall and crop production, and how strongly they are related. We use this information to create a rule (an equation) that predicts crop production based on rainfall. Then, we use this rule to guess the crop yield when the rainfall is 29 inches.

🎯 Exam Tip: When standard deviations and correlation coefficient are given, remember to use the formula \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \) or \( b_{xy} = r \frac{\sigma_x}{\sigma_y} \) to find the regression coefficients.

Question 6. The following data relate to advertisement expenditure (in lakh of rupees) and their corresponding sales (in crores of rupees)

Estimate the sales corresponding to advertising expenditure of Rs 30 lakh.
Answer:
Let X be Advertisement expenditure and Y be Sales. We need to estimate sales (Y) when advertisement expenditure (X) is given, so we will find the regression equation of Y on X.
First, calculate the mean of X and Y, and other sums. Let's create a table for calculations:

We have \( N = 7 \). From the table, we get the following sums:
\( \sum X = 338, \sum Y = 361, \sum X^2 = 17094, \sum Y^2 = 19773, \sum XY = 18160 \).
Now, let's find the means:
\( \overline{X} = \frac{\sum X}{N} = \frac{338}{7} = 48.2857 \)
Rounding to two decimal places, \( \overline{X} \approx 48.29 \).
\( \overline{Y} = \frac{\sum Y}{N} = \frac{361}{7} = 51.5714 \)
Rounding to two decimal places, \( \overline{Y} \approx 51.57 \).

Regression equation of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum X^2 - (\sum X)^2} \)
\( b_{yx} = \frac{7(18160) - (338)(361)}{7(17094) - (338)^2} \)
\( b_{yx} = \frac{127120 - 122018}{119658 - 114244} \)
\( b_{yx} = \frac{5102}{5414} = 0.94237 \)
Rounding to three decimal places, \( b_{yx} \approx 0.942 \).
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 51.57 = 0.942(X - 48.29) \)
\( Y = 0.942X - (0.942 \times 48.29) + 51.57 \)
\( Y = 0.942X - 45.48918 + 51.57 \)
\( Y = 0.942X + 6.08082 \)
Rounding to three decimal places, \( Y = 0.942X + 6.081 \).

To find the sales (Y) when the advertising expenditure (X) is Rs 30 lakh:
Substitute \( X = 30 \) into the regression equation:
\( Y = 0.942(30) + 6.081 \)
\( Y = 28.26 + 6.081 \)
\( Y = 34.341 \)
So, the estimated sales are Rs 34.34 crores.
In simple words: We gather data on advertising spending and sales, then calculate their averages. Using a specific formula, we find a relationship that helps us predict sales from advertising. Finally, we use this relationship to guess the sales if Rs 30 lakh is spent on advertising. More advertising generally leads to more sales.

🎯 Exam Tip: Remember to express the final answer in the correct units, such as 'crores of rupees' or 'lakh rupees', as specified in the question.

Question 7. You are given the following data:

If the Correlation coefficient between X and Y is 0.66, then find
(i) the two regression coefficients,
(ii) the most likely value of Y when X = 10.
Answer:
Given data:
Mean of X (\( \overline{X} \)) = 36
Standard Deviation of X (\( \sigma_x \)) = 11
Mean of Y (\( \overline{Y} \)) = 85
Standard Deviation of Y (\( \sigma_y \)) = 8
Coefficient of correlation (r) = 0.66

(i) The two regression coefficients:

Regression coefficient of Y on X (\(b_{yx}\)):
\( b_{yx} = r \times \frac{\sigma_y}{\sigma_x} \)
\( b_{yx} = 0.66 \times \frac{8}{11} \)
\( b_{yx} = 0.66 \times 0.7272... \)
\( b_{yx} = 0.48 \)

Regression coefficient of X on Y (\(b_{xy}\)):
\( b_{xy} = r \times \frac{\sigma_x}{\sigma_y} \)
\( b_{xy} = 0.66 \times \frac{11}{8} \)
\( b_{xy} = 0.66 \times 1.375 \)
\( b_{xy} = 0.9075 \)

(ii) The most likely value of Y when X = 10:
First, we need the regression equation of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 85 = 0.48(X - 36) \)
\( Y = 0.48X - (0.48 \times 36) + 85 \)
\( Y = 0.48X - 17.28 + 85 \)
\( Y = 0.48X + 67.72 \)
Now, substitute \( X = 10 \) into the equation to find Y:
\( Y = 0.48(10) + 67.72 \)
\( Y = 4.8 + 67.72 \)
\( Y = 72.52 \)
In simple words: We are given the average and spread of two things, plus how strongly they are connected. We first find two numbers that show how one changes with the other. Then, we use these numbers to make an equation that helps us guess the value of Y if X is 10.

