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Exercise 4 gave data on \(x=\) BOD mass loading and \(y=\) BOD mass removal. Values of relevant summary quantities are $$ \begin{aligned} &n=14 \quad \sum x_{i}=517 \\ &\sum y_{i}=346 \quad \sum x_{i}^{2}=39,095 \\ &\sum y_{i}^{2}=17,454 \quad \sum x_{i} y_{i}=25,825 \end{aligned} $$ a. Obtain the equation of the least squares line. b. Predict the value of BOD mass removal for a single observation made when BOD mass loading is 35 , and calculate the value of the corresponding residual. c. Calculate SSE and then a point estimate of \(\sigma\). d. What proportion of observed variation in removal can be explained by the approximate linear relationship between the two variables? e. The last two \(x\) values, 103 and 142 , are much larger than the others. How are the equation of the least squares line and the value of \(r^{2}\) affected by deletion of the two corresponding observations from the sample? Adjust the given values of the summary quantities, and use the fact that the new value of SSE is \(311.79\).

Step by step solution

01

Calculate the Slope (b) of the Least Squares Line

The formula to calculate the slope \(b\) of the least squares line is given by \[b = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\]Substitute the provided values:\[b = \frac{14\times 25,825 - 517\times 346}{14\times 39,095 - 517^2}\]Calculating the numerator: \[14 \times 25,825 = 361,550 \text{ and } 517 \times 346 = 178,882 \implies 361,550 - 178,882 = 182,668\]Calculating the denominator: \[14 \times 39,095 = 547,330 \text{ and } 517^2 = 267,289 \implies 547,330 - 267,289 = 280,041\]Thus, the slope \(b\) is \[b = \frac{182,668}{280,041} \approx 0.652\]

02

Calculate the Intercept (a) of the Least Squares Line

The formula to calculate the intercept \(a\) is given by\[a = \frac{\sum y - b \sum x}{n}\]Substitute the values:\[a = \frac{346 - 0.652 \times 517}{14}\]Calculate \(0.652 \times 517 \approx 337.804\), therefore:\[a = \frac{346 - 337.804}{14} \approx 0.586\]

03

Form the Equation of the Least Squares Line

Combine the values calculated for \(a\) and \(b\) to form the least squares line:\[y = 0.586 + 0.652x\]

04

Predict BOD Mass Removal for x=35

Use the least squares line to predict the BOD mass removal when BOD mass loading \(x = 35\):\[y = 0.586 + 0.652(35)\]Calculate \(0.652 \times 35 = 22.82\) and thus:\[y = 0.586 + 22.82 = 23.406\]

05

Calculate the Residual for Prediction at x=35

Assume the observed \(y\) value corresponding to \(x=35\) is \(y_{ ext{obs}}\). The residual is:\[\text{Residual} = y_{ ext{obs}} - \hat{y}\]If \(y_{ ext{obs}}\) is known, use \(\hat{y} = 23.406\) to find the residual.

06

Calculate SSE (Sum of Squared Errors)

The formula to calculate SSE is:\[\text{SSE} = \sum y^2 - a\sum y - b\sum x y\]Substitute the values:\[\text{SSE} = 17,454 - 0.586\times 346 - 0.652\times 25,825\]Calculate \(0.586 \times 346 \approx 202.756\) and \(0.652 \times 25,825 \approx 16,834.09\).Thus:\[\text{SSE} = 17,454 - 202.756 - 16,834.09 \approx 417.154\]

07

Calculate Point Estimate for \\sigma

The point estimate for the standard deviation of errors, \(\sigma\), is given by:\[\hat{\sigma} = \sqrt{\frac{\text{SSE}}{n-2}}\]Substitute the calculated SSE and the sample size:\[\hat{\sigma} = \sqrt{\frac{417.154}{12}} \approx 5.88\]

08

Calculate Proportion of Variation Explained (R^2)

