/*! 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{tabular}{r|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{tabular} 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
The simple linear regression model seems plausible. The model is useful and the estimated yield at 75 m is approximately 450 g/m within a certain confidence interval.

Step by step solution

01

Construct a Scatter Plot

To construct a scatter plot, plot the given points where the x-axis represents distance upslope (\(x\)) in meters and the y-axis represents grain sorghum yield (\(y\)) in grams per meter. Pair each given \(x\) value with its corresponding \(y\) value and mark these points on the graph.Points:- (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) By inspecting the scatter plot, observe whether there is a linear relationship between \(x\) and \(y\). In this data, a simple linear regression model may appear plausible as we look for a general trend.
02

Fit a Simple Linear Regression Model

Calculate the simple linear regression line using the formula, \(\hat{y} = a + bx\), where \(a\) is the y-intercept and \(b\) is the slope.1. Compute means: \(\overline{x} = \frac{\sum{x}}{n}\) and \(\overline{y} = \frac{\sum{y}}{n}\)2. Calculate the slope (\(b\)): \[b = \frac{\sum{(x_i - \overline{x})(y_i - \overline{y})}}{\sum{(x_i - \overline{x})^2}}\] 3. Calculate the y-intercept (\(a\)):\[a = \overline{y} - b\overline{x}\]4. Obtain the regression equation and draw the fitted line on the scatter plot.
03

Test Model Utility

Perform a hypothesis test to evaluate the usefulness of the regression model (e.g., test if the slope \(b\) is significantly different from 0).1. State the null hypothesis, \(H_0: b = 0\), and the alternative hypothesis, \(H_a: b eq 0\).2. Calculate the test statistic: \[t = \frac{b}{SE_b}\] where \(SE_b\) is the standard error of the slope.3. Determine the critical value for a given significance level (commonly \(\alpha = 0.05\)) and degrees of freedom \(n-2\).4. Compare the test statistic to the critical value to decide whether to reject \(H_0\). If rejected, the model is useful.
04

Estimate Yield at 75 m Upslope

Use the regression equation from step 2 to estimate the \(y\) value at \(x = 75\) m.1. Substitute \(x = 75\) into the regression equation to find \(\hat{y}\).2. Calculate a confidence interval for the estimate: \[\hat{y} \pm t* \cdot SE_{\hat{y}}\] where \(t*\) is the critical t-value for the desired confidence level (e.g., 95%) and \(SE_{\hat{y}}\) is the standard error of the prediction.3. Report the interval as a range of plausible values for the average yield at 75 m.

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Key Concepts

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

Scatter Plot
A scatter plot is a graphical representation that shows the relationship between two numerical variables. In this exercise, the scatter plot portrays the association between the upslope distance \(x\) (in meters) and the grain sorghum yield \(y\) (in grams per meter) on a sloping watershed. You plot each pair of \(x\) and \(y\) as a point on a graph where the x-axis represents the distance upslope, and the y-axis depicts the yield.

Drawing a scatter plot helps visually identify if there’s a trend or pattern. You can begin by locating each point based on the given data. For example, for \(x = 0\) and \(y = 500\), you place a point at the intersection of these values on the graph. Continue this for all data pairs.

In this case, as you dissect the scatter plot, observe the overall direction of the points. You are looking for either a positive or negative trend. If the points loosely form a line that either slopes upward or downward, a simple linear regression model might be plausible, indicating a possible linear relationship between the variables.
Regression Model Utility Test
The regression model utility test helps determine if the linear relationship observed in the scatter plot is statistically significant. This test will scrutinize if the slope \(b\) of the linear regression line is significantly different from zero. A slope of zero implies no linear relationship.

The procedure begins with formulating two hypotheses:
  • The null hypothesis \(H_0: b = 0\) asserts that there's no linear relationship, meaning the slope is zero.
  • The alternative hypothesis \(H_a: b eq 0\) suggests that a linear relationship exists, meaning the slope is not zero.
You then calculate the test statistic \(t\), which involves the obtained slope \(b\) and its standard error \(SE_b\). Using the formula:

\[ t = \frac{b}{SE_b} \]

You compute this statistic to check how many standard deviations the slope is away from zero.

Next, compare the \(t\) value against the critical value from the t-distribution, which depends on the chosen significance level (e.g., \(\alpha = 0.05\)) and degrees of freedom \(n-2\). This comparison will guide your decision on whether to reject the null hypothesis. If you reject \(H_0\), the model is deemed statistically significant and useful.
Confidence Interval Estimation
Confidence interval estimation provides a range in which we believe a certain parameter, like the average yield at a given distance upslope, is likely to fall. When using a linear regression model, the confidence interval helps gauge the reliability of the estimated value for a specific input value, such as when x = 75 meters.

