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Maths (commerce) Solve the Following Question.(4 Marks)

Linear Regression (p-2) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 13 questions.

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Solve the Following Question.(4 Marks)

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13 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
The equation of the two lines of regression are 3x + 2y – 26 = 0 and 6x + y – 31 = 0. Find
(i) Means of X and Y
(ii) Correlation coefficient between X on Y
(iii) Estimate of Y for X = 2
(iv) var (X) if var (Y) = 36
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Question 24 Marks
The equation of the line of regression of $y$ on $x$ is $v=\frac{2}{9} x$ and $x$ on $y$ is $x=\frac{y}{2}+\frac{7}{6}$. Find (i) $\mathrm{r}$ (ii) $\sigma_y^2$ if $\sigma_x^2=4$.
Answer
(i) Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is $y=\frac{2}{9} x$
$
\therefore b_{y x}=\frac{2}{9}
$
Regression equation of $\mathrm{X}$ on $\mathrm{Y}$ is $x=\frac{y}{2}+\frac{7}{6}$
$
\begin{aligned}
& \therefore b_{x y}=\frac{1}{2} \\
& r^2=b_{y x} \cdot b_{x y} \\
& =\frac{2}{9} \times \frac{1}{2}=\frac{1}{9} \\
& r= \pm \frac{1}{3} \quad \text { }
\end{aligned}
$
Since $b_{y x}$ and $b_{x y}$ are positive, $\therefore r=\frac{1}{3}$
(ii)
$
\begin{aligned}
& \sigma_x^2=4 \quad \therefore \sigma_x=2 \\
& b_{y x}=\frac{r \cdot \sigma_y}{\sigma_x} \\
& \frac{2}{9}=\frac{1}{3} \times \frac{\sigma_y}{2} \\
& \sigma_y=\frac{4}{3} \\
& \therefore \sigma_y{ }^2=\frac{16}{9}
\end{aligned}
$ />& \therefore \sigma_y{ }^2=\frac{16}{9}
\end{aligned}
$
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Question 34 Marks
The equations of two regression lines are $10 x-4 y=80$ and $10 y-9 x=-40$ Find:
(i) $\bar{x}$ and $y$
(ii) $b_{y x}$ and $b_{x y}$
(iii) If $\operatorname{var}(Y)=36$, obtain $\operatorname{var}(X)$
(iv) $\mathrm{r}$
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Question 44 Marks
For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean values of X and Y.
Answer
$\begin{array}{ll}10 y=3 x+170 & 5 x+70=6 y \\ y=\frac{3}{10} x+17 & x=\frac{6}{5} y-14 \\ b_{y x}=\frac{3}{10} & b_{y x}=\frac{6}{5}\end{array}$
$
b_{y x} \cdot b_{x y}=\frac{3}{10} \times \frac{6}{5}=\frac{18}{50}=\frac{9}{25} \in[0,1]
$
$\therefore$ Our assumption is correct
$
\begin{aligned}
& r^2=b_{y x} \cdot b_{x y} \text {} \\
& =\frac{9}{25} \\
& r= \pm \frac{3}{5}
\end{aligned}
$
Since byx and bxy are positive,
$
r =\frac{3}{5}=0.6
$
Since, $(\bar{x}, \bar{y})$ is the point of intersection of the regression lines
$
\begin{aligned}
& 3 x-10 y=-170 . . . . . . \text { (i) } \\
& 5 x-6 y=-70 \ldots \ldots . . \text { (ii) } \\
& 9 x-30 y=-510 \\
& 25 x-30 y=-350
\end{aligned}
$
on subtracting
$
\begin{aligned}
& -16 x=-160 \\
& x=10
\end{aligned}
$
Substituting $x=10$ in equation (i)
$
\begin{aligned}
& 3(10)-10 y =-170 \\
& 30+170=10 y \\
& 200=10 y \\
& y =20 \\
& \therefore \bar{x}=10, \bar{y}=20
\end{aligned}
$
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Question 54 Marks
Two lines of regression are $10 x+3 y-62=0$ and $6 x+5 y-50=0$ Identify the regression equation equation of $x$ on $y$. Hence find $\bar{x}, \bar{y}$, and $\mathrm{r}$.
Answer
$
\begin{array}{lr}
10 x+3 y=62 & 6 x+5 y=50 \\
10 x=-3 y+62 & 5 y=-6 x+50 \\
x=\frac{-3}{10} y+\frac{62}{10} & y=\frac{-6}{5} x+10 \\
b_{x y}=\frac{-3}{10} & \text { } \\
b_{x y} \cdot b_{y x}=\frac{-3}{10} \times \frac{-6}{5}=\frac{18}{50}=\frac{9}{25} \in[0,1]
\end{array}
$
$\therefore$ Our assumption is correct.
$\therefore$ Regression equation of $X$ on $Y$ is $10 x+3 y-62=0$
$
\begin{aligned}
& r^2=b_{y x} \cdot b_{x y} \\
& r^2=\frac{9}{25} \\
& r= \pm \frac{3}{5}
\end{aligned}
$
Since, byx and bxy are negative, $r=-\frac{3}{5}=-0.6$
Also $(\bar{x}, \bar{y})$ is the point of intersection of the regression lines
$
\begin{aligned}
& 50 x+15 y=310 \\
& 18 x+15 y=150
\end{aligned}
$
on subtracting
$
\begin{aligned}
& 32 x=160 \\
& x=5
\end{aligned}
