Question

The regression of savings $\left(s\right)$ of a family on income $y$may be expressed as $s=a+\left(\frac{y}{m}\right)$, 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

• A. $2$
• B. $5$
• C. $8$
• D. None of these

Answer 1

The correct option is B

$5$

Explanation for the correct answer:

Finding the value of $m$:

Given that, $s=a+\left(\frac{y}{m}\right)$,$s^2=\left(\frac{1}{4}\right)y^2$ and $r=0.4$

$\begin{array}{rcl}s& =& a+\left(\frac{y}{m}\right)\\ and\overline{s}& =& a+\left(\frac{\overline{y}}{m}\right)\end{array}$

On subtracting $s-\overline{s}$, we get

$$\begin{gathered} s-\overline{s}=\left(\frac{1}{m}\right)\left(y-\overline{y}\right)...\left(i\right) \\ \sum \left(s-\overline{s}\right)\left(y-\overline{y}\right)=\left(\frac{1}{m}\right)\sum {\left(y-\overline{y}\right)}^2 \end{gathered}$$

Also, $Cov(s,y)=\left(\frac{1}{m}\right)y^2...\left(ii\right)$

Finding the value of $m.$

$\begin{array}{rcl}r& =& \frac{cov(s,y)}{sy}\\ & \Rightarrow & 0.4=\frac{\left({\displaystyle \frac{1}{m}}\right)y^2}{sy}\\ & \Rightarrow & 0.4=\frac{\left({\displaystyle \frac{1}{m}}\right)}{\left({\displaystyle \frac{y}{s}}\right)}\\ & \Rightarrow & 0.2=\frac{1}{m}\\ & \Rightarrow & m=5\end{array}$

Therefore, the correct answer is option (B).