GSEB Class 12 Statistics Solutions Chapter 3 Linear Regression Exercise 3.2

Linear Regression Ex 3.2

Question 1. The following information is obtained from a study to know the effect of use of fertilizer on the yield of cotton:

Consumption of fertilizer (10 kg) x	28	35	25	24	20	25	20
Yield of cotton per hectare (Quintals) y	128	140	115	120	105	122	100

Obtain the regression line of Y on X and estimate the yield of cotton per hectare if 300 kg fertilizer is used.

Answer:

Here, we have 7 pairs of data (n = 7). X represents the amount of fertilizer used, and Y represents the cotton yield. We first calculate the average values for X and Y.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{177}{7} = 25.29 \] \[ \bar{y} = \frac{\Sigma y}{n} = \frac{830}{7} = 118.57 \]

Our goal is to find the regression line for cotton yield (Y) based on fertilizer consumption (X), which has the form \( \hat{y} = a + bx \). Since the average values of x and y are not whole numbers, we will use a shortcut method to find 'a' and 'b'. We define new variables, where A is taken as 25 and B as 120.

The table for calculation is prepared as follows:

x	y			uv	u2
		u = (x-A) A = 25	v = (y-B) B = 120
28	128	3	8	24	9
35	140	10	20	200	100
25	115	0	-5	0	0
24	120	-1	0	0	1
20	105	-5	-15	75	25
25	122	0	2	0	0
20	100	-5	-20	100	25
\( \Sigma x = 177 \)	\( \Sigma y = 830 \)	\( \Sigma u = 2 \)	\( \Sigma v = -10 \)	\( \Sigma uv = 399 \)	\( \Sigma u^2 = 160 \)

The formula to calculate the regression coefficient 'b' is:

\[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \]

Using the values: n = 7, \( \Sigma uv = 399 \), \( \Sigma u = 2 \), \( \Sigma v = -10 \), and \( \Sigma u^2 = 160 \).

\[ b = \frac{7(399)-(2)(-10)}{7(160)-(2)^2} \] \[ = \frac{2793+20}{1120-4} \] \[ = \frac{2813}{1116} \] \[ = 2.52 \]

Now, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 118.57 \), \( \bar{x} = 25.29 \), and \( b = 2.52 \).

\[ a = 118.57 - 2.52 (25.29) \] \[ = 118.57 - 63.73 \] \[ = 54.84 \]

The regression line for Y on X is obtained by putting the values of 'a' and 'b' into the equation \( \hat{y} = a + bx \).

So, the regression line is: \( \hat{y} = 54.84 + 2.52x \).

To estimate the yield of cotton when 300 kg of fertilizer is used, we note that 300 kg is equivalent to 30 units of x (since x is in 10 kg units).

We substitute \( x = 30 \) into the regression equation:

\[ \hat{y} = 54.84 + 2.52 (30) \] \[ = 54.84 + 75.6 \] \[ = 130.44 \]

Therefore, the estimated yield of cotton is 130.44 Quintals.

In simple words: We want to guess how much cotton will grow if we use a certain amount of fertilizer. We used past data to find a rule (a regression line). This rule helps us predict the cotton yield. When 300 kg of fertilizer is used, our prediction is 130.44 Quintals of cotton.

🎯 Exam Tip: Remember to correctly identify the units for 'x' (10 kg in this case) when performing estimations. Also, ensure accurate calculations for \( \Sigma u \), \( \Sigma v \), \( \Sigma uv \), and \( \Sigma u^2 \) as any error will propagate throughout the problem. Using the shortcut method is essential when \( \bar{x} \) and \( \bar{y} \) are not integers.

 

Question 2. To know the relationship between the heights of father and sons, obtain the regression line of height of son on the height of father from the following information of eight pairs of fathers and adult sons.

Height of father (cm) x	167	169	171	168	173	166	167	165
Height of son (cm) y	158	170	169	172	170	168	164	167

Estimate the height of a son whose father's height is 170 cm.

