/*! 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 63 A sample of \(n=61\) penguin bur... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 63

A sample of \(n=61\) penguin burrows was selected, and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one ("Effects of Substrate on the Distribution of Magellanic Penguin Burrows," The Auk [1991]: 923-933). The equation of the least-squares line was \(\hat{y}=11.607-\) \(1.4187 x\), and \(r^{2}=.386\). a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{e}=2.35, \bar{x}=4.5\), and \(\sum(x-\bar{x})^{2}=250\), predict trail length when soil hardness is \(6.0\) in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0\) ? Explain your

Short Answer

Expert verified
a. A t-test should be conducted to test the hypothesis that there is a linear relationship between the variables. b. The predicted trail length of a 6.0 hardness soil is obtained by substituting x=6 into the equation of the line, and the precision of the estimate can be gleaned from the standard error \(s_{e}\). c. Without knowing the range of hardness values in the dataset, it is difficult to definitively say whether the linear regression model is an appropriate model to use, as using it for values outside the dataset's range would involve extrapolation, which may not provide reliable predictions.

Step by step solution

01

- Hypothesis test

Given that the value of \(r^{2}=.386\), it tells us that approximately 38.6% of the variation in trail length is explained by its linear relationship with soil hardness. However, to carry out an appropriate test for the hypothesis that there is indeed a linear relationship, we should conduct a t-test. Here, the null hypothesis is that there is no relationship (i.e., the slope of the regression line equals 0), and the alternate hypothesis is that there is a relationship (the slope is not 0).
02

- Prediction

The equation of the least-squares line has been given as \(\hat{y}=11.607-1.4187x\), with \(s_{e}=2.35\), average soil hardness \(\bar{x}=4.5\), and the total variation of the soil hardness, \(\sum(x-\bar{x})^{2}=250\). To predict the trail length when soil hardness is 6.0, we use the given equation: \(\hat{y}=11.607-1.4187*(6)\). The standard error of the estimate shows the precision of the prediction. The smaller the standard error, the more confident we can be in our prediction.
03

- Model Suitability

To decide whether to use the linear regression model to predict trail length when hardness is 10.0, we need to consider whether hardness at this level falls within the range of hardness values in our dataset. If it does, it is reasonable to use the model. If it does not, caution should be used as the model may not make reliable predictions outside the range of the dataset. This is known as extrapolation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Least-Squares Line
The least-squares line, also known as the line of best fit, plays a crucial role in linear regression analysis. It's designed to minimize the sum of the squares of the vertical distances of the points from the line, hence the name 'least-squares'. In basic terms, it's the straight line that best represents the data on a scatter plot.

Let's consider an example related to penguin burrows. The linear equation given by \(\hat{y} = 11.607 - 1.4187x\) represents the least-squares line for the relationship between soil hardness (x) and trail length (y). This equation implies that for every unit increase in soil hardness, the trail length decreases by approximately 1.4187 meters. The negative slope indicates an inverse relationship; as soil hardness increases, the trail length tends to decrease.

In our specific context, the least-squares line is crucial for making predictions and interpreting the strength and direction of the relationship between trail length and soil hardness.
Hypothesis Testing in Regression
Hypothesis testing in regression analysis is a statistical method used to determine if there is a significant relationship between two variables. This process usually involves setting up two hypotheses: the null hypothesis \(H_0\), which proposes no effect or no relationship, and the alternative hypothesis \(H_1\) or \(H_a\), which suggests there is an effect or a relationship.

In the case of the penguin burrows study, the null hypothesis states that the slope of the regression line is zero, indicating no relationship between soil hardness and trail length. The alternative hypothesis suggests that the slope is not zero, thus implying a significant linear relationship. To determine the validity of these hypotheses, a t-test can be employed using the given coefficient of determination \(r^2 = 0.386\) and other values from the sample. This t-test assesses whether the observed relationship is likely to have occurred by chance, or if it's statistically significant.
Coefficient of Determination
The coefficient of determination, denoted as \(r^2\), is a key statistic in regression that measures the proportion of variability in the dependent variable that can be explained by the independent variable. The value of \(r^2\) ranges from 0 to 1, where 0 indicates no explanatory power and 1 indicates perfect explanation.

In our exercise, \(r^2\) is reported to be 0.386. This means that approximately 38.6% of the variation in trail length can be attributed to its linear relationship with soil hardness. Higher \(r^2\) values would show a stronger linear relationship between the variables. Knowing this value helps in understanding the strength of the model and in making more informed decisions when predicting new data points.
Standard Error in Regression
The standard error (SE) in regression analysis quantifies the amount of variability in the estimate of the regression coefficient or the prediction. It is a measure of the precision of the regression estimate: a smaller SE indicates more precise estimates.

