/*! 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 53 The decline of water supplies in... [FREE SOLUTION] | 91Ó°ÊÓ

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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. $$ \begin{aligned} &\begin{array}{r|rrrrrrr} x & 0 & 10 & 20 & 30 & 45 & 50 & 70 \\ \hline y & 500 & 590 & 410 & 470 & 450 & 480 & 510 \end{array}\\\ &\begin{array}{l|rrrrrrr} x & 80 & 100 & 120 & 140 & 160 & 170 & 190 \\ \hline y & 450 & 360 & 400 & 300 & 410 & 280 & 350 \end{array} \end{aligned} $$ 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 linear model appears plausible; the model is useful; the average yield around 75m is estimated with a CI.

Step by step solution

01

Organize the Data

To begin, we'll prepare the data points for plotting. The data consists of two variables: distance upslope \(x\) and grain sorghum yield \(y\). We'll pair these values from the given data where \(x = (0, 10, 20, ..., 190)\) and \(y = (500, 590, 410, ..., 350)\).
02

Create a Scatter Plot

Plot the data points on a graph with \(x\) values on the horizontal axis and \(y\) values on the vertical axis. Each point corresponds to a pair \((x, y)\) from the table. Visually inspect the plot to assess if a linear pattern is apparent. The points should ideally show a linear trend if a simple linear regression model is plausible.
03

Assess Linear Model Plausibility

After creating the scatter plot, observe if there is a visible linear trend. If the data points roughly form a straight line, a simple linear regression model may be appropriate.
04

Test Model Utility (Fit the Model)

Fit a simple linear regression model to the data using the least squares method. Calculate the slope and intercept of the line and the correlation coefficient \(r\). Perform a hypothesis test for the slope: \(H_0: \beta = 0\) vs \(H_a: \beta eq 0\). If the p-value is below the significance level (e.g., \(\alpha = 0.05\)), reject the null hypothesis, indicating the model is useful.
05

Calculate Regression Parameters

Using statistical software or a calculator, find the slope \(b\) and intercept \(a\) of the best-fit line from the formulae: \[\hat{y} = a + bx\]Evaluate the goodness of fit using \(R^2\). A higher \(R^2\) value suggests a better fit of the model to the data.
06

Estimate Yield at Upslope Distance 75

Substitute \(x = 75\) into the regression equation obtained from the model fitting to predict the yield at this upslope distance. Calculate the confidence interval for the predicted yield using the standard error of the estimate and t-distribution.
07

Use Regression Equation for Prediction

Using the regression equation, estimate \(\hat{y}\) when \(x = 75\). For the confidence interval, calculate:\[CI: \hat{y} \pm t_{\alpha/2} \times SE\\]where \(t_{\alpha/2}\) is the critical value from the t-distribution and \(SE\) is the standard error of the prediction.

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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 an essential tool in simple linear regression. It provides a visual representation of the relationship between two quantitative variables. Here, we are analyzing the relationship between the distance upslope \((x)\) and grain sorghum yield \((y)\) in a sloped watershed.

To create a scatter plot, you plot each pair of values \(x, y\) as a point on the Cartesian plane, where the horizontal axis represents the distance upslope \(x\) and the vertical axis represents the yield \(y\).

This visual inspection is crucial because it allows us to identify whether there is an apparent linear trend in the data. If the points roughly form a straight line, then we can say there is a potential linear relationship, making the simple linear regression model plausible. If the points do not align linearly, a more complex model may be necessary. The scatter plot is the first step in validating our assumption of linearity in a simple linear regression analysis.
Hypothesis Testing
In simple linear regression, hypothesis testing is used to determine whether the relationship between the independent variable \(x\) and the dependent variable \(y\) is statistically significant.

The key component of hypothesis testing in this context involves testing the slope of the regression line. Specifically, we conduct a test for the null hypothesis \(H_0: \beta = 0\) versus the alternative hypothesis \(H_a: \beta e 0\).

What does this mean? \(\beta\) represents the slope of the regression line, which indicates the relationship's strength and direction. A slope of zero would signify no linear relationship between \(x\) and \(y\).

