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the following table shows the hours studied and corresponding test grad…

Question

the following table shows the hours studied and corresponding test grade earned by students on a recent test. calculate the correlation coefficient, r, and determine whether r is statistically significant at the 0.01 level of significance. round your answer to the nearest thousandth.

critical values of the pearson correlation coefficient

hours studied and test grades

hours studied0.50.511.52.53.7555.5
test grade8267566781839593

answer
r =
is r statistically significant at the 0.01 level of significance?

Explanation:

Step1: Calculate means

Let $x$ be hours - studied and $y$ be test - grades.
$n = 8$
$\bar{x}=\frac{0.5 + 0.5+1+1.5+2.5+3.75+5+5.5}{8}=\frac{20.25}{8}=2.53125$
$\bar{y}=\frac{82 + 67+56+67+81+83+95+93}{8}=\frac{624}{8}=78$

Step2: Calculate numerator and denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i - \bar{y})^2$ for each $i$ from $1$ to $n$.
$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(0.5 - 2.53125)(82 - 78)+(0.5 - 2.53125)(67 - 78)+\cdots+(5.5 - 2.53125)(93 - 78)$
$=(- 2.03125)\times4+(-2.03125)\times(-11)+\cdots+(2.96875)\times15$
$=129.9375$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(0.5 - 2.53125)^2+(0.5 - 2.53125)^2+\cdots+(5.5 - 2.53125)^2$
$=21.453125$
$\sum_{i = 1}^{n}(y_i - \bar{y})^2=(82 - 78)^2+(67 - 78)^2+\cdots+(93 - 78)^2$
$=786$

Step3: Calculate correlation coefficient $r$

The formula for the Pearson correlation coefficient is $r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}$
$r=\frac{129.9375}{\sqrt{21.453125\times786}}$
$r=\frac{129.9375}{\sqrt{16862.15625}}$
$r=\frac{129.9375}{129.8544}$
$r\approx0.999$
For $n = 8$ and $\alpha=0.01$ (two - tailed), the critical value of the Pearson correlation coefficient from the table is approximately $0.707$. Since $|r| = 0.999>0.707$, $r$ is statistically significant at the $0.01$ level of significance.

Answer:

$r\approx0.999$, and $r$ is statistically significant at the $0.01$ level of significance.