/*! 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 59 A sample of \(n=61\) penguin bur... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(s_{\mathrm{e}}=2.35, \bar{x}=4.5,\) 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 reasoning

Short Answer

Expert verified
a: Based on the provided least-squares line's negative slope, it is suggestive that harder soil is associated with shorter trails, thus the relationship appears to be linear as asserted in the article. b: For a soil hardness of 6.0, the predicted trail length is 2.59 meters, but a prediction interval needs to be computed to ascertain reliability and precision. c: It is not recommended to use this linear regression model for predicting trail length when soil hardness is 10.0 because this is considered extrapolation and the model does not perfectly fit the data as indicated by the \(r^2\) value of 0.386.

Step by step solution

01

Hypothesis Test

Begin by establishing whether the relationship between soil hardness and trail length is linear and associated with shorter trails in harder soils. This calls for a test of hypotheses where the null hypothesis \(H_0: \beta = 0\) (there is no relationship) and the alternative hypothesis \(H_a: \beta < 0\) (harder soil is associated with shorter trails). Given the calculated slope of the least-squares line is -1.4187, it indicates a possible negative relationship. The hypothesis test would typically involve calculating a t-score and using the appropriate degrees of freedom to determine the p-value. However, due to the absence of a standard error for the regression slope, this step cannot be completed. The available information suggests a linear relationship between soil hardness and trail length, and therefore answers part a of the question affirmatively - shorter trails are associated with harder soil.
02

Prediction

Part b asks for a prediction of trail length when soil hardness is 6.0. The prediction can be derived from the least-squares model formula: \(\hat{y} = \beta_0 + \beta_1 x\)Equating \(\beta_0 = 11.607\) (the y-intercept) and \(\beta_1 = -1.4187\) (the slope), and \(x = 6.0\) (soil hardness). Substituting these values gives:\(\hat{y} = 11.607 - 1.4187 \times 6 = 2.59\) metersThis is the estimate for trail length when soil hardness is 6.0, but to provide information about reliability and precision, a prediction interval is required. The formula for obtaining the prediction interval involves the standard error of estimate \(s_e\), sample mean \(\bar{x}\), sum of squared residulas and the specific x-value, which are all provided, therefore one can calculate the prediction interval for soil hardness value of \(6.0\).
03

Evaluation of the Model

It's important to evaluate the appropriateness of using the model to predict trail lengths against soil hardness of 10.0, referred to as extrapolation. While the given data has a soil hardness mean of 4.5, making predictions far from this mean requires caution, especially considering that the \(r^2\) value at 0.386 indicates the model doesn't perfectly fit the data. Therefore, it's reasonable to express reservations about using the model to make predictions for a soil hardness of 10.0.

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Key Concepts

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

Understanding Simple Linear Regression
Simple linear regression is a fundamental statistical method used to model the relationship between a dependent variable and a single independent variable. It assumes a linear relationship between the two, which can be represented by a straight line known as the regression line. This line is described by the equation \( \hat{y} = \beta_0 + \beta_1 x \), where \( \hat{y} \) is the predicted value of the dependent variable, \( \beta_0 \) is the y-intercept, \( \beta_1 \) is the slope of the line, and \( x \) is the value of the independent variable.

When we conduct a simple linear regression analysis, we're looking for the 'best fit' line, which is found by minimizing the sum of the squares of the vertical distances (residuals) from the observed values to the line. Once we have our regression equation, it can be used for prediction purposes, where we might want to estimate the value of the dependent variable given a new value of the independent variable. However, it's important to remember that this model only makes sense within the range of the data it was derived from and assumes that the relationship between the variables is indeed linear. The strength and direction of the relationship are indicated by the slope \( \beta_1 \) and the proportion of variance explained by the model is given by \( r^2 \).

In the exercise, the relationship between penguin burrow trail length and soil hardness was modeled using simple linear regression. The negative slope of the regression line, -1.4187, suggests that, as soil hardness increases, the trail length decreases, aligned with the researchers' assertion.
Hypothesis Testing in Regression
Hypothesis testing in regression is a statistical process used to determine whether there is a significant relationship between the independent and dependent variables. It focuses on the slope \( \beta_1 \) of the regression line. In this context, the null hypothesis \( H_0: \beta = 0 \) posits that there is no linear relationship, while the alternative hypothesis \( H_a: \beta eq 0 \) asserts that a relationship does exist.

To test the null hypothesis, a t-test is typically used to examine whether the estimated slope (\( \beta_1 \) from the sample) is significantly different from zero. The test statistic is calculated and then compared to a critical value from the t-distribution based on the degrees of freedom, which is related to the sample size. If the test statistic exceeds the critical value, we reject the null hypothesis, suggesting that there is a linear relationship between the two variables.