🎯 Exam Tip: Be careful to use the correct standard deviation in the numerator and denominator when calculating \(b_{yx}\) and \(b_{xy}\) to avoid errors.

Question 8. Find the equation of the regression line of Y on X, if the observations (\(X_i\), \(Y_i\)) are the following (1, 4) (2, 8) (3, 2) (4, 12) ( 5, 10) (6, 14) (7, 16) (8, 6) (9, 18).
Answer:
Given observations are:
(1, 4), (2, 8), (3, 2), (4, 12), (5, 10), (6, 14), (7, 16), (8, 6), (9, 18).
We need to find the regression equation of Y on X. Let's create a table to calculate the necessary sums:

We have \( N = 9 \). From the table, we get the following sums:
\( \sum X = 45, \sum Y = 90, \sum X^2 = 285, \sum Y^2 = 1140, \sum XY = 530 \).
Now, let's find the means:
\( \overline{X} = \frac{\sum X}{N} = \frac{45}{9} = 5 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{90}{9} = 10 \)

Regression line of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum X^2 - (\sum X)^2} \)
\( b_{yx} = \frac{9(530) - (45)(90)}{9(285) - (45)^2} \)
\( b_{yx} = \frac{4770 - 4050}{2565 - 2025} \)
\( b_{yx} = \frac{720}{540} = 1.3333 \)
Rounding to two decimal places, \( b_{yx} \approx 1.33 \).
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 10 = 1.33(X - 5) \)
\( Y = 1.33X - (1.33 \times 5) + 10 \)
\( Y = 1.33X - 6.65 + 10 \)
\( Y = 1.33X + 3.35 \)
In simple words: We list all the given pairs of numbers and find their total sums and squares. From these, we calculate the average for each set of numbers. Then, using a special formula, we find a number that tells us how much Y changes for every change in X. This helps us write an equation to predict Y from X.

🎯 Exam Tip: Always double-check your sums and calculations for \( \sum XY \), \( \sum X^2 \), and \( (\sum X)^2 \) as small errors here can drastically affect the regression coefficients.

Question 9. A survey was conducted to study the relationship between expenditure on accommodation (X) and expenditure on Food and Entertainment (Y) and the following results were obtained:

Write down the regression equation and estimate the expenditure on Food and Entertainment, if the expenditure on accommodation is Rs 200.
Answer:
Let X be Expenditure on Accommodation and Y be Expenditure on Food and Entertainment.
Given data:
Mean of X (\( \overline{X} \)) = 178
Standard Deviation of X (\( \sigma_x \)) = 63.15
Mean of Y (\( \overline{Y} \)) = 47.8
Standard Deviation of Y (\( \sigma_y \)) = 22.98
Coefficient of correlation (r) = 0.43
We need to estimate Y when X is given, so we will find the regression equation of Y on X.

First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = r \times \frac{\sigma_y}{\sigma_x} \)
\( b_{yx} = 0.43 \times \frac{22.98}{63.15} \)
\( b_{yx} = 0.43 \times 0.364 \)
\( b_{yx} = 0.15652 \)
Rounding to four decimal places, \( b_{yx} \approx 0.1565 \).

Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 47.8 = 0.1565(X - 178) \)
\( Y = 0.1565X - (0.1565 \times 178) + 47.8 \)
\( Y = 0.1565X - 27.857 + 47.8 \)
\( Y = 0.1565X + 19.943 \)
Rounding to two decimal places, \( Y = 0.1565X + 19.94 \).

To estimate the expenditure on Food and Entertainment (Y) when the expenditure on accommodation (X) is Rs 200:
Substitute \( X = 200 \) into the regression equation:
\( Y = 0.1565(200) + 19.94 \)
\( Y = 31.3 + 19.94 \)
\( Y = 51.24 \)
So, the estimated expenditure on Food and Entertainment is Rs 51.24.
In simple words: We are given the average spending and how much it varies for accommodation and entertainment. We also know how these two spending types are connected. We use this to create a simple equation that predicts entertainment spending based on accommodation spending. Finally, we use this equation to guess the entertainment cost if accommodation costs Rs 200.

