Question

find the correlation coefficient, r, of the data described below. mr. baldwin wants to demonstrate the importance of proofreading to his english class. he had students read the same passage and mark all the spelling and grammar errors they could find. he recorded how many minutes each student had spent on the exercise, x, and how many errors that student had missed, y. minutes errors 6 27 7 20 9 11 9 17 12 13 round your answer to the nearest thousandth. r =

Explanation:

Step1: Calculate the means

Let $x = [6,7,9,9,12]$ and $y=[27,20,11,17,13]$.
The mean of $x$, $\bar{x}=\frac{6 + 7+9+9+12}{5}=\frac{43}{5} = 8.6$.
The mean of $y$, $\bar{y}=\frac{27+20 + 11+17+13}{5}=\frac{88}{5}=17.6$.

Step2: Calculate the numerator and denominator components

For the numerator of the correlation - coefficient formula $r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}$:
$(x_1-\bar{x})(y_1 - \bar{y})=(6 - 8.6)(27-17.6)=(-2.6)\times9.4=- 24.44$
$(x_2-\bar{x})(y_2 - \bar{y})=(7 - 8.6)(20 - 17.6)=(-1.6)\times2.4=-3.84$
$(x_3-\bar{x})(y_3 - \bar{y})=(9 - 8.6)(11 - 17.6)=0.4\times(-6.6)=-2.64$
$(x_4-\bar{x})(y_4 - \bar{y})=(9 - 8.6)(17 - 17.6)=0.4\times(-0.6)=-0.24$
$(x_5-\bar{x})(y_5 - \bar{y})=(12 - 8.6)(13 - 17.6)=3.4\times(-4.6)=-15.64$
$\sum_{i = 1}^{5}(x_{i}-\bar{x})(y_{i}-\bar{y})=-24.44-3.84 - 2.64-0.24-15.64=-46.8$.

For the denominator:
$(x_1-\bar{x})^2=(6 - 8.6)^2=(-2.6)^2 = 6.76$
$(x_2-\bar{x})^2=(7 - 8.6)^2=(-1.6)^2 = 2.56$
$(x_3-\bar{x})^2=(9 - 8.6)^2=(0.4)^2 = 0.16$
$(x_4-\bar{x})^2=(9 - 8.6)^2=(0.4)^2 = 0.16$
$(x_5-\bar{x})^2=(12 - 8.6)^2=(3.4)^2 = 11.56$
$\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=6.76+2.56 + 0.16+0.16+11.56=21.2$.

$(y_1-\bar{y})^2=(27-17.6)^2=(9.4)^2 = 88.36$
$(y_2-\bar{y})^2=(20 - 17.6)^2=(2.4)^2 = 5.76$
$(y_3-\bar{y})^2=(11 - 17.6)^2=(-6.6)^2 = 43.56$
$(y_4-\bar{y})^2=(17 - 17.6)^2=(-0.6)^2 = 0.36$
$(y_5-\bar{y})^2=(13 - 17.6)^2=(-4.6)^2 = 21.16$
$\sum_{i = 1}^{5}(y_{i}-\bar{y})^{2}=88.36+5.76+43.56+0.36+21.16=159.2$.

$\sqrt{\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{5}(y_{i}-\bar{y})^{2}}=\sqrt{21.2\times159.2}=\sqrt{3375.04}\approx58.095$.

Step3: Calculate the correlation coefficient

$r=\frac{-46.8}{58.095}\approx - 0.806$.

Answer:

$-0.806$