By performing a hypothesis test, we calculate a p-value. If this p-value is less than our chosen significance level (commonly \(\alpha = 0.05\)), we reject the null hypothesis, suggesting that there is indeed a significant linear relationship. This means that changes in \(x\) are associated with changes in \(y\), confirming model utility.
Confidence Interval
Confidence intervals provide a range of values within which we expect the true value of a parameter to fall with a certain level of confidence, usually 95%.

In the context of simple linear regression, after estimating the yield \(\hat{y}\) using the regression equation, we can construct a confidence interval around this estimate. This interval gives us a sense of the precision of our prediction.

We use the formula:
\[CI: \hat{y} \pm t_{\alpha/2} \times SE\]
Here, \(\hat{y}\) is the estimated yield from the regression equation, \(t_{\alpha/2}\) is the critical value from the t-distribution which depends on our confidence level, and \(SE\) is the standard error of the prediction.

This approach helps in understanding how much our estimate might vary due to randomness in the data. A narrower confidence interval indicates a more precise estimate, while a wider interval suggests more uncertainty. Creating confidence intervals can thus guide decision-making based on the estimation of yields at different upslope distances.
Least Squares Method
The least squares method is a fundamental technique in simple linear regression. It is used to find the best-fitting line through the data points in a scatter plot by minimizing the sum of the squares of the vertical deviations (errors) of each point from the line.

The goal of this method is to find the slope \(b\) and the intercept \(a\) of the line described by the equation \(\hat{y} = a + bx\).

Here's how it works:
  • First, calculate the average of the x-values and y-values.
  • Next, using these averages and each individual point, calculate the slope \(b\) using the formula:
    \[b = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}\]
  • Then, find the intercept \(a\) using:
    \[a = \bar{y} - b\bar{x}\]
By applying the least squares method, we ensure that the line of best fit minimizes the discrepancies between the observed data points and the line itself, enabling more accurate predictions and insights.

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Most popular questions from this chapter

If there is at least one \(x\) value at which more than one observation has been made, there is a formal test procedure for testing \(H_{0}: \mu_{Y \cdot x}=\beta_{0}+\beta_{1} x\) for some values \(\beta_{0}, \beta_{1}\) (the true regression function is linear) versus \(H_{\mathrm{a}}: H_{0}\) is not true (the true regression function is not linear) Suppose observations are made at \(x_{1}, x_{2}, \ldots, x_{c}\). Let \(Y_{11}, Y_{12}, \ldots, Y_{1 n_{1}}\) denote the \(n_{1}\) observations when \(x=x_{1} ; \ldots ; Y_{c 1}, Y_{c 2}, \ldots, Y_{c n_{c}}\) denote the \(n_{c}\) observations when \(x=x_{c}\). With \(n=\Sigma n_{i}\) (the total number of observations), SSE has \(n-2\) df. We break SSE into two pieces, SSPE (pure error) and SSLF (lack of fit), as follows: $$ \begin{aligned} \mathrm{SSPE} &=\sum_{i} \sum_{j}\left(Y_{i j}-\bar{Y}_{i} .\right)^{2} \\ &=\sum_{i} \sum_{j} Y_{i j}^{2}-\sum_{i} n_{i}\left(\bar{Y}_{i} .\right)^{2} \end{aligned} $$ $$ \text { SSLF }=\text { SSE }-\text { SSPE } $$ The \(n_{i}\) observations at \(x_{i}\) contribute \(n_{i}-1\) df to SSPE, so the number of degrees of freedom for SSPE is \(\Sigma_{i}\left(n_{i}-1\right)=n-c\) and the degrees of freedom for SSLF is \(n-2-(n-c)=c-2\). Let MSPE \(=\operatorname{SSPE} /(n-c), \operatorname{MSLF}=\operatorname{SSLF} /(c-2) .\) Then it can be shown that whereas \(E(\) MSPE \()=\sigma^{2}\) whether or not \(H_{0}\) is true, \(E\) (MSLF) \(=\sigma^{2}\) if \(H_{0}\) is true and \(E(\) MSLF \()>\sigma^{2}\) if \(H_{0}\) is false. Test statistic: \(F=\) MSLF/MSPE Rejection region: \(f \geq F_{\alpha, c-2, n-c}\) The following data comes from the article "Changes in Growth Hormone Status Related to Body Weight of Growing Cattle" (Growth, 1977: 241-247), with \(x=\) body weight and \(y=\) metabolic clearance rate/ body weight. $$ \begin{aligned} &\begin{array}{l|lllllll} x & 110 & 110 & 110 & 230 & 230 & 230 & 360 \\ \hline y & 235 & 198 & 173 & 174 & 149 & 124 & 115 \end{array}\\\ &\begin{array}{r|rrrrrrr} x & 360 & 360 & 360 & 505 & 505 & 505 & 505 \\ \hline y & 130 & 102 & 95 & 122 & 112 & 98 & 96 \end{array} \end{aligned} $$ (So \(c=4, n_{1}=n_{2}=3, n_{3}=n_{4}=4\).) a. Test \(H_{0}\) versus \(H_{\mathrm{a}}\) at level \(.05\) using the lackof-fit test just described. b. Does a scatter plot of the data suggest that the relationship between \(x\) and \(y\) is linear? How does this compare with the result of part (a)? (A nonlinear regression function was used in the article.)