For the computation of trail length, the given standard error is \(s_e = 2.35\). This suggests that the predicted trail lengths for the penguins' burrows are expected to vary from the least-squares line by an average of about 2.35 meters. The standard error plays an integral role when forming prediction intervals or when conducting hypothesis tests on regression coefficients, as it helps to quantify the uncertainty around these estimates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Occasionally an investigator may wish to compute a confidence interval for \(\alpha\), the \(y\) intercept of the true regression line, or test hypotheses about \(\alpha .\) The estimated \(y\) intercept is simply the height of the estimated line when \(x=0\), since \(a+b(0)=a .\) This implies that \(s_{a}\) the estimated standard deviation of the statistic \(a\), results from substituting \(x^{\prime \prime}=0\) in the formula for \(s_{a+b x^{+}} .\) The desired confidence interval is then \(a \pm(t\) critical value \() s_{a}\) \(-\) and a test statistic is 1 $$ t=\frac{a-\text { hypothesized value }}{s_{a}} $$ a. The article "Comparison of Winter-Nocturnal Geostationary Satellite Infrared-Surface Temperature with Shelter-Height Temperature in Florida" (Remote Sensing of the Emvironment \([1983]: 313-327\) ) used the simple linear regression model to relate surface temperature as measured by a satellite \((y)\) to actual air temperature \((x)\) as determined from a thermocouple placed on a traversing vehicle. Selected data are given (read from a scatterplot in the article). \(\begin{array}{rrrrrrrr}x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ y & -3.9 & -2.1 & -2.0 & -1.2 & 0.0 & 1.9 & 0.6 \\ x & 5 & 6 & 7 & & & & \\ y & 2.1 & 1.2 & 3.0 & & & & \end{array}\) Estimate the true regression line. b. Compute the estimated standard deviation \(s_{a}\). Carry out a test at level of significance \(.05\) to see whether the \(y\) intercept of the true regression line differs from zero. c. Compute a \(95 \%\) confidence interval for \(\alpha\). Does the result indicate that \(\alpha=0\) is plausible? Explain.

Exercise \(5.48\) described a regression situation in which \(y=\) hardness of molded plastic and \(x=\) amount of time elapsed since termination of the molding process. Summary quantities included \(n=15\), SSResid \(=\) \(1235.470\), and SSTo \(=25,321.368\). a. Calculate a point estimate of \(\sigma\). On how many degrees of freedom is the estimate based? b. What percentage of observed variation in hardness can be explained by the simple linear regression model relationship between hardness and elapsed time?

An investigation of the relationship between traffic flow \(x\) (thousands of cars per \(24 \mathrm{hr}\) ) and lead content \(y\) of bark on trees near the highway (mg/g dry weight) yielded the accompanying data. A simple linear regression model was fit, and the resulting estimated regression line was \(\hat{y}=28.7+33.3 x .\) Both residuals and standardized residuals are also given. \(\begin{array}{lrrrrr}\text { iduals are also given. } & & & & \\ x & 8.3 & 8.3 & 12.1 & 12.1 & 17.0 \\ y & 227 & 312 & 362 & 521 & 640 \\ \text { Residual } & -78.1 & 6.9 & -69.6 & 89.4 & 45.3 \\ \text { St. resid. } & -0.99 & 0.09 & -0.81 & 1.04 & 0.51\end{array}\) \(\begin{array}{lrrrrr}x & 17.0 & 17.0 & 24.3 & 24.3 & 24.3 \\ y & 539 & 728 & 945 & 738 & 759 \\ \text { Residual } & -55.7 & 133.3 & 107.2 & -99.8 & -78.8 \\\ \text { St. resid. } & -0.63 & 1.51 & 1.35 & -1.25 & -0.99\end{array}\) a. Plot the \((x\), residual \()\) pairs. Does the resulting plot suggest that a simple linear regression model is an appropriate choice? Explain your reasoning. b. Construct a standardized residual plot. Does the plot differ significantly in general appearance from the plot in Part (a)?

Let \(x\) be the size of a house (sq \(\mathrm{ft}\) ) and \(y\) be the amoun of natural gas used (therms) during a specified period. Suppose that for a particular community, \(x\) and \(y\) are related according to the simple linear regression model with \(\beta=\) slope of population regression line \(=.017\) \(\alpha=y\) intercept of population regression line \(=-5.0\) a. What is the equation of the population regression line? b. Graph the population regression line by first finding the point on the line corresponding to \(x=1000\) and then the point corresponding to \(x=2000\), and drawing a line through these points. c. What is the mean value of gas usage for houses with 2100 sq ft of space? d. What is the average change in usage associated with a 1 -sq-ft increase in size? e. What is the average change in usage associated with a 100 -sq-ft increase in size? f. Would you use the model to predict mean usage for a 500 -sq-ft house? Why or why not? (Note: There are no small houses in the community in which this model is valid.)

The accompanying data on \(x=\) U.S. population (millions) and \(y=\) crime index (millions) appeared in the article "The Normal Distribution of Crime" (Journal of Police Science and Administration \([1975]: 312-318)\). The author comments that "The simple linear regression analysis remains one of the most useful tools for crime prediction." When observations are made sequentially in time, the residuals or standardized residuals should be plotted in time order (that is, first the one for time \(t=1\) ( 1963 here), then the one for time \(t=2\), and so on ). Notice that here \(x\) increases with time, so an equivalent plot is of residuals or standardized residuals versus \(x\). Using \(\hat{y}=47.26+.260 x\), calculate the residuals and plot the \((x\), residual) pairs. Does the plot exhibit a pattern that casts doubt on the appropriateness of the simple linear regression model? Explain. \(\begin{array}{lrrrrrr}\text { Year } & 1963 & 1964 & 1965 & 1966 & 1967 & 1968 \\ x & 188.5 & 191.3 & 193.8 & 195.9 & 197.9 & 199.9 \\ y & 2.26 & 2.60 & 2.78 & 3.24 & 3.80 & 4.47 \\\ \text { Year } & 1969 & 1970 & 1971 & 1972 & 1973 & \\ x & 201.9 & 203.2 & 206.3 & 208.2 & 209.9 & \\ y & 4.99 & 5.57 & 6.00 & 5.89 & 8.64 & \end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.