Answer:

We have 8 data pairs (n = 8) for fathers' and sons' heights. X is the father's height, and Y is the son's height. First, we find the average height for fathers and sons.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{1346}{8} = 168.25 \text{ cm} \] \[ \bar{y} = \frac{\Sigma y}{n} = \frac{1338}{8} = 167.25 \text{ cm} \]

We want to find the regression line \( \hat{y} = a + bx \) that predicts a son's height (Y) from his father's height (X). Since the mean values are not integers, we will use a shortcut method by introducing new variables \( u = x - A \) and \( v = y - B \). We choose A = 168 and B = 167 for simpler calculations.

The table for calculation is prepared as follows:

x	y			uv	u2
		u = (x-A) A = 168	v = (y-B) B = 167
167	158	-1	-9	9	1
169	170	1	3	3	1
171	169	3	2	6	9
168	172	0	5	0	0
173	170	5	3	15	25
166	168	-2	1	-2	4
167	164	-1	-3	3	1
165	167	-3	0	0	9
\( \Sigma x = 1346 \)	\( \Sigma y = 1338 \)	\( \Sigma u = 2 \)	\( \Sigma v = 2 \)	\( \Sigma uv = 34 \)	\( \Sigma u^2 = 50 \)

The formula for the regression coefficient 'b' is:

\[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \]

Using n = 8, \( \Sigma uv = 34 \), \( \Sigma u = 2 \), \( \Sigma v = 2 \), and \( \Sigma u^2 = 50 \).

\[ b = \frac{8(34)-(2)(2)}{8(50)-(2)^2} \] \[ = \frac{272-4}{400-4} \] \[ = \frac{268}{396} \] \[ = 0.68 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substitute the known values: \( \bar{y} = 167.25 \), \( \bar{x} = 168.25 \), and \( b = 0.68 \).

\[ a = 167.25 - 0.68 (168.25) \] \[ = 167.25 - 114.41 \] \[ = 52.84 \]

By substituting 'a' and 'b' into \( \hat{y} = a + bx \), we get the regression line:

\[ \hat{y} = 52.84 + 0.68x \]

To estimate the height of a son when the father's height (X) is 170 cm, we put \( x = 170 \) into the regression equation:

\[ \hat{y} = 52.84 + 0.68 (170) \] \[ = 52.84 + 115.6 \] \[ = 168.44 \]

So, the estimated height of the son is 168.44 cm.

In simple words: We looked at how tall fathers and sons are to find a pattern. This pattern helps us predict a son's height if we know his father's height. For a father who is 170 cm tall, we estimate his son will be about 168.44 cm tall.

🎯 Exam Tip: When using the shortcut method, ensure careful selection of arbitrary origins A and B to simplify calculations. Always cross-check the sum of u, v, uv, and u² columns. Pay close attention to the decimal points in the final calculation of 'a' and 'b' and the estimation.

 

Question 3. From the following information of altitude and the amount of effective oxygen in air at the place, obtain the regression line of amount of effective oxygen (Y) on the altitude (X): (305 metre = 1000 feet)

Altitude (305 meter) x	0	1	2	3	4	5	6
Effective Oxygen (%) y	20.9	20.1	19.4	17.9	17.9	17.3	16.6

If the altitude of a place is 7 units (1 unit = 305 metre), estimate the percentage of effective oxygen in air at that place.

Answer:

In this problem, we have 7 data pairs (n = 7). X represents the altitude, and Y represents the percentage of effective oxygen. We first find the average values for X and Y.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{21}{7} = 3 \] \[ \bar{y} = \frac{\Sigma y}{n} = \frac{130.1}{7} = 18.59 \]

We aim to determine the regression line \( \hat{y} = a + bx \) for effective oxygen (Y) based on altitude (X). Since the mean values are not simple integers, and the Y values are multiples of 0.1, we'll use a shortcut method with new variables. We define \( u = \frac{x-A}{C_x} \) and \( v = \frac{y-B}{C_y} \). We choose A = 3, \( C_x = 1 \), B = 17.9, and \( C_y = 0.1 \).

The table for calculation is prepared as follows:

x	y			uv	u2
		u = \(\frac{x-A}{C_x}\) A=3, Cx=1	v = \(\frac{y-B}{C_y}\) B=17.9, Cy=0.1
0	20.9	-3	30	-90	9
1	20.1	-2	22	-44	4
2	19.4	-1	15	-15	1
3	17.9	0	0	0	0
4	17.9	1	0	0	1
5	17.3	2	-6	-12	4
6	16.6	3	-13	-39	9
\( \Sigma x = 21 \)	\( \Sigma y = 130.1 \)	\( \Sigma u = 0 \)	\( \Sigma v = 48 \)	\( \Sigma uv = -200 \)	\( \Sigma u^2 = 28 \)

The formula for the regression coefficient 'b' when using new variables is:

\[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \]

Using n = 7, \( \Sigma uv = -200 \), \( \Sigma u = 0 \), \( \Sigma v = 48 \), \( \Sigma u^2 = 28 \), \( C_y = 0.1 \), and \( C_x = 1 \).

\[ b = \frac{7(-200)-(0)(48)}{7(28)-(0)^2} \times \frac{0.1}{1} \] \[ = \frac{-1400-0}{196-0} \times \frac{0.1}{1} \] \[ = \frac{-1400 \times 0.1}{196} \] \[ = \frac{-140}{196} \] \[ = -0.71 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the known values: \( \bar{y} = 18.59 \), \( \bar{x} = 3 \), and \( b = -0.71 \).

\[ a = 18.59 - (-0.71) (3) \] \[ = 18.59 + 2.13 \] \[ = 20.72 \]

By substituting 'a' and 'b' into \( \hat{y} = a + bx \), we get the regression line:

\[ \hat{y} = 20.72 - 0.71x \]

To estimate the percentage of effective oxygen when the altitude (X) is 7 units, we put \( x = 7 \) into the regression equation:

\[ \hat{y} = 20.72 - 0.71 (7) \] \[ = 20.72 - 4.97 \] \[ = 15.75 \]

Therefore, the estimated percentage of effective oxygen in the air is 15.75%.

In simple words: We used data on altitude and oxygen levels to create a rule that helps us guess the oxygen level at different heights. This rule shows that oxygen tends to decrease as altitude increases. At an altitude of 7 units, we predict the oxygen level to be 15.75%.

🎯 Exam Tip: When using the shortcut method with scaling factors \( C_x \) and \( C_y \), remember to include the ratio \( \frac{C_y}{C_x} \) in the 'b' coefficient calculation. Be careful with negative signs in calculations, especially when subtracting a negative value. Ensure the final interpretation of units (percentage in this case) is correct.

 

Question 4. The following information is obtained to study the relation between the carpet area in a house and its monthly rent in a city:

Carpet area (square meter) x	55	60	75	80	100	120	140
Monthly rent (Rs.) y	18,000	19,000	20,000	20,000	25,000	30,000	50,000

Obtain the regression line of Y on X. Estimate the monthly rent of a house having carpet area of 110 square metre.

Answer:

We are given 7 pairs of data (n = 7). X represents the carpet area, and Y represents the monthly rent. We first calculate the average carpet area and average monthly rent.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{630}{7} = 90 \] \[ \bar{y} = \frac{\Sigma y}{n} = \frac{182000}{7} = 26000 \]

To find the regression line \( \hat{y} = a + bx \) for monthly rent (Y) based on carpet area (X), we will use a shortcut method. We introduce new variables \( u = \frac{x-A}{C_x} \) and \( v = \frac{y-B}{C_y} \). We select A = 80, \( C_x = 5 \), B = 25000, and \( C_y = 1000 \) to simplify computations.

