The linear correlation coefficient, \(r\), is a measure which tells us the strength and direction of a relationship between two variables. The correlation coefficient \(r\in \left[-1;1\right]\). When \(r=-1\), there is perfect negative correlation, when \(r=0\), there is no correlation and when \(r=1\) there is perfect positive correlation. NB. See the fifth bullet at the beginning of the chapter regarding the formula for the correlation coefficient.
The linear correlation coefficient \(r\) can be calculated using the formula where \(b\) is the gradient of the least squares regression line, \(\sigma_{x}\) is the standard deviation of the \(x\)-values and \(\sigma_{y}\) is the standard deviation of the \(y\)-values. This is known as the Pearson's product moment correlation coefficient. It is much easier to do on a calculator
where you simply follow the procedure for the regression equation, and go on to find \(r\). In general: Correlation does not imply causation! Just because two variables are correlated does not mean that they are causally linked, i.e. if A and B are correlated, that does not mean A causes B, or vice versa. This is a common mistake made by many people, especially journalists looking for their next juicy story. For example, ice cream sales and shark attacks are correlated. This does not mean that the sale of ice cream is somehow causing more shark attacks. Instead, a
simpler explanation is that the warmer it is, the more likely people are to buy ice cream and the more likely people are to go to the beach as well, thus increasing the likelihood of a shark attack. Video: 29DV A cardiologist wanted to test the relationship between resting heart rate and the peak heart rate during exercise. Heart rate is measured in beats per minute (bpm). The following set of data was generated from 12 study participants after they had run on a treadmill at \(\text{10}\) \(\text{km/h}\) for 10 minutes. Draw the scatter plot
 Calculate the equation of the line of best fitAs you learnt previously, use your calculator to determine the values for \(a\) and \(b\). \(a = \text{86,75}\) \(b = \text{0,96}\) Therefore, the equation for the line of best fit is \(y = \text{86,75} + \text{0,96}x\) Calculate the estimated value for \(y\)If \(x = 70\), using our equation, the estimated value for \(y\) is: \[y= \text{86,75} + \text{0,96} \times 70 = \text{153,95}\] Calculate the correlation co-efficientThe formula for \(r\) is: \[r=b\dfrac{\sigma_{x}}{\sigma_{y}}\] We already know the value of \(b\) and you know how to calculate \(b\) by hand from worked example 5, so we are just left to determine the value for \(\sigma_{x}\) and \(\sigma_{y}\). The formula for standard deviation is: \[\sigma_{x}= \frac{\sqrt{\sum\limits_{i=1}^{n}(x_i - \bar{x})^{2}}}{n}\] First, you need to determine \(\bar{x}\) and \(\bar{y}\) and then complete a table like the one below. \begin{align*} \bar{x} &= \frac{\sum\limits_{i=1}^{n}x_i}{n} = 71 \\ \bar{y} &= \frac{\sum\limits_{i=1}^{n}y_i}{n} = \text{154,83} \text{ (rounded to two decimal places)} \end{align*}
\begin{align*} \sigma_{x}&= \frac{\sqrt{\sum\limits_{i=1}^{n}(x_i - \bar{x})^{2}}}{n} = \frac{\sqrt{\text{1 534}}}{12} = \pm \text{3,26} \\ \sigma_{y}&= \frac{\sqrt{\sum\limits_{i=1}^{n}(y_i - \bar{y})^{2}}}{n} = \frac{\sqrt{\text{1 861,68}}}{12} = \pm \text{3,60} \\ b&=\text{0,96} \\ \therefore r&= \text{0,96} \times \frac{\text{3,26}}{\text{3,60}} \\ &= \text{0,87} \end{align*} Confirm your answer using your calculatorOnce you know the method for finding the equation of the best line of fit on your calculator, finding the value for \(r\) is trivial. After you have entered all your \(x\) and \(y\) values into your calculator, in STAT mode:
Comment on the correlation coefficient\[r = \text{0,87}\] Therefore, there is a strong, positive, linear relationship between resting heart rate and peak heart rate during exercise. This means that the higher your resting heart rate, the higher your peak heart rate during exercise is likely to be. Correlation coefficientTextbook Exercise 9.4
\begin{align*} r&= b\frac{\sigma_x}{\sigma_{y}} \\ b & = \frac{n\sum xy-\sum x\sum y}{n\sum {x}^{2}-{\left(\sum x\right)}^{2}}=\frac{10\times 479-125\times 41}{10\times \text{1 705}-{125}^{2}}=-\text{0,235} \\ \sigma_{x}&=\sqrt{\frac{\sum\left(x-\bar{x}\right)^{2}}{n}} = \sqrt{\frac{\text{142,5}}{10}}=\sqrt{\text{14,25}}=\pm \text{3,775}\\ \sigma_{y}&=\sqrt{\frac{\sum\left(y-\bar{y}\right)^{2}}{n}} = \sqrt{\frac{\text{94,9}}{10}}=\sqrt{\text{9,49}}= \pm \text{3,081}\\ \therefore r&= -\text{0,235} \times \frac{\text{3,775}}{\text{3,081}}\\ &= -\text{0,29} \end{align*} Therefore, the correlation between \(x\) and \(y\) is negative but weak.
