91Ó°ÊÓ

The catch basin in a storm-sewer system is the interface between surface runoff and the sewer. The catch-basin insert is a device for retrofitting catch basins to improve pollutantremoval properties. The article "An Evaluation of the Urban Stormwater Pollutant Removal Efficiency of Catch Basin Inserts" (Water Envir. Res., 2005: 500-510) reported on tests of various inserts under controlled conditions for which inflow is close to what can be expected in the field. Consider the following data, read from a graph in the article, for one particular type of insert on \(x=\) amount filtered (1000s of liters) and \(y=\%\) total suspended solids removed. $$ \begin{array}{l|cccccccccc} x & 23 & 45 & 68 & 91 & 114 & 136 & 159 & 182 & 205 & 228 \\ \hline y & 53.3 & 26.9 & 54.8 & 33.8 & 29.9 & 8.2 & 17.2 & 12.2 & 3.2 & 11.1 \end{array} $$ Summary quantities are $$ \begin{aligned} &\sum x_{i}=1251, \sum x_{i}^{2}=199,365, \sum y_{i}=250.6, \\ &\sum y_{i}^{2}=9249.36, \sum x_{i} y_{i}=21,904.4 \end{aligned} $$ a. Does a scatterplot support the choice of the simple linear regression model? Explain. b. Obtain the equation of the least squares line. c. What proportion of observed variation in \% removed can be attributed to the model relationship? d. Does the simple linear regression model specify a useful relationship? Carry out an appropriate test of hypotheses using a significance level of .05. e. Is there strong evidence for concluding that there is at least a \(2 \%\) decrease in true average suspended solid removal associated with a 10,000 liter increase in the amount filtered? Test appropriate hypotheses using \(\alpha=.05 .\) f. Calculate and interpret a \(95 \%\) CI for true average \(\%\) removed when amount filtered is 100,000 liters. How does this interval compare in width to a CI when amount filtered is 200,000 liters? g. Calculate and interpret a \(95 \%\) PI for \% removed when amount filtered is 100,000 liters. How does this interval compare in width to the CI calculated in (f) and to a PI when amount filtered is 200,000 liters?

Step by step solution

01

Scatterplot Visualization

To determine if a linear regression model is appropriate, we first create a scatterplot of the data points. By plotting the amount filtered, \(x\), against the percentage of total suspended solids removed, \(y\), we can visually assess the relationship between the variables. A linear pattern or trend (either positive or negative) would support a linear regression model. However, if the points are scattered randomly with no visible pattern, a linear model may not be suitable.

02

Calculate the Least Squares Line Equation

The equation of the least squares line is given by \(y = a + bx\), where \(b\) is the slope and \(a\) is the intercept of the line. To calculate \(b\), use the formula \(b = \frac{n\sum x_i y_i - \sum x_i \sum y_i}{n \sum x_i^2 - (\sum x_i)^2}\), and \(a = \frac{\sum y_i - b\sum x_i}{n}\). With given sums and \(n=10\), plug in the values to find \(b\) and \(a\).

03

Evaluate the Model Fit with R-squared

The coefficient of determination \(R^2\) quantifies the proportion of total variation in \(y\) that is explained by the model. Calculate \(R^2\) using \[(R^2 = \frac{\sum (\hat{y}_i - \bar{y})^2}{\sum (y_i - \bar{y})^2})\],\ where \(\hat{y}_i\) is the predicted value and \(\bar{y}\) is the mean of \(y\). Substitute values into the formula to get the \(R^2\) value.

04

Perform Hypothesis Testing

To test if the linear regression model specifies a useful relationship, we perform hypothesis testing for the slope \(b\). The null hypothesis \(H_0: b = 0\) (no effect), and alternative \(H_a: b eq 0\) (nonzero effect). Use the t-statistic \(t = \frac{b}{SE(b)}\), compare with critical t-value at \(\alpha = 0.05\) to accept or reject \(H_0\). If \(p < \alpha\), reject \(H_0\) indicating a significant relationship.

