/*! 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 24 Representative data read from a ... [FREE SOLUTION] | 91Ó°ÊÓ

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Representative data read from a plot that appeared in the paper "Effect of Cattle Treading on Erosion from Hill Pasture: Modeling Concepts and Analysis of Rainfall Simulator Data" (Australian Journal of Soil Research [2002]: 963-977) on runoff sediment concentration for plots with varying amounts of grazing damage, measured by the percentage of bare ground in the plot, are given for gradually sloped plots and for steeply sloped plots. \begin{tabular}{lrrrr} Gradually Sloped Plots & & & & \\ Bare ground (\%) & 5 & 10 & 15 & 25 \\ Concentration & 50 & 200 & 250 & 500 \\ Bare ground (\%) & 30 & 40 & & \\ Concentration & 600 & 500 & & \\ & & & \multicolumn{2}{r} { (contimued) } \end{tabular} Steeply Sloped Plots \(\begin{array}{lrrrr}\text { Bare ground (\%) } & 5 & 5 & 10 & 15 \\ \text { Concentration } & 100 & 250 & 300 & 600 \\ \text { Bare ground (\%) } & 20 & 25 & 20 & 30 \\ \text { Concentration } & 500 & 500 & 900 & 800 \\ \text { Bare ground (\%) } & 35 & 40 & 35 & \\ \text { Concentration } & 1100 & 1200 & 1000 & \end{array}\) a. Using the data for steeply sloped plots, find the equation of the least- squares line for predicting \(y=\) runoff sediment concentration using \(x=\) percentage of bare ground. b. What would you predict runoff sediment concentration to be for a steeply sloped plot with \(18 \%\) bare ground? c. Would you recommend using the least-squares equation from Part (a) to predict runoff sediment concentration for gradually sloped plots? If so, explain why it would be appropriate to do so. If not, provide an alternative way to make such predictions.

Short Answer

Expert verified
To solve this problem, it's necessary to calculate the least-squares line for the steeply sloped plots, use this line to predict runoff sediment concentration for a steeply sloped plot with 18% bare ground, and analyze whether this line would be appropriate for use with gradually sloped plot data.

Step by step solution

01

Calculation for least squares line

First, organize the data from the steeply sloped plots into two columns, X being percentage of bare ground and Y being concentration. Then, calculate the sums needed for the least squares formula: sum of X, Y, XY, and X squared. The formula for the slope, b, of the least squares line is \((n\sum{XY} - \sum{X}\sum{Y})/(n\sum{X^2}- (\sum{X})^2)\) and the formula for the y-intercept, a, is \((\sum{Y} - b\sum{X}) / n\). Use the gathered sums to calculate b and a.
02

Predicting runoff sediment concentration

To predict the runoff sediment concentration for 18\% bare ground in steeply sloped plots, use the derived least squares line equation from step 1. Replace X in the equation with 18 to find the predicted concentration, \(Y_predicted\).
03

Recommendation on the apply of the least-squares equation

The appropriateness of using the least-squares equation from data for steeply sloped plots to predict runoff sediment concentration for gradually sloped plots depends on how similar the trends in both datasets are. Visually compare the scatterplots of both datasets or perform a statistical test (like ANOVA) to affirm if the slopes are significantly different. If they are not significantly different, it could be appropriate to use the equation. If they are significantly different, it would be preferable to create a new least-squares line equation for the data on gradually sloped plots to predict the runoff sediment concentration for new percentages of bare ground.

