Gujarat Board Statistics Class 12 GSEB Solutions Part 1 Chapter 3 Linear Regression Ex 3.1 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Statistics Part 1 Chapter 3 Linear Regression Ex 3.1

Question 1.

From the following data of price (in â‚¹) and demand (in hundred units) of a commodity, obtain the regression line of demand on price. Also estimate the demand when price is â‚¹ 20.

Answer

Here, n = 6; X = Price and Y = Demand

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{96}{6}\) = â‚¹ 16, yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{60}{6}\) = 10 (â€™00) units.

To obtain the equation of regression line of Y on X,

yÌ‚ = a + bx, we calculate a and b.

xÌ„ and yÌ„ are integers. So the table of calculation is prepared as follows:

b = \(\frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}\)

Putting, Î£(x – xÌ„) (y – yÌ„) = – 67 and Î£(x – xÌ„)^{2} = 50 in the formula,

b = \(\frac{-67}{50}\)

= – 1.34

a = yÌ„ – bxÌ„

Putting yÌ„ = 10; xÌ„ = 16 and b = – 1.34, we get

a = 10 – (- 1.34) (16)

= 10 + 21.44

= 31.44

Regression line of demand (Y) on price (X) :

Putting a = 31.44 and b = – 1.34 in yÌ‚ = a + bx, we get

yÌ‚ = 31.44 – 1.34x

When price X = â‚¹ 20, the estimate of demand Y:

Putting x = 20 in yÌ‚ = 31.44 – 1.34x, we get

yÌ‚ = 31.44 – 1.34 (20)

= 31.44 – 26.8 = 4.64

Hence, when price is â‚¹ 20, the estimated demand Y obtained is yÌ‚ = 4.64 (â€™00) units.

Question 2.

To study the relationship between the time of usage of cars and its average annual maintenance cost, the following information is obtained:

Obtain the regression line of Y on X. Find an estimate of average annual maintenance cost when the usage time of a car is 5 years. Also find its error.

Answer:

Here, n = 6; X = Time of usage of a car and Y = Annual average maintenance cost

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{16}{6}\) = 2.67; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{51}{6}\) = 8.5

We obtain the regression line of average annual maintenance cost (Y) on the time of usage (X) yÌ‚ = a + bx.

The values of X and Y are not large. So we calculate the values of a and b by preparing the following table:

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

Putting n = 6, Î£xy = 154; Î£x = 16; Î£y = 15 and Î£x^{2} = 52 in the formula

= \(\frac{6(154)-(16)(51)}{6(52)-(16)^{2}}\)

= \(\frac{924-816}{312-256}\)

= \(\frac{108}{56}\) = 1.93

a = yÌ„ – bxÌ„

Putting yÌ„ = 8.5; xÌ„ = 2.67 and b = 1.93, we get,

a = 8.5 – 1.93 (2.67)

= 8.5 – 5.15

= 3.35

Regression line of Y on X:

Putting a = 3.35 and b = 1.93 in yÌ‚ = a + bx, we obtain yÌ‚ = 3.35 + 1.93x

The estimate of average annual maintenance cost Y when usage of car X = 5 years:

Putting x = 5 in yÌ‚ = 3.35 + 1.93x, we get

yÌ‚ = 3.35 + 1.93 (5)

= 3.35 + 9.65 = â‚¹ 13 thousand

Hence, the estimate of average annual maintenance cost obtained is y = â‚¹ 13 thousand.

Error: yÌ‚ = 13, y = 13

âˆ´ Error = yÌ‚ – y = 13 – 13 = 0

Hence, point (5, 13) is on he fitted regression line yÌ‚ = 3.35 + 1.93 x.

Question 3.

The information for a year regarding the average rainfall (in cm) and total production of crop (in tons) of five districts is given below:

Find the regression line of production of crop on rainfall and estimate the crop if average rainfall is 35 cm.

Answer:

Here, n = 5; X = Average rainfall and Y = Crop

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{155}{5}\) = 31 cm; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{450}{5}\) = 90 tons

We obtain the regression line of crop (Y) on the average rainfall (X) yÌ‚ = a + bx.

xÌ„ and yÌ„ are integers. So to calculate a and b, we prepare the following table:

b = \(\frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}\)

Putting, Î£(x – xÌ„) (y – yÌ„) = 75 and Î£(x – xÌ„)^{2} = 90 in the formula,

b = \(\frac{75}{50}\)

= 0.83

a = yÌ„ – bxÌ„

Putting yÌ„ = 90; xÌ„ = 31 and b = 0.83, we get

a = 90 – 0.83 (31)

= 90 – 25.73

= 64.27

Regression line of crop (Y) on average rainfall (X) :

Putting b = 0.83 and a = 64.27 in yÌ‚ = a + bx, we get

yÌ‚ = 64.27 + 0.83x.

Estimate of Y when X = 35 cm :

Putting x = 35 in yÌ‚ = 64.27 + 0.83x, we get

yÌ‚ = 64.27 + 0.83 (35) = 64.27 + 29.05 = 93.32

Hence, the estimate of crop obtained is yÌ‚ = 93.32 tons.

Question 4.

The following data gives the experience of machine operators and their performance ratings:

Calculate the regression line of performance ratings on the experience and estimate the performance rating of an operator having 7 years of experience.

Answer:

Here, n = 8; X = Experience and Y = Performance rating.

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{80}{8}\) = 10 year; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{648}{8}\) = 81

We obtain the regression line of Performance rating (Y) on experience (X) yÌ‚ = a + bx.

xÌ„ and yÌ„ are integers. So to calculate a and b we prepare the following table:

b = \(\frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}\)

Putting, Î£(x – xÌ„) (y – yÌ„) = 247 and Î£(x – xÌ„)^{2} = 218 in the formula,

b = \(\frac{247}{218}\)

= 1.13

a = yÌ„ – bxÌ„

Putting yÌ„ = 81; xÌ„ = 10 and b = 1.13, we get

a = 81 – 1.13 (31)

= 81 – 11.3

= 69.7

Regression line of Y on X:

Putting a = 69.7 and b = 1.13 in yÌ‚ = a + bx, we get

yÌ‚ = 69.7 + 1.13x

Estimate of Y when X = 7 years :

Putting x = 7 in yÌ‚ = 69.7 + 1.13x, we get

yÌ‚ = 69.7 + 1.13 (7)

= 69.7 + 7.91

= 77.61

Hence, the estimate of performance rating obtained is yÌ‚ = 77.61.