91Ó°ÊÓ

First, let's use the data for steeply sloped plots to calculate the equation of the least squares regression line for predicting runoff sediment concentration using the percentage of bare ground. We are given the following data: $$ \begin{array}{lrrrrrrr} \text { Bare ground }(\%) & 5 & 5 & 10 & 15 & 20 & 25 & 20 \\ \text { Concentration } & 100 & 250 & 300 & 600 & 500 & 500 & 900 \end{array} $$ To calculate the least squares regression line, we can use the equation: $$ y = a + bx $$ where: - \(y\) is the runoff sediment concentration - \(a\) is the y-intercept - \(b\) is the slope of the line - \(x\) is the percentage of bare ground The slope \(b\) can be calculated using the following formula: $$ b = \frac{N\sum(xy) - \sum x \sum y}{N\sum x^2 - (\sum x)^2} $$ and the y-intercept \(a\) can be calculated using: $$ a = \frac{\sum y - b\sum x}{N} $$ where \(N\) is the number of data points. Using the given data, we can calculate the necessary sums: $$ \begin{aligned} \sum x &= 5+5+10+15+20+25+20 = 100 \\ \sum y &= 100+250+300+600+500+500+900 = 3150 \\ \sum xy &= 5(100) + 5(250) + 10(300) + 15(600) + 20(500) + 25(500) + 20(900) = 25500 \\ \sum x^2 &= 5^2+5^2+10^2+15^2+20^2+25^2+20^2 = 2000 \\ N &= 7 \end{aligned} $$ Now we can find the slope \(b\) and y-intercept \(a\): $$ \begin{aligned} b &= \frac{7(25500) - 100(3150)}{7(2000) - (100)^2} = \frac{178500 - 315000}{14000 - 10000} = \frac{-136500}{4000} = -34.125 \\ a &= \frac{3150 - (-34.125)(100)}{7} = \frac{3150 + 3412.5}{7} = \frac{6562.5}{7} = 937.5 \end{aligned} $$ So the least squares regression line equation is: $$ y = 937.5 - 34.125x $$

Now we have the equation of the least squares regression line, we can use it to predict the runoff sediment concentration for a steeply sloped plot with 18% bare ground. Using the equation: $$ y = 937.5 - 34.125x $$ Let \(x = 18\): $$ y = 937.5 - 34.125(18) \approx 322.75 $$ Thus, the predicted runoff sediment concentration for a plot with 18% bare ground is approximately 322.75.

To check if the least squares regression line can be used to predict runoff sediment concentration for gradually sloped plots, we have to analyze the relationship between the data from steeply sloped plots and gradually sloped plots. The data for gradually sloped plots looks like this: $$ \begin{array}{lrrrrrr} \text { Bare ground }(\%) & 5 & 10 & 15 & 25 & 30 & 40 \\ \text { Concentration } & 50 & 200 & 250 & 500 & 600 & 500 \end{array} $$ Upon inspection, there is no clear linear relationship between the data for gradually sloped plots and steeply sloped plots, as the data points for gradually sloped plots do not seem to follow the same trend as the steeply sloped plots. Therefore, it is not recommended to use the least squares regression line from Part (a) to predict runoff sediment concentration for gradually sloped plots. The relationship between the two sets of data is not strong enough to make accurate predictions using the same regression line.