Solve the Following Question.(2 Marks)

Question 12 Marks

Find the equation of line regression of $Y$ on $X$ for the following data:
$
\mathrm{n}=8, \Sigma\left(\mathrm{x}_{\mathrm{i}}-\bar{x}\right)\left(\mathrm{y}_{\mathrm{i}}-\bar{y}\right)=120, \bar{x}=20, \bar{y}=36, \sigma_{\mathrm{x}}=2, \sigma_{\mathrm{y}}=3 .
$

Answer

$
\begin{aligned}
& b_{y x}=\frac{\operatorname{Cov}(x, y)}{\sigma_x{ }^2}=\frac{\frac{\left(\sum(x-\bar{x})(y-\bar{y})\right.}{n}}{\sigma_x{ }^2} \\
& =\frac{120}{8 \times 4}=3.75
\end{aligned}
$
Regression equation of $Y$ on $X$ is
$
\begin{aligned}
& (y-\bar{y})=b_{y x}(x-\bar{x}) \\
& (y-36)=3.75(x-20) \\
& (y-36)=3.75 x-75 \\
& y=3.75 x-39
\end{aligned}
$

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Question 22 Marks

(i) If for a bivariate data $b_{y x}=-1.2$ and $b_{x y}=-0.3$ then find $r$.
(ii) From the two regression equations $\mathrm{y}=4 \mathrm{x}-5$ and $3 \mathrm{x}=2 \mathrm{y}+5$, find $\bar{x}$ and $\bar{y}$.

Answer

$
\begin{aligned}
& r^2=b_{y x} \cdot b_{x y} \\
& r^2=(-1.2) \times(-0.3) \\
& r^2=0.36 \\
& r= \pm 0.6
\end{aligned}
$
Since, $b_{y x} \cdot b_{x y}$ are negative, $r=-0.6$
Also, $(\bar{x}, \bar{y})$ is the point of intersection of the regression lines
$
\begin{aligned}
& y=4 x-5,3 x=2 y+5 \\
& 8 x-2 y=10 \\
& 3 x-2 y=5
\end{aligned}
$
on subtracting,
$
\begin{aligned}
& 5 x=5 \\
& x=1
\end{aligned}
$
Substituting $x=1$ in $y=4 x-5$
$
\begin{aligned}
& \mathrm{y}=4(1)-5 \\
& \mathrm{y}=-1 \\
& \therefore \bar{x}=1, \bar{y}=-1
\end{aligned}
$

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Question 32 Marks

The following result was obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men.

	X	Y
MEAN	50	140
VARIANCE	160	165

and $\Sigma\left(x_i-\bar{x}\right)\left(y_i-\bar{x}\right)=1120$. Find the Prediction of blood pressure of a man of age 40 years.

Answer

$
\begin{aligned}
& \text { Given, } n=10, \bar{x}=50, \bar{y}=140, \sigma_x^2=150, \\
& \sigma_y^2=16 \\
& b_{y x}=\frac{\operatorname{Cov}(x, y)}{\sigma_x{ }^2}=\frac{\frac{\left(\sum(x-\bar{x})(y-\bar{y})\right)}{n}}{150} \\
& =\frac{1120}{10 \times 150}=0.7
\end{aligned}
$
Regression equation of $Y$ on $X$ is
$
\begin{aligned}
& (y-\bar{y})=b_{y x}(x-\bar{x}) \\
& (y-140)=0.7(40-50) \\
& y-140=0.7(-10) \\
& y-140=-7 \\
& \therefore y=133
\end{aligned}
$

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Question 42 Marks

The equation of two regression lines are $2 x+3 y-6=0$ and $3 x+2 y-12=0$ Find (i) Correlation coefficient (ii) $\frac{\sigma_x}{\sigma_y}$