The table for calculation is prepared as follows:

x	y			uv	u2
		u = \(\frac{x-A}{C_x}\) A=80, Cx=5	v = \(\frac{y-B}{C_y}\) B=25000, Cy=1000
55	18000	-5	-7	35	25
60	19000	-4	-6	24	16
75	20000	-1	-5	5	1
80	20000	0	-5	0	0
100	25000	4	0	0	16
120	30000	8	5	40	64
140	50000	12	25	300	144
\( \Sigma x = 630 \)	\( \Sigma y = 182000 \)	\( \Sigma u = 14 \)	\( \Sigma v = 7 \)	\( \Sigma uv = 404 \)	\( \Sigma u^2 = 266 \)

The formula for the regression coefficient 'b' is:

\[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \]

Using n = 7, \( \Sigma uv = 404 \), \( \Sigma u = 14 \), \( \Sigma v = 7 \), \( \Sigma u^2 = 266 \), \( C_x = 5 \), and \( C_y = 1000 \).

\[ b = \frac{7(404)-(14)(7)}{7(266)-(14)^2} \times \frac{1000}{5} \] \[ = \frac{2828-98}{1862-196} \times 200 \] \[ = \frac{2730}{1666} \times 200 \] \[ = \frac{546000}{1666} \] \[ = 327.73 \]

Now, we calculate 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 26000 \), \( b = 327.73 \), and \( \bar{x} = 90 \).

\[ a = 26000 - 327.73 (90) \] \[ = 26000 - 29495.7 \] \[ = -3495.7 \]

The regression line for Y on X is formed by substituting 'a' and 'b' into \( \hat{y} = a + bx \).

\[ \hat{y} = -3495.7 + 327.73x \]

To estimate the monthly rent for a house with a carpet area (X) of 110 square meters, we substitute \( x = 110 \) into the regression equation:

\[ \hat{y} = -3495.7 + 327.73 (110) \] \[ = -3495.7 + 36050.3 \] \[ = 32554.6 \]

Therefore, the estimated monthly rent is Rs. 32554.6.

In simple words: We studied how carpet area affects house rent. Using this data, we made a formula to predict rent based on area. For a house with 110 square meters of carpet area, we estimate the monthly rent to be Rs. 32554.6.

🎯 Exam Tip: When dealing with large numerical values like rent, using scaling factors \( C_x \) and \( C_y \) in the shortcut method is highly beneficial to manage calculations. Ensure correct application of the \( \frac{C_y}{C_x} \) ratio and proper handling of negative intercepts ('a' value).

 

Question 5. The following sample data is obtained to study the relation between the number of customers visiting a mall per day and the sales (ten thousand Rs.) :

No. of customers x	50	70	100	70	150	120
Sales (ten thousand Rs.) y	2.0	2.0	2.5	1.4	4.0	2.5

Obtain the regression line of Y on X. Estimate the sales of a mall if 80 customers have visited the mall on a particular day.

Answer:

We have 6 data pairs (n = 6). X represents the number of customers, and Y represents the sales in ten thousand Rs. We first calculate the average number of customers and average sales.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{560}{6} = 93.33 \] \[ \bar{y} = \frac{\Sigma y}{n} = \frac{14.4}{6} = 2.4 \]

To find the regression line \( \hat{y} = a + bx \) for sales (Y) based on customer count (X), we use a shortcut method. We define new variables \( u = \frac{x-A}{C_x} \) and \( v = \frac{y-B}{C_y} \). We choose A = 100, \( C_x = 10 \), B = 2.5, and \( C_y = 0.1 \) to make calculations easier.

The table for calculation is prepared as follows:

x	y			uv	u2
		u = \(\frac{x-A}{C_x}\) A=100, Cx=10	v = \(\frac{y-B}{C_y}\) B=2.5, Cy=0.1
50	2.0	-5	-5	25	25
70	2.0	-3	-5	15	9
100	2.5	0	0	0	0
70	1.4	-3	-11	33	9
150	4.0	5	15	75	25
120	2.5	2	0	0	4
\( \Sigma x = 560 \)	\( \Sigma y = 14.4 \)	\( \Sigma u = -4 \)	\( \Sigma v = -6 \)	\( \Sigma uv = 148 \)	\( \Sigma u^2 = 72 \)

The formula for the regression coefficient 'b' is:

\[ b = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2} \times \frac{C_y}{C_x} \]

