GSEB Class 12 Statistics Solutions Chapter 3 Linear Regression Exercise 3

GSEB Solutions Class 12 Statistics Part 1 Chapter 3 Linear Regression Ex 3

Section A

Answer the following questions by selecting a correct option from the given options:

 

Question 1.
Which of the following indicates the functional relation between the two variables ?
(a) Correlation
(b) Regression
(c) Mean
(d) Variance
Answer: (b) Regression
In simple words: Regression shows how two variables are related and how one changes with the other.

🎯 Exam Tip: Understanding the basic definitions of statistical terms like correlation and regression is crucial for scoring well.

 

Question 2.
The best fitted line of regression can be obtained by which method ?
(c) Maximum Square Method
(d) Bowley's Method
Answer: (a) Least Square Method
In simple words: The Least Square Method finds the best line that fits the data by making the errors as small as possible.

🎯 Exam Tip: Remember the specific method used for finding the best fit line, as it's a fundamental concept in regression analysis.

 

Question 3.
In usual notation, what is \(b_{yx}\)?
(a) Intercept
(b) Dependent Variable
(c) The approximate change in the value of Y for a unit change in the value of X.
(d) The approximate change in the value of X for a unit change in the value of Y.
Answer: (c) The approximate change in the value of Y for a unit change in the value of X.
In simple words: \(b_{yx}\) tells us how much Y is expected to change when X increases by one unit.

🎯 Exam Tip: The regression coefficient \(b_{yx}\) indicates the rate of change of Y with respect to X; understand its practical interpretation.

 

Question 4.
Which of the following is correct ?
(a) \(b_{yx} = r \frac{s_{x}}{s_{y}}\)
(b) \(b_{yx} = r \frac{s_{y}^2}{s_{x}^2}\)
(c) \(b_{yx} = \frac{Cov(x, y)}{s_{y}^2}\)
(d) \(b_{yx} = r \frac{s_{y}}{s_{x}}\)
Answer: (d) \(b_{yx} = r \frac{s_{y}}{s_{x}}\)
In simple words: This formula shows how the regression coefficient is calculated using the correlation coefficient and the standard deviations of X and Y.

🎯 Exam Tip: Memorize the standard formula for the regression coefficient, as it is frequently tested in calculations and theoretical questions.

 

Question 5.
The regression line always passes which point ?
(a) \((\bar{x}, \bar{y})\)
(b) \((0, \bar{y})\)
(c) \((\bar{x}, 0)\)
(d) \((0, 0)\)
Answer: (a) \((\bar{x}, \bar{y})\)
In simple words: A regression line will always go through the point where the average of X and the average of Y meet.

🎯 Exam Tip: Always remember that the mean point \((\bar{x}, \bar{y})\) lies on the regression line, which is a key property.

 

Question 6.
What is error e in estimation in case of line of regression of Y on X?
(a) \(y - \hat{y}\)
(c) \(x - \bar{x}\)
(d) \(\hat{y} - \bar{x}\)
Answer: (a) \(y - \hat{y}\)
In simple words: The error is the difference between the actual observed value and the value predicted by the regression line.

🎯 Exam Tip: The error term (residual) is the vertical distance between an observed data point and the regression line.

 

Question 7.
Which regression line is used if the sale of a commodity depends on its advertisement cost?
(a) Regression line of advertisement cost on sale
(b) Regression line of advertisement cost on advertisement cost
(c) Regression line of sales on advertisement cost
(d) Regression line of sales on sales
Answer: (c) Regression line of sales on advertisement cost
In simple words: When sales depend on advertising cost, we predict sales based on advertising cost.

🎯 Exam Tip: Identify the dependent (effect) and independent (cause) variables correctly to choose the right regression line (Y on X or X on Y).

 

Question 8.
Which of the following is a regression line of Y on X?
(a) \(\hat{y} = a + bx + cx^2\)
(b) \(x = c + by\)
(c) \(\hat{y} = a + bx\)
(d) \(\hat{y} = a + bx^2\)
Answer: (c) \(\hat{y} = a + bx\)
In simple words: This is the standard form for a simple linear regression line, where Y is predicted using X.

🎯 Exam Tip: Recognize the standard linear regression equation, where '\(a\)' is the intercept and '\(b\)' is the slope (regression coefficient).

 

Question 9.
For which value of the correlation coefficient (r), the regression coefficient becomes zero?
(a) 1
(b) -1
(c) \(\frac{1}{2}\)
(d) 0
Answer: (d) 0
In simple words: If there is no correlation between X and Y, then changing X will not affect Y, so the regression coefficient will be zero.

🎯 Exam Tip: Remember that if the correlation coefficient \(r\) is zero, there is no linear relationship, and thus the regression coefficient \(b\) will also be zero.

 

Question 10.
Coefficient of determination in the study of regression for two variables ?
(a) Product of two standard deviations
(b) Square of correlation coefficient
(c) Square of covariance
(d) Product of two variances
Answer: (b) Square of correlation coefficient
In simple words: The coefficient of determination is the square of the correlation coefficient, showing how much variation in one variable is explained by the other.

🎯 Exam Tip: The coefficient of determination \(R^2\) is a measure of how well the regression model explains the variability of the dependent variable. It is the square of the correlation coefficient \(r\).

 

Question 11.
If the regression line is \(\hat{y} = 10 + 3x\), what is the estimate of Y for X = 20 ?
(a) 13
(b) 60
(c) 70
(d) 203
Answer: (c) 70
In simple words: To find Y, put the value of X into the given equation: \(10 + 3 \times 20 = 10 + 60 = 70\).

🎯 Exam Tip: To estimate Y for a given X, simply substitute the value of X into the regression equation.

 

Question 12.
What is the value of \(b_{yx}\) if the regression line is \(2x + 3y - 50 = 0\)?
(a) \(\frac{3}{2}\)
(b) \(-\frac{3}{2}\)
(c) \(-\frac{2}{3}\)
(d) \(2\)
Answer: (c) \(-\frac{2}{3}\)
In simple words: Rewrite the equation to find Y in terms of X, like \(\hat{y} = a + bx\). Here, \(3y = -2x + 50\), so \(y = -\frac{2}{3}x + \frac{50}{3}\). The coefficient of x is \(-\frac{2}{3}\).

🎯 Exam Tip: To find \(b_{yx}\) from a linear equation, rearrange it into the form \(\hat{y} = a + bx\), and the coefficient of X will be \(b_{yx}\).

 

Question 13.
The regression line of Y on X is \(\hat{y} = 30 - 1.5x\). What is the value of \(\bar{y}\) if \(\bar{x} = 10\)?
(a) 28.5
(b) 20
(c) 15
(d) 45
Answer: (c) 15
In simple words: Since the regression line passes through the mean point \((\bar{x}, \bar{y})\), substitute \(\bar{x} = 10\) into the equation to find \(\bar{y}\). So, \(\bar{y} = 30 - 1.5(10) = 30 - 15 = 15\).

🎯 Exam Tip: Remember that the point of means \((\bar{x}, \bar{y})\) always lies on the regression line of Y on X.

 

Question 14.
If \(u = \frac{x-5}{10}\) and \(v = \frac{y-50}{2}\) and \(b_{yx} = 7.5\), what is the value of \(b_{vu}\)?
(a) 7.5
(b) 1.5
(c) 37.5
(d) 150
Answer: (c) 37.5
In simple words: The regression coefficient \(b_{vu}\) is found by multiplying \(b_{yx}\) by the ratio of the scale changes for Y and X, which are \(C_y = 2\) and \(C_x = 10\). So, \(b_{vu} = b_{yx} \times \frac{C_x}{C_y} = 7.5 \times \frac{10}{2} = 7.5 \times 5 = 37.5\).

🎯 Exam Tip: Understand how changes in origin and scale affect the regression coefficient. While a change in origin does not affect it, a change in scale does, specifically \(b_{vu} = b_{yx} \times \frac{C_x}{C_y}\).

 

Question 15.
If r = 0.8, how much part of the total variation in the dependent variable can be explained by the regression model ?
(a) 80%
(b) 64%
(c) 36%
(d) 20%
Answer: (b) 64%
In simple words: The explained variation is found by squaring the correlation coefficient \(r\). So, \(R^2 = (0.8)^2 = 0.64\), which is 64%.

🎯 Exam Tip: The coefficient of determination \(R^2 = r^2\) quantifies the proportion of the variance in the dependent variable that is predictable from the independent variable(s).

Section B

Answer the following questions in one sentence:

 

Question 1.
Define : Linear Regression
Answer: A mathematical connection between two related variables that helps predict the value of a dependent variable for a given independent variable is called Linear Regression.
In simple words: Linear regression finds a straight line that best predicts one variable using another.

🎯 Exam Tip: Provide a concise and accurate definition, highlighting the predictive nature and linearity of the relationship.

 

Question 2.
Define: Regression Coefficient
Answer: In the regression line of Y on X, which is \(\hat{y} = a + bx\), the coefficient 'b' of the variable x is known as the regression coefficient for the regression line of Y on X.
In simple words: The regression coefficient 'b' in the equation \(\hat{y} = a + bx\) tells us the average change in Y for every one-unit change in X.

🎯 Exam Tip: Clearly state that 'b' represents the slope of the regression line and signifies the expected change in the dependent variable for a unit change in the independent variable.

 

Question 3.
State the Linear Regression model.
Answer: The linear regression model is written as \(Y = \alpha + \beta X + u\); where \(\alpha\) and \(\beta\) are constant values, and \(u\) represents the error variable.
In simple words: The linear regression model is an equation that describes how a dependent variable Y is related to an independent variable X, plus a random error.

🎯 Exam Tip: Include all components of the model-dependent variable, intercept, slope, independent variable, and error term-for a complete answer.

 

Question 4.
What is an error in context with a regression line?
Answer: The error in the context of a regression line is the difference between the observed value \(y\) of the variable Y and its estimated value \(\hat{y}\) obtained from the regression model. It is denoted by 'e', so \(e = y - \hat{y}\).
In simple words: Error in regression means how much the actual data point differs from the value predicted by the regression line.

🎯 Exam Tip: Define the error (residual) as the vertical distance between observed and predicted values, and provide its mathematical notation.

