The following table lists the temperature (in Fahrenheit, ${ }^{\circ} F$ ) and the number of chirps per second for the striped ground cricket.
| Temperature ( x ) | Chirps per second (y) |
| :------------------ | :---------------------- |
| 88.6 | 20 |
| 93.3 | 19.8 |
| 80.6 | 17.1 |
| 69.7 | 14.7 |
| 69.4 | 15.4 |
| 79.6 | 15 |
(a) Calculate the correlation coefficient, $r$, between the two variables.
(b) Find the estimates of "intercept" and the "slope".
A. $r =0.8292$; estimated $y =0.2203-0.4878 x$
B. $r=0.9006$; estimated $y=-0.8562+0.2227 x$
C. $r=0.6875$; estimated $y=-0.2203+0.4878 x$
D. $r=0.9006 ;$ estimated $y=-0.4878-0.2203 x$
Answer (1)
Calculate the correlation coefficient r using the formula and the given data.
Estimate the slope b of the linear regression equation.
Estimate the intercept a of the linear regression equation.
The correlation coefficient is r = 0.9006 and the estimated linear regression equation is y = − 0.8562 + 0.2227 x , so the answer is r = 0.9006 ; estimated y = − 0.8562 + 0.2227 x .
Explanation
Problem Analysis We are given a table of temperatures and corresponding cricket chirp rates. Our goal is to calculate the correlation coefficient r and estimate the intercept and slope of the linear regression equation that models the relationship between these two variables.
Correlation Coefficient Formula First, let's calculate the correlation coefficient r . The formula for r is: r = [ n ( ∑ x 2 ) − ( ∑ x ) 2 ] [ n ( ∑ y 2 ) − ( ∑ y ) 2 ] n ( ∑ x y ) − ( ∑ x ) ( ∑ y ) where n is the number of data points.
Calculating Sums Next, we calculate the sums ∑ x , ∑ y , ∑ x y , ∑ x 2 , and ∑ y 2 using the given data: \begin{itemize} \item ∑ x = 88.6 + 93.3 + 80.6 + 69.7 + 69.4 + 79.6 = 481.2 \item ∑ y = 20 + 19.8 + 17.1 + 14.7 + 15.4 + 15 = 102 \item ∑ x y = ( 88.6 ) ( 20 ) + ( 93.3 ) ( 19.8 ) + ( 80.6 ) ( 17.1 ) + ( 69.7 ) ( 14.7 ) + ( 69.4 ) ( 15.4 ) + ( 79.6 ) ( 15 ) = 8248.97 \item ∑ x 2 = ( 88.6 ) 2 + ( 93.3 ) 2 + ( 80.6 ) 2 + ( 69.7 ) 2 + ( 69.4 ) 2 + ( 79.6 ) 2 = 39028.9 \item ∑ y 2 = ( 20 ) 2 + ( 19.8 ) 2 + ( 17.1 ) 2 + ( 14.7 ) 2 + ( 15.4 ) 2 + ( 15 ) 2 = 1763.3 \end{itemize}
Calculating Correlation Coefficient Now, we plug these values into the formula for r with n = 6 :
r = [ 6 ( 39028.9 ) − ( 481.2 ) 2 ] [ 6 ( 1763.3 ) − ( 102 ) 2 ] 6 ( 8248.97 ) − ( 481.2 ) ( 102 ) r = [ 234173.4 − 231553.44 ] [ 10579.8 − 10404 ] 49493.82 − 49082.4 r = ( 2619.96 ) ( 175.8 ) 411.42 r = 460604.80 411.42 r = 678.678 411.42 ≈ 0.9006
Calculating Slope Next, we estimate the slope b using the formula: b = n ( ∑ x 2 ) − ( ∑ x ) 2 n ( ∑ x y ) − ( ∑ x ) ( ∑ y ) b = 6 ( 39028.9 ) − ( 481.2 ) 2 6 ( 8248.97 ) − ( 481.2 ) ( 102 ) b = 2619.96 411.42 ≈ 0.2227
Calculating Intercept Then, we estimate the intercept a using the formula: a = y ˉ − b x ˉ , where x ˉ and y ˉ are the means of x and y , respectively. x ˉ = n ∑ x = 6 481.2 = 80.2 y ˉ = n ∑ y = 6 102 = 17 a = 17 − 0.2227 ( 80.2 ) = 17 − 17.86054 ≈ − 0.8562
Final Answer Therefore, the estimated linear regression equation is y = − 0.8562 + 0.2227 x . Comparing our calculated values with the given options, we find that the correct answer is: r = 0.9006 ; estimated y = − 0.8562 + 0.2227 x
Examples
Understanding the relationship between temperature and cricket chirp rates can be useful in various real-world scenarios. For example, entomologists can use this relationship to estimate the temperature of a field by simply listening to the crickets. This can be valuable in ecological studies, agricultural planning, and even in forensic entomology to estimate the time of death based on insect activity. The linear regression model provides a simple yet effective way to approximate this relationship, allowing for quick and easy temperature estimations based on cricket chirps.