Using n = 6, \( \Sigma uv = 148 \), \( \Sigma u = -4 \), \( \Sigma v = -6 \), \( \Sigma u^2 = 72 \), \( C_x = 10 \), and \( C_y = 0.1 \).

\[ b = \frac{6(148)-(-4)(-6)}{6(72)-(-4)^2} \times \frac{0.1}{10} \] \[ = \frac{888-24}{432-16} \times \frac{0.1}{10} \] \[ = \frac{864 \times 0.1}{416 \times 10} \] \[ = \frac{86.4}{4160} \] \[ = 0.02 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 2.4 \), \( \bar{x} = 93.33 \), and \( b = 0.02 \).

\[ a = 2.4 - 0.02 (93.33) \] \[ = 2.4 - 1.87 \] \[ = 0.53 \]

The regression line for Y on X is obtained by placing 'a' and 'b' into \( \hat{y} = a + bx \).

\[ \hat{y} = 0.53 + 0.02x \]

To estimate the sales when 80 customers visit the mall, we substitute \( x = 80 \) into the regression equation:

\[ \hat{y} = 0.53 + 0.02 (80) \] \[ = 0.53 + 1.6 \] \[ = 2.13 \]

So, the estimated sales are Rs. 2.13 (ten thousand).

In simple words: We used past data to find a connection between how many customers visit a mall and how much they buy. This connection helps us predict sales. If 80 customers visit the mall, we expect the sales to be Rs. 2.13 (which means Rs. 21,300).

🎯 Exam Tip: Pay close attention to the units of Y (ten thousand Rs.) when interpreting the final estimated value. Careful handling of negative signs during shortcut calculations is vital. Always re-verify the sum of products (Σuv) and squared terms (Σu²) to avoid errors.

 

Question 6. The following information is given for ten firms running business of clothes in a city regarding their average annual profit (in lakh Rs.) and average annual administrative cost (in lakh Rs.) :

Particulars	Profit (in lakh Rs.) x	Administrative Cost (in lakh Rs.) y
Mean	60	25
Standard Deviation	6	3
Covariance	10.4

Obtain the regression line of Administrative Cost (Y) on Profit (X). Estimate the administrative cost if the average annual profit is 70 lakh Rs.

Answer:

We are provided with the following statistics: mean profit \( \bar{x} = 60 \) lakh Rs., mean administrative cost \( \bar{y} = 25 \) lakh Rs., standard deviation of profit \( S_x = 6 \), standard deviation of administrative cost \( S_y = 3 \), and the covariance between profit and administrative cost \( Cov(x, y) = 10.4 \).

Our goal is to find the regression line \( \hat{y} = a + bx \) for administrative cost (Y) based on profit (X).

The formula for the regression coefficient 'b' using covariance and standard deviation is:

\[ b = \frac{Cov(x, y)}{S_x^2} \]

Substituting the given values:

\[ b = \frac{10.4}{(6)^2} \] \[ = \frac{10.4}{36} \] \[ = 0.29 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 25 \), \( b = 0.29 \), and \( \bar{x} = 60 \).

\[ a = 25 - 0.29 (60) \] \[ = 25 - 17.4 \] \[ = 7.6 \]

By substituting 'a' and 'b' into \( \hat{y} = a + bx \), we get the regression line:

\[ \hat{y} = 7.6 + 0.29x \]

To estimate the administrative cost when the average annual profit (X) is 70 lakh Rs., we substitute \( x = 70 \) into the regression equation:

\[ \hat{y} = 7.6 + 0.29 (70) \] \[ = 7.6 + 20.3 \] \[ = 27.9 \]

Therefore, the estimated administrative cost is 27.9 lakh Rs.

In simple words: We used information about profit and administrative costs to create a formula. This formula helps us guess the administrative cost based on how much profit a business makes. If a business makes 70 lakh Rs. profit, we expect its administrative cost to be about 27.9 lakh Rs.

🎯 Exam Tip: When given mean, standard deviation, and covariance, use the specific formula \( b = \frac{Cov(x, y)}{S_x^2} \) directly to find the regression coefficient 'b'. Ensure all units (lakh Rs. in this case) are consistently carried through the problem and stated in the final answer.

