claudia records the hours she spent studying and her test scores for 5 tests. what is the correlation…

Explanation:

Step1: Recall correlation coefficient formula

The formula for the correlation coefficient $r$ is $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}$. Let $x$ be the hours spent studying ($x = 1,2,3,4,5$) and $y$ be the test - scores ($y=72,80,90,82,95$). First, calculate the necessary sums: $\sum x=1 + 2+3+4+5=15$. $\sum y=72 + 80+90+82+95=419$. $\sum xy=1\times72+2\times80 + 3\times90+4\times82+5\times95=72+160+270+328+475 = 1305$. $\sum x^{2}=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=1 + 4+9+16+25 = 55$. $\sum y^{2}=72^{2}+80^{2}+90^{2}+82^{2}+95^{2}=5184+6400+8100+6724+9025=35433$. $n = 5$.

Step2: Substitute values into the formula

$n(\sum xy)=5\times1305 = 6525$. $(\sum x)(\sum y)=15\times419 = 6285$. $n\sum x^{2}=5\times55 = 275$. $(\sum x)^{2}=15^{2}=225$. $n\sum y^{2}=5\times35433=177165$. $(\sum y)^{2}=419^{2}=175561$. The denominator is $\sqrt{(275 - 225)(177165-175561)}=\sqrt{50\times1604}=\sqrt{80200}\approx283.196$. The numerator is $6525-6285 = 240$. So, $r=\frac{240}{283.196}\approx0.85$. The strength of the linear - correlation is determined by the absolute value of the correlation coefficient. Since $|r|\approx0.85$, the strength of the model is strong.

Answer:

The correlation coefficient is approximately $0.85$. The strength of the model is strong.