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Key Concepts

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

Least Squares Method
The least squares method is a statistical technique used to find the line of best fit for a set of data points. It minimizes the sum of the squared differences between the observed values and the values predicted by the line. For simple linear regression, we aim to fit a line to the data in the form of \( y = a + bx \), where \( a \) is the y-intercept and \( b \) is the slope. In the original exercise, this method helps to model the relationship between the percentage of bare ground (as \( x \)) and the runoff sediment concentration (as \( y \)).
To calculate the slope \( b \), we use:
  • Calculate products and sums: \( n\sum{XY}, \sum{X}, \sum{Y}, \sum{X^2}, \sum{XY} \)
  • Use the formula: \( b = \frac{n\sum{XY} - \sum{X}\sum{Y}}{n\sum{X^2} - (\sum{X})^2} \)
The y-intercept \( a \) is then calculated as:
  • \( a = \frac{\sum{Y} - b\sum{X}}{n} \)
These calculations provide the equation needed to make predictions on future data points.
Scatterplots
Scatterplots are graphical representations of data points for two variables. They allow us to visualize relationships or trends between them. Each point on the scatterplot corresponds to a pair of values, such as the percentage of bare ground and the runoff sediment concentration in our exercise.
When creating a scatterplot:
  • Place one variable along the x-axis (e.g., percentage of bare ground)
  • Place the other variable along the y-axis (e.g., runoff sediment concentration)
This visualization helps in identifying whether a linear relationship exists and assists in deciding where to apply linear regression. By plotting the points for both steeply and gradually sloped plots, one can visually assess if a linear trend holds and if separate regression lines are needed for each set of conditions.
Statistical Prediction
Statistical prediction involves using a model (like a least-squares regression line) to forecast unknown values based on known data. Once you have the regression equation \( y = a + bx \), you can use it to estimate the runoff sediment concentration for any given percentage of bare ground.
In the exercise:
  • The least-squares line derived for steeply sloped plots helps predict concentration for particular bare ground percentages
  • For 18% bare ground, substitute \( x = 18 \) into the equation to find the corresponding \( y \)
The precision of these predictions depends on the goodness of fit of the model. If the scatterplot shows a strong linear relationship and the residual variance is low, predictions will likely be accurate.
Slope and Y-intercept
The slope and y-intercept are critical components of the regression line equation. They define the line's angle and where it crosses the y-axis, respectively.

What is the slope?

The slope \( b \) measures the line’s steepness. It quantifies the change in the runoff sediment concentration for each one-unit increase in the percentage of bare ground. A higher slope implies a steeper increase in concentration as bare ground increases.

And the y-intercept?

The y-intercept \( a \) is the value of \( y \) when \( x = 0 \). It represents the starting point of the regression line on the y-axis. In environmental studies like in our exercise, interpreting the y-intercept outside the context of meaningful data can be complex since having zero percent bare ground may not be realistic.
Understanding these concepts aids in making sense of the regression equation and the nature of the relationship it represents.

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Most popular questions from this chapter

The following quote is from the paper "Evaluation of the Accuracy of Different Methods Used to Estimate Weights in the Pediatric Population" (Pediatrics [2009 ]: el045-elo51): As expected, the model demonstrated that weight increased with age, but visual inspection of an age versus weight plot demonstrated a nonlinear relationship unless infants and children were analyzed separately. The linear coefficient for age as a predictor of weight was \(6.93\) in infants and \(3.1\) to \(3.48\) in children. This quote suggests that when a scatterplot of weight versus age was constructed for all 1011 children in the study described in the paper, the relationship between \(y=\) weight and \(x=\) age was not linear. When the 1011 children were separated into two groups-infants (age birth to 1\. year) and children (age 1 to 10 years) - and separate scatterplots were constructed, the relationship between weight and age appeared linear in each scatterplot. The slopes reported in the given quote (referred to as "the linear coefficient") are expressed in kg/year. Briefly explain why the relationship between weight and age in the scatterplot for the combined group would appear nonlinear.

The paper "Developmental and Individual Differences in Pure Numerical Estimation" (Developmental Psychology [2006]: \(189-201)\) describes a study of how young children develop the ability to estimate lengths. Children were shown a piece of paper with two lines. One line was a short line labeled as having length zip. The second line was a much longer line labeled as having length 1000 zips. The child was then asked to draw a line that had a length of a specified number of zips, such as 438 zips. The data in the accompanying table gives the length requested and the average of the actual lengths of the lines drawn by 30 second graders. \begin{tabular}{cc} Requested Length & Second Grade Average Length Drawn \\ \hline 3 & \(37.15\) \\ 7 & \(92.88\) \\ 19 & \(207.43\) \\ 52 & \(272.45\) \\ 103 & \(458.20\) \\ 158 & \(442.72\) \\ 240 & \(371.52\) \\ 297 & \(467.49\) \\ 346 & \(487.62\) \\ 391 & \(530.96\) \\ 438 & \(482.97\) \\ 475 & \(544.89\) \\ 502 & \(515.48\) \\ 586 & \(595.98\) \\ 613 & \(575.85\) \\ 690 & \(605.26\) \\ 721 & \(637.77\) \\ 760 & \(674.92\) \\ 835 & \(701.24\) \\ 874 & \(662.54\) \\ 907 & \(758.51\) \\ 962 & \(749.23\) \\ \hline \end{tabular} a. Construct a scatterplot of \(y=\) second grade average length drawn versus \(x=\) requested length. b. Based on the scatterplot in Part (a), would you suggest using a line, a quadratic curve, or a cubic curve to describe the relationship between \(x\) and \(y\) ? Explain choice c. Using a statistical software package or a graphing calculator, fit a cubic curve to this data and use it to predict average length drawn for a requested length of 500 zips.

