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The article "The Epiphytic Lichen Hypogymnia physodes as a Bioindicator of Atmospheric Nitrogen and Sulphur Deposition in Norway" (Environmental Monitoring and Assessment \([1993]: 27-47\) ) gives the following data (read from a graph in the paper) on \(x=\mathrm{NO}_{3}\) wet deposition (in grams per cubic meter) and \(y=\) lichen (\% dry weight): $$ \begin{array}{llllll} x & 0.05 & 0.10 & 0.11 & 0.12 & 0.31 \\ y & 0.48 & 0.55 & 0.48 & 0.50 & 0.58 \\ x & 0.37 & 0.42 & 0.58 & 0.68 & 0.68 \\ y & 0.52 & 1.02 & 0.86 & 0.86 & 1.00 \\ x & 0.73 & 0.85 & 0.92 & & \\ y & 0.88 & 1.04 & 1.70 & & \end{array} $$ a. What is the equation of the least-squares regression line? b. Predict lichen dry weight percentage for an \(\mathrm{NO}_{3}\) deposition of \(0.5 \mathrm{~g} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The answer includes the equation of the least-squares regression line in Step 4 and the predicted lichen dry weight percentage for an \(\mathrm{NO}_{3}\) deposition of \(0.5 \mathrm{~g} / \mathrm{m}^{3}\) in Step 5.

Step by step solution

01

Calculate the means of x and y

First calculate the means of the \(x\) and \(y\) values. These will be used in later formulas. The mean (average) of a set of numbers is the sum of the numbers divided by the count of numbers. Let's represent mean of \(x\) as \(\bar{x}\) and mean of \(y\) as \(\bar{y}\).
02

Calculate slope (b1)

The slope of regression line (also known as b1 or the regression coefficient) is calculated as \[ b1 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})(y_i-\bar{y})}{\sum_{i=1}^{n} (x_i - \bar{x})^2}\] where \(x_i\) and \(y_i\) represent each value of \(x\) and \(y\), \(\bar{x}\) and \(\bar{y}\) are the means of \(x\) and \(y\) respectively, and \(n\) is the number of observations.
03

Calculate intercept (b0)

The y-intercept (also known as b0) is calculated using the formula \[ b0 = \bar{y} - b1 \cdot \bar{x}\] where \(\bar{y}\) and \(\bar{x}\) are the means of \(y\) and \(x\) respectively, and \(b1\) is the slope calculated in the previous step.
04

Formulate the regression equation

Now, put the values of \(b0\) and \(b1\) into the equation \[y = b0 + b1 \cdot x\] It gives the least squares regression line.
05

Predict the value of y for given x

For \(x = 0.5 \mathrm{~g} / \mathrm{m}^{3}\), put it into the equation derived in the previous step to get the predicted value of \(y\) (lichen dry weight percentage).

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Key Concepts

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

Statistical Analysis
Statistical analysis is a component of data science that involves collecting, examining, interpreting, presenting, and modeling data. It is the process that allows us to explore and infer from data, providing a way to look beyond the raw numbers and find meaningful patterns and trends.

In the context of the given exercise, statistical analysis is used to understand the relationship between two variables: the NO3 wet deposition (grams per cubic meter) and the lichen's dry weight percentage. These variables are quantitative, and statistical tools, specifically least-squares regression, are applied to quantify the associated or trend between them.

The least-squares regression line, which is central to this analysis, is computed based on minimizing the sum of the squares of the vertical deviations from each data point to the line. By using the calculated equation of the regression line, predictions for unobserved values can be made, which is a practical application of statistical analysis in predicting outcomes based on historical data.

The computation steps in the provided solution involve finding the means of both x and y values, calculating the slope and intercept of the regression line, formulating the regression equation, and finally making predictions.
Bivariate Data
Bivariate data involves two sets of related data—pairs of observations taken together usually representing two aspects of the same event. For example, in the given exercise, the pairs are NO3 wet deposition and lichen dry weight percentage. Bivariate data is crucial in research and analysis, because it lets us explore possible connections between two variables.

