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Question 35 of 42
Q.

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

Car123456
Time of usage of a car (Years) xx312253
Average annual maintenance cost (thousand ₹) yy10587138

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

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2025Subjective· 5mImportance★★★★★
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byx=10856=1.9286b_{yx} = \tfrac{108}{56} = 1.9286, a=3.3571a = 3.3571; y^=3.3571+1.9286x\hat{y} = 3.3571 + 1.9286x; at x=5x = 5, y^=13.0\hat{y} = 13.0; error =13−13=0= 13 - 13 = 0.

GSEB Class-12 Statistics, Linear Regression (line of YY on XX):

Let xx = years of usage, yy = maintenance cost, n=6n = 6.

xxyyx2x^2xyxy
310930
1515
28416
27414
5132565
38924
Σx=16\Sigma x = 16Σy=51\Sigma y = 51Σx2=52\Sigma x^2 = 52Σxy=154\Sigma xy = 154

xˉ=166=2.6667\bar{x} = \dfrac{16}{6} = 2.6667, yˉ=516=8.5\bar{y} = \dfrac{51}{6} = 8.5.

Regression coefficient of YY on XX:

byx=nΣxy−Σx ΣynΣx2−(Σx)2=6(154)−16(51)6(52)−(16)2=924−816312−256=10856=1.9286b_{yx} = \frac{n\Sigma xy - \Sigma x \, \Sigma y}{n\Sigma x^2 - (\Sigma x)^2} = \frac{6(154) - 16(51)}{6(52) - (16)^2} = \frac{924 - 816}{312 - 256} = \frac{108}{56} = 1.9286

Intercept:

a=yˉ−byx xˉ=8.5−1.9286×2.6667=8.5−5.1429=3.3571a = \bar{y} - b_{yx}\,\bar{x} = 8.5 - 1.9286 \times 2.6667 = 8.5 - 5.1429 = 3.3571

Regression line of YY on XX: …

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