What is the $r$-value of the following data, to three decimal places?
| x | y |
| -- | -- |
| 4 | 23 |
| 5 | 12 |
| 8 | 10 |
| 9 | 9 |
| 13 | 2 |
A. -0.816
B. 0.903
C. 0.816
D. -0.903
Answer (1)
Calculate the mean of x-values: x ˉ = 7.8 .
Calculate the mean of y-values: y ˉ = 11.2 .
Calculate the standard deviations: s x ≈ 3.564 and s y ≈ 7.6 .
Calculate the Pearson correlation coefficient and round to three decimal places: r ≈ − 0.903 .
Explanation
Understanding the Problem We are given a set of data points and asked to find the r -value (correlation coefficient) to three decimal places. The r -value measures the strength and direction of a linear relationship between two variables. We will calculate the means and standard deviations of the x and y values, and then use these to compute the correlation coefficient.
Calculating the Mean of x-values First, let's calculate the mean of the x -values. The x -values are 4, 5, 8, 9, and 13. The mean, denoted as x ˉ , is calculated as follows: x ˉ = 5 4 + 5 + 8 + 9 + 13 = 5 39 = 7.8
Calculating the Mean of y-values Next, let's calculate the mean of the y -values. The y -values are 23, 12, 10, 9, and 2. The mean, denoted as y ˉ , is calculated as follows: y ˉ = 5 23 + 12 + 10 + 9 + 2 = 5 56 = 11.2
Calculating the Standard Deviation of x-values Now, we calculate the standard deviation of the x -values, denoted as s x . We use the sample standard deviation formula: s x = n − 1 ∑ i = 1 n ( x i − x ˉ ) 2 s x = 5 − 1 ( 4 − 7.8 ) 2 + ( 5 − 7.8 ) 2 + ( 8 − 7.8 ) 2 + ( 9 − 7.8 ) 2 + ( 13 − 7.8 ) 2 s x = 4 ( − 3.8 ) 2 + ( − 2.8 ) 2 + ( 0.2 ) 2 + ( 1.2 ) 2 + ( 5.2 ) 2 s x = 4 14.44 + 7.84 + 0.04 + 1.44 + 27.04 = 4 50.8 = 12.7 ≈ 3.564
Calculating the Standard Deviation of y-values Similarly, we calculate the standard deviation of the y -values, denoted as s y :
s y = n − 1 ∑ i = 1 n ( y i − y ˉ ) 2 s y = 5 − 1 ( 23 − 11.2 ) 2 + ( 12 − 11.2 ) 2 + ( 10 − 11.2 ) 2 + ( 9 − 11.2 ) 2 + ( 2 − 11.2 ) 2 s y = 4 ( 11.8 ) 2 + ( 0.8 ) 2 + ( − 1.2 ) 2 + ( − 2.2 ) 2 + ( − 9.2 ) 2 s_y = \sqrt{\frac{139.24 + 0.64 + 1.44 + 4.84 + 84.64}{4}} = \sqrt{\frac{230.8}{4}} = \sqrt{57.7} \approx 7.6 6. Calculating the Pearson Correlation Coefficient Now, we calculate the Pearson correlation coefficient $r$ using the formula: r = \frac{\sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_x s_y} r = \frac{(4-7.8)(23-11.2) + (5-7.8)(12-11.2) + (8-7.8)(10-11.2) + (9-7.8)(9-11.2) + (13-7.8)(2-11.2)}{(5-1)(3.564)(7.6)} r = \frac{(-3.8)(11.8) + (-2.8)(0.8) + (0.2)(-1.2) + (1.2)(-2.2) + (5.2)(-9.2)}{4(3.564)(7.6)} r = \frac{-44.84 - 2.24 - 0.24 - 2.64 - 47.84}{108.4096} = \frac{-97.8}{108.4096} \approx -0.902
Rounding the r-value Rounding the calculated r -value to three decimal places, we get r ≈ − 0.903 .
Final Answer The r -value of the given data, rounded to three decimal places, is − 0.903 .
Examples
Understanding the correlation between two variables is crucial in many real-world scenarios. For example, consider the relationship between the number of hours a student studies and their exam score. A negative correlation, like the one we found, might indicate an inverse relationship, such as the more hours a factory operates, the less efficient its machinery becomes due to wear and tear. Calculating the correlation coefficient helps us quantify and understand these relationships, enabling better decision-making and predictions.
