Exam 10: Bivariate Measures of Association

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You are given the following pairs of sample scores on X and Y. Determine the Pearson correlation coefficient using SPSS.

$\begin{array}{ccc}X && Y \\\hline 20 & & 400 \\30 & & 300 \\25 & & 250 \\40 & & 500 \\50 & & 100 \\\hline\end{array}$
Using SPSS, we compute the Pearson correlation to be ?.397.

Correct Answer:

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You are given the following pairs of sample scores on X and Y. Determine the Pearson correlation coefficient.

$\begin{array}{ccc}X & Y \\\hline 8 & 7 \\6 & 6 \\6 & 8 \\9 & 6 \\1 & 5\\\hline\end{array}$
$\begin{array}{cccccc}X & X^{2} & & Y & Y^{2} & X Y \\\hline8 & 64 & & 7 & 49 & 56 \\4 & 16 & & 6 & 36 & 24 \\6 & 36 & & 8 & 64 & 48 \\9 & 81 & & 6 & 36 & 54 \\1 & 1 & & 5 & 25 & 5\\\\\sum=28&\Sigma=198 &&\sum=32&\sum=210 & \Sigma=187\\\hline\end{array}$

The Pearson product-moment correlation coefficient is computed as follows.
$r_{X Y}=\frac{n\left(\sum_{i=1}^{n} X_{i} Y_{i}\right)-\left(\sum_{i=1}^{n} X_{i}\right)\left(\sum_{i=1}^{n} Y_{i}\right)}{\sqrt{\left[n\left(\sum_{i=1}^{n} X_{i}^{2}\right)-\left(\sum_{i=1}^{n} X_{i}\right)^{2}\left[n\left(\sum_{i=1}^{n} Y_{i}^{2}\right)-\left(\sum_{i=1}^{n} Y_{i}\right)^{2}\right]\right.}}$ $r_{X Y}=\frac{5(187)-(28)(32)}{\sqrt{\left[5(198)-(28)^{2}\right]\left[5(210)-(32)^{2}\right]}}=\frac{935-896}{\sqrt{(990-784)(1050-1024)}}=\frac{39}{73.185}=.533$