Before conducting any advanced analyses or predictive modeling, it is essential to visualize bivariate data. The usual method is constructing a scatter plot, where one variable is plotted along the x-axis (independent variable) and the other along the y-axis (dependent variable). The pattern of points on the scatter plot can provide initial insights into the type of relationship – for instance, linear, exponential, or no definite pattern.

By visual examination of this plot, we can hypothesize whether a linear model, like the least-squares regression line, is appropriate for the data. This line aims to explain the variation in the dependent variable (y) with the variation in the independent variable (x), and the exercise demonstrates the numerical method of finding this line through calculation.
Predictive Modeling
Predictive modeling is a form of data analysis technique used to create models that predict future outcomes based on historical data. It is widely used in various fields such as finance, marketing, meteorology, and healthcare. Predictive models are built using statistical techniques that consider current and historical facts to make forecasts about future or otherwise unknown events.

In the case of the exercise, predictive modeling involves using the least-squares regression line as a predictive model. After establishing the equation of the regression line from the dataset, it becomes possible to predict the lichen dry weight percentage for an unknown NO3 deposition value. We assume that the underlying relationship established from the historical data will hold for new data points.

This process involves plugging in the NO3 deposition value into the regression equation to estimate the expected lichen dry weight percentage. The quality and reliability of these predictions hinge upon the representativeness of the sample data, the accuracy of the modeling process, and the assumption that the future will follow patterns found in the past data.

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Most popular questions from this chapter

An auction house released a list of 25 recently sold paintings. Eight artists were represented in these sales. The sale price of each painting appears on the list. Would the correlation coefficient be an appropriate way to summarize the relationship between artist \((x)\) and sale price \((y)\) ? Why or why not?

The following data on degree of exposure to \({ }^{242} \mathrm{Cm}\) alpha particles \((x)\) and the percentage of exposed cells without aberrations \((y)\) appeared in the paper "Chromosome Aberrations Induced in Human Lymphocytes by D-T Neutrons" (Radiation Research \([1984]: 561-573)\) : $$ \begin{array}{rrrrr} x & 0.106 & 0.193 & 0.511 & 0.527 \\ y & 98 & 95 & 87 & 85 \\ x & 1.08 & 1.62 & 1.73 & 2.36 \\ y & 75 & 72 & 64 & 55 \\ x & 2.72 & 3.12 & 3.88 & 4.18 \\ y & 44 & 41 & 37 & 40 \end{array} $$ Summary quantities are $$ \begin{gathered} n=12 \quad \sum x=22.027 \quad \sum y=793 \\ \sum x^{2}=62.600235 \quad \sum x y=1114.5 \quad \sum y^{2}=57,939 \end{gathered} $$ a. Obtain the equation of the least-squares line. b. Construct a residual plot, and comment on any interesting features.

The paper cited in Exercise \(5.69\) gave a scatterplot in which \(x\) values were for anterior teeth. Consider the following representative subset of the data: $$ \begin{aligned} &\begin{array}{lllrrl} x & 15 & 19 & 31 & 39 & 41 \\ y & 23 & 52 & 65 & 55 & 32 \\ x & 44 & 47 & 48 & 55 & 65 \\ y & 60 & 78 & 59 & 61 & 60 \\ \sum x=404 & \sum x y=18,448 & \sum y=545 & \end{array}\\\ &\begin{array}{rl} \sum x=404 & \sum x y=18,448 \quad \sum y=545 \\ a=32.080888 & b=0.554929 \end{array} \end{aligned} $$ a. Calculate the predicted values and residuals. b. Use the results of Part (a) to obtain SSResid and \(r^{2}\). c. Does the least-squares line appear to give accurate predictions? Explain your reasoning.

The sales manager of a large company selected a random sample of \(n=10\) salespeople and determined for each one the values of \(x=\) years of sales experience and \(y=\) annual sales (in thousands of dollars). A scatterplot of the resulting \((x, y)\) pairs showed a marked linear pattern. a. Suppose that the sample correlation coefficient is \(r=\) \(.75\) and that the average annual sales is \(\bar{y}=100\). If a particular salesperson is 2 standard deviations above the mean in terms of experience, what would you predict for that person's annual sales? b. If a particular person whose sales experience is \(1.5\) standard deviations below the average experience is predicted to have an annual sales value that is 1 standard deviation below the average annual sales, what is the value of \(r ?\)

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