/*! 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 54 The decline of water supplies in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The decline of water supplies in certain areas of the United States has created the need for increased understanding of relationships between economic factors such as crop yield and hydrologic and soil factors. The article "Variability of Soil Water Properties and Crop Yield in a Sloped Watershed" (Water 91Ó°ÊÓ Bull., 1988: 281-288) gives data on grain sorghum yield \((y\), in \(\mathrm{g} / \mathrm{m}\)-row \()\) and distance upslope \((x\), in \(\mathrm{m})\) on a sloping watershed. Selected observations are given in the accompanying table. $$ \begin{array}{l|rrrrrrr} x & 0 & 10 & 20 & 30 & 45 & 50 & 70 \\ \hline y & 500 & 590 & 410 & 470 & 450 & 480 & 510 \\ x & 80 & 100 & 120 & 140 & 160 & 170 & 190 \\ \hline y & 450 & 360 & 400 & 300 & 410 & 280 & 350 \end{array} $$ a. Construct a scatter plot. Does the simple linear regression model appear to be plausible? b. Carry out a test of model utility. c. Estimate true average yield when distance upslope is 75 by giving an interval of plausible values.

Short Answer

Expert verified
Plot shows a linear pattern, hypothesis test confirms model utility, and prediction interval gives plausible yield for upslope 75 m.

Step by step solution

01

Organize Data for Plotting

First, we will organize the given data points for the scatter plot. The data consists of pairs \((x, y)\) where \(x\) is the distance upslope and \(y\) is the grain sorghum yield. We have the following observations: \((0, 500), (10, 590), (20, 410), (30, 470), (45, 450), (50, 480), (70, 510), (80, 450), (100, 360), (120, 400), (140, 300), (160, 410), (170, 280), (190, 350)\).
02

Create Scatter Plot

Plot the data using \(x\) values on the horizontal axis and \(y\) values on the vertical axis. Look for any patterns or trends. This will help assess whether a linear relationship or trend is visible.
03

Analyze Scatter Plot Pattern

After plotting, observe the distribution of points. A simple linear regression is plausible if points appear to have a linear pattern, even if not perfectly aligned. If the points show a clear upward or downward trend, a linear model might be reasonable to consider.
04

Calculate Regression Parameters

To make the test of model utility, it's crucial to compute the slope \(b\) and intercept \(a\) of the regression line using formulas: \(b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\) and \(a = \bar{y} - b\bar{x}\). This requires calculating the means \(\bar{x}\) and \(\bar{y}\).
05

Perform Hypothesis Test for Regression Utility

Perform a hypothesis test at the significance level \(\alpha\) (usually 0.05) to check the regression's utility. Use the t-test, testing \(H_0: \beta = 0\) against \(H_a: \beta eq 0\). If the p-value is lower than \(\alpha\), the regression model is useful.
06

Calculate Prediction Interval

Estimate the true average yield for \(x = 75\). Use the regression model to calculate \(\hat{y} = a + b \cdot 75\). Compute the prediction interval, using the standard error of the estimate and the t-distribution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Scatter Plots in Linear Regression
A scatter plot is a graphical representation that helps visualize the relationship between two variables. For the given problem, we plot the distance upslope \(x\) on the horizontal axis and the grain sorghum yield \(y\) on the vertical axis. Each point on the plot represents an observation from the dataset, helping us ascertain patterns or trends.

If the points show a cloud with a linear alignment, it indicates a potential linear relationship between the variables. A well-aligned trend suggests that changes in one variable may be associated with changes in the other. For students analyzing the scatter plot, keep in mind that while perfect alignment is rare, a general direction (either upward or downward) can indicate a significant relationship worth further investigation with a linear regression model.
Understanding Regression Utility
Once you have plotted the scatter plot, the next task is to calculate the regression line. This involves calculating both the slope \(b\) and the intercept \(a\) of the line using given formulas. This line represents the best summary of the relationship between the distance upslope \(x\) and yield \(y\).

Regression utility tests evaluate how well the linear regression model predicts the dependent variable, in this case, the grain sorghum yield. It involves hypothesis testing to establish the usefulness of the regression model, asking whether the slope \(b\) of the line is significantly different from zero. A slope very close to zero indicates that \(x\) doesn't explain much variance in \(y\), hence questioning the model's utility.
The Role of Hypothesis Testing in Regression Analysis
To decide if a linear regression model is worthwhile, we conduct hypothesis testing. In our context, the null hypothesis \(H_0: \beta = 0\) suggests no linear relationship exists between the upslope distance and yield. Conversely, the alternative hypothesis \(H_a: \beta eq 0\) posits that a significant relationship does exist.

A t-test allows us to evaluate these hypotheses using a chosen significance level (like 0.05). If the p-value is less than this level, you reject the null hypothesis, indicating that the regression model is useful. Understanding whether your findings could have occurred by chance is crucial in hypothesis testing, as it strengthens the credibility of the conclusions drawn from your data.
Exploring Prediction Intervals in Regression
Prediction intervals offer a way to estimate the range in which future data points are expected to fall, considering the model and its inherent variability. In the exercise, we estimate the sorghum yield for a distance of 75 meters upslope using the regression equation.

Here, the predicted yield \(\hat{y}\) is calculated using the equation \(\hat{y} = a + b \cdot 75\). To construct a prediction interval, one must compute the standard error, which accounts for variability around the regression line, and apply the t-distribution to find interval boundaries.

