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In Exercise 3.18 , \(\mathrm{EX}_{1223}\) we found that 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: $$ \begin{array}{l|llllllllll} \text { Chirps per Second } & 20 & 16 & 19 & 18 & 18 & 16 & 14 & 17 & 15 & 16 \\ \hline \text { Temperature } & 88 & 73 & 91 & 85 & 82 & 75 & 69 & 82 & 69 & 83 \end{array} $$ a. Use the formulas given in this chapter to find the least-squares regression line relating the number of chirps to temperature. Compare to the results obtained in Exercise 3.18 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?

Step by step solution

01

Calculate the mean of both variables

To find the mean of chirps per second and temperature, sum the values and divide by the total number of observations (10 in this case). Mean of chirps per second: \(\bar{x} = \frac{20+16+19+18+18+16+14+17+15+16}{10} = \frac{169}{10} = 16.9\) Mean of temperature: \(\bar{y} = \frac{88+73+91+85+82+75+69+82+69+83}{10} = \frac{797}{10} = 79.7\)

02

Calculate the slope of the least-squares regression line

To calculate the slope (b) of the least-squares regression line, we use the following formula: \(b = \frac{\sum_{i=1}^{n} (x_{i} - \bar{x})(y_{i} - \bar{y})}{\sum_{i=1}^{n} (x_{i} - \bar{x})^{2}}\) We will need to calculate both the numerator and the denominator. Since we have 10 data points, we need to calculate the expression for i=1 to 10, and then sum them up. Denominator: \(\sum_{i=1}^{n} (x_{i} - \bar{x})^{2} = (-3.1)^{2} + (-0.9)^{2} + (2.1)^{2} + (1.1)^{2} + (1.1)^{2} + (-0.9)^{2} + (-2.9)^{2} + (0.1)^{2} + (-1.9)^{2} + (-0.9)^{2} = 35.3\) Numerator: \(\sum_{i=1}^{n} (x_{i} - \bar{x})(y_{i} - \bar{y}) = (-3.1)(8.3) + (-0.9)(-6.7) + (2.1)(11.3) + (1.1)(5.3) + (1.1)(2.3) + (-0.9)(-4.7) + (-2.9)(-10.7) + (0.1)(2.3) + (-1.9)(-10.7) + (-0.9)(3.3) = 121.59\) Now we can find the slope b: \(b = \frac{121.59}{35.3} = 3.45\)

03

Calculate the intercept of the least-squares regression line

To calculate the intercept (a) of the least-squares regression line, we can use the following formula: \(a = \bar{y} - b\bar{x}\) Using the mean values of both variables (16.9 chirps per second and 79.7 degrees) and the slope (3.45), we can find the intercept: \(a = 79.7 - 3.45(16.9) = 21.19\)

04

Find the least-squares regression line

Now that we have the slope and intercept of the least-squares regression line, we can write the equation for the line: \(y = 3.45x + 21.19\) This is the least-squares regression line relating the number of chirps to the temperature.

05

Check for linear relationship

To check if the data provides sufficient evidence for a linear relationship, we can calculate the correlation coefficient (r): \(r = \frac{\sum_{i=1}^{n} (x_{i} - \bar{x})(y_{i} - \bar{y})}{\sqrt{[\sum_{i=1}^{n} (x_{i} - \bar{x})^2][\sum_{i=1}^{n} (y_{i} - \bar{y})^2]}}\) \(r = \frac{121.59}{\sqrt{(35.3)(68)}} \approx 0.83\) The correlation coefficient is 0.83, which indicates a strong positive linear relationship between the number of chirps and the temperature.

06

Calculate \(r^{2}\)

Now, we will calculate the coefficient of determination, \(r^{2}\), which tells us the proportion of the total variation in the dependent variable (temperature) that can be explained by the independent variable (number of chirps): \(r^{2} = (0.83)^2 = 0.6889\) The coefficient of determination, \(r^2\), is approximately 0.689. This tells us that about 68.9% of the variation in the temperature can be explained by the number of chirps. This implies that the linear regression analysis is moderately effective in describing the relationship between the number of chirps and the temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Least-Squares Regression

The concept of Least-Squares Regression plays a crucial role in finding the best linear fit for a set of data points. This process involves determining a straight line that minimizes the sum of the squared differences between the observed data and the values predicted by the line.
The line is described by the equation:\[ y = bx + a \]where:

• \( b \) is the slope of the line, indicating the change in temperature for each chirp per second.
• \( a \) is the y-intercept, representing the temperature when there are zero chirps.

To find these coefficients, we calculate:

• The slope \( b \) as:\[ b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \]
• And the intercept \( a \) using:\[ a = \bar{y} - b\bar{x} \]

Through this method, the relationship between crickets' chirps and the ambient temperature can be represented effectively, providing insights into biological and environmental interactions.

Correlation Coefficient

The Correlation Coefficient, represented as \( r \), is a statistical measure that expresses the extent of a linear relationship between two variables. The value of \( r \) ranges from -1 to 1.

• An \( r \) value close to 1 suggests a strong positive linear relationship, indicating that as the number of chirps increases, so does the temperature.
• Conversely, an \( r \) value near -1 would imply a strong negative linear relationship.
• An \( r \) around 0 indicates a weak or no linear relationship.

In the context of our exercise, an \( r \) of approximately 0.83 signals a strong positive relationship between the number of cricket chirps and temperature. This means that chirps can reliably predict temperature changes. This measure not only provides insights into biological patterns but also underlines the predictive power of statistical analysis.

Coefficient of Determination

The Coefficient of Determination, denoted as \( r^2 \), is integral to understanding how well our model explains the observed outcomes. It reflects the proportion of variance in the dependent variable, which can be accounted for by the independent variable.

• A high \( r^2 \) value, such as 0.689 in this example, indicates that 68.9% of the temperature variation can be explained by the changes in chirps per second.
• Consequently, a higher \( r^2 \) suggests a better-fitting model.
• An \( r^2 \) value closer to 1 is ideal, as almost all the variation in temperatures is captured by the number of chirps.

This coefficient provides a measure of model effectiveness, reinforcing the significance of the linear relationship between the number of chirps and the temperature adjustments.

Statistical Analysis

Statistical Analysis encompasses a suite of techniques used to interpret, describe, and infer patterns from data. In the context of the relationship between chirps and temperature, statistical analysis serves multiple purposes.

• It allows us to estimate the parameters (slope and intercept) of the least-squares regression line.
• Evaluates the strength of the association using the correlation coefficient (\( r \)).
• Assesses the explanatory power of the model with the coefficient of determination (\( r^2 \)).
• Tests hypotheses, such as the existence of a linear relationship.

By applying statistical analysis, we can move beyond raw data to meaningful interpretations, enabling the prediction and understanding of phenomena, such as the environmental impact on cricket behavior. These insights not only facilitate ecological studies but also enhance our comprehension of statistical methods.