🎯 Exam Tip: When dealing with monetary units, clearly state the units (e.g., Rs, lakh, crores) in your final answer to avoid ambiguity.

Question 10. For 5 observations of pairs of (X, Y) of variables X and Y the following results are obtained.
\( \sum X = 15, \sum Y = 25, \sum X^2 = 55, \sum Y^2 = 135, \sum XY = 83 \). Find the equation of the lines of regression and estimate the values of X and Y if Y = 8; X = 12.
Answer:
Given data:
\( N = 5, \sum X = 15, \sum Y = 25, \sum X^2 = 55, \sum Y^2 = 135, \sum XY = 83 \).
First, calculate the means:
\( \overline{X} = \frac{\sum X}{N} = \frac{15}{5} = 3 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{25}{5} = 5 \)

Regression line of Y on X:
First, calculate the regression coefficient \(b_{yx}\):
\( b_{yx} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum X^2 - (\sum X)^2} \)
\( b_{yx} = \frac{5(83) - (15)(25)}{5(55) - (15)^2} \)
\( b_{yx} = \frac{415 - 375}{275 - 225} \)
\( b_{yx} = \frac{40}{50} = 0.8 \)
Now, use the formula for the regression line of Y on X:
\( Y - \overline{Y} = b_{yx}(X - \overline{X}) \)
\( Y - 5 = 0.8(X - 3) \)
\( Y = 0.8X - (0.8 \times 3) + 5 \)
\( Y = 0.8X - 2.4 + 5 \)
\( Y = 0.8X + 2.6 \)

Regression line of X on Y:
First, calculate the regression coefficient \(b_{xy}\):
\( b_{xy} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum Y^2 - (\sum Y)^2} \)
\( b_{xy} = \frac{5(83) - (15)(25)}{5(135) - (25)^2} \)
\( b_{xy} = \frac{415 - 375}{675 - 625} \)
\( b_{xy} = \frac{40}{50} = 0.8 \)
Now, use the formula for the regression line of X on Y:
\( X - \overline{X} = b_{xy}(Y - \overline{Y}) \)
\( X - 3 = 0.8(Y - 5) \)
\( X = 0.8Y - (0.8 \times 5) + 3 \)
\( X = 0.8Y - 4 + 3 \)
\( X = 0.8Y - 1 \)

Estimate the values:
When \( X = 12 \), find Y using the regression line of Y on X:
\( Y = 0.8(12) + 2.6 \)
\( Y = 9.6 + 2.6 \)
\( Y = 12.2 \)

When \( Y = 8 \), find X using the regression line of X on Y:
\( X = 0.8(8) - 1 \)
\( X = 6.4 - 1 \)
\( X = 5.4 \)
In simple words: We are given basic statistical information about two variables. We use these facts to find their average values and then calculate how they relate to each other. This helps us create two special equations, one to guess Y if we know X, and another to guess X if we know Y. We then use these equations to predict specific values.

🎯 Exam Tip: Always calculate both regression equations when asked to find the lines of regression. If estimations are also required, make sure to use the correct equation for the estimation (Y on X for Y given X, and X on Y for X given Y).

Question 11. The two regression lines were found to be \(4X - 5Y + 33 = 0\) and \(20X - 9Y - 107 = 0\). Find the mean values and coefficient of correlation between X and Y.
Answer:
To find the mean values (\( \overline{X} \) and \( \overline{Y} \)), we solve the two given regression equations simultaneously, as the point of intersection of the regression lines represents the mean values of X and Y.

The given equations are:
1) \( 4X - 5Y + 33 = 0 \implies 4X - 5Y = -33 \)
2) \( 20X - 9Y - 107 = 0 \implies 20X - 9Y = 107 \)

Multiply equation (1) by 5 to make the X coefficients equal:
\( 5 \times (4X - 5Y) = 5 \times (-33) \)
\( 20X - 25Y = -165 \)
Now, subtract this modified equation from equation (2):
\( (20X - 9Y) - (20X - 25Y) = 107 - (-165) \)
\( 20X - 9Y - 20X + 25Y = 107 + 165 \)
\( 16Y = 272 \)
\( Y = \frac{272}{16} \)
\( Y = 17 \)
So, the mean value of Y is \( \overline{Y} = 17 \).

