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in Fish In Exercise 9.21 , we see that the conditions are met for using the \(\mathrm{pH}\) of a lake in Florida to predict the mercury level of fish in the lake. The data are given in FloridaLakes. Computer output is shown for the linear model with several values missing: The regression equation is AvgMercury \(=1.53-0.152 \mathrm{pH}\) Predictor Constant \(\mathrm{pH}\) \(\begin{array}{rrrr}\text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ 1.5309 & 0.2035 & 7.52 & 0.000 \\ -0.15230 & \text { * (c) }^{* *} & -5.02 & 0.000\end{array}\) stant \(S=* *(b)^{* *} \quad R-S q=* *(a)^{* x}\) Analysis of Variance Source Regression Residual Erri Total \(\begin{array}{rrrrrr} & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { ion } & 1 & 2.0024 & 2.0024 & 25.24 & 0.000 \\ \text { Error } & 51 & 4.0455 & 0.0793 & & \\ & 52 & 6.0479 & & & \end{array}\) (a) Use the information in the ANOVA table to compute and interpret the value of \(R^{2}\). (b) Show how to estimate the standard deviation of the error term, \(s_{\epsilon}\). (c) Use the result from part (b) and the summary statistics below to compute the standard error of the slope, \(S E,\) for this model: $$ \begin{array}{lrrrrr} \text { Variable } & \text { N } & \text { Mean } & \text { StDev } & \text { Minimum } & \text { Maximum } \\ \text { pH } & 53 & 6.591 & 1.288 & 3.600 & 9.100 \\ \text { AvgMercury } & 53 & 0.5272 & 0.3410 & 0.0400 & 1.3300 \end{array} $$

Short Answer

Expert verified
To get the necessary answers, you follow three main steps. Firstly, compute \(R^{2}\) which is approximately 0.3315. Secondly, calculate the standard deviation of the error term, \(s_{\epsilon}\) which is found to be approximately 0.2816. Lastly, compute the standard error of the slope, which is found to be approximately 0.0303

Step by step solution

01

Compute \(R^{2}\)

To compute the big R-squared (\(R^{2}\)), we use the formula \[R^{2} = \frac {SS_{Regression}} {SS_{Total}}\] Here, \(SS_{Regression}\) is 2.0024 and \(SS_{Total}\) is 6.0479. Substitute these values into the formula and calculate \(R^{2}\).
02

Estimate the standard deviation of the error term, \(s_{\epsilon}\)

\[s_{\epsilon}\] is calculated as the squareroot of the Mean Square Error (MSE). The MSE is obtained from ANOVA table and equals 0.0793. Calculate s by taking the squareroot of the Mean Square Error.
03

Compute the standard error of the slope (SE)

The formula to calculate the standard error of the slope is \[SE_{slope} = \frac{s_{\epsilon}}{\sqrt{SSX}}\] where \(s_{\epsilon}\) is the standard deviation of the error term found in step 2, and \(SSX\) is the sum of squares of the differences from the mean value of pH. The latter, \(SSX\), is calculated as follows: \[\sum{(x_i - \overline{x})^2}\]where \(x_i\) is each individual value of pH, and \(\overline{x}\) is the mean of pH. Since we are missing individual values for pH, we make use of the fact that \(SSX = (n-1) \times \text{{StDev}}^2\), where n is the number of observations of pH, and StDev is the standard deviation of pH. Substitutes these values into the formula for SE_{slope} and calculate it.

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Key Concepts

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

Understanding R-Squared (R^2) in Linear Regression Analysis
The metric R-Squared, symbolized as R2, is a statistical measure in linear regression that determines the proportion of the variation in the dependent variable that can be explained by the independent variable(s). Imagine you are trying to predict the weight of a group of people based on their heights. An R2 close to 1 indicates that the regression model explains a large portion of the variation in weight, while an R2 near 0 suggests that height provides little information about weight in this scenario.

It acts as a goodness-of-fit measure, with higher values usually indicating a better model, but beware - it doesn't mean the model is the right one! It's possible for a model with a high R2 to be misleading, especially if it's based on a nonsensical relationship or if we've just got too many predictors throwing in their two cents.