 

Question 5.
Give the name of a method to obtain the best fitted regression line.
Answer: The method used to find the best fitted regression line is called the 'Least Square Method'.
In simple words: The best regression line is found using the Least Square Method.

🎯 Exam Tip: Naming the "Least Square Method" directly is sufficient for this type of question.

 

Question 6.
The regression coefficient is independent of which transformation?
Answer: The regression coefficient is not affected by a change of origin.
In simple words: Changing where the graph starts (origin) does not change the regression coefficient.

🎯 Exam Tip: Clearly distinguish between transformations that affect the regression coefficient (scale) and those that do not (origin).

 

Question 7.
The regression coefficient is not independent of which transformation?
Answer: The regression coefficient is not independent of a change of scale.
In simple words: Changing the units of measurement (scale) will change the regression coefficient.

🎯 Exam Tip: Remember that scale transformations (e.g., changing units from meters to centimeters) will alter the value of the regression coefficient.

 

Question 8.
What is the value of error if a sample point is on the fitted line?
Answer: If a sample point lies exactly on the fitted line, the value of the error will be zero.
In simple words: If a data point is on the line, there is no error.

🎯 Exam Tip: A zero error implies a perfect fit for that specific data point on the regression line.

 

Question 9.
Will the regression coefficient change if the values of both the variables are doubled with the help of transformation of scale?
Answer: If both variable values are doubled, the regression coefficient will not change, because \(C_x = \frac{1}{2}\) and \(C_y = \frac{1}{2}\).
In simple words: Doubling both X and Y values using scale transformation does not change the regression coefficient.

🎯 Exam Tip: Understand the formula for scale transformation's effect on the regression coefficient \(b_{vu} = b_{yx} \times \frac{C_x}{C_y}\). If both \(C_x\) and \(C_y\) are the same (e.g., both 1/2 for doubling), they cancel out, leaving \(b_{vu} = b_{yx}\).

 

Question 10.
If r = 0.5, \(S_x = 2\), \(S_y = 4\), what is the value of \(b_{yx}\)?
Answer:
Given: \(r = 0.5\), \(S_x = 2\), \(S_y = 4\)
Now, \(b_{yx} = r \frac{S_y}{S_x}\)

\( \implies b_{yx} = 0.5 \times \frac{4}{2}\)
\( \implies b_{yx} = 0.5 \times 2\)
\( \implies b_{yx} = 1\)
In simple words: We use the formula \(b_{yx} = r \times (S_y / S_x)\) and put in the given numbers to get the answer.

🎯 Exam Tip: Always write down the given values and the formula before starting the calculation to avoid errors.

 

Question 11.
If a regression line is \(\hat{y} = 31.5 + 1.85x\). estimate Y for X = 10.
Answer:
Given regression line: \(\hat{y} = 31.5 + 1.85x\)
Substitute \(x = 10\) into the equation:
\(\hat{y} = 31.5 + 1.85 (10)\)
\(\hat{y} = 31.5 + 18.5\)
\(\hat{y} = 50\)
Thus, the estimated value of Y for X = 10 is 50.
In simple words: We put the value of X (10) into the regression equation to find the predicted Y value.

🎯 Exam Tip: Direct substitution is the method for estimating Y. Ensure careful arithmetic to get the correct result.

 

Question 12.
If Y and X have the relation \(y = a + bx\), where \(b > 0\), then what is the value of r?
Answer:
If \(y = a + bx\) and \(b > 0\), then the value of the correlation coefficient \(r\) is 1.
In simple words: When the relationship is perfectly linear and the slope is positive, the correlation coefficient is 1.

🎯 Exam Tip: A regression line with a positive slope (\(b > 0\)) indicates a perfect positive linear relationship between X and Y if all points lie on the line, resulting in \(r=1\).

 

Question 13.
If \(y = 5 - 3x\) is the relation between Y and X then what is the value of r?
Answer:
Given relation: \(y = 5 - 3x\)
Here, the regression coefficient \(b = -3\).
Since \(b\) is negative, the correlation coefficient \(r\) must also be negative. In a perfect linear relationship where all points lie on the line, \(r\) would be -1.
Thus, the value of \(r\) is -1.
In simple words: If the line perfectly describes Y and has a negative slope, the correlation is -1.

🎯 Exam Tip: If the relationship is a perfect straight line, the absolute value of the correlation coefficient \(|r|\) is 1. The sign of \(r\) matches the sign of the regression coefficient \(b\).

Section C

Answer the following questions as required:

 

Question 1.
Explain constants a and b in the regression line \(\hat{y} = a + bx\)?
Answer: In the regression line \(\hat{y} = a + bx\), the constant '\(a\)' is known as the intercept of the regression line of Y on X. The constant '\(b\)' is called the regression coefficient of the regression line of Y on X. It is also referred to as the slope of the regression line of Y on X.
In simple words: 'a' is where the line crosses the Y-axis, and 'b' shows how steep the line is and how much Y changes for each unit change in X.

🎯 Exam Tip: Clearly define 'a' as the intercept (value of Y when X is 0) and 'b' as the slope (change in Y per unit change in X), emphasizing their roles.

 

Question 2.
The fitted regression line of Y on X is \(\hat{y} = 23.2 - 1.2x\) and one of the observations used in fitting of the line is (6, 17). Find the error In estimating Y for \(x = 6\).
Answer:
The given regression line is \(\hat{y} = 23.2 - 1.2x\).
One observation point is \((6, 17)\), where observed \(x = 6\) and observed \(y = 17\).
To find the estimated \(\hat{y}\) for \(x = 6\):
Substitute \(x = 6\) into the regression equation:
\(\hat{y} = 23.2 - 1.2 (6)\)
\(\hat{y} = 23.2 - 7.2\)
\(\hat{y} = 16\)
Now, calculate the error \((e)\):
\(e = y - \hat{y}\)
Substitute \(y = 17\) and \(\hat{y} = 16\):
\(e = 17 - 16\)
\(e = 1\)
Therefore, the error when estimating Y for \(X = 6\) is 1.
In simple words: We first find the predicted Y value using the given X, then subtract this predicted value from the actual observed Y value to find the error.

🎯 Exam Tip: Remember that error is always the observed value minus the estimated value \((e = y - \hat{y})\). Show all calculation steps clearly.

 

Question 3.
If \(\bar{x} = 30\), \(\bar{y} = 20\) and \(b = 0.6\), find the intercept of the regression line of Y on X and write equation of the line.
Answer:
Given: \(\bar{x} = 30\), \(\bar{y} = 20\), and \(b = 0.6\).
The formula for the intercept \(a\) is: \(a = \bar{y} - b\bar{x}\)
Substitute the given values:
\(a = 20 - 0.6(30)\)
\(a = 20 - 18\)
\(a = 2\)
The equation of the regression line of Y on X is \(\hat{y} = a + bx\).
Substitute the values of \(a = 2\) and \(b = 0.6\):
\(\hat{y} = 2 + 0.6x\)
In simple words: We find the intercept 'a' using the mean values and the slope, then use 'a' and 'b' to write the complete regression equation.

🎯 Exam Tip: The intercept 'a' can be calculated using the mean values of X and Y, and the regression coefficient 'b', via the formula \(a = \bar{y} - b\bar{x}\).

 

Question 4.
Interpret \(b_{yx} = 5\).
Answer: \(b_{yx} = 5\) indicates that for every one-unit increase in the value of variable X, there is an estimated increase of 5 units in the value of variable Y.
In simple words: If X goes up by 1, Y is expected to go up by 5.

🎯 Exam Tip: Interpretation questions require you to explain the meaning of the coefficient in practical terms related to the variables.

 

Question 5.
If \(b = 1.5\), \(r = 0.8\), and the standard deviation of X is 1.6, find the standard deviation of Y.
Answer:
Given: \(b = 1.5\), \(r = 0.8\), \(S_x = 1.6\). We need to find \(S_y\).
The formula relating these values is: \(b = r \frac{S_y}{S_x}\)
Substitute the given values into the formula:
\(1.5 = 0.8 \times \frac{S_y}{1.6}\)
To solve for \(S_y\), first multiply \(0.8\) by \(\frac{1}{1.6}\):
\(1.5 = \frac{0.8}{1.6} \times S_y\)
\(1.5 = \frac{1}{2} \times S_y\)
Now, multiply both sides by 2:
\(1.5 \times 2 = S_y\)
\(S_y = 3\)
Therefore, the standard deviation of Y is 3.
In simple words: We use the formula that connects regression coefficient, correlation coefficient, and standard deviations to calculate the missing standard deviation of Y.

🎯 Exam Tip: Be comfortable manipulating the formula \(b_{yx} = r \frac{S_y}{S_x}\) to solve for any of the variables, including \(S_x\) or \(S_y\).

 

Question 6.
If the regression coefficient of the regression line of Y on X is 0.6 and the standard deviations of X and Y are 5 and 3 respectively, find the coefficient of determination.
Answer:
Given: Regression coefficient \(b = 0.6\), standard deviation of X \(S_x = 5\), standard deviation of Y \(S_y = 3\). We need to find the coefficient of determination \(R^2\).
First, find the correlation coefficient \(r\) using the formula: \(b = r \frac{S_y}{S_x}\)
Substitute the known values:
\(0.6 = r \times \frac{3}{5}\)
\(0.6 = r \times 0.6\)
Divide both sides by 0.6 to find \(r\):
\(r = \frac{0.6}{0.6}\)
\(r = 1\)
Now, the coefficient of determination \(R^2\) is the square of the correlation coefficient:
\(R^2 = (r)^2\)
\(R^2 = (1)^2\)
\(R^2 = 1\)
In simple words: First, we use the given numbers to find the correlation coefficient. Then, we square it to get the coefficient of determination.

🎯 Exam Tip: This question requires two steps: first calculate \(r\) using \(b, S_x, S_y\), then calculate \(R^2\) using \(r^2\). A value of \(r=1\) means perfect positive linear correlation, and \(R^2=1\) means the model explains 100% of the variation.