Substitute \( Y = 17 \) into equation (1) to find X:
\( 4X - 5(17) = -33 \)
\( 4X - 85 = -33 \)
\( 4X = -33 + 85 \)
\( 4X = 52 \)
\( X = \frac{52}{4} \)
\( X = 13 \)
So, the mean value of X is \( \overline{X} = 13 \).

The mean values are \( \overline{X} = 13 \) and \( \overline{Y} = 17 \).

Now, we find the regression coefficients. We assume one equation is Y on X and the other is X on Y. If \( 4X - 5Y + 33 = 0 \) is the regression line of Y on X:
\( 5Y = 4X + 33 \)
\( Y = \frac{4}{5}X + \frac{33}{5} \)
So, \( b_{yx} = \frac{4}{5} = 0.8 \)

If \( 20X - 9Y - 107 = 0 \) is the regression line of X on Y:
\( 20X = 9Y + 107 \)
\( X = \frac{9}{20}Y + \frac{107}{20} \)
So, \( b_{xy} = \frac{9}{20} = 0.45 \)

Since both \( b_{yx} \) and \( b_{xy} \) are positive, our assumption is correct. If one was negative, we would swap the assignments of the equations. Also, \( |b_{yx}| \times |b_{xy}| = 0.8 \times 0.45 = 0.36 \le 1 \), which is valid.

The coefficient of correlation \(r\) is given by:
\( r = \pm \sqrt{b_{yx} \times b_{xy}} \)
Since both regression coefficients are positive, \(r\) must also be positive.
\( r = \sqrt{0.8 \times 0.45} \)
\( r = \sqrt{0.36} \)
\( r = 0.6 \)
In simple words: We are given two equations that show how two things, X and Y, are related. First, we solve these equations together to find the average values of X and Y. Then, we look at how X changes with Y and how Y changes with X, giving us two special numbers called regression coefficients. Finally, we use these two numbers to find a single number, called the correlation coefficient, which tells us how strongly X and Y move together.

🎯 Exam Tip: To find the mean values, always solve the two regression equations simultaneously. For the correlation coefficient, remember its sign must match the sign of both regression coefficients.

Question 12. The equations of two lines of regression obtained in a correlation analysis are the following 2X = 8 - 3Y and 2Y = 5 - X . Obtain the value of the regression coefficients and correlation coefficient.
Answer: We are given two regression equations:
1. \( 2Y = 5 - X \)
2. \( 2X = 8 - 3Y \)

First, let's find the regression line of Y on X from the first equation:
\( 2Y = 5 - X \)
To get Y by itself, we divide both sides by 2:
\( Y = \frac{5 - X}{2} \)
\( Y = 2.5 - 0.5X \)
So, the regression coefficient of Y on X, which is \( b_{yx} \), is the number multiplied by X:
\( b_{yx} = -0.5 \)

Next, let's find the regression line of X on Y from the second equation:
\( 2X = 8 - 3Y \)
To get X by itself, we divide both sides by 2:
\( X = \frac{8 - 3Y}{2} \)
\( X = 4 - 1.5Y \)
So, the regression coefficient of X on Y, which is \( b_{xy} \), is the number multiplied by Y:
\( b_{xy} = -1.5 \)

Now, we calculate the coefficient of correlation, \( r \). We use the formula that connects \( r \) with \( b_{yx} \) and \( b_{xy} \):
\( r = \pm \sqrt{b_{yx} \times b_{xy}} \)
Substitute the values we found:
\( r = \pm \sqrt{(-0.5) \times (-1.5)} \)
\( r = \pm \sqrt{0.75} \)
\( r \approx \pm 0.866 \)
Since both \( b_{yx} \) and \( b_{xy} \) are negative, the correlation coefficient \( r \) must also be negative. The sign of the correlation coefficient always matches the sign of both regression coefficients.
Therefore,
\( r = -0.866 \)
In simple words: We rearranged each equation to find Y in terms of X and X in terms of Y. The numbers we got for \( b_{yx} \) and \( b_{xy} \) were both negative. Since both are negative, the correlation coefficient \( r \) is also negative, which we found by taking the square root of their product.

🎯 Exam Tip: Always ensure that the sign of the correlation coefficient \( r \) matches the common sign of both regression coefficients \( b_{yx} \) and \( b_{xy} \). If both are negative, \( r \) is negative.