05

Test Hypothesis for Decrease in Removal Efficiency

We wish to test if a 10,000-liter increase results in at least a 2% decrease. The null hypothesis is \(H_0: b = -0.2\) (no decrease) and \(H_a: b < -0.2\) (a decrease). Compute the t-statistic \(t = \frac{b + 0.2}{SE(b)}\) and compare with the critical value from a t-distribution at \(\alpha = 0.05\). Reject \(H_0\) if \(p < \alpha\).

06

Construct a 95% Confidence Interval for True Mean

For a 95% CI, use the formula \((a + bx_0) \pm t_{\alpha/2, n-2} \cdot SE_{\text{mean}}(\hat{y})\), where \(x_0 = 100\). Calculate standard error \(SE_{\text{mean}}(\hat{y})\) using predicted values and variance. Compare CI width for 100k and 200k to observe changes.

07

Calculate a 95% Prediction Interval

The prediction interval (PI) is given by \((a + bx_0) \pm t_{\alpha/2, n-2} \cdot SE_{\text{pred}}(\hat{y})\) where \(SE_{\text{pred}}(\hat{y})\) includes additional model and residual uncertainty terms. Calculate PI for both 100k and 200k filtering amounts to compare with the confidence intervals obtained in Step 6.

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Key Concepts

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

Scatterplot Interpretation

When working with linear regression analysis, one of the first steps involves creating a scatterplot of the data you're analyzing. This visual tool helps to identify the relationship between two variables. In this particular context, we are looking at the amount of water filtered and the percentage of total suspended solids removed. By plotting these variables against each other as points on a graph, students can explore whether there is any obvious pattern or trend present.

A scatterplot that exhibits a linear trend—meaning the data points roughly align along a straight line—suggests that a linear regression model could be appropriate. However, if the points are very scattered with no discernable pattern, this could indicate that a linear model might not fit well.

Being able to interpret scatterplots is crucial in determining whether to proceed with linear regression. Look for a "running curve". This could be an upward or downward trend that indicates correlation. If the data appears scattered randomly, consider exploring other types of models.

Least Squares Method

The Least Squares Method is a fundamental technique used to find the equation of the line of best fit for a set of data points. This line minimizes the sum of the squares of the vertical distances of the points from the line. Essentially, this means you are trying to find the line that best represents the data available, providing a clearer picture of the relationship between variables.

To find this line, often described in the form \( y = a + bx \), you must calculate the slope \( b \) and the intercept \( a \). The slope determines how steep the line is, while the intercept is where the line crosses the y-axis. Using formulas involving the sums of the products of data values and their squares, you can solve for \( b \) and \( a \). Here is a simplified view:

• Calculate the slope \( b \) using the provided formula involving sums and sums of products.
• Determine the intercept \( a \) by rearranging the equation for the best fit.

Once you have these values, you can construct the equation of the line of best fit. This equation can then be used to make predictions, providing insights into the potential outcomes of varying conditions.

Coefficient of Determination

After obtaining the equation of the best fit line, the next question to address is how well this line explains the variability in your data. This is where the coefficient of determination, denoted as \( R^2 \), comes into play.

The \( R^2 \) value ranges between 0 and 1, and it represents the proportion of the variance for the dependent variable (in this case, the percentage of solids removed) that's predicted from the independent variable (the amount filtered). More simply, it tells you how much of the change in the output variable can be predicted by changes in the input variable.

A higher \( R^2 \) value signifies that the model provides a good fit to the data. For instance, an \( R^2 \) value of 0.8 would suggest that 80% of the variation in the percentage of solids removed is explained by the model. It's an important measure because it gives an immediate visual indication of how reliable the predictions made by your line of best fit are likely to be. An \( R^2 \) closer to 1 suggests a better fit, while a value closer to 0 suggests a weaker model.