Example 4

EXAMPLE 4 Calculating the correlation coefficient $r$

highlowtemp

Find the value of the correlation coefficient $r$ for the temperature data in Table 3.

Solution

We will outline the steps used in calculating the value of $r$ using the temperature data.

Step 1 Calculate the respective sample means, $\bar{x}$ and $\bar{y}$ .
$\begin{matrix} \bar{x} = \frac{\sum x}{n} = 40, & \bar{y} = \frac{\sum y}{n} = 64 \end{matrix}$

Step 2 Construct a table, as shown here in Table 4.

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Table 4.4: Table 4 Calculation table for the correlation coefficient

$r$

City	$x$	$y$	$(x - \bar{x})$	${(x - \bar{x})}^{2}$	$(y - \bar{y})$	${(y - \bar{y})}^{2}$	$(x - \bar{x})$ $(y - \bar{y})$
Boston	30	50	−10	100	−14	196	140
Chicago	35	55	−5	25	−9	81	45
Philadelphia	40	70	0	0	6	36	0
Washington, DC	45	65	5	25	1	1	5
Dallas	50	80	10	100	16	256	160
				$\sum {(x - \bar{x})}^{2} = 250$		$\sum {(y - \bar{y})}^{2} = 570$	$\sum (x - \bar{x}) (y - \bar{y}) = 350$

Note on Rounding: Whenever you calculate a quantity that will be needed for later calculations, do not round. Round only when you arrive at the final answer. Here, because the quantities $s_{x}$ and $s_{y}$ are used to calculate the correlation coefficient $r$ , neither of them is rounded until the end of the calculation.

Step 3 Calculate the respective sample standard deviations $s_{x}$ and $s_{y}$ . Using the sums calculated from Table 4, we have
$\begin{array}{l} s_{x} & = & \sqrt{\frac{\sum {(x - \bar{x})}^{2}}{n - 1}} & = & \sqrt{\frac{250}{5 - 1}} & \approx & 7.90569415 & and \\ s_{y} & = & \sqrt{\frac{\sum {(y - \bar{y})}^{2}}{n - 1}} & = & \sqrt{\frac{570}{5 - 1}} & \approx & 11.93733639 \end{array}$
Step 4 Put these values all together in the formula for the correlation coefficient $r$ :
$r = \frac{\sum (x - \bar{x}) (y - \bar{y})}{(n - 1) s_{x} s_{y}} = \frac{350}{(4) (7.90569415) (11.93733639)} \approx 0.92717265 \approx 0.9272$

The correlation coefficient $r$ for the high and low temperatures is 0.9272.

NOW YOU CAN DO

Exercises 13c–20c.