 

Question 7. The following information is obtained to study the relationship between average rainfall (in cm) and the yield of maize (in quintal per hectare) in different talukaa of Gujarat:

Particulars	Rainfall (cm) x	Yield of Maize (Quintal per Hectare) y
Mean	82	180
Variance	64	225
Correlation coefficient	0.82

Estimate the yield of maize when the rainfall is 60 cm.

Answer:

We are given the following information: mean rainfall \( \bar{x} = 82 \) cm, mean maize yield \( \bar{y} = 180 \) quintals per hectare, correlation coefficient \( r = 0.82 \). We are also given variance of rainfall \( S_x^2 = 64 \), which means its standard deviation \( S_x = \sqrt{64} = 8 \). Similarly, variance of maize yield \( S_y^2 = 225 \), so its standard deviation \( S_y = \sqrt{225} = 15 \).

To predict the maize yield (Y) based on rainfall (X), we will find the regression line \( \hat{y} = a + bx \).

The formula for the regression coefficient 'b' using the correlation coefficient and standard deviations is:

\[ b = r \frac{S_y}{S_x} \]

Substituting \( r = 0.82 \), \( S_y = 15 \), and \( S_x = 8 \).

\[ b = 0.82 \frac{15}{8} \] \[ = \frac{12.3}{8} \] \[ = 1.54 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 180 \), \( b = 1.54 \), and \( \bar{x} = 82 \).

\[ a = 180 - 1.54 (82) \] \[ = 180 - 126.28 \] \[ = 53.72 \]

By substituting 'a' and 'b' into \( \hat{y} = a + bx \), we get the regression line:

\[ \hat{y} = 53.72 + 1.54x \]

To estimate the yield of maize when the rainfall (X) is 60 cm, we substitute \( x = 60 \) into the regression equation:

\[ \hat{y} = 53.72 + 1.54 (60) \] \[ = 53.72 + 92.40 \] \[ = 146.12 \]

Therefore, the estimated yield of maize is 146.12 quintals per hectare.

In simple words: We used data about rainfall and maize yield to find a rule that links them. This rule helps us predict how much maize will grow based on the rainfall. If the rainfall is 60 cm, we expect about 146.12 quintals of maize per hectare.

🎯 Exam Tip: When variances are given, remember to calculate the standard deviations by taking the square root before using them in the formula for 'b'. This formula using correlation coefficient and standard deviations is a quick way to find 'b' when direct covariance or sum of products are not provided.

 

Question 8. The following results are obtained to study the relation between the price of battery (cell) of wrist watch in rupees (X) and its supply in hundred units (Y) : n = 10, \( \Sigma x = 130 \), \( \Sigma y = 220 \), \( \Sigma x^2 = 2288 \) and \( \Sigma xy = 3467 \). Obtain the regression line of Y on X and estimate the supply when price is Rs. 16.

Answer:

We are given the following information for 10 firms (n = 10): sum of prices \( \Sigma x = 130 \), sum of supplies \( \Sigma y = 220 \), sum of squared prices \( \Sigma x^2 = 2288 \), and sum of product of price and supply \( \Sigma xy = 3467 \). First, we calculate the average price and average supply.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{130}{10} = 13 \] \[ \bar{y} = \frac{\Sigma y}{n} = \frac{220}{10} = 22 \]

Our goal is to determine the regression line \( \hat{y} = a + bx \) for supply (Y) based on price (X).

