Question

a teacher recorded the number of hours students studied for a test and their final test scores. the data is shown in the table.
study time vs. test score
find the correlation coefficient between the variables x, hours studied, and y, test score.
use the drop - down menus to choose the response to make each statement true.
the correlation coefficient is approximately choose your answer... this means there is choose your answer... correlation between x and y.

Explanation:

Step1: Calculate the means

The mean of \(x\) values: \(\bar{x}=\frac{1 + 2+3+5+6}{5}=\frac{17}{5} = 3.4\)
The mean of \(y\) values: \(\bar{y}=\frac{70+78+82+94+98}{5}=\frac{422}{5}=84.4\)

Step2: Calculate the numerator and denominator of the correlation coefficient formula

The formula for the correlation coefficient \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)

For \((x_{i}-\bar{x})(y_{i}-\bar{y})\):
When \(x = 1,y = 70\): \((1 - 3.4)(70 - 84.4)=(- 2.4)\times(-14.4)=34.56\)
When \(x = 2,y = 78\): \((2 - 3.4)(78 - 84.4)=(-1.4)\times(-6.4) = 8.96\)
When \(x = 3,y = 82\): \((3 - 3.4)(82 - 84.4)=(-0.4)\times(-2.4)=0.96\)
When \(x = 5,y = 94\): \((5 - 3.4)(94 - 84.4)=1.6\times9.6 = 15.36\)
When \(x = 6,y = 98\): \((6 - 3.4)(98 - 84.4)=2.6\times13.6=35.36\)
\(\sum_{i = 1}^{5}(x_{i}-\bar{x})(y_{i}-\bar{y})=34.56+8.96 + 0.96+15.36+35.36=95.2\)

For \((x_{i}-\bar{x})^{2}\):
When \(x = 1\): \((1 - 3.4)^{2}=(-2.4)^{2}=5.76\)
When \(x = 2\): \((2 - 3.4)^{2}=(-1.4)^{2}=1.96\)
When \(x = 3\): \((3 - 3.4)^{2}=(-0.4)^{2}=0.16\)
When \(x = 5\): \((5 - 3.4)^{2}=1.6^{2}=2.56\)
When \(x = 6\): \((6 - 3.4)^{2}=2.6^{2}=6.76\)
\(\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=5.76+1.96+0.16+2.56+6.76 = 17.2\)

For \((y_{i}-\bar{y})^{2}\):
When \(y = 70\): \((70 - 84.4)^{2}=(-14.4)^{2}=207.36\)
When \(y = 78\): \((78 - 84.4)^{2}=(-6.4)^{2}=40.96\)
When \(y = 82\): \((82 - 84.4)^{2}=(-2.4)^{2}=5.76\)
When \(y = 94\): \((94 - 84.4)^{2}=9.6^{2}=92.16\)
When \(y = 98\): \((98 - 84.4)^{2}=13.6^{2}=184.96\)
\(\sum_{i = 1}^{5}(y_{i}-\bar{y})^{2}=207.36+40.96+5.76+92.16+184.96=531.2\)

Then \(r=\frac{95.2}{\sqrt{17.2\times531.2}}=\frac{95.2}{\sqrt{9136.64}}=\frac{95.2}{95.6}=0.996\approx1\)

Answer:

The correlation coefficient is approximately \(1\). This means there is a strong positive correlation between \(x\) and \(y\).