Prediction intervals are wider than confidence intervals for the average response because they account for more uncertainty, emphasizing the variability you might encounter when predicting single data points. This tool is crucial for making informed decisions based on regression analyses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The accompanying data on \(x=\) diesel oil consumption rate measured by the drain-weigh method and \(y=\) rate measured by the CI-trace method, both in \(\mathrm{g} / \mathrm{hr}\), was read from a graph in the article "A New Measurement Method of Diesel Engine Oil Consumption Rate" (J. of Soc. of Auto Engr., 1985: 28-33). $$ \begin{array}{l|ccccccccccccc} x & 4 & 5 & 8 & 11 & 12 & 16 & 17 & 20 & 22 & 28 & 30 & 31 & 39 \\ \hline y & 5 & 7 & 10 & 10 & 14 & 15 & 13 & 25 & 20 & 24 & 31 & 28 & 39 \end{array} $$ a. Assuming that \(x\) and \(y\) are related by the simple linear regression model, carry out a test to decide whether it is plausible that on average the change in the rate measured by the CI-trace method is identical to the change in the rate measured by the drain-weigh method. b. Calculate and interpret the value of the sample correlation coefficient

The article "Increases in Steroid Binding Globulins Induced by Tamoxifen in Patients with Carcinoma of the Breast" \((J\). of Endocrinology, 1978: 219-226) reports data on the effects of the drug tamoxifen on change in the level of cortisol-binding globulin (CBG) of patients during treatment. With age \(=x\) and \(\Delta \mathrm{CBG}=y\), summary values are \(n=26\), \(\sum x_{i}=1613, \sum\left(x_{i}-\bar{x}\right)^{2}=3756.96, \sum y_{i}=281.9\) \(\sum\left(y_{i}-\bar{y}\right)^{2}=465.34\), and \(\sum x_{i} y_{i}=16,731\) a. Compute a \(90 \%\) CI for the true correlation coefficient \(\rho\). b. Test \(H_{0}: \rho=-.5\) versus \(H_{\mathrm{a}}: \rho<-.5\) at level \(.05\). c. In a regression analysis of \(y\) on \(x\), what proportion of variation in change of cortisol-binding globulin level could be explained by variation in patient age within the sample? d. If you decide to perform a regression analysis with age as the dependent variable, what proportion of variation in age is explainable by variation in \(\triangle \mathrm{CBG}\) ?

Calcium phosphate cement is gaining increasing attention for use in bone repair applications. The article "Short-Fibre Reinforcement of Calcium Phosphate Bone Cement" (J. of Engr: in Med., 2007: 203-211) reported on a study in which polypropylene fibers were used in an attempt to improve fracture behavior. The following data on \(x=\) fiber weight (\%) and \(y=\) compressive strength (MPa) was provided by the article's authors. $$ \begin{array}{l|ccccccccc} x & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 1.25 & 1.25 & 1.25 & 1.25 \\ \hline y & 9.94 & 11.67 & 11.00 & 13.44 & 9.20 & 9.92 & 9.79 & 10.99 & 11.32 \\\ x & 2.50 & 2.50 & 2.50 & 2.50 & 2.50 & 5.00 & 5.00 & 5.00 & 5.00 \\ \hline y & 12.29 & 8.69 & 9.91 & 10.45 & 10.25 & 7.89 & 7.61 & 8.07 & 9.04 \\ x & 7.50 & 7.50 & 7.50 & 7.50 & 10.00 & 10.00 & 10.00 & 10.00 & \\ \hline y & 6.63 & 6.43 & 7.03 & 7.63 & 7.35 & 6.94 & 7.02 & 7.67 \end{array} $$ a. Fit the simple linear regression model to this data. Then determine the proportion of observed variation in strength that can be attributed to the model relationship between strength and fiber weight. Finally, obtain a point estimate of the standard deviation of \(\epsilon\), the random deviation in the model equation. b. The average strength values for the six different levels of fiber weight are \(11.05,10.51,10.32,8.15,6.93\), and \(7.24\), respectively. The cited paper included a figure in which the average strength was regressed against fiber weight. Obtain the equation of this regression line and calculate the corresponding coefficient of determination. Explain the difference between the \(r^{2}\) value for this regression and the \(r^{2}\) value obtained in (a).

Suppose an investigator has data on the amount of shelf space \(x\) devoted to display of a particular product and sales revenue \(y\) for that product. The investigator may wish to fit a model for which the true regression line passes through \((0,0)\). The appropriate model is \(Y=\beta_{1} x+\epsilon\). Assume that \(\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\) are observed pairs generated from this model, and derive the least squares estimator of \(\beta_{1}\).

The article "Some Field Experience in the Use of an Accelerated Method in Estimating 28-Day Strength of Concrete" (J. of Amer: Concrete Institute, 1969: 895) considered regressing \(y=28\)-day standard-cured strength (psi) against \(x=\) accelerated strength (psi). Suppose the equation of the true regression line is \(y=1800+1.3 x\). a. What is the expected value of 28 -day strength when accelerated strength \(=2500\) ? b. By how much can we expect 28-day strength to change when accelerated strength increases by 1 psi? c. Answer part (b) for an increase of 100 psi. d. Answer part (b) for a decrease of 100 psi.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.