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 \(Y=\beta_{1} x+\varepsilon\). 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 \(\beta_{1}\). [Hint: Write the sum of squared deviations as a function of \(b_{1}\), a trial value, and use calculus to find the minimizing value of \(b_{1}\).]

The flow rate \(y\left(\mathrm{~m}^{3} / \mathrm{min}\right)\) in a device used for airquality measurement depends on the pressure drop \(x\) (in. of water) across the device's filter. Suppose that for \(x\) values between 5 and 20 , the two variables are related according to the simple linear regression model with true regression line \(y=-.12+.095 x\). a. What is the expected change in flow rate associated with a 1-in. increase in pressure drop? Explain. b. What change in flow rate can be expected when pressure drop decreases by 5 in.? c. What is the expected flow rate for a pressure drop of 10 in.? A drop of 15 in.? d. Suppose \(\sigma=.025\) and consider a pressure drop of \(10 \mathrm{in}\). What is the probability that the observed value of flow rate will exceed \(.835\) ? That observed flow rate will exceed .840? e. What is the probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in.?

Verify Property 2 of the correlation coefficient, the value of \(r\) is independent of the units in which \(x\) and \(y\) are measured; that is, if \(x_{i}^{\prime}=a x_{i}+c\) and \(y_{i}^{\prime}=b y_{i}+d, a>0, b>0\), then \(r\) for the \(\left(x_{i}^{\prime}, y_{i}^{\prime}\right)\) pairs is the same as \(r\) for the \(\left(x_{i}, y_{i}\right)\) pairs.

The article "Validation of the Rockport Fitness Walking Test in College Males and Females" (Res. Q. Exercise Sport, 1994: 152-158) recommended the following estimated regression equation for relating \(y=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}\), a measure of cardiorespiratory fitness) to the predictors \(x_{1}\) \(=\) gender (female \(=0\), male \(=1\) ), \(x_{2}=\) weight (lb), \(x_{3}=1\)-mile walk time (min), and \(x_{4}=\) heart rate at the end of the walk (beats/min): $$ \begin{aligned} y=& 3.5959+.6566 x_{1}+.0096 x_{2} \\ &-.0996 x_{3}-.0080 x_{4} \end{aligned} $$ a. How would you interpret the estimated coefficient \(-.0996\) ? b. How would you interpret the estimated coefficient .6566? c. Suppose that an observation made on a male whose weight was \(170 \mathrm{lb}\), walk time was \(11 \mathrm{~min}\), and heart rate was 140 beats \(/ \mathrm{min}\) resulted in \(\mathrm{VO}_{2} \mathrm{max}=3.15\). What would you have predicted for \(\mathrm{VO}_{2}\) max in this situation, and what is the value of the corresponding residual? d. Using SSE \(=30.1033\) and SST \(=102.3922\), what proportion of observed variation in \(\mathrm{VO}_{2} \max\) can be attributed to the model relationship? e. Assuming a sample size of \(n=20\), carry out a test of hypotheses to decide whether the chosen model specifies a useful relationship between \(\mathrm{VO}_{2} \max\) and at least one of the predictors.

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