Question

x, fertility rate | y, average life expectancy
1.7 | 77.2
2.9 | 72.1
2.6 | 68.9
2.1 | 76.4
2.8 | 70.8
1.5 | 76
2.2 | 69.9
2 | 78.2
6.8 | 54.4
2.4 | 67.1
1.4 | 80.7
4.2 | 66
1.8 | 75
2.6 | 73
5.2 | 55.9
1.5 | 81
4.7 | 62.9
3.9 | 72.3
1.5 | 73.3
6.9 | 54.2
1.4 | 82.6
5.8 | 55.4
2.1 | 78.3
2.5 | 74.2
a) state the random variables.
rv x = the fertility rate of a randomly selected country
rv y = the average life expectancy of a randomly selected country
b) the symbol and value of the correlation coefficient are as follows:
round final answer to 3 decimal places.
r =

Explanation:

Step1: Recall correlation - coefficient formula

The formula for the correlation coefficient $r$ between two variables $X$ and $Y$ with $n$ data - points is $r=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{\sqrt{[n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}][n\sum_{i = 1}^{n}y_i^{2}-(\sum_{i = 1}^{n}y_i)^{2}]}}$. First, calculate the necessary sums. Let $n = 25$.
Let $x_i$ be the fertility - rate values and $y_i$ be the average life - expectancy values.
$\sum_{i = 1}^{25}x_i=1.7 + 2.9+2.6+\cdots+2.5=84.8$
$\sum_{i = 1}^{25}y_i=77.2 + 72.1+68.9+\cdots+74.2 = 1739.4$
$\sum_{i = 1}^{25}x_i^{2}=1.7^{2}+2.9^{2}+2.6^{2}+\cdots+2.5^{2}=379.96$
$\sum_{i = 1}^{25}y_i^{2}=77.2^{2}+72.1^{2}+68.9^{2}+\cdots+74.2^{2}=123979.94$
$\sum_{i = 1}^{25}x_iy_i=1.7\times77.2+2.9\times72.1+2.6\times68.9+\cdots+2.5\times74.2 = 5749.38$

Step2: Substitute values into the formula

$n = 25$.
$n\sum_{i = 1}^{n}x_iy_i=25\times5749.38 = 143734.5$
$\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i=84.8\times1739.4=147591.12$
$n\sum_{i = 1}^{n}x_i^{2}=25\times379.96 = 9499$
$(\sum_{i = 1}^{n}x_i)^{2}=84.8^{2}=7191.04$
$n\sum_{i = 1}^{n}y_i^{2}=25\times123979.94 = 3099498.5$
$(\sum_{i = 1}^{n}y_i)^{2}=1739.4^{2}=3025412.36$

$r=\frac{143734.5 - 147591.12}{\sqrt{(9499 - 7191.04)(3099498.5 - 3025412.36)}}$
$=\frac{- 3856.62}{\sqrt{2307.96\times74086.14}}$
$=\frac{-3856.62}{\sqrt{170959799.7}}$
$=\frac{-3856.62}{13075.16}$
$=- 0.295$

Answer:

$r=-0.295$