 

Question 7.
If the regression line of Y on X is \(\hat{y} = 35 + 2x\) and \(Cov(x, y) = 50\), find the standard deviation of X.
Answer:
Given regression line: \(\hat{y} = 35 + 2x\). From this, we know that the regression coefficient \(b = 2\).
Also given: Covariance \(Cov(x, y) = 50\). We need to find \(S_x\).
The formula for the regression coefficient using covariance and standard deviation of X is: \(b = \frac{Cov(x, y)}{S_x^2}\)
Substitute the known values:
\(2 = \frac{50}{S_x^2}\)
Rearrange the equation to solve for \(S_x^2\):
\(2S_x^2 = 50\)
\(S_x^2 = \frac{50}{2}\)
\(S_x^2 = 25\)
Now, take the square root to find \(S_x\):
\(S_x = \sqrt{25}\)
\(S_x = 5\)
Therefore, the standard deviation of X is 5.
In simple words: We use the slope from the line equation and the given covariance to find the standard deviation of X.

🎯 Exam Tip: Remember the alternative formula for the regression coefficient \(b_{yx} = \frac{Cov(x, y)}{S_x^2}\) and use it when covariance is provided.

 

Question 8.
For the regression line given in the previous question (7), if the value of Y is to be increased by 10 units, how many units should be increased in the value of X?
Answer:
From Question 7, the regression line is \(\hat{y} = 35 + 2x\), and the regression coefficient is \(b = 2\).
The regression coefficient \(b\) represents the change in Y for a one-unit change in X.
Here, \(b = 2\) means that Y increases by 2 units for every 1-unit increase in X.
If Y needs to be increased by 10 units, then the required increase in X can be calculated as:
Increase in X = \(\frac{\text{Desired increase in Y}}{\text{Regression coefficient (b)}}\)
Increase in X = \(\frac{10}{2}\)
Increase in X = \(5\) units
Therefore, X should be increased by 5 units to increase Y by 10 units.
In simple words: Since Y increases by 2 units for every 1 unit of X, to make Y increase by 10 units, X must increase by 5 units.

🎯 Exam Tip: The regression coefficient \(b\) is key for predicting the effect of changes in X on Y. Always ensure the units of change are consistent.

 

Question 9.
If \(\bar{x} = 10\), \(\bar{y} = 25\), \(\Sigma(x - 10)(y - 25) = 120\) and \(\Sigma(x - 10)^2 = 100\), find the values of \(a\) and \(b\) for the regression line of Y on X.
Answer:
Given: \(\bar{x} = 10\), \(\bar{y} = 25\)
We are also given: \(\Sigma(x - 10)(y - 25) = 120\) and \(\Sigma(x - 10)^2 = 100\).
Since \(\bar{x} = 10\) and \(\bar{y} = 25\), we can write:
\(\Sigma(x - \bar{x})(y - \bar{y}) = 120\)
\(\Sigma(x - \bar{x})^2 = 100\)
The formula for the regression coefficient \(b\) is: \(b = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2}\)
Substitute the given sums:
\(b = \frac{120}{100}\)
\(b = 1.2\)
Now, find the intercept \(a\) using the formula: \(a = \bar{y} - b\bar{x}\)
Substitute the values of \(\bar{y}\), \(b\), and \(\bar{x}\):
\(a = 25 - 1.2(10)\)
\(a = 25 - 12\)
\(a = 13\)
Therefore, for the regression line of Y on X, the value of \(a\) is 13 and the value of \(b\) is 1.2.
In simple words: We calculate the slope 'b' using the provided sums and mean values. Then, we use 'b' and the means to calculate the intercept 'a'.

🎯 Exam Tip: When deviation sums from the mean are given, recognize that \((x - 10)\) corresponds to \((x - \bar{x})\) if \(\bar{x} = 10\). This simplifies the calculation for \(b\) and then \(a\).

 

Question 10.
If \(b_{yx} = 0.75\), \(u = 6(x - 20)\) and \(v = 2(y - 15)\) for the data in the study of a regression line then find the value of \(b_{vu}\).
Answer:
Given: \(b_{yx} = 0.75\)
The transformations are: \(u = 6(x - 20)\) and \(v = 2(y - 15)\)
From the transformation \(u = 6(x - 20)\), we can write \(u = \frac{x - 20}{1/6}\). So, the scale factor for X is \(C_x = \frac{1}{6}\).
From the transformation \(v = 2(y - 15)\), we can write \(v = \frac{y - 15}{1/2}\). So, the scale factor for Y is \(C_y = \frac{1}{2}\).
The relationship between \(b_{vu}\) and \(b_{yx}\) for scale transformations is: \(b_{vu} = b_{yx} \times \frac{C_y}{C_x}\)
Substitute the known values:
\(b_{vu} = 0.75 \times \frac{1/2}{1/6}\)
\(b_{vu} = 0.75 \times \frac{1}{2} \times \frac{6}{1}\)
\(b_{vu} = 0.75 \times 3\)
\(b_{vu} = 2.25\)
Therefore, the value of \(b_{vu}\) is 2.25.
In simple words: We use the given regression coefficient and the scale factors from the 'u' and 'v' equations to find the new regression coefficient.

🎯 Exam Tip: Carefully determine the scale factors \(C_x\) and \(C_y\) from the given transformations. Remember that \(u = \frac{x-A}{C_x}\) and \(v = \frac{y-B}{C_y}\), and the formula is \(b_{vu} = b_{yx} \times \frac{C_y}{C_x}\).

Section D

Answer the following questions as required:

 

Question 1.
Explain the statement, "There is a cause and effect relationship between two variables" by giving a suitable example. Also define independent variable and dependent variable.
Answer:
When a connection exists between two random variables such that changes in one variable lead to changes in the other, and we can predict the type and extent of these changes, this relationship is called a cause and effect relationship. This allows us to estimate the value of one variable given the value of the other.
For instance, if we know the relationship between income and expenditure, we can estimate how much expenditure will increase due to a certain rise in income. We can also estimate expenditure for a specific income level.
Here, income is the cause variable, and expenditure is the effect variable. Thus, there is a cause and effect relationship between income and expenditure.

**Independent Variable:** Out of two related variables, the one that acts as the 'cause' is called the independent variable. It is usually represented by the symbol X.

**Dependent Variable:** Out of two related variables, the one that represents the 'effect' is called the dependent variable. It is usually represented by the symbol Y.
In simple words: A cause and effect relationship means one thing makes another thing change. For example, more income causes more spending. The thing that causes the change is the independent variable (X), and the thing that changes is the dependent variable (Y).

🎯 Exam Tip: Use clear examples to illustrate cause and effect. Definitions of independent and dependent variables should be precise, linking them to cause and effect respectively.

 

Question 2.
Explain the method of scatter diagram for fitting a line of regression and state its limitation.
Answer:
**Method of Scatter Diagram for Fitting a Regression Line:**
Imagine you have 'n' pairs of observations \((x_1, y_1), (x_2, y_2), ..., (x_n, y_n)\) from a bivariate population. To fit a regression line, we first create a scatter diagram by plotting these 'n' pairs of observations on a graph. Then, we draw a straight line on this diagram that lies as close as possible to all the plotted points. This line aims to best represent the relationship between Y and X. Such a line is known as the regression line of Y on X.
We can find the equation of this regression line by selecting any two points on the line. This method is straightforward, quick, and doesn't require complex calculations.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक स्कैटर डायग्राम दिखाता है, जिसमें डेटा पॉइंट्स \( (x_i, y_i) \) प्लॉट किए गए हैं। एक सीधी रेखा इन पॉइंट्स के बीच से गुज़र रही है, जो रिग्रेशन लाइन को दर्शाती है। प्रत्येक पॉइंट से रेखा तक की ऊर्ध्वाधर दूरी \( e_i \) (त्रुटि या अवशिष्ट) को दर्शाती है, जो वास्तविक और अनुमानित मान के बीच का अंतर है।

**Limitations:**
* The regression line obtained by this method might not provide the most accurate estimate of the relationship between Y and X.
* Different people might draw slightly different regression lines from the same data, leading to varied conclusions, because they might choose different points to define the line.
* This method is subjective, meaning it relies on personal judgment rather than objective calculation.
In simple words: A scatter diagram method involves plotting data points and drawing a line that looks like it fits best. It's easy, but different people might draw slightly different lines, so it's not always the most accurate or objective way.

🎯 Exam Tip: For the scatter diagram method, describe the process of plotting points and visually fitting a line. Crucially, emphasize its subjective nature as its main limitation. Ensure to follow the diagram explanation rule with Hindi text.

 

Question 3.
Explain the method of least square for fitting a regression line.
Answer:
**Method of Least Squares for Fitting a Regression Line:**
The best-fitted regression line of Y on X can be found using the method of least squares. Suppose we have 'n' ordered pairs of sample observations \((x_1, y_1), (x_2, y_2), ..., (x_n, y_n)\) for two correlated variables X and Y. We use the least square method to draw the scatter diagram for this data.
If the equation of the best-fitted line representing the linear regression between Y and X is \(\hat{y} = a + bx\), then the values of constants \(a\) and \(b\) are found by minimizing the sum of the squares of the errors. These constants are obtained as follows:
Let \(\hat{y}_1, \hat{y}_2, ..., \hat{y}_n\) be the estimated values of Y for the observed values \(y_1, y_2, ..., y_n\), obtained from the regression equation for the corresponding X values \(x_1, x_2, ..., x_n\).
For a given value \(x_i\) of X, the estimated value of Y is \(\hat{y}_i = a + bx_i\).
The error (residual) for each observation is \(e_i = y_i - \hat{y}_i\).
The least squares method determines \(a\) and \(b\) such that the sum of the squared errors, \(\Sigma e_i^2\), is minimized. This means we minimize \(\Sigma(y_i - \hat{y}_i)^2 = \Sigma(y_i - (a + bx_i))^2\).
The line \(\hat{y} = a + bx\) obtained using this method passes closest to most points on the scatter diagram because it minimizes the sum of squared errors. This is why it is called the method of least squares.
In simple words: The Least Square Method finds the best-fit line by making the total squared distance between the actual data points and the line as small as possible. This mathematically ensures the line is the "best" fit.

🎯 Exam Tip: The core idea of the least squares method is minimizing \(\Sigma e_i^2 = \Sigma(y_i - \hat{y}_i)^2\). Mentioning this minimization principle and the resulting optimal values for 'a' and 'b' is crucial.

