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LINEAR REGRASSION — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsLINEAR REGRASSION5 Marks
Question
The following information is obtained to study the relation between the carpet area in a house and its monthly rent in a city:

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

Answer

Here, $n = 7; X = $ Carpet area and $Y =$ Monthly rent.
Now, $x̄ = \frac{\Sigma x}{n} = \frac{630}{7} = 90; ȳ = \frac{\Sigma y}{n} = \frac{182000}{7} = 26000$
To obtain the regression line $ŷ = a + bx$ of monthly rent $(Y)$ on the carpet area $(X),$ we calculate
the values of $a$ and $b$ by shortcut method, taking new variables $u = \frac{x-\mathrm{A}}{\mathrm{C}_{x}}, A = 80, C_x = 5$ and $v = \frac{y-\mathrm{B}}{\mathrm{C}_{y}}, B = 25000, C_y = 1000$
The table for calculation is prepared as follows :
Image
$b = b_{vu} = \frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{\mathrm{C}_{y}}{\mathrm{C}_{x}}$
Putting $n = 7; \sum uv = 404; \sum u = 14; \sum v = 7; \sum u^2 = 266; C_x = 5$ and $C_y = 1000$ in the formula,
$b = \frac{7(404)-(14)(7)}{7(266)-(14)^{2}} \times \frac{1000}{5}$
$= \frac{2828-98}{1862-196} \times 200$
$= \frac{2730 \times 200}{1666}$
$= \frac{546000}{1666}$
$= 327.73$
$a = ȳ – bx̄$
Putting $ȳ = 26000; b = 327.73$ and $x̄ = 90,$ we get
$a = 26000 – 327.73 (90)$
$= 26000 – 29495.7$
$= – 3495.7$
Regression line of $Y$ on $X :$
Putting $a = – 3495.7 $ and $b = 327.73$ in $ŷ = a + bx,$ we get
$ŷ = – 3495.7 + 327.73x$
Estimate of monthly rent $(Y)$ when $X=110 :$
Putting $x = 110$ in $ŷ = – 3495.7 + 327.73x,$ we get
$ŷ = – 3495.7 + 327.73 (110)$
$= – 3495.7 + 36050.3$
$= 32554.6$
Hence, the estimate of monthly rate obtained is $ŷ = ₹ 32554.6$

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