/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Utilization of sucrose as a carb... [FREE SOLUTION] | 91Ó°ÊÓ

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Utilization of sucrose as a carbon source for the production of chemicals is uneconomical. Beet molasses is a readily available and low-priced substitute. The article "Optimization of the Production of \(\beta\)-Carotene from Molasses by Blakeslea Trispora" \((J\). of Chem. Tech. and Biotech., 2002: 933-943) carried out a multiple regression analysis to relate the dependent variable \(y=\) amount of \(\beta\)-carotene \(\left(\mathrm{g} / \mathrm{dm}^{3}\right)\) to the three predictors amount of lineolic acid, amount of kerosene, and amount of antioxidant \(\left(\right.\) all \(\mathrm{g} / \mathrm{dm}^{3}\) ). a. Fitting the complete second-order model in the three predictors resulted in \(R^{2}=987\) and adjusted \(R^{2}=\) \(.974\), whereas fitting the first-order model gave \(R^{2}=\) .016. What would you conclude about the two models? b. For \(x_{1}=x_{2}=30, x_{3}=10\), a statistical software package reported that \(\hat{y}=.66573, s_{\hat{y}}=.01785\), based on the complete second-order model. Predict the amount of \(\beta\)-carotene that would result from a single experimental run with the designated values of the independent variables, and do so in a way that conveys information about precision and reliability.

Short Answer

Expert verified
The second-order model fits significantly better than the first-order model. For \((x_1, x_2, x_3) = (30, 30, 10)\), predict \(y = 0.66573\) with a 95% CI of \((0.63075, 0.70071)\).

Step by step solution

01

Analyze Model Fit

Start by assessing the quality of the model fits. The second-order model has an \( R^2 \) of 0.987 and an adjusted \( R^2 \) of 0.974, indicating that the model explains approximately 97.4% of the variability in the dependent variable when adjusted for the number of predictors. In contrast, the first-order model has an \( R^2 \) of only 0.016, showing it explains just 1.6% of variability and is probably not a useful model.
02

Compare the Models

Given the significant differences in \( R^2 \) values, the second-order model provides a much better fit than the first-order model. This suggests that the interaction and quadratic terms included in the second-order model are crucial for explaining the variability in the response variable, and that the relationship between predictors and response is not simply linear.
03

Predict Using Second-Order Model

To predict \( y \) using the second-order model, use the provided prediction \( \hat{y} = 0.66573 \) for \( x_1 = x_2 = 30, x_3 = 10 \). This prediction indicates the expected amount of \( \beta \)-carotene based on the model when each predictor is set to the given value.
04

Evaluate Prediction Precision

Assess the precision of the prediction by considering the standard error of the prediction, \( s_{\hat{y}} = 0.01785 \). A smaller standard error suggests more precise predictions. To express this precision, construct a confidence interval (CI) around \( \hat{y} \). For a 95% CI, use \( \hat{y} \pm 1.96 \, s_{\hat{y}} \).
05

Calculate Confidence Interval

Calculate the 95% confidence interval as follows: \( 0.66573 \pm 1.96 \times 0.01785 = 0.66573 \pm 0.03498 \). Thus, the interval is \( (0.63075, 0.70071) \), indicating there's a 95% chance the true amount of \( \beta \)-carotene lies within this range under the specified conditions.

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Key Concepts

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

Second-Order Model
In multiple regression analysis, a second-order model is one where the relationship between the independent variables (predictors) and the dependent variable includes both linear and interaction effects. This type of model allows for the inclusion of squared terms and products of different predictors. By doing so, the model can capture more complex relationships between variables than a first-order model, which only considers linear main effects.