 

Question 4.
State the utility of regression.
Answer:
The utility of regression can be described as follows:
* We can identify the functional relationship between two correlated variables.
* We can estimate the unknown value of Y for a given value of the independent variable X.
* We can determine the estimated change in Y for a unit change in X using regression.
* We can calculate the error made when finding the estimated value of the dependent variable using the regression line.
* It is a valuable tool for economists, planners, businessmen, administrators, and researchers.
In simple words: Regression helps us understand how variables are connected, predict future values, measure changes, find errors in predictions, and is useful for many professionals.

🎯 Exam Tip: Focus on the predictive and explanatory powers of regression. List practical applications to demonstrate its importance.

 

Question 5.
State properties of regression coefficient. Also state the point through which a regression line always passes.
Answer:
**Properties of Regression Coefficient:**
* The signs of the correlation coefficient \(r\) and the regression coefficient \(b\) are always the same.
* The value of the regression coefficient \(b\) is independent of a change of origin but is dependent on a change of scale.
* From the regression coefficient, we can determine the estimated change in the value of Y for a unit change in the value of X.
* The value of \(b\) can be less than 1 or greater than 1.
* The regression coefficient is a relative measure.
* The sign of '\(b\)' depends on the sign of \(Cov(x, y)\).
* If \(b > 0\), a unit increase in X implies an estimated increase of '\(b\)' units in Y.
* If \(b < 0\), a unit increase in X implies an estimated decrease of '\(b\)' units in Y.

**Point through which a regression line always passes:**
A regression line always passes through the mean point \((\bar{x}, \bar{y})\).
In simple words: The regression coefficient's sign matches the correlation's sign, it's not affected by shifting the data (origin) but is affected by changing measurement units (scale), and it tells us how much Y changes with X. The regression line always goes through the average point of X and Y.

🎯 Exam Tip: List key properties concisely, especially independence from origin and dependence on scale. Crucially, mention that the regression line always passes through \((\bar{x}, \bar{y})\).

 

Question 6.
Explain: coefficient of determination
Answer:
The square of the correlation coefficient between the observed value \(y\) of the dependent variable Y and its corresponding estimated value \(\hat{y}\) from the regression line \(\hat{y} = a + bx\) is called the coefficient of determination. It is denoted by \(R^2\).
Thus, \(R^2 = [Cov(y, \hat{y})]^2\)
\(\implies R^2 = [Cov(y, a + bx)]^2\)
\(\implies R^2 = [Cov(y, x)]^2\)
\(\implies R^2 = r^2\)

**Uses:**
(i) **To determine the reliability of estimates obtained from the regression lines:**
* If \(R^2 = 1\), the estimates obtained from the regression line are 100 % reliable. This indicates a perfect linear correlation between variables Y and X.
* If \(R^2 = 0\), the estimates obtained from the regression line are not reliable. This shows a lack of linear correlation between variables Y and X.

(ii) **For the interpretation regarding the assumption of linear regression between two random variables X and Y:**
* If the obtained value of \(R^2\) is close to 1 (i.e., \(0.5 \le R^2 < 1\)), it suggests that the correlation between Y and X is close to a perfect linear correlation. Therefore, the assumption of linear regression between X and Y is appropriate.
* If the obtained value of \(R^2\) is close to zero (0) (i.e., \(0 \le R^2 < 0.5\)), it suggests that the correlation between Y and X is far from a perfect linear correlation. In this case, the assumption of linear regression between X and Y is not proper.
In simple words: The coefficient of determination (\(R^2\)) tells us how well the regression line explains the changes in the dependent variable. If it's close to 1, the model is very good; if it's close to 0, it's not.

🎯 Exam Tip: Define \(R^2\) as \(r^2\) and explain its interpretation in terms of explained variance. Discuss how its value indicates the reliability and appropriateness of the linear regression model.

 

Question 7. State precautions which are necessary while using the regression.

Answer: To use regression effectively, some important steps are needed:

• First, always check if the linear regression assumption holds true by looking at the coefficient of determination (R²).
• Second, the regression line found from a scatter diagram or least squares should not be used for predicting values that are very different from the original independent variable data.
• Third, conclusions from the regression line are reliable only if the correlation between Y and X is very strong.
• Finally, if the chosen linear regression model is not suitable, use proper statistical methods to find a better model before making any estimations or conclusions.

In simple words: When using regression, make sure your model is appropriate for the data. Avoid making predictions too far outside your data range, and ensure there's a strong linear relationship before trusting your results.

🎯 Exam Tip: Pay attention to the conditions for applying regression models; misapplication, especially extrapolation, is a common error and can lead to incorrect conclusions. Understanding R² and correlation strength is key for reliability checks.

 

Question 8. For two related variables X and Y, \( \Sigma(x - \bar{x})^2 = 80 \), \( \Sigma(x - \bar{x})(y - \bar{y}) = 60 \), \( \bar{x} = 8 \), \( \bar{y} = 10 \). Obtain the regression line of Y on X.

Answer: Given: sum of squared deviations for X is 80, sum of products of deviations for X and Y is 60. The mean of X is 8, and the mean of Y is 10.
First, calculate the regression coefficient 'b' using the formula: \( b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}} \)
Substitute the given values: \( b = \frac{60}{80} = 0.75 \).
Next, calculate the intercept 'a' using the formula: \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b' value: \( a = 10 - 0.75(8) = 10 - 6 = 4 \).
Therefore, the regression line of Y on X is \( \hat{y} = a + bx \), which means \( \hat{y} = 4 + 0.75x \).In simple words: We used the given sums and means to calculate the regression coefficient 'b' as 0.75 and the intercept 'a' as 4. This gives us the regression line equation \( \hat{y} = 4 + 0.75x \).

🎯 Exam Tip: Remember the formulas for 'b' and 'a' in terms of deviations and means. Accurate substitution and calculation are crucial for obtaining the correct regression line.

 

Question 9. If \( \bar{x} = 30 \), \( \bar{y} = 50 \), \( r = 0.8 \) and the standard deviations of X and Y are 2 and 5 respectively, obtain the regression line of Y on X.

Answer: Given values are: mean of X is 30, mean of Y is 50, correlation coefficient (r) is 0.8. Standard deviation of X (\( S_x \)) is 2, and standard deviation of Y (\( S_y \)) is 5.
First, find the regression coefficient 'b' using the formula: \( b = r \frac{S_y}{S_x} \).
Plug in the numbers: \( b = 0.8 \times \frac{5}{2} = 0.8 \times 2.5 = 2 \).
Next, find the intercept 'a' using the formula: \( a = \bar{y} - b\bar{x} \).
Substitute the values: \( a = 50 - 2(30) = 50 - 60 = -10 \).
So, the regression line equation for Y on X, \( \hat{y} = a + bx \), becomes \( \hat{y} = -10 + 2x \).In simple words: We used the correlation coefficient and standard deviations to find 'b' (2), then used the means to find 'a' (-10). The final regression line equation is \( \hat{y} = -10 + 2x \).

🎯 Exam Tip: When given 'r' and standard deviations, use the formula \( b = r \frac{S_y}{S_x} \). Ensure correct substitution and calculation for 'b' and then 'a'.

 

Question 10. If the regression line of Y on X is \( \hat{y} = 11 + 3x \) and \( S_x: S_y = 3 : 10 \), find the coefficient of determination.

Answer: The regression line is \( \hat{y} = 11 + 3x \), which tells us that the regression coefficient 'b' is 3.
The ratio of standard deviations of X and Y is given as \( S_x : S_y = 3 : 10 \). This means \( S_x = 3 \) and \( S_y = 10 \).
Now, use the formula for 'b': \( b = r \frac{S_y}{S_x} \).
Substitute the known values: \( 3 = r \times \frac{10}{3} \).
Solve for 'r': \( r = 3 \times \frac{3}{10} = \frac{9}{10} = 0.9 \).
The coefficient of determination \( R^2 \) is simply the square of the correlation coefficient 'r'.
So, \( R^2 = (0.9)^2 = 0.81 \).In simple words: From the regression line, we found 'b' is 3. Using the ratio of standard deviations, we calculated the correlation coefficient 'r' as 0.9. Squaring 'r' gave us the coefficient of determination \( R^2 \), which is 0.81.

🎯 Exam Tip: Remember that \( R^2 \) is the square of 'r' and indicates how much variation in Y is explained by X. Always identify 'b' from the given regression equation first.

 

Question 11. In usual notations, \( n = 7 \); \( \Sigma u = 2 \); \( \Sigma v = 25 \); \( \Sigma u^2 = 160 \) and \( \Sigma uv = 409 \). Obtain the regression coefficient of a regression line of Y on X and interpret it.

Answer: Given are: number of observations (n) = 7, sum of u (\( \Sigma u \)) = 2, sum of v (\( \Sigma v \)) = 25, sum of u-squared (\( \Sigma u^2 \)) = 160, and sum of uv (\( \Sigma uv \)) = 409.
We need to find the regression coefficient 'b' using the formula: \( b = \frac{n\Sigma uv - (\Sigma u)(\Sigma v)}{n\Sigma u^2 - (\Sigma u)^2} \).
Substitute the values into the formula:
\( b = \frac{7(409) - (2)(25)}{7(160) - (2)^2} \)
\( b = \frac{2863 - 50}{1120 - 4} \)
\( b = \frac{2813}{1116} \)
\( b \approx 2.52 \).
This means that for every one-unit increase in the variable X, the estimated value of Y will increase by approximately 2.52 units.In simple words: We calculated the regression coefficient 'b' using the provided sums and the formula, getting 2.52. This means if X increases by 1, Y is expected to increase by 2.52.

🎯 Exam Tip: Be careful with the signs and order of operations in the formula for 'b'. The interpretation of 'b' as the change in Y for a unit change in X is very important.

 

Question 12. If \( b_{yx} = 0.8 \), then find the value of \( b_{vu} \) for the following u and v :
(i) \( u = x - 105 \) and \( v = y - 90 \)
(ii) \( u = \frac{x-1400}{100} \) and \( v = \frac{y-750}{50} \)
(iii) \( u = 10 (x - 4.6) \) and \( v = y - 75 \)

Answer: Given that \( b_{yx} = 0.8 \).
(i) If \( u = x - 105 \) and \( v = y - 90 \):
Here, the new variables 'u' and 'v' are formed by just subtracting constants from 'x' and 'y'. This is called a change of origin.
Since the regression coefficient 'b' is not affected by a change in origin, the value of \( b_{vu} \) will be the same as \( b_{yx} \).
Therefore, \( b_{vu} = 0.8 \).