To calculate R2, you take the sum of squares due to regression (SSRegression) divided by the total sum of squares (SSTotal). It's like figuring out what percentage of your pizza has been eaten by comparing the eaten slices (SSRegression) to the whole pie (SSTotal).
Deciphering the Standard Deviation of the Error Term (s_epsilon)
The standard deviation of error term, denoted as sε, is the estimate of the variability or dispersion of the observed values around the regression line. Think of throwing darts at a dartboard - sε tells you how spread out those darts are from the bullseye, which represents the predictions made by your regression model.

To get this value, we take the square root of the Mean Square Error (MSE) from the ANOVA table. MSE is like an average of the squares of the 'darts' distances from the bullseye. Actually calculating sε is like measuring the average distance itself, not just the squared distance, making it easier to understand and more tangible, a bit like comparing apples to apples instead of apples to apple squares. It's a crucial component for assessing model performance and for calculating other important statistics such as SEslope.
Standard Error of the Slope (SE_slope): Nailing Down the Precision
The standard error of the slope, or SEslope, measures the precision of the estimate of the slope coefficient in your regression model. In simpler terms, it tells you how accurate your estimation of the relationship between the independent and dependent variable is - are you just in the ballpark, or did you hit it out of the park?

An easy way to think about SEslope is like this: If you recorded the heights and shoe sizes of everyone in your class to find out the general trend, SEslope would tell you how confident you can be about the strength of the connection between those two measurements. Small values of SEslope suggest a more precise estimate, meaning you can be pretty confident about the trend you've observed. To calculate SEslope, you need the standard deviation of the error term (sε) and the total sum of squares for the independent variable, which is like the variability of the class heights in this metaphor.
ANOVA Table in Linear Regression: Breaking Down Variability
The ANOVA (Analysis of Variance) table in the context of linear regression analysis breaks down the variability in the data into components that can help us assess how well our regression model works. It's like checking which ingredients in a recipe are making your cake rise.

The ANOVA table typically includes figures like the degrees of freedom (DF), sum of squares (SS), mean squares (MS), and an F-statistic. Here's a quick bite of what those ingredients are serving up:
  • DF: Like telling you how many independent pieces of information went into making your cake.
  • SS: The sum of squares is like quantifying how much sugar and flour you've added to the mix - for regression, how much the model and the errors contribute to the total variability.
  • MS: Mean Squares are like an average - if you divided all that sugar and flour evenly into every bite of your cake. For our analysis, MS is SS divided by DF.
  • F-statistic: It compares the model to a scenario where there's no relationship (plain cake with no rising agent). A high F-statistic means your model adds significant lift, just like baking powder in the cake.
Understanding the ANOVA table arms you with the info to say how much faith to put in your regression model's ability to predict outcomes - like knowing if that cake will be a showstopper.
Sum of Squares (SS): Measuring Variability in Data
Sum of Squares (SS) is a measure used to describe how much the values in a data set vary around their mean. It's like measuring the roughness of a road by checking out how many potholes there are and how deep they go.

In linear regression, we deal with two main types of sum of squares:
  • The SSRegression represents how much variance in your dependent variable can be explained by the model. It's like showing how many of those bumps in the road are because of potholes versus something else.
  • The SSResidual (or Error) measures the variance that the model fails to explain, which is the remaining bumpiness after you've patched up the potholes.
  • The SSTotal is the overarching measurement of variability, or in our road analogy, the total bumpiness including both potholes and other flaws in the road.
The SS is a foundational figure in many statistical tests and procedures, such as ANOVA, allowing you to judge whether the potholes (your model's predictions) are the main reason for the road's condition, or if other factors are also at play.

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Most popular questions from this chapter