The benefits of using a second-order model are significant. It can reveal nuances in data that linear models overlook, which is especially useful in a scientific context where interactions between components matter. For instance, in the given example about \( \beta \)-Carotene production from molasses: the second-order model provides an excellent fit with an \( R^2 \) of 0.987 and an adjusted \( R^2 \) of 0.974, indicating that this model explains approximately 97.4% of the variability in the amount of \( \beta \)-Carotene produced. In contrast, the first-order model, with an \( R^2 \) of only 0.016, failed to capture these complex interactions. This suggests that the true relationship among the predictors and the output is not merely additive or linear but involves interactions and potentially curvilinear trends that are appropriately modeled with a second-order approach.
Prediction Precision
Prediction precision refers to how close a prediction by the model is likely to be to the actual future observations. In statistical terms, it can be quantified by examining metrics such as the standard error of the prediction. A smaller standard error indicates higher precision and greater confidence in the prediction made by the model.

In the context of the example, the prediction for \( y \) (amount of \( \beta \)-Carotene) using the second-order model is given by \( \hat{y} = 0.66573 \). The standard error of this prediction, \( s_{\hat{y}} = 0.01785 \), is relatively small. This indicates that the predicted value of \( \beta \)-Carotene is precise and likely to be very close to the true value, assuming the model is correct.

A common way to assess prediction precision and reliability is to construct a confidence interval. For a 95% confidence interval, we calculate \( \hat{y} \pm 1.96 \times s_{\hat{y}} \), which in this example becomes: \( 0.66573 \pm 0.03498 \). This results in an interval of \([0.63075, 0.70071]\), suggesting that there's a 95% probability the actual amount of \( \beta \)-Carotene will fall within this range if the assumptions of the model hold.
Model Comparison
Comparing different models is a critical step in regression analysis as it helps determine which model best explains the variability in the dataset. There are several criteria used for such comparisons, including the \( R^2 \) and adjusted \( R^2 \) values.

The example illustrates this with a comparison between a first-order and a second-order model. The first-order model, characterized by an \( R^2 \) of 0.016, explains a negligible 1.6% of the variance in \( \beta \)-Carotene production. This suggests that the model fails to capture key variability and interactions in the data, making it an inadequate representation for this context.

On the other hand, the second-order model achieves a significantly higher \( R^2 \) of 0.987, demonstrating its ability to account for 98.7% of the variance in the data. The considerable difference between these \( R^2 \) values indicates the importance of including non-linear and interaction terms, caught by the second-order model, to accurately model the underlying biological process in \( \beta \)-Carotene production from molasses. Understanding this distinction and choosing the model with appropriate complexity is crucial for accurate data representation and prediction.

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Most popular questions from this chapter

The article "Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns" (Textile Res. \(J ., 1997: 39-44\) ) reported on a study carried out to relate yarn tenacity ( \(y\), in \(\mathrm{g} /\) tex \()\) to yarn count \(\left(x_{1}\right.\), in tex), percentage polyester \(\left(x_{2}\right)\), first nozzle pressure \(\left(x_{3}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right)\), and second nozzle pressure \(\left(x_{4}\right.\), in \(\mathrm{kg} /\) \(\left.\mathrm{cm}^{2}\right)\). The estimate of the constant term in the corresponding multiple regression equation was \(6.121\). The estimated coefficients for the four predictors were \(-.082\), \(.113, .256\), and \(-.219\), respectively, and the coefficient of multiple determination was 946 . a. Assuming that the sample size was \(n=25\), state and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors. b. Again using \(n=25\), calculate the value of adjusted \(R^{2}\). c. Calculate a \(99 \%\) confidence interval for true mean yarn tenacity when yarn count is \(16.5\), yarn contains \(50 \%\) polyester, first nozzle pressure is 3 , and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is . 350 .

