From the data given below, calculate Karl Pearson's coefficient o…

Question

From the data given below, calculate Karl Pearson's coefficient of correlation between density of population and death rate by step deviation method.

Region	Area (in sq km)	Population	Death
A	200	40000	480
B	150	75000	1200
C	120	72000	1080
D	80	20000	280

✓

Answer

First of all, we shall compute density of population i.e., population per sq km and death rate per 1000.$\text{Density of Population}=\frac{\text{Population}}{\text{Area}},$
$\text{Death Rate}=\frac{\text{Number of Deaths}}{\text{Population}}\times100 $

Region	Density(X)	dx (X - A), A = 500	$\text{dx}'\bigg(\frac{\text{dx}}{\text{c}_1}\bigg),\text{c}_1=50$	$\text{dx}^2$	Death Rate (Y)	dy (Y - A), A = 16	$\text{dy}'\bigg(\frac{\text{dx}}{\text{c}_2}\bigg),\text{c}_2=1$	$\text{dy}^2$	dx'dy'
A	200	-300	-6	36	12	-4	-4	16	24
B	500	0	0	0	16	0	0	0	0
C	600	100	2	4	15	-1	-1	1	-2
D	250	-250	-5	25	14	-2	-2	4	10
			$\Sigma\text{dx}'=-9$	$\Sigma\text{dy}^2=65$			$\Sigma\text{dy}'=-7$	$\Sigma\text{dy}^2=21$	$\Sigma\text{dx}'\text{dy}'=32$

Here, $\text{dx}'=-9,\Sigma\text{dx}'^2=65,\Sigma\text{dy}'=-7,\Sigma\text{dy}'^2=21\ \text{and}\Sigma\text{dx}'\text{dy}'=32$$\therefore\ \text{r}=\frac{\Sigma\text{dx}'\text{dy}'-\frac{\Sigma\text{dx}'\times\Sigma\text{dy}'}{\text{n}}}{\sqrt{\Sigma\text{dx}'^2-\frac{(\Sigma\text{dx}')^2}{\text{n}}\times}\sqrt{\Sigma\text{dx}'^2-\frac{(\Sigma\text{dy}')^2}{\text{n}}}} $
$\frac{32-\frac{(-9\times-7)}{4}}{\sqrt{65-\frac{(-9)^2}{4}\times\sqrt{21-\frac{(-7)^2}{4}}}}=\frac{32-15.75}{\sqrt{65-20.25\times\sqrt{21-12.25}}{}} $
$=\frac{16.25}{\sqrt{44.75}\times\sqrt{8.75}}=\frac{16.25}{6.96\times2.96}=\frac{16.25}{19.80}=0.82 $
There is high degree of positive correlation between density of population and death rate.