Question

find the correlation coefficient, r, of the data described below.
a graduate school rates entrance applications on a scale of 1 to 10, where 10 is excellent. administrators are curious to know whether the application rating can be used to predict the number of - years it will take an admitted student to graduate.
for the students graduating this year, the administrators note the ratings of their applications, x, as well as the number of years it took them to graduate, y.
application rating|years to graduate
5|7
6|8
7|9
9|6
10|5
10|6
round your answer to the nearest thousandth

Explanation:

Step1: Calculate the means

Let $x = [5,6,7,9,10,10]$ and $y=[7,8,9,6,5,6]$.
The mean of $x$, $\bar{x}=\frac{5 + 6+7+9+10+10}{6}=\frac{47}{6}\approx7.833$.
The mean of $y$, $\bar{y}=\frac{7 + 8+9+6+5+6}{6}=\frac{41}{6}\approx6.833$.

Step2: Calculate the numerator and denominator components

For the numerator of the correlation - coefficient formula $r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}$:

• Calculate $(x_{i}-\bar{x})(y_{i}-\bar{y})$ for each $i$:
• For $i = 1$: $(5 - 7.833)(7 - 6.833)=(-2.833)\times0.167=-0.473$.
• For $i = 2$: $(6 - 7.833)(8 - 6.833)=(-1.833)\times1.167=-2.139$.
• For $i = 3$: $(7 - 7.833)(9 - 6.833)=(-0.833)\times2.167=-1.795$.
• For $i = 4$: $(9 - 7.833)(6 - 6.833)=(1.167)\times(-0.833)=-0.972$.
• For $i = 5$: $(10 - 7.833)(5 - 6.833)=(2.167)\times(-1.833)=-3.972$.
• For $i = 6$: $(10 - 7.833)(6 - 6.833)=(2.167)\times(-0.833)=-1.806$.
• $\sum_{i = 1}^{6}(x_{i}-\bar{x})(y_{i}-\bar{y})=-0.473-2.139 - 1.795-0.972-3.972-1.806=-11.157$.
• Calculate $(x_{i}-\bar{x})^{2}$ for each $i$:
• $(5 - 7.833)^{2}=(-2.833)^{2}=8.026$.
• $(6 - 7.833)^{2}=(-1.833)^{2}=3.36$.
• $(7 - 7.833)^{2}=(-0.833)^{2}=0.694$.
• $(9 - 7.833)^{2}=(1.167)^{2}=1.362$.
• $(10 - 7.833)^{2}=(2.167)^{2}=4.696$.
• $(10 - 7.833)^{2}=(2.167)^{2}=4.696$.
• $\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=8.026 + 3.36+0.694+1.362+4.696+4.696=22.834$.
• Calculate $(y_{i}-\bar{y})^{2}$ for each $i$:
• $(7 - 6.833)^{2}=(0.167)^{2}=0.028$.
• $(8 - 6.833)^{2}=(1.167)^{2}=1.362$.
• $(9 - 6.833)^{2}=(2.167)^{2}=4.696$.
• $(6 - 6.833)^{2}=(-0.833)^{2}=0.694$.
• $(5 - 6.833)^{2}=(-1.833)^{2}=3.36$.
• $(6 - 6.833)^{2}=(-0.833)^{2}=0.694$.
• $\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}=0.028+1.362+4.696+0.694+3.36+0.694=10.834$.
• The denominator is $\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}=\sqrt{22.834\times10.834}=\sqrt{247.327}\approx15.727$.

Step3: Calculate the correlation coefficient

$r=\frac{-11.157}{15.727}\approx - 0.709$.

Answer:

$-0.709$