(ii) If \( u = \frac{x-1400}{100} \) and \( v = \frac{y-750}{50} \):
In this case, the new variables 'u' and 'v' involve both a change of origin and a change of scale.
From the given equations, we can identify the scale factors: \( C_x = 100 \) (for x) and \( C_y = 50 \) (for y).
The relationship between \( b_{yx} \) and \( b_{vu} \) with scale changes is: \( b_{yx} = b_{vu} \cdot \frac{C_y}{C_x} \).
To find \( b_{vu} \), we rearrange the formula: \( b_{vu} = b_{yx} \cdot \frac{C_x}{C_y} \).
Substitute the known values: \( b_{vu} = 0.8 \times \frac{100}{50} = 0.8 \times 2 = 1.6 \).

(iii) If \( u = 10 (x - 4.6) \) and \( v = y - 75 \):
From \( u = 10 (x - 4.6) \), we can write it as \( u = \frac{x - 4.6}{1/10} \). So, the scale factor for x is \( C_x = \frac{1}{10} \).
From \( v = y - 75 \), the scale factor for y is \( C_y = 1 \).
The formula to convert \( b_{yx} \) to \( b_{vu} \) is: \( b_{vu} = b_{yx} \cdot \frac{C_x}{C_y} \).
Substitute the values: \( b_{vu} = 0.8 \times \frac{1/10}{1} \).
This simplifies to: \( b_{vu} = 0.8 \times \frac{1}{10} = 0.08 \).In simple words: The regression coefficient 'b' is unaffected by shifting the origin (adding/subtracting constants). However, it changes when the scale is changed (multiplying/dividing by constants). We used the scale factors to adjust \( b_{yx} \) for the new variables 'u' and 'v'.

🎯 Exam Tip: Remember that regression coefficients are independent of the change of origin but dependent on the change of scale. Understand how \( C_x \) and \( C_y \) affect the transformation of 'b'.

 

Question 13. The following results are obtained for a bivariate data.

Particulars	x	y
No. of observations		8
Mean	100	100
The sum of squares of deviations taken from mean	130	145
The sum of product of deviations taken from mean	115

Obtain the regression line of Y on X.

Answer: Given the following data: number of observations (n) = 8, mean of X (\( \bar{x} \)) = 100, mean of Y (\( \bar{y} \)) = 100.
Also, the sum of squared deviations for X is \( \Sigma(x-\bar{x})^2 = 130 \), the sum of squared deviations for Y is \( \Sigma(y-\bar{y})^2 = 145 \), and the sum of products of deviations is \( \Sigma(x-\bar{x})(y-\bar{y}) = 115 \).
First, calculate the regression coefficient 'b' using the formula: \( b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}} \).
Substitute the values: \( b = \frac{115}{130} \approx 0.88 \).
Next, calculate the intercept 'a' using the formula: \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b' value: \( a = 100 - 0.88(100) = 100 - 88 = 12 \).
Thus, the regression line of Y on X, \( \hat{y} = a + bx \), is \( \hat{y} = 12 + 0.88x \).In simple words: We used the given statistical sums and means to calculate the regression coefficient 'b' as 0.88 and the intercept 'a' as 12. This leads to the regression line \( \hat{y} = 12 + 0.88x \).

🎯 Exam Tip: When summary statistics like sum of squares and product of deviations are provided, use the direct formulas for 'b' and 'a'. Be careful with rounding in intermediate steps.

 

Section E

 

Question 1. A manager of an I.T. company has collected the following information regarding the years of job and monthly income of seven marketing executives;

Years of job	10	6	8	5	9	7	11
Monthly income (ten thousand Rs.)	11	7	9	5	6	8	10

Obtain the regression line of the monthly income on the years of job of the marketing executives.

Answer: Given data for 7 marketing executives: X represents years of job, and Y represents monthly income.
Calculate the mean of X: \( \bar{x} = \frac{\Sigma x}{n} = \frac{56}{7} = 8 \).
Calculate the mean of Y: \( \bar{y} = \frac{\Sigma y}{n} = \frac{56}{7} = 8 \).
To find the regression line \( \hat{y} = a + bx \), we first need to calculate 'b' and then 'a'.
The table for calculating the values of a and b is prepared as follows:

X	y	xy	\( x^2 \)
10	11	110	100
6	7	42	36
8	9	72	64
5	5	25	25
9	6	54	81
7	8	56	49
11	10	110	121
\( \Sigma x = 56 \)	\( \Sigma y = 56 \)	\( \Sigma xy = 469 \)	\( \Sigma x^2 = 476 \)

Now, use the formula for 'b': \( b = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{n \Sigma x^2 - (\Sigma x)^2} \).
Substitute the values:
\( b = \frac{7(469) - (56)(56)}{7(476) - (56)^2} \)
\( b = \frac{3283 - 3136}{3332 - 3136} \)
\( b = \frac{147}{196} = 0.75 \).
Next, calculate the intercept 'a': \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 8 - 0.75(8) = 8 - 6 = 2 \).
Therefore, the regression line for monthly income (Y) on years of job (X) is \( \hat{y} = 2 + 0.75x \).In simple words: We found the average job years and income. Then, using a table to sum necessary values, we calculated the regression coefficient 'b' as 0.75 and the intercept 'a' as 2. The regression line predicting income from job years is \( \hat{y} = 2 + 0.75x \).

🎯 Exam Tip: Carefully construct the calculation table, especially for \( xy \) and \( x^2 \), as errors here propagate. Remember to calculate both 'b' and 'a' accurately to form the complete regression equation.

 

Question 2. The information collected regarding price (in Rs.) of a commodity and its supply (in hundred units) is as follows:

Price (Rs.)	59	60	61	62	64	57	58	59
Supply (hundred units)	78	82	82	79	81	77	78	75

Obtain the regression line of the supply on the price.

Answer: Given data for 8 observations: X is price (in Rs.), and Y is supply (in hundred units).
Calculate the mean of X: \( \bar{x} = \frac{\Sigma x}{n} = \frac{480}{8} = 60 \).
Calculate the mean of Y: \( \bar{y} = \frac{\Sigma y}{n} = \frac{632}{8} = 79 \).
To find the regression line \( \hat{y} = a + bx \), we use a calculation table to find the sum of products of deviations \( \Sigma(x-\bar{x})(y-\bar{y}) \) and the sum of squared deviations for X \( \Sigma(x-\bar{x})^2 \).
The table for calculating the values of a and b is prepared as follows:

x	y	\( (x-\bar{x}) \) \( \bar{x} = 60 \)	\( (y-\bar{y}) \) \( \bar{y} = 79 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)
59	78	-1	-1	1	1
60	82	0	3	0	0
61	82	1	3	3	1
62	79	2	0	0	4
64	81	4	2	8	16
57	77	-3	-2	6	9
58	78	-2	-1	2	4
59	75	-1	-4	4	1
\( \Sigma x = 480 \)	\( \Sigma y = 632 \)	\( \Sigma(x-\bar{x}) = 0 \)	\( \Sigma(y-\bar{y}) = 0 \)	\( \Sigma(x-\bar{x})(y-\bar{y}) = 24 \)	\( \Sigma(x-\bar{x})^2 = 36 \)

Now, calculate 'b': \( b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}} \).
Substitute the values: \( b = \frac{24}{36} \approx 0.67 \).
Next, calculate the intercept 'a': \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 79 - 0.67(60) = 79 - 40.2 = 38.8 \).
Therefore, the regression line for supply (Y) on price (X) is \( \hat{y} = 38.8 + 0.67x \).In simple words: We found the average price and supply. By calculating deviations from the mean, we determined 'b' as 0.67 and 'a' as 38.8. The regression line to predict supply from price is \( \hat{y} = 38.8 + 0.67x \).

🎯 Exam Tip: When using the deviation method for 'b', ensure \( \Sigma(x-\bar{x}) \) and \( \Sigma(y-\bar{y}) \) sum to zero, which is a good check for calculations of deviations.

 

Question 3. The following information is obtained for monthly advertisement cost and the sales of the last year for a company providing online shopping:

Particulars	Advertisement cost (ten thousand Rs.)	Sales (lakh Rs.)
Mean	10	90
Standard Deviation	3	12
r	0.8

Obtain the regression line of the sales on the advertisement cost.

Answer: Given: X is advertisement cost, and Y is sales.
We have: mean of X (\( \bar{x} \)) = 10, mean of Y (\( \bar{y} \)) = 90. Standard deviation of X (\( S_x \)) = 3, standard deviation of Y (\( S_y \)) = 12, and the correlation coefficient (r) = 0.8.
First, calculate the regression coefficient 'b' using the formula: \( b = r \frac{S_y}{S_x} \).
Substitute the values: \( b = 0.8 \times \frac{12}{3} = 0.8 \times 4 = 3.2 \).
Next, calculate the intercept 'a' using the formula: \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 90 - 3.2(10) = 90 - 32 = 58 \).
Therefore, the regression line for sales (Y) on advertisement cost (X) is \( \hat{y} = 58 + 3.2x \).In simple words: Using the given means, standard deviations, and correlation coefficient, we calculated 'b' as 3.2 and 'a' as 58. The regression line for sales based on advertisement cost is \( \hat{y} = 58 + 3.2x \).

🎯 Exam Tip: When standard deviations and correlation coefficient are provided, this method is usually faster than constructing a full table. Ensure you correctly identify \( S_x \) and \( S_y \) to avoid errors in the 'b' calculation.

 

Question 4. The following results are obtained from the information of average rain and yield of a crop per acre in the last ten years of an arid region:

Particulars	Rainfall (cm)	Yield of crop (kg)
Mean	18	970
Standard Deviation	2	38
Correlation Coefficient	0.6

Estimate the yield of the crop if it rains 20 cms.

