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?

Question

Accepted Answer

# Explanation:
## Step1: Calculate the means
Let $x$ be hours spent studying and $y$ be test score.
$\bar{x}=\frac{1 + 2+3+4+5}{5}=\frac{15}{5} = 3$
$\bar{y}=\frac{72 + 80+90+82+95}{5}=\frac{419}{5}=83.8$

## Step2: Calculate numerator and denominator for $r$
$$
\begin{align*}
\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})&=(1 - 3)(72-83.8)+(2 - 3)(80 - 83.8)+(3 - 3)(90-83.8)+(4 - 3)(82-83.8)+(5 - 3)(95-83.8)\
&=(- 2)(-11.8)+(-1)(-3.8)+0\times6.2+1\times(-1.8)+2\times11.2\
&=23.6 + 3.8+0-1.8+22.4\
&=48
\end{align*}
$$

$$
\begin{align*}
\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}&=(1 - 3)^{2}+(2 - 3)^{2}+(3 - 3)^{2}+(4 - 3)^{2}+(5 - 3)^{2}\
&=4 + 1+0+1+4\
&=10
\end{align*}
$$

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

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

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

Since $|r|\approx0.85$ and $0.7<|r|\leq1$, the strength of the model is strong.

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

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the means

Step2: Calculate numerator and denominator for \(r\)

Answer: