Questions: Find the value of the linear correlation coefficient (r). The paired data below consist of the temperatures (in degrees Fahrenheit) on randomly chosen days and the amount a certain kind of plant grew (in millimeters). Temperature 62 76 50 51 71 46 51 44 79 Growth 36 39 50 13 33 33 17 6 16 0 0.256 -0.210 0.196

Find the value of the linear correlation coefficient (r). The paired data below consist of the temperatures (in degrees Fahrenheit) on randomly chosen days and the amount a certain kind of plant grew (in millimeters).
Temperature 62 76 50 51 71 46 51 44 79
Growth 36 39 50 13 33 33 17 6 16
0
0.256
-0.210
0.196
Transcript text: Find the value of the linear correlation coefficient $r$. The paired data below consist of the temperatures (in degrees Fahrenheit) on randomly chosen days and the amount a certain kind of plant grew (in millimeters). \begin{tabular}{l|lllllllll} Temperature & 62 & 76 & 50 & 51 & 71 & 46 & 51 & 44 & 79 \\ \hline Growth & 36 & 39 & 50 & 13 & 33 & 33 & 17 & 6 & 16 \end{tabular} 0 0.256 $-0.210$ 0.196
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Solution

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Solution Steps

Step 1: Calculate Covariance

The covariance between the temperature \( X \) and the growth \( Y \) is calculated as follows:

\[ \text{Cov}(X,Y) = 38.125 \]

Step 2: Calculate Standard Deviations

The standard deviation of the temperature \( X \) is given by:

\[ \sigma_X = 13.439 \]

The standard deviation of the growth \( Y \) is given by:

\[ \sigma_Y = 14.509 \]

Step 3: Calculate Correlation Coefficient

The correlation coefficient \( r \) is calculated using the formula:

\[ r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} \]

Substituting the values:

\[ r = \frac{38.125}{13.439 \times 14.509} \approx 0.196 \]

Final Answer

The correlation coefficient \( r \) is

\[ \boxed{0.196} \]

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