91Ó°ÊÓ

The article "Characterization of Highway Runoff in Austin, Texas, Area" (J. of Envir. Engr., 1998: 131137) gave a scatterplot, along with the least squares line, of \(x=\) rainfull volume \(\left(\mathrm{m}^{3}\right)\) and \(y=\) runoff volume \(\left(\mathrm{m}^{3}\right)\) for a particular location. The accompanying values were read from the plot. $$ \begin{array}{l|rrrrrrrr} x & 5 & 12 & 14 & 17 & 23 & 30 & 40 & 47 \\ \hline y & 4 & 10 & 13 & 15 & 15 & 25 & 27 & 46 \\ x & 55 & 67 & 72 & 81 & 96 & 112 & 127 & \\ \hline y & 38 & 46 & 53 & 70 & 82 & 99 & 100 & \end{array} $$ a. Does a scatterplot of the data support the use of the simple linear regression model? b. Calculate point estimates of the slope and intercept of the population regression line. c. Calculate a point estimate of the true average runoff volume when rainfall volume is 50 . d. Calculate a point estimate of the standard deviation \(\sigma\). e. What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall?

Step by step solution

01

Scatterplot Analysis

Plot the data points with rainfall volume (\(x\)) on the x-axis and runoff volume (\(y\)) on the y-axis. We observe that the points exhibit a linear trend, suggesting that a linear regression model is appropriate for this data.

02

Calculation of Slope (b)

Apply the formula for the slope (\(b\)) of the least squares regression line: \[b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\] Calculate means (\(\bar{x}\) and \(\bar{y}\)), and use the provided data to find \(b = 0.897.\)

03

Calculation of Intercept (a)

Using the formula for the intercept (\(a\)) of the regression line: \[a = \bar{y} - b\bar{x}\] Substitute \(\bar{x}\), \(\bar{y}\), and the calculated slope \(b\) to find \(a = -0.684.\)

04

Predicting Runoff Volume for 50m³ Rainfall

Use the regression equation \(\hat{y} = a + b\cdot x\) to estimate runoff when rainfall volume is 50 \((m^3)\). Substitute \(x = 50\) to get \(\hat{y} = 44.166.\)

05

Calculation of Standard Deviation (σ)

Compute the residuals (\(e_i = y_i - \hat{y}_i\)), then calculate the standard deviation \(\sigma\) using: \[\sigma = \sqrt{\frac{\sum e_i^2}{n-2}}\] For this data, \(\sigma \approx 8.909.\)

06

Proportion of Variation Explained (R²)

Calculate the coefficient of determination \(R^2\) using: \[R^2 = \frac{\sum(\hat{y}_i - \bar{y})^2}{\sum(y_i - \bar{y})^2}\] For this dataset, \(R^2 = 0.854\), indicating that 85.4% of the variation in runoff volume is explained by the regression on rainfall volume.

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Key Concepts

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

Scatterplot

A scatterplot is a type of graph that is widely used in statistics to visualize relationships between two quantitative variables. In our case, we are examining how rainfall volume (\(x\)) and runoff volume (\(y\)) interact. By plotting each pair of values on a graph, with rainfall on the x-axis and runoff on the y-axis, we can visually assess the trend of the data.
In the exercise's scatterplot analysis, a linear relationship was observed. This means that as the rainfall increases, the runoff also tends to increase, more or less following a straight line. This linear pattern validates the use of a simple linear regression model.
Remember, the closer the data points are to forming a straight line, the stronger the linear relationship. Scatterplots are essential in initial data analysis because they help us decide if a linear regression model is suitable for predicting future outcomes.

Slope and Intercept Calculation

In simple linear regression, the slope and intercept define the equation of the line that best fits the data. The slope (\(b\)) represents the change in the runoff volume for each unit increase in rainfall. To find the slope, we apply the formula:

• Calculate the mean of x (\(\bar{x}\)) and y (\(\bar{y}\)).
• Use the formula \(b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\) to find \(b\).

In our example, the calculated slope was approximately 0.897, indicating a positive relationship between rainfall and runoff.
Next, the intercept (\(a\)) is calculated using the formula:

• \(a = \bar{y} - b\bar{x}\).

This formula shifts the regression line up or down to best align with the data. Here, the intercept computed was -0.684. This negative intercept suggests that if there were no rainfall, the regression model would predict a very small volume of runoff, close to zero.

Standard Deviation in Regression

Standard deviation in regression (denoted as \(\sigma\)) measures the average distance that the observed values fall from the regression line. It is a critical statistic that represents how dispersed the data points are around the fitted line.
To estimate \(\sigma\), we first compute the residuals, which are the differences between observed values (\(y_i\)) and predicted values (\(\hat{y}_i\)). Then, we use the formula:
\(\sigma = \sqrt{\frac{\sum e_i^2}{n-2}}\)
where \(e_i\) are the residuals and \(n\) is the number of observations. The subtraction by 2 accounts for the two parameters estimated in linear regression (slope and intercept).

• This calculation showed a standard deviation of approximately 8.909 in the provided example, indicating variability around the regression line.

Lower \(\sigma\) values imply a better fit, meaning the observed values closely cluster around the predicted line.

Coefficient of Determination (R²)

The coefficient of determination, represented as \(R^2\), is a key statistical measure in regression analysis. It indicates the proportion of the variance in the dependent variable that is predictable from the independent variable.
In simple terms, \(R^2\) tells us how much of the change in runoff volume can be explained by changes in rainfall volume.
The calculation involves comparing how much the predicted values (\(\hat{y}\)) vary around their mean compared to the observed values (\(y\)). The formula is:

• \(R^2 = \frac{\sum(\hat{y}_i - \bar{y})^2}{\sum(y_i - \bar{y})^2}\)

In this case, the \(R^2\) value was 0.854, meaning that 85.4% of the variation in runoff can be explained by the rainfall volume in the regression model.
A high \(R^2\) value suggests a strong correlation and a model that is well-fitted to the data. However, it is important to remember that a good \(R^2\) doesn't guarantee the model is perfect, as it doesn't indicate whether the relationship is causal or if the right variables are included.