The viscosity \((y)\) of an oil was measured by a cone and plate viscometer at six different cone speeds \((x)\). It was assumed that a quadratic regression model was appropriate, and the estimated regression function resulting from the \(n=6\) observations was $$ y=-113.0937+3.3684 x-.01780 x^{2} $$ a. Estimate \(\mu_{\gamma \cdot 75}\), the expected viscosity when speed is \(75 \mathrm{rpm} .\) b. What viscosity would you predict for a cone speed of \(60 \mathrm{rpm}\) ? c. If \(\Sigma y_{i}^{2}=8386.43, \Sigma y_{i}=210.70, \Sigma x_{i} y_{i}=17,002.00\), and \(\Sigma x_{i}^{2} y_{i}=1,419,780\), compute SSE \(\left[=\Sigma y_{i}^{2}-\right.\) \(\left.\hat{\beta}_{0} \Sigma y_{i}-\hat{\beta}_{1} \Sigma x_{i} y_{i}-\hat{\beta}_{2} \Sigma x_{i}^{2} y_{i}\right]\) and \(s\). d. From part (c), SST \(=8386.43-(210.70)^{2} / 6=987.35\). Using SSE computed in part (c), what is the computed value of \(R^{2}\) ? e. If the estimated standard deviation of \(\hat{\beta}_{2}\) is \(s_{\hat{\beta}_{2}}=.00226\), test \(H_{0}: \beta_{2}=0\) versus \(H_{2}: \beta_{2} \neq 0\) at level \(.01\), and interpret the result.

A trucking company considered a multiple regression model for relating the dependent variable \(y=\) total daily travel time for one of its drivers (hours) to the predictors \(x_{1}=\) distance traveled (miles) and \(x_{2}=\) the number of deliveries made. Suppose that the model equation is $$ Y=-.800+.060 x_{1}+.900 x_{2}+\epsilon $$ a. What is the mean value of travel time when distance traveled is 50 miles and three deliveries are made? b. How would you interpret \(\beta_{1}=.060\), the coefficient of the predictor \(x_{1}\) ? What is the interpretation of \(\beta_{2}=.900 ?\) c. If \(\sigma=.5\) hour, what is the probability that travel time will be at most 6 hours when three deliveries are made and the distance traveled is 50 miles?

An aeronautical engineering student carried out an experiment to study how \(y=\) lift/drag ratio related to the variables \(x_{1}=\) position of a certain forward lifting surface relative to the main wing and \(x_{2}=\) tail placement relative to the main wing, obtaining the following data (Statistics for Engineering Problem Solving, p. 133\():\)a. Fitting the first-order model gives \(\mathrm{SSE}=5.18\), whereas including \(x_{3}=x_{1} x_{2}\) as a predictor results in \(\mathrm{SSE}=\) 3.07. Calculate and interpret the coefficient of multiple determination for each model. b. Carry out a test of model utility using \(\alpha=.05\) for each of the models described in part (a). Does either result surprise you?

The article cited in Exercise 49 of Chapter 7 gave summary information on a regression in which the dependent variable was power output \((\mathrm{W})\) in a simulated 200 -m race and the predictors were \(x_{1}=\) arm girth \((\mathrm{cm}), x_{2}=\) excess post-exercise oxygen consumption \((\mathrm{ml} / \mathrm{kg})\), and \(x_{3}=\) immediate posttest lactate (mmol/L). The estimated regression equation was reported as $$ \begin{aligned} &y=-408.20+14.06 x_{1}+.76 x_{2}-3.64 x_{3} \\ &\left(n=11, R^{2}=.91\right) \end{aligned} $$ a. Carry out the model utility test using a significance level of .01. [Note: All three predictors were judged to be important.] b. Interpret the estimate \(14.06\). c. Predict power output when arm girth is \(36 \mathrm{~cm}\), excess oxygen consumption is \(120 \mathrm{ml} / \mathrm{kg}\), and lactate is \(10.0\). d. Calculate a point estimate for true average power output when values of the predictors are as given in (c). e. Obtain a point estimate for the true average change in power output associated with a \(1 \mathrm{mmol} / \mathrm{L}\) increase in lactate while arm girth and oxygen consumption remain fixed.

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