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Two separate tests are designed to measure a students ability to solve problems. Several students are randomly selected to take both tests and the results are:

$$ \begin{matrix} \text{Test A}(x) & 43 & 65 & 73 & 34 & 99 & 78 & 65 \\ \text{Test B}(y) & 39 & 60 & 62 & 20 & 85 & 70 & 54 \end{matrix} $$

Calculate $r$, the linear correlation coefficient.

My answer is always $1.07$ after I solve using the formula.

$$r = \frac{n \sum x y - \sum x \sum y}{\sqrt{ n (\sum x^2) - (\sum x)^2} \cdot \sqrt{n (\sum y^2) - (\sum y)^2}}$$

I don't have a linear calculator, so my numbers might be off due to that. If anyone needs my calculations for the formula that I have gotten, I can provide.

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It is good that you recognize that 1.07 can't be right. Go ahead and show your calculations. – Doug Chatham Dec 4 '11 at 17:55

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