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.
hours studied and test grades

hours studied	0.25	1	1.25	1.25	2	2.5	2.75	3.75	4	4.5	5	5.25
test grade	68	72	85	62	76	82	91	84	79	93	95	97

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

Explanation:

Step1: Calculate the means

Let $x$ be hours - studied and $y$ be test - grade.
$n = 12$
$\bar{x}=\frac{0.25 + 1+1.25+1.25+2+2.5+2.75+3.75+4+4.5+5+5.25}{12}=\frac{33.5}{12}\approx2.792$
$\bar{y}=\frac{68 + 72+85+62+76+82+91+84+79+93+95+97}{12}=\frac{994}{12}\approx82.833$

Step2: Calculate the numerator of the correlation - coefficient formula

$\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})$
$(0.25 - 2.792)(68 - 82.833)+(1 - 2.792)(72 - 82.833)+(1.25 - 2.792)(85 - 82.833)+(1.25 - 2.792)(62 - 82.833)+(2 - 2.792)(76 - 82.833)+(2.5 - 2.792)(82 - 82.833)+(2.75 - 2.792)(91 - 82.833)+(3.75 - 2.792)(84 - 82.833)+(4 - 2.792)(79 - 82.833)+(4.5 - 2.792)(93 - 82.833)+(5 - 2.792)(95 - 82.833)+(5.25 - 2.792)(97 - 82.833)$
$=(-2.542)(-14.833)+(-1.792)(-10.833)+(-1.542)(2.167)+(-1.542)(-20.833)+(-0.792)(-6.833)+(-0.292)(-0.833)+(-0.042)(8.167)+(0.958)(1.167)+(1.208)(-3.833)+(1.708)(10.167)+(2.208)(12.167)+(2.458)(14.167)$
$=37.705+19.413 - 3.342+32.125+5.312+0.243 - 0.343+1.118 - 4.630+17.365+26.865+34.823$
$=162.893$

Step3: Calculate the denominator of the correlation - coefficient formula

$\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}$
First, calculate $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$:
$(0.25 - 2.792)^{2}+(1 - 2.792)^{2}+(1.25 - 2.792)^{2}+(1.25 - 2.792)^{2}+(2 - 2.792)^{2}+(2.5 - 2.792)^{2}+(2.75 - 2.792)^{2}+(3.75 - 2.792)^{2}+(4 - 2.792)^{2}+(4.5 - 2.792)^{2}+(5 - 2.792)^{2}+(5.25 - 2.792)^{2}$
$=(-2.542)^{2}+(-1.792)^{2}+(-1.542)^{2}+(-1.542)^{2}+(-0.792)^{2}+(-0.292)^{2}+(-0.042)^{2}+(0.958)^{2}+(1.208)^{2}+(1.708)^{2}+(2.208)^{2}+(2.458)^{2}$
$=6.462+3.211+2.378+2.378+0.627+0.085+0.002+0.918+1.460+2.917+4.875+6.042$
$=31.355$
Next, calculate $\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}$:
$(68 - 82.833)^{2}+(72 - 82.833)^{2}+(85 - 82.833)^{2}+(62 - 82.833)^{2}+(76 - 82.833)^{2}+(82 - 82.833)^{2}+(91 - 82.833)^{2}+(84 - 82.833)^{2}+(79 - 82.833)^{2}+(93 - 82.833)^{2}+(95 - 82.833)^{2}+(97 - 82.833)^{2}$
$=(-14.833)^{2}+(-10.833)^{2}+(2.167)^{2}+(-20.833)^{2}+(-6.833)^{2}+(-0.833)^{2}+(8.167)^{2}+(1.167)^{2}+(-3.833)^{2}+(10.167)^{2}+(12.167)^{2}+(14.167)^{2}$
$=220.025+117.361+4.696+434.018+46.690+0.694+66.694+1.362+14.694+103.361+148.025+200.694$
$=1458.414$
$\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}=\sqrt{31.355\times1458.414}=\sqrt{45734.377}\approx213.856$

Step4: Calculate the correlation coefficient $r$

$r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}=\frac{162.893}{213.856}\approx0.762$

Step5: Determine statistical significance

The degrees of freedom $df=n - 2=12 - 2 = 10$.
For a two - tailed test at $\alpha = 0.01$ and $df = 10$, the critical value of the Pearson correlation coefficient is approximately $\pm0.708$.
Since $|r|=|0.762|>0.708$, $r$ is statistically significant at the $0.01$ level of significance.

Answer:

$r = 0.762$, and $r$ is statistically significant at the $0.01$ level of significance.