Question

Regression of savings \[\left( s \right)\] of a family on income $\left( y \right)$ may be expressed as $s = a + \left( {\dfrac{y}{m}} \right)$ , where $a$ and $m$ are constants. In a random sample of 100 families, the variance of savings is one-quarter of the variance of incomes and the correlation is found to be 0.4 the value of $m$ is
1. $2$
2. $5$
3. $8$
4. None of these

Answer 1

Hint: In this question, we are given the value of correlation and the regression of $s$ on $y$, from the regression we have the value of regression coefficient (co-efficient of $y$). Also, used the linear regression co-efficient formula and compare them. You’ll get the formula and put the values. Also, for the variance use of the given condition the variance of savings is one-quarter of the variance of incomes.

Formula Used:
Equation of straight line $y = mx + c$
Linear regression co-efficient formula ${b_{xy}} = r\dfrac{{{\sigma _x}}}{{{\sigma _y}}}$($x$ on $y$)

Complete step by step Solution:
Given that,
Regression equation of $s$ on $y$ is $s = a + \left( {\dfrac{y}{m}} \right)$,
Also written as $s = a + \left( {\dfrac{1}{m}} \right)y - - - - - \left( 1 \right)$
Equation (1) is the equation of straight line i.e., $y = mx + c$
Therefore, the regression co-efficient of $s$ on $y$ will be the co-efficient of $y$ from equation (1),
It implies that, ${b_{sy}} = \dfrac{1}{m} - - - - - \left( 2 \right)$
Also, ${b_{sy}} = r\dfrac{{{\sigma _s}}}{{{\sigma _y}}} - - - - - \left( 3 \right)$(Linear regression co-efficient formula)
Now, the variance of savings is one-quarter of the variance of incomes and the correlation is found to be 0.4 (given)
$ \Rightarrow r = 0.4$, ${\sigma _s}^2 = \left( {\dfrac{1}{4}} \right){\sigma _y}^2$ Or $\dfrac{{{\sigma _s}}}{{{\sigma _y}}} = \dfrac{1}{2}$
From equation (2) and (3),
$r\dfrac{{{\sigma _s}}}{{{\sigma _y}}} = \dfrac{1}{m}$
$0.4\left( {\dfrac{1}{2}} \right) = \dfrac{1}{m}$
Cross-multiplying both the sides,
$m = \dfrac{1}{{0.2}}$
$m = 5$


Hence, the correct option is 2.

Note:The key concept involved in solving this problem is a good knowledge of linear regression. Students must remember that linear regression analysis predicts the value of one variable based on the value of another. The variable you wish to predict is referred to as the dependent variable. The variable you are using to predict the value of the other variable is known as the independent variable.