\begin{align*} r&= b\frac{\sigma_x}{\sigma_{y}} \\ b & = \frac{n\sum xy-\sum x\sum y}{n\sum {x}^{2}-{\left(\sum x\right)}^{2}}=\frac{10\times \text{5 138}-87\times 529}{10\times \text{863}-{87}^{2}}=\text{5,049} \\ \sigma_{x}&=\sqrt{\frac{\sum\left(x-\bar{x}\right)^{2}}{n}} = \sqrt{\frac{\text{106,1}}{10}}=\sqrt{\text{10,61}}=\pm \text{3,26}\\ \sigma_{y}&=\sqrt{\frac{\sum\left(y-\bar{y}\right)^{2}}{n}} = \sqrt{\frac{\text{5 862,9}}{10}}=\sqrt{\text{586,29}}=\pm \text{24,21}\\ \therefore r&= \text{5,049} \times \frac{\text{3,26}}{\text{24,21}}\\ &= \text{0,68} \end{align*} Therefore, the correlation between \(x\) and \(y\) is positive and moderate.
\begin{align*} r&= b\frac{\sigma_x}{\sigma_{y}} \\ b & = \frac{n\sum xy-\sum x\sum y}{n\sum {x}^{2}-{\left(\sum x\right)}^{2}}=\frac{10\times \text{8 012}-122\times 541}{10\times \text{1 774}-{122}^{2}}=\text{4,943} \\ \sigma_{x}&=\sqrt{\frac{\sum\left(x-\bar{x}\right)^{2}}{n}} = \sqrt{\frac{\text{285,6}}{10}}=\sqrt{\text{28,56}}=\pm \text{5,344}\\ \sigma_{y}&=\sqrt{\frac{\sum\left(y-\bar{y}\right)^{2}}{n}} = \sqrt{\frac{\text{8 372,9}}{10}}=\sqrt{\text{837,29}}= \pm \text{28,936}\\ \therefore r&= \text{4,943} \times \frac{\text{5,344}}{\text{28,936}}\\ &= \text{0,91} \end{align*} Therefore, the correlation between \(x\) and \(y\) is positive and very strong.
\(r=-\text{0,95}\), negative, very strong.
\(r=-\text{0,40}\), negative, weak.
\(r=\text{0,00}\), no correlation
\(r=\text{0,14}\), positive, very weak.
\(r=\text{0,83}\), positive, strong. \(b = -\text{1,88}; \enspace \sigma^{2}_x = \text{48,62}; \enspace \sigma^{2}_y = \text{736,54}.\) \[r=-\text{1,88} \times \sqrt{\frac{\text{48,62}}{\text{736,54}}} = -\text{0,48}\] \(a = \text{32,19}; \enspace x = \text{4,3}; \enspace \bar{y} = \text{36,6}; \enspace \sum\limits_{i=1}^{n}(x_i-\bar{x})^{2} = \text{620,1};\enspace \sum\limits_{i=1}^{n}(y_i-\bar{y})^{2}= \text{2 636,4}.\) \begin{align*} a&= \bar{y} - b\bar{x} \\ \therefore b&=\frac{\hat{y}-a}{x} = \frac{\text{36,6} - \text{32,19}}{\text{4,3}} = \text{1,03}\\ \therefore r&= \text{1,03} \times \sqrt{\frac{\text{620,1}}{\text{2 636,4}}} = \text{0,50} \end{align*} Copy and complete the table.