No tortilla chip lover likes soggy chips, so it is important to find characteristics of the production process that produce chips with an appealing texture. The accompanying data on \(x=\) frying time (in seconds) and \(y=\) moisture content \((\%)\) appeared in the paper, "Thermal and Physical Properties of Tortilla Chips as a Function of Frying Time" (journal of Food Processing and Preservation [1995]: \(175-189\) ): \(\begin{array}{lrrrrrrrr}\text { Frying time }(x): & 5 & 10 & 15 & 20 & 25 & 30 & 45 & 60 \\ \text { Moisture } & 16.3 & 9.7 & 8.1 & 4.2 & 3.4 & 2.9 & 1.9 & 1.3\end{array}\) content \((y)\) : a. Construct a scatterplot of these data. Does the relationship between moisture content and frying time appear to be linear? b. Transform the \(y\) values using \(y^{\prime}=\log (y)\) and construct a scatterplot of the \(\left(x, y^{\prime}\right)\) pairs. Does this scatterplot look more nearly linear than the one in Part (a)? c. Find the equation of the least-squares line that describes the relationship between \(y^{\prime}\) and \(x\). d. Use the least-squares line from Part (c) to predict moisture content for a frying time of 35 minutes.

The accompanying data are \(x=\) cost (cents per serving) and \(y=\) fiber content (grams per serving) for 18 high-fiber cereals rated by Consumer Reports (www .consumerreports.org/health). $$ \begin{array}{cc} \begin{array}{c} \text { Cost per } \\ \text { serving } \end{array} & \begin{array}{c} \text { Fiber per } \\ \text { serving } \end{array} \\ \hline 33 & 7 \\ 46 & 10 \\ 49 & 10 \\ 62 & 7 \\ 41 & 8 \\ 19 & 7 \\ 77 & 12 \\ 71 & 12 \\ 30 & 8 \\ 53 & 13 \\ 53 & 10 \\ 67 & 8 \\ 43 & 12 \\ 48 & 7 \\ 28 & 14 \\ 54 & 7 \\ 27 & 8 \\ 58 & 8 \\ \hline \end{array} $$ a. Compute and interpret the correlation coefficient for this data set. b. The serving size differed for the different cereals, with serving sizes varying from \(1 / 2\) cup to \(1^{1 / 4}\) cups. Converting price and fiber content to "per cup" rather than "per serving" results in the accompanying data. Is the correlation coefficient for the per cup data greater than or less than the correlation coefficient for the per serving data? $$ \begin{array}{cc} \text { Cost per Cup } & \text { Fiber per Cup } \\ \hline 9.3 & 44 \\ 10 & 46 \\ 10 & 49 \\ 7 & 62 \\ 6.4 & 32.8 \\ 7 & 19 \\ 12 & 77 \\ 9.6 & 56.8 \\ 8 & 30 \\ 13 & 53 \\ 10 & 53 \\ 8 & 67 \\ 12 & 43 \\ 7 & 48 \\ 28 & 56 \\ 7 & 54 \\ 16 & 54 \\ 10.7 & 77.3 \\ \hline \end{array} $$

The article "That's Rich: More You Drink, More You Earn" (Calgary Herald, April 16. 2002) reported that there was a positive correlation between alcohol consumption and income. Is it reasonable to conclude that increasing alcohol consumption will increase income? Give at least two reasons or examples to support your answer.

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