The formula for the regression coefficient 'b' is:

\[ b = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{n \Sigma x^2 - (\Sigma x)^2} \]

Substituting the given values:

\[ b = \frac{10(3467)-(130)(220)}{10(2288)-(130)^2} \] \[ = \frac{34670-28600}{22880-16900} \] \[ = \frac{6070}{5980} \] \[ = 1.02 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 22 \), \( b = 1.02 \), and \( \bar{x} = 13 \).

\[ a = 22 - 1.02 (13) \] \[ = 22 - 13.26 \] \[ = 8.74 \]

By substituting 'a' and 'b' into \( \hat{y} = a + bx \), we get the regression line:

\[ \hat{y} = 8.74 + 1.02x \]

To estimate the supply (Y) when the price (X) is Rs. 16, we substitute \( x = 16 \) into the regression equation:

\[ \hat{y} = 8.74 + 1.02 (16) \] \[ = 8.74 + 16.32 \] \[ = 25.06 \]

Therefore, the estimated supply is 25.06 hundred units.

In simple words: We used data on battery price and supply to find a rule that helps us predict how many batteries will be supplied at a certain price. If the price of a battery is Rs. 16, we expect a supply of 25.06 hundred units (which means 2506 units).

🎯 Exam Tip: When given sums directly (Σx, Σy, Σx², Σxy), use the direct formula for 'b' which is efficient. Double-check calculations involving multiplication and subtraction, especially to avoid common arithmetic errors. Clearly state the units (hundred units) in the final estimated answer.

 

Question 9. The information regarding maximum temperature (X) and sale of ice cream (Y) of six different days in summer for a city is given below:

Maximum temperature = X (in celsius).

Sale of ice cream = Y (in lakh Rs.)

Given: n = 6, \( \bar{x} = 40 \), \( \bar{y} = 1.2 \), \( \Sigma xy = 306 \), and \( S_x^2 = 20 \). Obtain the regression line of sale of ice cream on maximum temperature. Estimate the sale of ice cream if the maximum temperature on a day is 42 Celsius.

Answer:

We are provided with data for 6 days (n = 6), where X is the maximum temperature and Y is the sale of ice cream in lakh Rs. The given statistics are: mean temperature \( \bar{x} = 40 \), mean sales \( \bar{y} = 1.2 \), sum of product of X and Y \( \Sigma xy = 306 \), and variance of temperature \( S_x^2 = 20 \).

We need to find the regression line \( \hat{y} = a + bx \) to predict ice cream sales (Y) from the maximum temperature (X).

The formula for the regression coefficient 'b' when mean values and \( \Sigma xy \) are given, along with \( S_x^2 \), is:

\[ b = \frac{\Sigma xy - n \bar{x} \bar{y}}{n \cdot S_x^2} \]

Substituting the values: n = 6, \( \Sigma xy = 306 \), \( \bar{x} = 40 \), \( \bar{y} = 1.2 \), and \( S_x^2 = 20 \).

\[ b = \frac{306 - 6(40)(1.2)}{6 \times 20} \] \[ = \frac{306 - 288}{120} \] \[ = \frac{18}{120} \] \[ = 0.15 \]

Next, we find 'a' using the formula:

\[ a = \bar{y} - b\bar{x} \]

Substituting the values: \( \bar{y} = 1.2 \), \( b = 0.15 \), and \( \bar{x} = 40 \).

\[ a = 1.2 - 0.15 (40) \] \[ = 1.2 - 6 \] \[ = -4.8 \]

By substituting 'a' and 'b' into \( \hat{y} = a + bx \), we get the regression line:

\[ \hat{y} = -4.8 + 0.15x \]

To estimate the sale of ice cream when the maximum temperature (X) is 42 Celsius, we substitute \( x = 42 \) into the regression equation:

\[ \hat{y} = -4.8 + 0.15 (42) \] \[ = -4.8 + 6.3 \] \[ = 1.5 \]

Therefore, the estimated sale of ice cream is Rs. 1.5 lakh.

In simple words: We examined the connection between daily temperature and ice cream sales. We created a formula to predict ice cream sales based on temperature. If the temperature reaches 42 Celsius, we expect ice cream sales to be around Rs. 1.5 lakh.

🎯 Exam Tip: This problem provides summary statistics directly, so choose the appropriate formula for 'b' that uses \( \bar{x}, \bar{y}, \Sigma xy, \) and \( S_x^2 \). Be meticulous with negative values during calculations to ensure accuracy in the regression equation and final estimate. Always include units in your final answer.