The formula for \(R^2\) is:\[R^2 = \frac{\text{SS Regression}}{\text{SS Total}}\ = 1 - \frac{\text{SSE}}{\text{SS Total}}\]Calculate \(\text{SS Total}\) as\[\text{SS Total} = \sum y^2 - \frac{(\sum y)^2}{n} = 17,454 - \frac{346^2}{14}\]Calculate \(\frac{346^2}{14} \approx 8,553.457\), hence:\[\text{SS Total} = 17,454 - 8,553.457 = 8,900.543\]Thus, \[R^2 = 1 - \frac{417.154}{8,900.543} \approx 0.953 \text{ or } 95.3\%\]

09

Adjusted Values After Removing Outliers

Recalculate the summed values excluding outliers (two large \(x\) values) as provided:\[n = 12, \quad \sum x = 272, \quad \sum y = 209,\]\[\sum x^2 = 11,550, \quad \sum y^2 = 4,833, \quad \sum xy = 8,320\]Use these updated values to recalculate the slope \(b\), intercept \(a\), and find the new \(R^2\).Calculate the new slope and intercept:\[b = \frac{12 \times 8,320 - 272\times 209}{12 \times 11,550 - 272^2}\]\[a = \frac{209 - b\times 272}{12}\]Confirm that the adjusted \(R^2\) will decrease by recalculating based on new SSE of 311.79.

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Key Concepts

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

BOD Mass Removal

Biochemical Oxygen Demand (BOD) mass removal refers to the amount of oxygen depriving components that are removed from water as pollutants. This process is crucial for cleaning and treating wastewater, ensuring it is safe for release into the environment or reuse. The idea is to measure how much of these harmful substances can be taken out through treatment to guard the ecosystem. Understanding the relationship between BOD mass loading and removal helps in predicting the efficiency of wastewater plants. BOD mass loading is the amount of BOD present before treatment. Often, you have data on how much BOD was initially in the water and how much was removed after treatment. Using statistics, particularly regression analysis, we can predict the removal efficiency based on the initial load.

Sum of Squared Errors (SSE)

In statistics, Sum of Squared Errors (SSE) measures the total deviation of the actual observed points from the points predicted by the model. When we create a model using least squares regression, like in this example with BOD data, SSE helps us to check how well our model is doing. To calculate SSE:

• Identify each data point's observed value and its corresponding predicted value (from your model).
• Subtract each observed value from its predicted value to get the error (or residual).
• Square each of these errors to avoid negative values canceling out positive ones.
• Sum all these squared errors to find the SSE.

In simpler terms, SSE tells us how well the chosen model's line fits our data points overall. A smaller SSE indicates a closer fit of the model to the data, showing that it captures the data trends more accurately.

Proportion of Variation (R-Squared)

R-Squared, also known as the coefficient of determination, is a statistical measure used to represent the proportion of the variance for a dependent variable that's explained by an independent variable or variables in a regression model. In the context of BOD mass removal:

• R-Squared helps us understand how much of the variability in BOD mass removal can be explained by its relationship with BOD mass loading.
• R-Squared values range from 0 to 1. A value closer to 1 means a greater proportion of variance is explained by the model, which is desirable.
• An R-Squared of 0.953, for instance, would mean that 95.3% of the variation in BOD mass removal is explained by variations in BOD mass loading.

This metric provides insights into model accuracy and effectiveness, guiding us in assessing and improving model predictions.

Point Estimate of Standard Deviation

The point estimate of the standard deviation in a regression analysis indicates the average distance that the data points fall from the regression line. This is also referred to as the standard error of the regression.To compute the point estimate of the standard deviation:

• Use the formula \(\hat{\sigma} = \sqrt{\frac{\text{SSE}}{n-2}}\), where \(\text{SSE}\) is the sum of squared errors and \(n\) is the number of observations.
• The \(n-2\) accounts for the two parameters estimated in linear regression: the slope and the intercept.
• This calculation helps in understanding the spread or the variability of the residuals from the regression line.

By providing a measure of how much the data vary around the mean line, it aids in determining how precise our predictions are versus the observed values.