A common (and hotly debated) saying among sports fans is "Defense wins championships." Is offensive scoring ability or defensive stinginess a better indicator of a team's success? To investigate this question we'll use data from the \(2015-2016\) National Basketball Association (NBA) regular season. The data \(^{6}\) stored in NBAStandings2016 include each team's record (wins, losses, and winning percentage) along with the average number of points the team scored per game (PtsFor) and average number of points scored against them ( PtsAgainst). (a) Examine scatterplots for predicting \(\operatorname{WinPct}\) using PtsFor and predicting WinPct using PtsAgainst. In each case, discuss whether conditions for fitting a linear model appear to be met. (b) Fit a model to predict winning percentage (WinPct) using offensive ability (PtsFor). Write down the prediction equation and comment on whether PtsFor is an effective predictor. (c) Repeat the process of part (b) using PtsAgainst as the predictor. (d) Compare and interpret \(R^{2}\) for both models. (e) The Golden State Warriors set an NBA record by winning 73 games in the regular season and only losing 9 (WinPct \(=0.890\) ). They scored an average of 114.9 points per game while giving up an average of 104.1 points against. Find the predicted winning percentage for the Warriors using each of the models in (b) and (c). (f) Overall, does one of the predictors, PtsFor or PtsAgainst, appear to be more effective at explaining winning percentages for NBA teams? Give some justification for your answer.

How well does a student's Verbal SAT score (on an 800 -point scale) predict future college grade point average (on a four-point scale)? Computer output for this regression analysis is shown, using the data in StudentSurvey: The regression equation is \(\mathrm{GPA}=2.03+0.00189\) VerbalSAT Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 6.8029 & 6.8029 & 48.84 & 0.000 \\ \text { Residual Error } & 343 & 47.7760 & 0.1393 & & \\ \text { Total } & 344 & 54.5788 & & & \end{array}\) (a) What is the predicted grade point average of a student who receives a 550 on the Verbal SAT exam? (b) Use the information in the ANOVA table to determine the number of students included in the dataset. (c) Use the information in the ANOVA table to compute and interpret \(R^{2}\). (d) Is the linear model effective at predicting grade point average? Use information from the computer output and state the conclusion in context.

In Exercises 9.1 to \(9.4,\) use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{aligned} &\text { The regression equation is } Y=29.3+4.30 \mathrm{X}\\\ &\begin{array}{lrrrr} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 29.266 & 6.324 & 4.63 & 0.000 \\ \text { X } & 4.2969 & 0.6473 & 6.64 & 0.000 \end{array} \end{aligned} $$

Exercise A .97 on page 189 , we introduce a study about mating activity of water striders. The dataset is available as WaterStriders and includes the variables FemalesHiding, which gives the proportion of time the female water striders were in hiding, and MatingActivity, which is a measure of mean mating activity with higher numbers meaning more mating. The study included 10 groups of water striders. (The study also included an examination of the effect of hyper-aggressive males and concludes that if a male wants mating success, he should not hang out with hyper-aggressive males.) Computer output for a model to predict mating activity based on the proportion of time females are in hiding is shown below, and a scatterplot of the data with the least squares line is shown in Figure 9.12 . The regression equation is MatingActivity \(=0.480-0.323\) FemalesHiding \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 0.48014 & 0.04213 & 11.40 & 0.000 \\ \text { FemalesHiding } & -0.3232 & 0.1260 & -2.56 & 0.033\end{array}\) \(\begin{array}{lll}S=0.101312 & \text { R-Sq }=45.1 \% & \text { R-Sq(adj) }=38.3 \%\end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 0.06749 & 0.06749 & 6.58 & 0.033 \\ \text { Residual Error } & 8 & 0.08211 & 0.01026 & & \\ \text { Total } & 9 & 0.14960 & & & \end{array}\) (a) While it is hard to tell with only \(n=10\) data points, determine whether we should have any serious concerns about the conditions for fitting a linear model to these data. (b) Write down the equation of the least squares line and use it to predict the mating activity of water striders in a group in which females spend \(50 \%\) of the time in hiding (FemalesHiding = 0.50) (c) Give the hypotheses, t-statistic, p-value, and conclusion of the t-test of the slope to determine whether time in hiding is an effective predictor of mating activity. (d) Give the hypotheses, F-statistic, p-value, and conclusion of the ANOVA test to determine whether the regression model is effective at predicting mating activity. (e) How do the two p-values from parts (c) and (d) compare? (f) Interpret \(R^{2}\) for this model.

Hantavirus is carried by wild rodents and causes severe lung disease in humans. A study \(^{5}\) on the California Channel Islands found that increased prevalence of the virus was linked with greater precipitation. The study adds "Precipitation accounted for \(79 \%\) of the variation in prevalence." (a) What notation or terminology do we use for the value \(79 \%\) in this context? (b) What is the response variable? What is the explanatory variable? (c) What is the correlation between the two variables?

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