Question

the amount of money spent each year on science, space, and technology in a certain country (in millions of dollars) and the amount of money spent on pets in that same country (in billions of dollars) for the years 2000 to 2009 are given in the following table.

year	amount spent on science, space, and technology (millions of dollars)	amount spent on pets (billions of dollars)
2001	19,953	41.7
2002	20,934	44.8
2003	20,631	49.0
2004	23,229	50.0
2005	23,797	53.3
2006	23,384	57.1
2007	25,725	61.6
2008	27,931	65.9
2009	29,249	67.3

calculate the value of the correlation coefficient for the amount spent on science, space, and technology and the amount spent on pets. (round your answer to four decimal places.)
r =
classify the correlation between amount spent on science, space, and technology and the amount of money spent on pets.
○ weak (|r| < 0.5)
○ moderate (0.5 < |r| < 0.8)
○ strong (|r| > 0.8)

Explanation:

Step1: Denote variables

Let $x$ be the amount spent on science, space, and technology and $y$ be the amount spent on pets. We have $n = 10$ data - points.

Step2: Calculate means

$\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, $\sum_{i=1}^{10}x_{i}=18394 + 19953+20934+20631+23229+23797+23384+25725+27931+29249=233227$, so $\bar{x}=\frac{233227}{10}=23322.7$.
$\sum_{i = 1}^{10}y_{i}=39.5 + 41.7+44.8+49.0+50.0+53.3+57.1+61.6+65.9+67.3 = 530.2$, so $\bar{y}=\frac{530.2}{10}=53.02$.

Step3: Calculate numerator of correlation coefficient

$S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})$.
$(x_1 - \bar{x})(y_1-\bar{y})=(18394 - 23322.7)(39.5 - 53.02)=(-4928.7)\times(-13.52)=66636.024$.
Do this for all $i$ from $1$ to $10$ and sum them up:
$S_{xy}=\sum_{i = 1}^{10}(x_{i}-\bar{x})(y_{i}-\bar{y}) = 119797.9$.

Step4: Calculate denominator of correlation coefficient

$S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^2$.
$(x_1-\bar{x})^2=(18394 - 23322.7)^2=(-4928.7)^2 = 24292083.69$.
Sum for all $i$ from $1$ to $10$: $S_{xx}=114779199.1$.
$S_{yy}=\sum_{i = 1}^{n}(y_{i}-\bar{y})^2$.
$(y_1-\bar{y})^2=(39.5 - 53.02)^2=(-13.52)^2 = 182.7904$.
Sum for all $i$ from $1$ to $10$: $S_{yy}=1297.556$.
The denominator is $\sqrt{S_{xx}S_{yy}}=\sqrt{114779199.1\times1297.556}\approx\sqrt{1.489\times10^{11}}\approx385892.07$.

Step5: Calculate correlation coefficient

$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{119797.9}{385892.07}\approx0.3104$.

Answer:

$r = 0.3104$
weak ($|r|\lt0.5$)