From the data given below, calculate Karl Pearson’s coefficient o… - Vidyadip

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	80
D	80	20000	280

✓

Answer

Regio n	Density(X)	dx(X A), A = 500	$\begin{array}{c} d x \left(\frac{d x}{c_1}\right), \\ c _1= 5 0 \end{array}$	$dx ^{\prime 2}$	Death Rate(Y)	dy(Y A), A = 16	$\begin{array}{c} d y ^{\prime}\left(\frac{d y}{c_2}\right) \\ , c _2=1\end{array}$	$d y^{\prime 2}$	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
			$\Sigma dx ^{\prime}=-9$	$\begin{array}{c}\Sigma dx x^{\prime 2} =65\end{array}$			$\Sigma dy ^{\prime}=-7$	$\begin{array}{l} \Sigma dy ^{\prime 2} =21\end{array}$	$\begin{array}{l}\Sigma dx ^{\prime} dy ^{\prime} =32\end{array}$

Density is calculated as $\frac{\text { population }}{\text { area }}$
Death Rate is calculated as $\frac{\text { death }}{\text { population }} \times 100$
Here, $\Sigma d x { }^{\prime}=-9, \Sigma dx ^{\prime 2}=65, \Sigma dy ^{\prime}=-7, \Sigma dy ^{\prime 2}=21, \Sigma dx ^{\prime} dy ^{\prime}=32$
Now, $r =\frac{\sum d x^{\prime} d y^{\prime}-\frac{\Sigma d x^{\prime} \times \Sigma d y^{\prime}}{n}}{\sqrt{\sum d x^n-\frac{\left(\Sigma d d^{\prime}\right)^2}{n}} \times \sqrt{\Sigma d y^a-\frac{(\Sigma d)^2}{n}}}$
$=\frac{32-\frac{(-9 \times-7)}{4}}{\sqrt{65-\frac{(-9)^2}{4}} \times \sqrt{21-\frac{(-7)^2}{4}}}$
$\begin{array}{l}
=\frac{32-15.75}{\sqrt{65-20.25} \times \sqrt{21-12.25}} \\
=\frac{16.25}{\sqrt{44.75} \times \sqrt{8.75}}=\frac{16.25}{6.69 \times 2.96}=\frac{16.25}{19.80}=0.82
\end{array}$
- Therefore, Karl Pearson's coefficient of correlation between density of population and death rate is 0.82 .
- Interpretation of r: There is a high degree of positive correlation between density of population and death rate.

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