Answer

(i)
$
\begin{array}{ll}
2 x+3 y=6, & 3 x+2 y=12 \\
3 y=-2 x+6 & 3 x=-2 y+12 \\
y=\frac{-2}{3} x+2 & x=\frac{-2}{3} y+4 \\
b_{y x}=\frac{-2}{3} & b_{x y}=\frac{-2}{3} \\
b_{y x} \cdot b_{x y}=\frac{-2}{3} \times \frac{-2}{3}=\frac{4}{9} \in[0,1]
\end{array}
$
$\therefore$ Our assumption is correct.
$\therefore r^2=b_{y x} \cdot b_{x y}$
$r^2=\frac{4}{9}$
$
r= \pm \frac{2}{3}
$
Since $b_{y x}$ and $b_{x y}$ are negative $\therefore r=\frac{-2}{3}$
(ii)
$
\begin{aligned}
& b_{x y}=\frac{r \cdot \sigma_y}{\sigma_x} \\
& \frac{-2}{3}=\frac{-2}{3} \cdot \frac{\sigma_x}{\sigma_y} \\
& \therefore \frac{\sigma_x}{\sigma_y}=1
\end{aligned}
$

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Question 52 Marks

For bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of the variance of Y if the variance of X is 9.

Answer

Given, $b_{y x}=0.4, b_{x y}=0.9, \sigma_x^2=9, \sigma_x=3$
$
\begin{aligned}
& r^2=b_{y x} \cdot b_{x y} \\
& r^2=0.4 \times 0.9 \\
& r^2=0.36 \\
& r= \pm 0.6
\end{aligned}
$
Since $b_{y x}$ and $b_{x y}$ are positive $\therefore r=0.6$
$
b_{y x}=\frac{r \cdot \sigma_y}{\sigma_x}
$
$
\begin{aligned}
& 0.4=0.6 \times \frac{\sigma_y}{3} \\
& \frac{4}{10}=\frac{6}{10} \times \frac{\sigma_y}{3} \\
& \sigma_y=2 \\
& \therefore \sigma_y{ }^2=4
\end{aligned}
$
$\therefore$ Variance of $y$ is 4

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Question 62 Marks

Given the following data, obtain linear regression estimate of X for Y = 10

Answer

$
\begin{aligned}
& \bar{x}=7.6, \bar{y}=14.8, \sigma_{\mathrm{x}}=3.2, \sigma_{\mathrm{y}}=16 \text { and } \mathrm{r}=0.7 \\
& \mathrm{~b}_{\mathrm{xy}}=\frac{r_{\sigma_x}}{\sigma_y}=0.7 \times \frac{3.2}{16}=0.14
\end{aligned}
$
Regression equation of $\mathrm{X}$ on $\mathrm{Y}$ is
$
\begin{aligned}
& (X-\bar{y})=b_{x y}(Y-\bar{y}) \\
& (X-7.6)=0.14(y-14.8) \\
& X-7.6=0.14 y-2.072 \\
& X=0.14 y+5.528
\end{aligned}
$
When $y=10$
$
\begin{aligned}
& x=0.14(10)+5.528 \\
& =1.4+5.528 \\
& =6.928
\end{aligned}
$

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Question 72 Marks

An inquiry of 50 families to study the relationship between expenditure on accommodation (₹ x) and expenditure on food and entertainment ₹ gave the following result:
$
\Sigma x=8500, \Sigma y=9600, \sigma_x=60, \sigma_y=20, r=0.6
$
Estimate the expenditure on food and entertainment when expenditure on accommodation is ₹ 200

Answer

$
\begin{aligned}
\mathrm{n}=50 \text { (given) } \\
\quad b_{y x}=\frac{r \sigma_y}{\sigma_x}=0.6 \times \frac{20}{60}=0.2 \\
\bar{x}=\frac{\sum x_i}{n}=\frac{8500}{50}=170 \\
\bar{y}=\frac{\sum y_i}{n}=\frac{9600}{n}=192
\end{aligned}
$
Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
\begin{aligned}
& Y-\bar{y}=b_{y x}(X-\bar{x}) \\
& (Y-192)=0.2(200-170) \\
& Y-192=0.2(30) \\
& Y=192+6 \\
& Y=198
\end{aligned}
$

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Maths (commerce) Solve the Following Question.(2 Marks)

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