$
Substituting $x=5$ in $10 x+3 y=62$
$
10(5)+3 y=62
$
$3 y=12$
$
\therefore \mathrm{y}=4
$
$\therefore \bar{x}=5, \bar{y}=4$
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Question 64 Marks
For a certain bivariate data
  XY
MEN2520
S.D.43
And r = 0.5 estimate y when x = 10 and estimate x when y = 16
Answer
Given, $\bar{x}=25, \bar{y}=20, \sigma_{\mathrm{x}}=4, \sigma_{\mathrm{y}}=3, \mathrm{r}=0.5$
$
\mathrm{b}_{\mathrm{yx}}=\frac{r \sigma_y}{\sigma_y}=0.5 \times \frac{3}{4}=0.375
$
Regression equation of $Y$ on $X$ is
$
\begin{aligned}
& (Y-\bar{y})=b_{y x}(X-\bar{x}) \\
& (Y-20)=0.375(x-25) \\
& Y-20=0.375 x-9.375 \\
& Y=0.375 x+10.625
\end{aligned}
$
When, $x=10$
$
\begin{aligned}
& Y=0.375(10)+10.625 \\
& =3.75+10.625 \\
& =14.375 \\
& \mathrm{~b}_{\mathrm{xy}}=\frac{r \sigma_x}{\sigma_y}=0.5 \times \frac{4}{3}=0.67
\end{aligned}
$
Regression equation of $X$ on $Y$ is
$
\begin{aligned}
& (X-\bar{x})=b_{y x}(Y-\bar{y}) \\
& (X-25)=0.67(y-20) \\
& (X-25)=0.67 y-13.4 \\
& X=0.67 y+11.6 \\
& \text { When, } Y=16 \\
& X=0.67(16)+11.6 \\
& =10.72+11.6 \\
& =22.32
\end{aligned}
$
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Question 74 Marks
Bring out inconsistency if any, in the following:
(i) $b_{y x}+b_{x y}=1.30$ and $r=0.75$
(ii) $b_{y x}=b_{x y}=1.50$ and $r=-0.9$
(iii) $b_{y x}=1.9$ and $b_{x y}=-0.25$
(iv) $\mathrm{b}_{\mathrm{yx}}=2.6$ and $\mathrm{b}_{\mathrm{xy}}=\frac{1}{2.6}$
Answer
(i) Given, $b_{y x}+b_{x y}=1.30$ and $r=0.75$
$\frac{b_{y x}+b_{x y}}{2}=\frac{1.30}{2}=0.65$
But for regression coefficients $b_{\mathrm{yx}}$ and $b_{\mathrm{xy}}$
$\left|\frac{b_{y x}+b_{x y}}{2}\right| \geq r$
Here, $0.65<r=0.75$
$\therefore$ The data is inconsistent
(ii) The signs of $b_{y x}, b_{x y}$ and $r$ must be same (all three positive or all three negative)
$\therefore$ The data is inconsistent.
(iii) The signs of $b_{y x}$ and $b_{x y}$ should be same (either both positive or both negative)
$\therefore$ The data is consistent.
(iv) $b_{y x} \cdot b_{x y}=2.6 \times \frac{1}{2.6}=1$
$\therefore 0 \leq r^2 \leq 1$
$\therefore$ The data is consistent.
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Question 84 Marks
The following data about the sales and advertisement expenditure of a firms is given below (in ₹ crores)
SALESADV.EXP.
MEAN406
S.D.1015
Also correlation coefficient between X and Y is 0.9
(i) Estimate the likely sales for a proposed advertisement expenditure of ₹ 10 crores.
(ii) What should be the advertisement expenditure if the firm proposes a sales target ₹ 60 crores
Let the sales be X and advertisement expenditure be Y
Answer
Given, $\bar{x}=40, y=6, \sigma_{\mathrm{x}}=10, \sigma_{\mathrm{y}}=1.5, \mathrm{r}=0.9$
$
\begin{aligned}
& \mathrm{b}_{\mathrm{yx}}=\frac{r \sigma_y}{\sigma_x}=0.9 \times \frac{1.5}{10}=0.135 \\
& \mathrm{~b}_{\mathrm{xy}}=\frac{r_{\sigma_x}}{\sigma_y}=0.9 \times \frac{10}{1.5}=6
\end{aligned}
$
(i) Regression equation of $X$ on $Y$ is
$
\begin{aligned}
& (X-\bar{x})=b_{x y}(Y-\bar{y}) \\
& (X-40)=6(y-6) \\
& X-40=6 y-36 \\
& X=6 y+4
\end{aligned}
$
When $y=10$
$
\begin{aligned}
& x=6(10)+4 \\
& =60+4 \\
& =64 \text { crores }
\end{aligned}
$
(ii) Regression equation $Y$ on $X$ is
$
\begin{aligned}
& (Y-\bar{y})=b_{y x}(X-\bar{x}) \\
& (Y-6)=0.135(x-40) \\
& Y-6=0.135 x-5.4 \\
& Y=0.135 x+0.6
\end{aligned}
$
When $\mathrm{x}=60$
$
\begin{aligned}
& Y=0.135(60)+0.6 \\
& =8.1+0.6 \\
& =8.7 \text { crores }
\end{aligned}
$
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Question 114 Marks
The following are the marks obtained by the students in Economic (X) and Mathematics (Y)
X 59 60 61 62 63
Y 78 82 82 79 81
Find the regression equation of Y and X.
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Question 134 Marks
For a certain bivariate data on 5 pairs of observations given
$
\Sigma x=20, \Sigma y=20, \Sigma x^2=90, \Sigma y^2=90, \Sigma x y=76
$
Calculate (i) $\operatorname{cov}(x, y)$, (ii) $b_{y x}$ and $b_{x y}$, (iii) $r$
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