Answer: Given data: X is rainfall, and Y is crop yield.
We have: mean rainfall (\( \bar{x} \)) = 18 cm, mean yield (\( \bar{y} \)) = 970 kg. Standard deviation of X (\( S_x \)) = 2, standard deviation of Y (\( S_y \)) = 38, and correlation coefficient (r) = 0.6.
First, calculate the regression coefficient 'b' using the formula: \( b = r \frac{S_y}{S_x} \).
Substitute the values: \( b = 0.6 \times \frac{38}{2} = 0.6 \times 19 = 11.4 \).
Next, calculate the intercept 'a' using the formula: \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 970 - 11.4(18) = 970 - 205.2 = 764.8 \).
The regression line of crop yield (Y) on rainfall (X) is \( \hat{y} = 764.8 + 11.4x \).
To estimate the crop yield when rainfall (X) is 20 cm, substitute \( x = 20 \) into the equation:
\( \hat{y} = 764.8 + 11.4(20) \)
\( \hat{y} = 764.8 + 228 \)
\( \hat{y} = 992.8 \).
So, the estimated crop yield for 20 cm rainfall is 992.8 kg.In simple words: Using the given averages, standard deviations, and correlation, we found the regression line for crop yield. Then, we used this line to predict that 20 cm of rain would lead to an estimated crop yield of 992.8 kg.

🎯 Exam Tip: After finding the regression line, remember to use it to answer the estimation part of the question. Clearly state the units for your final estimated value.

 

Question 5. The information of investment (in lakh Rs.) and its market price (in lakh Rs.) after six months in share market in the last seven years for a Mutual Fund Company is obtained as follows:

Particulars	Investment (lakh Rs.) x	Market price after six months (lakh Rs.) y
Mean	40	50
Variance	100	256
Covariance	80

Obtain the regression line of Y on X and estimate the market price in the share market after six months if there is an investment of Rs. 45 lakh in a year.

Answer: Given data: X is investment (in Rs. lakh), and Y is market price after six months (in Rs. lakh).
We have: mean investment (\( \bar{x} \)) = 40, mean market price (\( \bar{y} \)) = 50. Variance of X (\( S_x^2 \)) = 100, variance of Y (\( S_y^2 \)) = 256. Covariance of X and Y (\( Cov(x, y) \)) = 80.
First, calculate the regression coefficient 'b' using the formula: \( b = \frac{Cov(x, y)}{S_x^2} \).
Substitute the values: \( b = \frac{80}{100} = 0.8 \).
Next, calculate the intercept 'a' using the formula: \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 50 - 0.8(40) = 50 - 32 = 18 \).
The regression line of Y on X is \( \hat{y} = 18 + 0.8x \).
To estimate the market price (Y) for an investment (X) of Rs. 45 lakh, substitute \( x = 45 \) into the equation:
\( \hat{y} = 18 + 0.8(45) \)
\( \hat{y} = 18 + 36 \)
\( \hat{y} = 54 \).
So, the estimated market price is Rs. 54 lakh.In simple words: We used the given means, variances, and covariance to find 'b' (0.8) and 'a' (18). The regression line for market price based on investment is \( \hat{y} = 18 + 0.8x \). Using this, an investment of Rs. 45 lakh is estimated to result in a market price of Rs. 54 lakh.

🎯 Exam Tip: When covariance and variance are given, use the formula \( b = \frac{Cov(x, y)}{S_x^2} \). Be careful not to confuse variance with standard deviation in the formula.

 

Section F

 

Question 1. Obtain the regression line of the demand on the price using the following information collected for the demand and the price of a commodity. Estimate the demand of the commodity if price is 40.

Price (Rs.)	38	36	37	37	36	38	39	36	38
Demand (hundred units)	12	18	15	12	17	13	13	15	12

Answer: Given data for 9 observations: X is price (in Rs.), and Y is demand (in hundred units).
Calculate the mean of X: \( \bar{x} = \frac{\Sigma x}{n} = \frac{335}{9} \approx 37.22 \).
Calculate the mean of Y: \( \bar{y} = \frac{\Sigma y}{n} = \frac{127}{9} \approx 14.11 \).
To find the regression line \( \hat{y} = a + bx \), we use a shortcut method with new variables \( u = x - A \) and \( v = y - B \), where \( A = 36 \) and \( B = 15 \).
The table for calculation is prepared as follows:

x	y	\( u=(x-A) \) \( A = 36 \)	\( v=(y-B) \) \( B = 15 \)	uv	\( u^2 \)
38	12	2	-3	-6	4
36	18	0	3	0	0
37	15	1	0	0	1
37	12	1	-3	-3	1
36	17	0	2	0	0
38	13	2	-2	-4	4
39	13	3	-2	-6	9
36	15	0	0	0	0
38	12	2	-3	-6	4
\( \Sigma x = 335 \)	\( \Sigma y = 127 \)	\( \Sigma u = 11 \)	\( \Sigma v = -8 \)	\( \Sigma uv = -25 \)	\( \Sigma u^2 = 23 \)

Now, calculate the regression coefficient 'b' using the formula: \( b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \).
Substitute the values:
\( b = \frac{9(-25) - (11)(-8)}{9(23) - (11)^2} \)
\( b = \frac{-225 + 88}{207 - 121} \)
\( b = \frac{-137}{86} \approx -1.59 \).
Next, calculate the intercept 'a': \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 14.11 - (-1.59)(37.22) \)
\( a = 14.11 + 59.18 = 73.29 \).
The regression line for demand (Y) on price (X) is \( \hat{y} = 73.29 - 1.59x \).
To estimate demand when the price (X) is Rs. 40, substitute \( x = 40 \) into the equation:
\( \hat{y} = 73.29 - 1.59(40) \)
\( \hat{y} = 73.29 - 63.6 \)
\( \hat{y} = 9.69 \).
So, the estimated demand is 9.69 (in hundred units).In simple words: We calculated the average price and demand. Using a shortcut method with transformed variables, we found the regression coefficient 'b' to be -1.59 and the intercept 'a' to be 73.29. The regression line is \( \hat{y} = 73.29 - 1.59x \). If the price is Rs. 40, the estimated demand is 9.69 hundred units.

🎯 Exam Tip: When dealing with many data points, using the shortcut method with 'u' and 'v' can simplify calculations. Ensure the correct transformation and formula are applied for 'b'.

 

Question 2. The information regarding the experience (in years) of eight workers on a machine and their performance ratings based on the nondefective units they manufactured in every 100 units is as follows:

Experience of worker (years)	5	12	15	8	20	18	22	25
Performance rating	80	82	85	81	90	90	95	97

Obtain the regression line of the performance rating on the experience and estimate the performance rating if a worker has an experience of 17 years.

Answer: Given data for 8 workers: X is years of experience, and Y is performance rating.
Calculate the mean of X: \( \bar{x} = \frac{\Sigma x}{n} = \frac{125}{8} = 15.63 \).
Calculate the mean of Y: \( \bar{y} = \frac{\Sigma y}{n} = \frac{700}{8} = 87.5 \).
To simplify calculations for the regression line \( \hat{y} = a + bx \), new variables \( u = x - A \) and \( v = y - B \) are used, with \( A = 15 \) and \( B = 90 \).
The table for calculation is prepared as follows:

x	y	\( u=(x-A) \) \( A = 15 \)	\( v=(y-B) \) \( B = 90 \)	uv	\( u^2 \)
5	80	-10	-10	100	100
12	82	-3	-8	24	9
15	85	0	-5	0	0
8	81	-7	-9	63	49
20	90	5	0	0	25
18	90	3	0	0	9
22	95	7	5	35	49
25	97	10	7	70	100
\( \Sigma x = 125 \)	\( \Sigma y = 700 \)	\( \Sigma u = 5 \)	\( \Sigma v = -20 \)	\( \Sigma uv = 292 \)	\( \Sigma u^2 = 341 \)

Now, calculate the regression coefficient 'b': \( b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \).
Substitute the values:
\( b = \frac{8(292) - (5)(-20)}{8(341) - (5)^2} \)
\( b = \frac{2336 + 100}{2728 - 25} \)
\( b = \frac{2436}{2703} \approx 0.9 \).
Next, calculate the intercept 'a': \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 87.5 - 0.9(15.63) \)
\( a = 87.5 - 14.07 = 73.43 \).
The regression line for performance rating (Y) on experience (X) is \( \hat{y} = 73.43 + 0.9x \).
To estimate performance rating when a worker has 17 years of experience, substitute \( x = 17 \) into the equation:
\( \hat{y} = 73.43 + 0.9(17) \)
\( \hat{y} = 73.43 + 15.3 \)
\( \hat{y} = 88.73 \).
So, the estimated performance rating for 17 years of experience is 88.73.In simple words: We found the average experience and performance rating. Using transformed variables for simpler calculations, we determined 'b' to be 0.9 and 'a' to be 73.43. The regression line is \( \hat{y} = 73.43 + 0.9x \). For a worker with 17 years of experience, the estimated performance rating is 88.73.

🎯 Exam Tip: The shortcut method using 'u' and 'v' is highly efficient for larger datasets. Ensure the calculation of \( \Sigma uv \) and \( \Sigma u^2 \) is accurate, as these are critical for 'b'.

 

Question 3. The information regarding daily income (in Rs.) and expenditure (in Rs.) of five labour families earning by daily work:

Daily income (Rs.)	200	300	400	600	900
Expenditure (Rs.)	180	270	320	480	700

Obtain the regression line of the expenditure on the daily income. Estimate the expenditure of a family having daily income of Rs. 500.

