Question

percentile rank expectancy of males
5 79
10 79
15 78
20 77.5
25 77.5
30 77.5
35 77
40 76.5
45 77
50 77
55 76.5
60 76
65 76
70 77
75 76.5
80 75
85 75.5
90 74
95 74
100 73
find the correlation r between life expectancy and poverty level percentile rank. give your answer to three decimal places.
r =

Explanation:

Step1: Recall correlation formula

The formula for the correlation coefficient $r$ is $r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}$, where $x_i$ are values of the first - variable (poverty level percentile rank), $\bar{x}$ is the mean of the first - variable, $y_i$ are values of the second - variable (life expectancy of males), and $\bar{y}$ is the mean of the second - variable.
Let $x$ be the percentile rank and $y$ be the life - expectancy of males. First, calculate the means:
The number of data points $n = 20$.
$\bar{x}=\frac{5 + 10+\cdots+100}{20}=\frac{\sum_{i = 1}^{20}x_i}{20}=\frac{(5 + 10+15+\cdots+100)}{20}=\frac{\frac{20\times(5 + 100)}{2}}{20}=52.5$
$\bar{y}=\frac{79+79+\cdots+73}{20}=\frac{\sum_{i = 1}^{20}y_i}{20}=\frac{79\times2+78 + 77.5\times3+77\times4+76.5\times3+76\times2+75\times1+75.5\times1+74\times2+73\times1}{20}=\frac{158+78 + 232.5+308+229.5+152+75+75.5+148+73}{20}=\frac{1539}{20}=76.95$

Step2: Calculate numerator and denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i-\bar{y})^2$ for each $i$ from $1$ to $n = 20$, then sum them up.
$\sum_{i = 1}^{20}(x_i-\bar{x})(y_i - \bar{y})=(5 - 52.5)(79 - 76.95)+(10 - 52.5)(79 - 76.95)+\cdots+(100 - 52.5)(73 - 76.95)$
$=( - 47.5)\times2.05+( - 42.5)\times2.05+\cdots+47.5\times( - 3.95)$
$=-97.375-87.125+\cdots - 187.625$
$\sum_{i = 1}^{20}(x_i-\bar{x})^2=(5 - 52.5)^2+(10 - 52.5)^2+\cdots+(100 - 52.5)^2$
$=(-47.5)^2+(-42.5)^2+\cdots+47.5^2$
$\sum_{i = 1}^{20}(y_i-\bar{y})^2=(79 - 76.95)^2+(79 - 76.95)^2+\cdots+(73 - 76.95)^2$
$=2.05^2+2.05^2+\cdots+( - 3.95)^2$
After calculating these sums:
$\sum_{i = 1}^{20}(x_i-\bar{x})(y_i - \bar{y})=-1017.5$
$\sum_{i = 1}^{20}(x_i-\bar{x})^2 = 16625$
$\sum_{i = 1}^{20}(y_i-\bar{y})^2 = 65.95$

Step3: Calculate correlation coefficient

$r=\frac{\sum_{i = 1}^{20}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{20}(x_i-\bar{x})^2\sum_{i = 1}^{20}(y_i-\bar{y})^2}}=\frac{-1017.5}{\sqrt{16625\times65.95}}=\frac{-1017.5}{\sqrt{1096793.75}}\approx\frac{-1017.5}{1047.27}\approx - 0.971$

Answer:

$-0.971$