Question

determining the strength of a correlation
claudia records the hours she spent studying and her
test scores for 5 tests.
what is the correlation coefficient?
what is the strength of the model?
strong negative correlation
moderate negative correlation
strong positive correlation
moderate positive correlation

Explanation:

Step1: Calculate the means

Let \(x\) be hours spent studying (\(x = [1,2,3,4,5]\)) and \(y\) be test scores (\(y=[72,80,90,82,95]\)).
The mean of \(x\), \(\bar{x}=\frac{1 + 2+3+4+5}{5}=\frac{15}{5} = 3\)
The mean of \(y\), \(\bar{y}=\frac{72 + 80+90+82+95}{5}=\frac{419}{5}=83.8\)

Step2: Calculate numerator and denominator for correlation coefficient formula

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

• Calculate \((x_{i}-\bar{x})(y_{i}-\bar{y})\):

\((1 - 3)(72-83.8)=(- 2)(-11.8) = 23.6\)
\((2 - 3)(80 - 83.8)=(-1)(-3.8)=3.8\)
\((3 - 3)(90 - 83.8)=0\times6.2 = 0\)
\((4 - 3)(82 - 83.8)=1\times(-1.8)=-1.8\)
\((5 - 3)(95 - 83.8)=2\times11.2 = 22.4\)
\(\sum_{i = 1}^{5}(x_{i}-\bar{x})(y_{i}-\bar{y})=23.6+3.8 + 0-1.8+22.4=48\)

• Calculate \((x_{i}-\bar{x})^{2}\):

\((1 - 3)^{2}=4\), \((2 - 3)^{2}=1\), \((3 - 3)^{2}=0\), \((4 - 3)^{2}=1\), \((5 - 3)^{2}=4\)
\(\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=4 + 1+0+1+4=10\)

• Calculate \((y_{i}-\bar{y})^{2}\):

\((72-83.8)^{2}=(-11.8)^{2}=139.24\)
\((80 - 83.8)^{2}=(-3.8)^{2}=14.44\)
\((90 - 83.8)^{2}=6.2^{2}=38.44\)
\((82 - 83.8)^{2}=(-1.8)^{2}=3.24\)
\((95 - 83.8)^{2}=11.2^{2}=125.44\)
\(\sum_{i = 1}^{5}(y_{i}-\bar{y})^{2}=139.24+14.44+38.44+3.24+125.44 = 320.8\)

Step3: Calculate \(r\)

\(r=\frac{48}{\sqrt{10\times320.8}}=\frac{48}{\sqrt{3208}}\approx\frac{48}{56.64}\approx0.85\)

Since \(r\approx0.85\) (close to \(1\)), it is a strong positive correlation.

Answer:

The correlation coefficient is approximately \(0.85\). The strength of the model is a strong positive correlation.