Answer: Given data for 5 families: X is daily income (in Rs.), and Y is expenditure (in Rs.).
Calculate the mean of X: \( \bar{x} = \frac{\Sigma x}{n} = \frac{2400}{5} = 480 \).
Calculate the mean of Y: \( \bar{y} = \frac{\Sigma y}{n} = \frac{1950}{5} = 390 \).
To simplify finding the regression line \( \hat{y} = a + bx \), new variables \( u = \frac{x-A}{C_x} \) and \( v = \frac{y-B}{C_y} \) are used, with \( A = 400 \), \( C_x = 100 \), \( B = 320 \), and \( C_y = 10 \).
The table for calculation is prepared as follows:

x	y	\( u=\frac{x-A}{C_x} \) \( A = 400 \) \( C_x = 100 \)	\( v=\frac{y-B}{C_y} \) \( B = 320 \) \( C_y = 10 \)	uv	\( u^2 \)
200	180	-2	-14	28	4
300	270	-1	-5	5	1
400	320	0	0	0	0
600	480	2	16	32	4
900	700	5	38	190	25
\( \Sigma x = 2400 \)	\( \Sigma y = 1950 \)	\( \Sigma u = 4 \)	\( \Sigma v = 35 \)	\( \Sigma uv = 255 \)	\( \Sigma u^2 = 34 \)

Now, calculate the regression coefficient 'b' using the formula: \( b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \).
Substitute the values:
\( b = \frac{5(255) - (4)(35)}{5(34) - (4)^2} \times \frac{10}{100} \)
\( b = \frac{1275 - 140}{170 - 16} \times \frac{1}{10} \)
\( b = \frac{1135}{154} \times \frac{1}{10} = \frac{1135}{1540} \approx 0.74 \).
Next, calculate the intercept 'a': \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 390 - 0.74(480) \)
\( a = 390 - 355.2 = 34.8 \).
The regression line for expenditure (Y) on daily income (X) is \( \hat{y} = 34.8 + 0.74x \).
To estimate expenditure when daily income (X) is Rs. 500, substitute \( x = 500 \) into the equation:
\( \hat{y} = 34.8 + 0.74(500) \)
\( \hat{y} = 34.8 + 370 \)
\( \hat{y} = 404.8 \).
So, the estimated expenditure is Rs. 404.8.In simple words: We found average income and expenditure. Using a shortcut method with scaled variables, we calculated 'b' as 0.74 and 'a' as 34.8. The regression line is \( \hat{y} = 34.8 + 0.74x \). For a daily income of Rs. 500, the estimated expenditure is Rs. 404.8.

🎯 Exam Tip: When using the shortcut method with scale changes, ensure you correctly identify and apply \( C_x \) and \( C_y \) in the 'b' formula. This is a common point of error.

 

Question 4. The following information is collected by a firm to know the effect of an advertisement campaign:

Year	1	2	3	4	5	6	7	8
Advertisement cost (ten thousand Rs.)	12	15	15	23	24	38	42	48
Sales (crore Rs.)	5	5.6	5.8	7	7.2	8.8	9.2	9.5

Obtain the regression line of sales on the advertisement cost. Estimate the sales when the advertisement cost is Rs. 5,00,000.

Answer: Given data for a firm's advertisement campaign: X is advertisement cost (in ten thousand Rs.), and Y is sales (in crore Rs.). There are \( n = 8 \) observations.
Calculate the mean of X: \( \bar{x} = \frac{\Sigma x}{n} = \frac{217}{8} \approx 27.13 \).
Calculate the mean of Y: \( \bar{y} = \frac{\Sigma y}{n} = \frac{58.1}{8} \approx 7.26 \).
To simplify calculating the regression line \( \hat{y} = a + bx \), new variables \( u = x - A \) and \( v = \frac{y-B}{C_y} \) are used, with \( A = 24 \), \( B = 7 \), and \( C_y = 0.1 \). The scale factor for x (\( C_x \)) is 1.
The table for calculation is prepared as follows:

Year	x	y	\( u = (x-A) \) \( A = 24 \)	\( v = \frac{y-B}{C_y} \) \( B = 7 \) \( C_y = 0.1 \)	uv	\( u^2 \)
1	12	5	-12	-20	240	144
2	15	5.6	-9	-14	126	81
3	15	5.8	-9	-12	108	81
4	23	7	-1	0	0	1
5	24	7.2	0	2	0	0
6	38	8.8	14	18	252	196
7	42	9.2	18	22	396	324
8	48	9.5	24	25	600	576
\( \Sigma x = 217 \)	\( \Sigma y = 58.1 \)	\( \Sigma u = 25 \)	\( \Sigma v = 21 \)	\( \Sigma uv = 1722 \)	\( \Sigma u^2 = 1403 \)

Now, calculate the regression coefficient 'b' using the formula: \( b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \).
Substitute the values:
\( b = \frac{8(1722) - (25)(21)}{8(1403) - (25)^2} \times \frac{0.1}{1} \)
\( b = \frac{13776 - 525}{11224 - 625} \times 0.1 \)
\( b = \frac{13251}{10599} \times 0.1 = 1.25 \times 0.1 \approx 0.13 \).
Next, calculate the intercept 'a': \( a = \bar{y} - b\bar{x} \).
Substitute the means and 'b': \( a = 7.26 - 0.13(27.13) \)
\( a = 7.26 - 3.53 = 3.73 \).
The regression line for sales (Y) on advertisement cost (X) is \( \hat{y} = 3.73 + 0.13x \).
To estimate sales when advertisement cost is Rs. 5,00,000 (which is 50 ten thousand Rs.), substitute \( x = 50 \) into the equation:
\( \hat{y} = 3.73 + 0.13(50) \)
\( \hat{y} = 3.73 + 6.5 \)
\( \hat{y} = 10.23 \).
So, the estimated sales are Rs. 10.23 crore.In simple words: We calculated the average advertisement cost and sales. By transforming variables for easier calculations, we found 'b' to be 0.13 and 'a' to be 3.73. The regression line is \( \hat{y} = 3.73 + 0.13x \). For an advertisement cost of Rs. 5,00,000, the estimated sales are Rs. 10.23 crore.

🎯 Exam Tip: Pay close attention to the units of the variables, especially when converting an estimated value like Rs. 5,00,000 into the 'ten thousand Rs.' scale used in the problem (i.e., x = 50).

 

Question 4. The following information is collected by a firm to know the effect of an advertisement campaign:

Year	1	2	3	4	5	6	7	8
Advertisement cost (ten thousand Rs.)	12	15	15	23	24	38	42	48
Sales (crore Rs.)	5	5.6	5.8	7	7.2	8.8	9.2	9.5

Obtain the regression line of sales on the advertisement cost. Estimate the sales when the advertisement cost is Rs 5,00,000.
Answer:Here, we have 8 observations, so \(n = 8\). X represents the advertisement cost, and Y represents the sales. First, we find the mean of X and Y: \(x = \frac{\Sigma x}{n} = \frac{217}{8} = 27.13\) \(y = \frac{\Sigma y}{n} = \frac{58.1}{8} = 7.26\) We need to find the regression line \(ŷ = a + bx\) for sales (Y) based on advertisement cost (X). To simplify the calculations for 'a' and 'b', we introduce new variables: \(u = x - A\), where \(A = 24\) \(v = \frac{y-B}{C_y}\), where \(B = 7\) and \(C_y = 0.1\) The table below helps in calculating the values:

Year	X	y	\(u = (x-A)\) \(A = 24\)	\(v = \frac{(y-B)}{C_y}\) \(B = 7\) \(C_y = 0.1\)	uv	\(u^2\)
1	12	5	-12	-20	240	144
2	15	5.6	-9	-14	126	81
3	15	5.8	-9	-12	108	81
4	23	7	-1	0	0	1
5	24	7.2	0	2	0	0
6	38	8.8	14	18	252	196
7	42	9.2	18	22	396	324
8	48	9.5	24	25	600	576
Sum	\( \Sigma x = 217 \)	\( \Sigma y = 58.1 \)	\( \Sigma u = 25 \)	\( \Sigma v = 21 \)	\( \Sigma uv = 1722 \)	\( \Sigma u^2 = 1403 \)

Now, we calculate the regression coefficient \(b\): \[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \] Given \(n = 8\), \( \Sigma uv = 1722 \), \( \Sigma u = 25 \), \( \Sigma v = 21 \), \( \Sigma u^2 = 1403 \). Also, \( C_y = 0.1 \) and \( C_x = 1 \) (since \(u = x-24\), \(C_x\) is 1). Substitute these values into the formula: \[ b = \frac{8(1722)-(25)(21)}{8(1403)-(25)^2} \times \frac{0.1}{1} \]
\( \implies b = \frac{13776-525}{11224-625} \times \frac{0.1}{1} \)
\( \implies b = \frac{13251}{10599} \times 0.1 \)
\( \implies b = 1.2502 \times 0.1 \)
\( \implies b \approx 0.13 \) Next, we find the intercept \(a\): \( a = \bar{y} - b\bar{x} \) Using \( \bar{y} = 7.26 \), \( \bar{x} = 27.13 \), and \( b = 0.13 \): \( a = 7.26 - 0.13(27.13) \) \( a = 7.26 - 3.5269 \) \( a \approx 3.73 \) The regression line for sales (Y) on advertisement cost (X) is: \( ŷ = a + bx \) Substituting the values of \(a\) and \(b\): \( ŷ = 3.73 + 0.13x \) Now, we estimate sales (Y) when the advertisement cost (X) is Rs 5,00,000, which is 50 ten thousand Rs. So, \(x = 50\). \( ŷ = 3.73 + 0.13(50) \) \( ŷ = 3.73 + 6.5 \) \( ŷ = 10.23 \) Therefore, the estimated sales are Rs 10.23 crore when the advertisement cost is Rs 5,00,000.
In simple words: This question asks us to find a formula that predicts sales based on advertisement cost and then use that formula to guess sales for a specific advertisement amount. We first found the average advertisement cost and sales, then calculated a special number (regression coefficient 'b') that tells us how much sales change for each unit change in advertisement. We also found another number ('a') which is the starting sales when advertisement is zero. Using these numbers, we built a prediction formula and then put in the given advertisement cost to find the predicted sales.

🎯 Exam Tip: When dealing with large numbers, using transformed variables (u and v) can simplify calculations. Remember to correctly apply the scaling factors \(C_x\) and \(C_y\) when finding 'b' and ensure all steps for calculating 'a' and 'ŷ' are shown clearly for full marks.