Using your table, determine the equation of the least squares regression line. Round \(a\) and \(b\) to two decimal places in your final answer. \begin{align*} b & = \frac{n{\sum }_{i=1}^{n}{x}_{i}{y}_{i}-{\sum }_{i=1}^{n}{x}_{i}{\sum }_{i=1}^{n}{y}_{i}}{n{\sum }_{i=1}^{n}{\left({x}_{i}\right)}^{2}-{\left({\sum }_{i=1}^{n}{x}_{i}\right)}^{2}} \\ & = \frac{10 \times \text{8 492} - 391 \times \text{228}}{10 \times \text{16 851} - 391^{2}} = -\text{0,2705227462} \\ \\ a&= \bar{y}-b\bar{x} = \frac{\text{228}}{\text{10}} - -\text{0,2705227462} \times \frac{391}{10} = \text{33,37743938} \\ \\ \therefore \hat{y}&= \text{33,38} -\text{0,27}x \end{align*} Use your calculator to confirm your equation for the least squares regression line. Answer should be as above. Using your table, determine the value of the correlation coefficient to two decimal places. \begin{align*} r&=b\left(\frac{\sigma_{x}}{\sigma_{y}}\right) \\ &=-\text{0,27}\left(\frac{\sqrt{\frac{\text{1 562,9}}{10}}}{\sqrt{\frac{\text{173,6}}{10}}}\right)\\ &=-\text{0,81} \end{align*} What can you deduce about the relationship between how far north a city is and its average temperature? There is a strong, negative, linear correlation between how far north a city is (latitude) and average temperature. Estimate the latitude of Paris if it has an average temperature of \(\text{25}\)\(\text{°C}\) \begin{align*} 25&= \text{33,38} + -\text{0,27}(x) \\ \therefore x&=\frac{25-\text{33,38}}{-\text{0,27}} \\ &=\text{31,04} \text{ degrees North} \end{align*} Draw a scatter plot of the data. Use your calculator to determine the equation of the least squares regression line and draw this line on your scatter plot. Round \(a\) and \(b\) to two decimal places in your final answer. \[\hat{y}=\text{2,67} + -\text{0,02}x\] Using your calculator, determine the correlation coefficient to two decimal places. \(r = -\text{0,92}\) Describe the relationship between the distance travelled per trip and the fuel cost per kilometre. There is a very strong, negative, linear relationship between distance travelled per trip and the fuel cost per kilometre. Predict the distance travelled if the cost per kilometre is \(\text{R}\,\text{1,75}\). \begin{align*} \text{1,75}&=\text{2,67} -\text{0,02}x \\ \therefore x &= \frac{\text{1,75}-\text{2,67}}{-\text{0,02}} = 46 \text{ kilometres} \end{align*} Draw a scatter plot of the data. What is the influence of more time taken to complete the task on the number of errors made? When more time is taken to complete the task, the learners make fewer errors. OR When less time is taken to complete the task, the learners make more errors. Determine the equation of the least squares regression line and draw this line on your scatter plot. Round \(a\) and \(b\) to two decimal places in your final answer. \begin{align*} a&= \text{14,71} \\ b&= -\text{0,53} \\ \hat{y}&=\text{14,71} - \text{0,53}x \end{align*} Determine the correlation coefficient to two decimal places. \(r = -\text{0,96}\) Predict the number of errors that will be made by a learner who takes 13 seconds to complete this task. \begin{align*} \hat{y}&=\text{14,71} -\text{0,53}(13) \\ &\approx \text{7,82} \\ &\approx 8 \end{align*} Comment on the strength of the relationship between the variables. There is a strong negative relationship between the variables. Draw a scatter plot of the data. Determine the equation of the least squares regression line. \begin{align*} a&=\text{293,06} \\ b&=\text{74,28} \\ \hat{y} &=\text{293,06} + \text{74,28}x \end{align*} Calculate the correlation coefficient. \(r=\text{0,95}\) Predict, correct to the nearest 50, the weekly sales for a CD that was played 45 times by the radio station in the previous week. \begin{align*} \hat{y}&= \text{293,06} + \text{74,28}(45) \\ &= \text{3 635,66} \\ &\approx \text{3 650} \text{ (to the nearest 50)} \end{align*} Comment on the strength of the relationship between the variables. There is a very strong positive relationship between the number of times that a CD was played and the sales of that CD in the following week. Which of the following correlation coefficient values indicate the strongest relationship between two variables?The strongest linear relationship is indicated by a correlation coefficient of -1 or 1. The weakest linear relationship is indicated by a correlation coefficient equal to 0. A positive correlation means that if one variable gets bigger, the other variable tends to get bigger.
Is .75 a strong correlation coefficient?As a rule of thumb, a correlation greater than 0.75 is considered to be a “strong” correlation between two variables.
Is 0.8 A strong correlation?A correlation coefficient of -0.8 indicates an exceptionally strong negative correlation, meaning that the two variables tend to move in opposite directions. The closer the coefficient is to -1.0, the stronger the negative relationship will be.
Why is 0.7 A strong correlation?Values between 0.7 and 1.0 (−0.7 and −1.0) indicate a strong positive (negative) linear relationship through a firm linear rule. It is the correlation coefficient between the observed and modelled (predicted) data values. It can increase as the number of predictor variables in the model increases; it does not decrease.
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Which of the following correlation coefficient values indicate the strongest relationship between two variables A 74 B 68 C 69 D 71?
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