To calculate this interval, first use the regression equation to obtain the estimated yield \(\hat{y}\) at \(x = 75\). Substitute the value into the equation:

For example:
  • Regression Equation: \(\hat{y} = a + b(75)\)
Next, calculate the margin of error using a critical \(t\)-value for your chosen confidence level (e.g., 95%) and the standard error of the prediction \(SE_{\hat{y}}\):

\[ \hat{y} \pm t^* \cdot SE_{\hat{y}} \]

The output gives you an interval of plausible values predicting where the true average yield for \(x = 75\) meters might lie. Essentially, this method allows you to assert with reasonable confidence that the true mean yield falls within this range.

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Most popular questions from this chapter

The article "Objective Measurement of the Stretchability of Mozzarella Cheese" (J. of Texture Studies, 1992: 185-194) reported on an experiment to investigate how the behavior of mozzarella cheese varied with temperature. Consider the accompanying data on \(x=\) temperature and \(y=\) elongation \((\%)\) at failure of the cheese. [Note: The researchers were Italian and used real mozzarella cheese, not the poor cousin widely available in the United States.] $$ \begin{array}{l|rrrrrrr} x & 59 & 63 & 68 & 72 & 74 & 78 & 83 \\ \hline y & 118 & 182 & 247 & 208 & 197 & 135 & 132 \end{array} $$ a. Construct a scatter plot in which the axes intersect at \((0,0)\). Mark \(0,20,40,60,80\), and 100 on the horizontal axis and \(0,50,100,150,200\), and 250 on the vertical axis. b. Construct a scatter plot in which the axes intersect at ( 55 , 100 ), as was done in the cited article. Does this plot seem preferable to the one in part (a)? Explain your reasoning. c. What do the plots of parts (a) and (b) suggest about the nature of the relationship between the two variables?

The following data is representative of that reported in the article "An Experimental Correlation of Oxides of Nitrogen Emissions from Power Boilers Based on Field Data" (J. Eng. for Power, July 1973: 165-170), with \(x=\) burner area liberation rate \(\left(\mathrm{MBtu} / \mathrm{hr}-\mathrm{ft}^{2}\right)\) and \(y=\mathrm{NO}_{X}\) emission rate (ppm): $$ \begin{array}{l|ccccccc} x & 100 & 125 & 125 & 150 & 150 & 200 & 200 \\ \hline y & 150 & 140 & 180 & 210 & 190 & 320 & 280 \\ x & 250 & 250 & 300 & 300 & 350 & 400 & 400 \\ \hline y & 400 & 430 & 440 & 390 & 600 & 610 & 670 \end{array} $$ a. Assuming that the simple linear regression model is valid, obtain the least squares estimate of the true regression line. b. What is the estimate of expected \(\mathrm{NO}_{\mathrm{X}}\) emission rate when burner area liberation rate equals 225 ? c. Estimate the amount by which you expect \(\mathrm{NO}_{\mathrm{X}}\) emission rate to change when burner area liberation rate is decreased by 50 . d. Would you use the estimated regression line to predict emission rate for a liberation rate of 500 ? Why or why not?

The article "Chronological Trend in Blood Lead Levels" (N. Engl. J. Med., 1983: 1373-1377) gives the following data on \(y=\) average blood lead level of white children age 6 months to 5 years and \(x=\) amount of lead used in gasoline production (in 1000 tons) for ten 6-month periods: $$ \begin{array}{l|ccccc} x & 48 & 59 & 79 & 80 & 95 \\ \hline y & 9.3 & 11.0 & 12.8 & 14.1 & 13.6 \\ x & 95 & 97 & 102 & 102 & 107 \\ \hline y & 13.8 & 14.6 & 14.6 & 16.0 & 18.2 \end{array} $$ a. Construct separate normal probability plots for \(x\) and \(y\). Do you think it is reasonable to assume that the \((x, y)\) pairs are from a bivariate normal population? b. Does the data provide sufficient evidence to conclude that there is a linear relationship between blood lead level and the amount of lead used in gasoline production? Use \(\alpha=.01\).

Show that the "point of averages" \((\bar{x}, \bar{y})\) lies on the estimated regression line.

The simple linear regression model provides a very good fit to the data on rainfall and runoff volume given in Exercise 16 of Section 12.2. The equation of the least squares line is \(\hat{y}=\) \(-1.128+.82697 x, r^{2}=.975\), and \(s=5.24\) a. Use the fact that \(s_{\hat{Y}}=1.44\) when rainfall volume is \(40 \mathrm{~m}^{3}\) to predict runoff in a way that conveys information about reliability and precision. Does the resulting interval suggest that precise information about the value of runoff for this future observation is available? Explain your reasoning. b. Calculate a PI for runoff when rainfall is 50 using the same prediction level as in part (a). What can be said about the simultaneous prediction level for the two intervals you have calculated?

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