 

Question 5. The information of eight construction companies regarding the number of contracts received in a year and the annual profit is as follows:

No. of contracts	2	5	9	12	6	4	8	10
Annual profit (lakh Rs.)	100	300	700	1000	350	250	700	750

Obtain the regression line of the annual profit on the number of contracts. Verify the reliability of the regression model.
Answer:Here, we have 8 companies, so \(n = 8\). X represents the number of contracts, and Y represents the annual profit. First, we find the mean of X and Y: \( \bar{x} = \frac{\Sigma x}{n} = \frac{56}{8} = 7 \) \( \bar{y} = \frac{\Sigma y}{n} = \frac{4150}{8} = 518.75 \) We want to find the regression line \( ŷ = a + bx \) for annual profit (Y) based on the number of contracts (X). To simplify calculations for 'a' and 'b', we use new variables: \( u = x - A \), where \( A = 8 \) \( v = \frac{y-B}{C_y} \), where \( B = 700 \) and \( C_y = 50 \) (since \(u = x-8\), \(C_x\) is 1). The table below shows the calculations:

X	y	\(u = (x-A)\) \(A = 8\)	\(v = \frac{(y-B)}{C_y}\) \(B = 700\) \(C_y = 50\)	uv	\(u^2\)	\(v^2\)
2	100	-6	-12	72	36	144
5	300	-3	-8	24	9	64
9	700	1	0	0	1	0
12	1000	4	6	24	16	36
6	350	-2	-7	14	4	49
4	250	-4	-9	36	16	81
8	700	0	0	0	0	0
10	750	2	1	2	4	1
\( \Sigma x = 56 \)	\( \Sigma y = 4150 \)	\( \Sigma u = -8 \)	\( \Sigma v = -29 \)	\( \Sigma uv = 172 \)	\( \Sigma u^2 = 86 \)	\( \Sigma v^2 = 375 \)

Now, we calculate the regression coefficient \(b\): \[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \] Given \(n = 8\), \( \Sigma uv = 172 \), \( \Sigma u = -8 \), \( \Sigma v = -29 \), \( \Sigma u^2 = 86 \). Also, \( C_y = 50 \) and \( C_x = 1 \). Substitute these values into the formula: \[ b = \frac{8(172)-(-8)(-29)}{8(86)-(-8)^2} \times \frac{50}{1} \]
\( \implies b = \frac{1376-232}{688-64} \times 50 \)
\( \implies b = \frac{1144}{624} \times 50 \)
\( \implies b = 1.8333 \times 50 \)
\( \implies b \approx 91.67 \) Next, we find the intercept \(a\): \( a = \bar{y} - b\bar{x} \) Using \( \bar{y} = 518.75 \), \( \bar{x} = 7 \), and \( b = 91.67 \): \( a = 518.75 - 91.67(7) \) \( a = 518.75 - 641.69 \) \( a \approx -122.94 \) The regression line for annual profit (Y) on the number of contracts (X) is: \( ŷ = -122.94 + 91.67x \) To check the reliability of the regression model, we calculate the coefficient of determination \(R^2\). \(R^2\) is the square of the correlation coefficient (\(r^2\)). \[ R^2 = \left[\frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^2 - (\Sigma u)^2} \cdot \sqrt{n \Sigma v^2 - (\Sigma v)^2}}\right]^2 \] Given \(n = 8\), \( \Sigma u = -8 \), \( \Sigma v = -29 \), \( \Sigma uv = 172 \), \( \Sigma u^2 = 86 \), and \( \Sigma v^2 = 375 \). Substitute these values into the formula: \[ R^2 = \left[\frac{8(172)-(-8)(-29)}{\sqrt{8(86)-(-8)^2} \cdot \sqrt{8(375)-(-29)^2}}\right]^2 \]
\( \implies R^2 = \left[\frac{1376 - 232}{\sqrt{688-64} \cdot \sqrt{3000-841}}\right]^2 \)
\( \implies R^2 = \left[\frac{1144}{\sqrt{624} \cdot \sqrt{2159}}\right]^2 \)
\( \implies R^2 = \left[\frac{1144}{\sqrt{1347216}}\right]^2 \)
\( \implies R^2 = \left[\frac{1144}{1160.7}\right]^2 \)
\( \implies R^2 = (0.9856)^2 \)
\( \implies R^2 \approx 0.97 \) Since \(R^2 = 0.97\) (which is very close to 1), the regression model is highly reliable.
In simple words: We calculated a formula to predict a company's profit based on how many contracts it gets. We found the 'b' and 'a' values using a special method to make the calculations easier. After getting the prediction formula, we checked how good it is using a number called \(R^2\). Since \(R^2\) is very close to 1, it means our prediction formula is very accurate and trustworthy.

🎯 Exam Tip: Always show all intermediate calculation steps, especially for 'b' and 'a', as they carry marks. When asked to verify reliability, calculate \(R^2\) and interpret its value (close to 1 means reliable, close to 0 means less reliable). Remember to handle negative signs correctly in calculations.

 

Question 6. Obtain the regression line of Y on X from the following data and estimate Y for X = 30 : n = 10, Σx = 250, Σy = 300, Σxy = 7900, Σx² = 6500
Answer:The given data includes: \(n = 10\), \( \Sigma x = 250 \), \( \Sigma y = 300 \), \( \Sigma xy = 7900 \), \( \Sigma x^2 = 6500 \). First, calculate the means for X and Y: \( \bar{x} = \frac{\Sigma x}{n} = \frac{250}{10} = 25 \) \( \bar{y} = \frac{\Sigma y}{n} = \frac{300}{10} = 30 \) We need to find the regression line \( ŷ = a + bx \) of Y on X. First, calculate the regression coefficient \(b\): \[ b = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{n \Sigma x^2 - (\Sigma x)^2} \] Substitute the given values into the formula: \[ b = \frac{10(7900)-(250)(300)}{10(6500)-(250)^2} \]
\( \implies b = \frac{79000-75000}{65000-62500} \)
\( \implies b = \frac{4000}{2500} \)
\( \implies b = 1.6 \) Next, calculate the intercept \(a\): \( a = \bar{y} - b\bar{x} \) Substitute the values of \( \bar{y} = 30 \), \( b = 1.6 \), and \( \bar{x} = 25 \): \( a = 30 - 1.6(25) \) \( a = 30 - 40 \) \( a = -10 \) The regression line of Y on X is: \( ŷ = a + bx \) Substituting the values of \(a\) and \(b\): \( ŷ = -10 + 1.6x \) Now, we need to estimate the value of Y when \(X = 30\). Substitute \(x = 30\) into the regression line equation: \( ŷ = -10 + 1.6(30) \) \( ŷ = -10 + 48 \) \( ŷ = 38 \) Thus, the estimated value of Y when X is 30 is 38.
In simple words: This problem gave us summarized data to find a line that predicts Y from X. We calculated the average values for X and Y, then used a formula to find the slope (b) of the prediction line, which shows how much Y changes for a unit change in X. After that, we found the starting point (a) of this line. With 'a' and 'b', we formed our prediction equation. Finally, we used this equation to estimate Y when X is 30.

🎯 Exam Tip: Pay close attention to the formula for 'b' and 'a'. Ensure correct substitution of given aggregated sums (\(\Sigma x\), \(\Sigma y\), \(\Sigma xy\), \(\Sigma x^2\)). Double-check arithmetic, especially during the final estimation step.

 

Question 7. The following results are obtained for a data : n = 12, Σx = 30, Σy = 5, Σx² = 670, Σxy = 344 Later on, it was known that one pair (10, 14) was wrongly taken as (11, 4). By correcting the above measures, obtain the regression line of Y on X. Estimate Y for X = 5.
Answer:We are given initial data: \(n = 12\), \( \Sigma x = 30 \), \( \Sigma y = 5 \), \( \Sigma x^2 = 670 \), \( \Sigma xy = 344 \). A mistake was found: the observation (10, 14) was wrongly recorded as (11, 4). Let's correct the sums: Original \( \Sigma x = 30 \) True x-value to add = 10 False x-value to remove = 11 Correct \( \Sigma x = 30 + 10 - 11 = 29 \) Original \( \Sigma y = 5 \) True y-value to add = 14 False y-value to remove = 4 Correct \( \Sigma y = 5 + 14 - 4 = 15 \) Original \( \Sigma x^2 = 670 \) True x-squared value to add = \(10^2 = 100\) False x-squared value to remove = \(11^2 = 121\) Correct \( \Sigma x^2 = 670 + 100 - 121 = 649 \) Original \( \Sigma xy = 344 \) True xy-value to add = \(10 \times 14 = 140\) False xy-value to remove = \(11 \times 4 = 44\) Correct \( \Sigma xy = 344 + 140 - 44 = 440 \) Now, using the corrected values, we find the means: \( \bar{x} = \frac{\text{Correct } \Sigma x}{n} = \frac{29}{12} \approx 2.42 \) \( \bar{y} = \frac{\text{Correct } \Sigma y}{n} = \frac{15}{12} = 1.25 \) We will find the regression line \( ŷ = a + bx \) of Y on X using these corrected sums. First, calculate the regression coefficient \(b\): \[ b = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{n \Sigma x^2 - (\Sigma x)^2} \] Substitute the corrected values into the formula: \[ b = \frac{12(440)-(29)(15)}{12(649)-(29)^2} \]
\( \implies b = \frac{5280-435}{7788-841} \)
\( \implies b = \frac{4845}{6947} \)
\( \implies b \approx 0.70 \) Next, calculate the intercept \(a\): \( a = \bar{y} - b\bar{x} \) Substitute the corrected values of \( \bar{y} = 1.25 \), \( b = 0.70 \), and \( \bar{x} = 2.42 \): \( a = 1.25 - 0.70(2.42) \) \( a = 1.25 - 1.694 \) \( a \approx -0.44 \) The regression line of Y on X is: \( ŷ = a + bx \) Substituting the values of \(a\) and \(b\): \( ŷ = -0.44 + 0.70x \) Finally, we need to estimate the value of Y when \(X = 5\). Substitute \(x = 5\) into the regression line equation: \( ŷ = -0.44 + 0.70(5) \) \( ŷ = -0.44 + 3.50 \) \( ŷ = 3.06 \) Thus, the estimated value of Y when X is 5 is 3.06.
In simple words: This question first asked us to fix some mistakes in the given data totals. We removed the incorrect data point and added the correct one to get new, accurate totals for sums of X, Y, X-squared, and XY. Using these corrected totals, we then found the regression line's slope ('b') and intercept ('a'). Finally, we used this new, corrected prediction formula to estimate the value of Y when X is 5.

🎯 Exam Tip: Data correction problems require meticulous attention to detail. Ensure you correctly subtract the incorrect values and add the true values for all sums (\(\Sigma x\), \(\Sigma y\), \(\Sigma x^2\), \(\Sigma xy\)) before proceeding. Any error in correction will propagate through the entire calculation.