Question

11\. Chirping Crickets Male crickets chirp by rubbing their front wings together, and their chirping is temperature dependent. The table below shows the number of chirps per second for a cricket, recorded at 10 different temperatures: $$ \begin{array}{l|llllllllll} \text { Chirps per Second } & 20 & 16 & 19 & 18 & 18 & 16 & 14 & 17 & 15 & 16 \\\ \hline \text { Temperature } & 31 & 22 & 32 & 29 & 27 & 23 & 20 & 27 & 20 & 28 \end{array} $$ a. Find the least-squares regression line relating the number of chirps to temperature. b. Do the data provide sufficient evidence to indicate that there is a linear relationship between number of chirps and temperature? c. Calculate \(r^{2}\). What does this value tell you about the effectiveness of the linear regression analysis?

Short answer

What does the coefficient of determination tell us about the relationship between the number of chirps and temperature? Answer: The equation of the regression line is \(y = 4.205 + 0.450x\), the correlation coefficient (r) is approximately 0.829, and the coefficient of determination (\(r^2\)) is approximately 0.687. The coefficient of determination tells us that about 68.7% of the variation in the number of chirps can be explained by the linear relationship with temperature, whereas the remaining 31.3% of the variability may be due to other factors.

Step-by-step solution

Calculate the mean of temperature and number of chirps

To find the least squares regression line, we first calculate the mean of the temperatures and the mean of the number of chirps. Mean temperature, \(\overline{T}= \frac{31+22+32+29+27+23+20+27+20+28}{10}=25.9\) Mean number of chirps, \(\overline{C}= \frac{20+16+19+18+18+16+14+17+15+16}{10}=16.9\)

Calculate the covariance between temperature and chirps

Now, let's find the covariance between the temperatures and the number of chirps: $$ \begin{aligned} \text{Cov}(T,C) &= \frac{1}{n} \sum_{i=1}^{n}(T_i - \overline{T})(C_i - \overline{C}) \\ &= \frac{1}{10}[(31-25.9)(20-16.9)+(22-25.9)(16-16.9)\\ &\enspace\enspace\enspace+(32-25.9)(19-16.9)+(29-25.9)(18-16.9)\\ &\enspace\enspace\enspace+(27-25.9)(18-16.9)+(23-25.9)(16-16.9)\\ &\enspace\enspace\enspace+(20-25.9)(14-16.9)+(27-25.9)(17-16.9)\\ &\enspace\enspace\enspace+(20-25.9)(15-16.9)+(28-25.9)(16-16.9)]\\ &=12.28 \end{aligned} $$

Calculate the variance of temperature

We need to find the variance of the temperatures in order to calculate the regression coefficients. $$ \begin{aligned} \text{Var}(T) & = \frac{1}{n} \sum_{i=1}^{n}(T_i - \overline{T})^2\\ & = \frac{1}{10}[(31-25.9)^2+(22-25.9)^2+(32-25.9)^2\\ &\enspace\enspace\enspace+(29-25.9)^2+(27-25.9)^2+(23-25.9)^2\\ &\enspace\enspace\enspace+(20-25.9)^2+(27-25.9)^2+(20-25.9)^2+(28-25.9)^2]\\ &= 27.29 \end{aligned} $$

Calculate the slope and the intercept

Now, we will find the coefficients for the best fit line, in the form \(y = a + bx\) Slope (b) = \(\frac{\text{Cov}(T,C)}{\text{Var}(T)} = \frac{12.28}{27.29} \approx 0.450\) Intercept (a) = \(\overline{C} - b\overline{T} = 16.9-0.450(25.9) \approx 4.205\) So, the regression line equation is \(y = 4.205 + 0.450x\).

Assess the linear relationship

To determine if there is a linear relationship between temperature and number of chirps, we need to look at the correlation coefficient (\(r\)). \(r = \frac{\text{Cov}(T,C)}{\sqrt{\text{Var}(T) \cdot \text{Var}(C)}}\) But since we weren't given the variance of number of chirps, we need to calculate that first: Mean square of chirps, \(\overline{C^2}=\frac{1}{10}(20^2+16^2+19^2+18^2+18^2+16^2+14^2+17^2+15^2+16^2)=287.8\) The variance of chirps, \(\text{Var}(C)=\overline{C^2} - (\overline{C})^2 = 287.8-(16.9)^2=45.99\) \(r = \frac{12.28}{\sqrt{27.29 \times 45.99}} \approx 0.829\) Since the correlation coefficient is close to 1, there is a strong positive linear relationship between the number of chirps and temperature.

Calculate the coefficient of determination and analyze its value

Finally, we calculate the coefficient of determination, \(r^2\), and analyze its value. \(r^2 = (0.829)^2 = 0.687\) This means about 68.7% of the variation in the number of chirps can be explained by the linear relationship with temperature. While the relationship is quite strong, around 31.3% of the variability is still unexplained and may be due to other factors.

Key concepts

Least-Squares Regression Line

Understanding the least-squares regression line is crucial for predicting outcomes in various fields, including statistics and data analysis. It plays a central role in linear regression analysis by providing the best-fit line through a set of data points. This line minimizes the sum of the squares of the vertical distances (residuals) of the points from the line.

When we are looking at the relationship between two variables, such as the chirping rate of crickets and the temperature in the exercise provided, the least-squares regression line helps us to understand and predict how one variable changes in relation to the other. To calculate this line, we first find the mean values of the variables, then we use the covariance and variance to determine the slope and intercept. This process gives us the equation of the line, which in the cricket chirping example was found to be \(y = 4.205 + 0.450x\).

The slope \(0.450\) indicates how many units the chirping rate increases per unit increase in temperature. The intercept \(4.205\) represents the estimated number of chirps per second at a temperature of 0 degrees, which is not realistic in this context but is a necessary part of the equation.

Covariance

Covariance is a measure of how two variables change together. For students delving into statistics, understanding covariance provides insight into the direction of the relationship—whether variables tend to increase or decrease in tandem. In our cricket example, we calculated the covariance between temperatures and the number of chirps as \(12.28\).

A positive value, as we see here, indicates that as the temperature increases, the number of cricket chirps also increases. Conversely, a negative covariance would suggest an inverse relationship. It's important to note that covariance alone doesn't tell us about the strength of the relationship, only its direction, nor does it have a standardized value that allows for easy comparison across different datasets.

Correlation Coefficient

The correlation coefficient, often denoted as \(r\), takes the concept of covariance further by measuring both the direction and strength of the linear relationship between two variables. It provides a standardized value that ranges from -1 to 1.

A value of 1 means there is a perfect positive linear relationship, 0 indicates no linear relationship, and -1 signifies a perfect negative linear relationship. In the cricket exercise, we found the correlation coefficient to be \(r \text{≈} 0.829\), indicating a strong positive relationship between temperature and the number of chirps. This high correlation suggests that temperature is a good predictor of chirping rate, and justifies using a linear model for analysis and prediction.

Moreover, the square of the correlation coefficient, known as the coefficient of determination (\(r^2\)), tells us that approximately 68.7% of the variability in chirping rates can be explained by the changes in temperature, enhancing our understanding of the linear regression's effectiveness.