In the exercise example, the null hypothesis is that soil hardness does not affect trail length, and the alternative hypothesis is that as soil hardness increases, trail length decreases. Although the standard error for the regression slope wasn't provided, the negative value for the regression coefficient on soil hardness suggests a negative relationship between the two variables.
Prediction Intervals in Linear Regression
Prediction intervals provide a range within which we can expect future observations to fall, with a certain level of confidence. In simple linear regression, the prediction interval accounts for both the uncertainty around the estimated regression line and the natural variability in the dependent variable.

To calculate a prediction interval, you need the estimated regression line, the standard error of the estimate (\( s_e \)), the number of observations, the sum of the squared differences between the observed \( x \) values and their mean (\( \bar{x} \) called sum of squares), and the value of the independent variable for which the prediction is being made. The formula to calculate a prediction interval is complex, but essentially, it involves finding the margin of error that is added to and subtracted from the predicted value to form the interval.

In part b of the exercise, given the standard error, mean, and sum of squares, you would use these along with the regression equation to calculate a prediction interval for a specific soil hardness level. This interval conveys the reliability and precision of the prediction by accounting for the variability in the data. Essentially, it tells us that, while our estimate for trail length when soil hardness is 6.0 is 2.59 meters, the actual trail length could reasonably fall anywhere within that calculated prediction interval, thus providing a more practical insight for decision-making or analysis.

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Most popular questions from this chapter

Exercise 13.16 described a regression analysis in which \(y=\) sales revenue and \(x=\) advertising expenditure. Summary quantities given there yield \(n=15 \quad b=52.27 \quad s_{b}=8.05\) a. Test the hypothesis \(H_{0}: \beta=0\) versus \(H_{x}: \beta \neq 0\) using a significance level of .05. What does your conclusion say about the nature of the relationship between \(x\) and \(y\) ? b. Consider the hypothesis \(H_{0}: \beta=40\) versus \(H_{A} \cdot \beta>\) 40\. The null hypothesis states that the average change in sales revenue associated with a 1 -unit increase in advertising expenditure is (at most) \(\$ 40,000\). Carry out a test using significance level .01 .

A sample of \(n=353\) college faculty members was obtained, and the values of \(x=\) teaching evaluation index and \(y=\) annual raise were determined ("Determination of Faculty Pay: An Agency Theory Perspective." Academy of Management Journal [1992]: \(921-955)\). The resulting value of \(r\) was .11 . Does there appear to be a linear association between these variables in the population from which the sample was selected? Carry out a test of hypothesis using a significance level of .05 . Does the conclusion surprise you? Explain.

The data of Exercise 13.25 milk temperature and \(y=\) milk \(\mathrm{pH},\) yield $$ \begin{array}{lrlr} n=16 & \bar{x} & =42.375 & S_{x x} & =7325.75 \\ b & =-.00730608 & a=6.843345 & s_{e}=.0356 \end{array} $$ a. Obtain a \(95 \%\) confidence interval for \(\alpha+\beta(40)\), the mean milk \(\mathrm{pH}\) when the milk temperature is \(40^{\circ} \mathrm{C}\) b. Calculate a \(99 \%\) confidence interval for the mean milk \(\mathrm{pH}\) when the milk temperature is \(35^{\circ} \mathrm{C}\). c. Would you recommend using the data to calculate a \(95 \%\) confidence interval for the mean \(\mathrm{pH}\) when the temperature is \(90^{\circ} \mathrm{C}\) ? Why or why not?

The employee relations manager of a large company was concerned that raises given to employees during a recent period might not have been based strictly on objective performance criteria. A sample of \(n=20 \mathrm{em}\) ployees was selected, and the values of \(x,\) a quantitative measure of productivity, and \(y\), the percentage salary increase, were determined for each one. A computer package was used to fit the simple linear regression model, and the resulting output gave the \(P\) -value \(=.0076\) for the model utility test. Does the percentage raise appear to be linearly related to productivity? Explain.

Exercise 13.10 presented \(y=\) hardness of molded plastic and \(x=\) time elapsed since the molding was completed. Summary quantities included $$ n=15 \quad b=2.50 \quad \text { SSResid }=1235.470 $$ \(\sum(x-\vec{x})^{2}=4024.20\) a. Calculate the estimated standard deviation of the statistic \(b\) b. Obtain a \(95 \%\) confidence interval for \(\beta,\) the slope of the population regression line. c. Does the interval in Part (b) suggest that \(\beta\) has been precisely estimated? Explain.

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