/*! 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 41 The ability of ecologists to ide... [FREE SOLUTION] | 91Ó°ÊÓ

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The ability of ecologists to identify regions of greatest species richness could have an impact on the preservation of genetic diversity, a major objective of the World Conservation Strategy. The article "Prediction of Rarities from Habitat Variables: Coastal Plain Plants on Nova Scotian Lakeshores" (Ecology, 1992: 1852-1859) used a sample of \(n=37\) lakes to obtain the estimated regression equation $$ \begin{aligned} y=3.89+.033 x_{1}+.024 x_{2}+.023 x_{3} \\ &-.0080 x_{4}-.13 x_{5}-.72 x_{6} \end{aligned} $$ where \(y=\) species richness, \(x_{1}=\) watershed area, \(x_{2}=\) shore width, \(x_{3}=\) poor drainage \((\%), x_{4}=\) water color (total color units), \(x_{5}=\) sand (\%), and \(x_{6}=\) alkalinity. The coefficient of multiple determination was reported as \(R^{2}=.83\). Carry out a test of model utility.

Short Answer

Expert verified
Reject the null hypothesis; the model is useful.

Step by step solution

01

State the Hypotheses

In regression, we test the null hypothesis that none of the predictor variables have any effect on the response variable. Formally, the null and alternative hypotheses are:- Null Hypothesis (H_0): \(\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5 = \beta_6 = 0\)- Alternative Hypothesis (H_a): At least one \(\beta_i eq 0\)
02

Calculate the Test Statistic

The test statistic for the utility of a regression model is based on the F-test. The formula is:\[F = \frac{(R^2 / k)}{((1 - R^2) / (n - k - 1))}\]where \(R^2 = 0.83\), \(k = 6\) (number of predictors), and \(n = 37\).Substituting the values:\[F = \frac{(0.83 / 6)}{((1 - 0.83) / (37 - 6 - 1))}\]
03

Compute the F-value

Calculate the value of the F-statistic:\[F = \frac{0.1383}{0.0066} \approx 20.95\]
04

Determine the Critical F-value

To find the critical F-value, use \(\alpha = 0.05\) and the F-distribution table with \(df_1 = k = 6\) and \(df_2 = n - k - 1 = 30\). For these degrees of freedom, the critical F-value is approximately 2.49.
05

Compare F-values and Decision Making

Compare the calculated F-value (20.95) with the critical F-value (2.49). Since 20.95 > 2.49, we reject the null hypothesis.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method that helps us make decisions based on data. In the context of multiple linear regression, we often want to understand if the independent variables (predictors) have any true effect on the dependent variable (response). For this, we start by setting up two statements:
  • **Null Hypothesis (\(H_0\)):** Assumes that none of the predictors affects the response variable, meaning their coefficients are all zero (\(\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5 = \beta_6 = 0\)).
  • **Alternative Hypothesis (\(H_a\)):** Suggests that at least one predictor has a non-zero effect (at least one \(\beta_i eq 0\)).
By rejecting or failing to reject the null hypothesis, we confirm or dismiss the utility of the regression model. Hypothesis testing ensures the robustness and usefulness of our models by statistically validating the relationships we observe.
Regression Equation
A regression equation models the relationship between a dependent variable and one or more independent variables. In multiple linear regression, this equation takes the form:\[ y = \beta_0 + \beta_1x_1 + \beta_2x_2 + \cdots + \beta_kx_k \]where \(y\) denotes the dependent variable, \(\beta_0\) is the intercept, and \(\beta_1x_1, \beta_2x_2, \ldots , \beta_kx_k\) are the effects of the predictors.In this exercise, species richness is modeled as a function of various lakeshore properties such as watershed area and shore width, among others. The given equation:\[ y = 3.89 + 0.033x_{1} + 0.024x_{2} + 0.023x_{3} - 0.0080x_{4} - 0.13x_{5} - 0.72x_{6} \]depicts the combined effect of these variables on species richness. Each coefficient represents the change in species richness per unit change in that particular variable, holding all others constant. This equation helps predict species diversity based on measurable environmental factors.
F-test
The F-test plays a crucial role in hypothesis testing for regression analysis. It tells us if there is a significant relationship between the dependent variable and the set of independent variables.In our case, we use the F-statistic to test the null hypothesis that all regression coefficients are zero. The F-statistic formula is:\[ F = \frac{(R^2 / k)}{((1 - R^2) / (n - k - 1))} \]where:
  • \(R^2\): Coefficient of determination.
  • \(k\): Number of predictors.
  • \(n\): Number of observations.
Substituting the given values, our F-statistic was approximately 20.95. We compare this against a critical F-value (2.49 at a 0.05 significance level) from statistical tables. Since the calculated F-value is greater, we reject the null hypothesis, confirming that at least one of our predictors is statistically significant in explaining the variation in species richness.
Coefficient of Determination
The coefficient of determination, denoted as $R^2$ , measures how well the regression model explains the variability of the dependent variable. In simple terms, it tells us what percentage of the total variation in the response variable is accounted for by the predictors in the model. For instance, an $R^2$ value of 0.83 means that 83% of the variation in species richness is explained by the different habitat variables like watershed area and water color. An $R^2$ value close to 1 indicates a strong relationship between the predictors and the response variable, suggesting that the model fits the data well. However, it's important to note that a high $R^2$ does not imply causation but simply indicates a good fit. Hence, we use it along with other statistical tests, like the F-test, to validate the model's utility effectively.

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Most popular questions from this chapter

High-alumina refractory castables have been extensively investigated in recent years because of their significant advantages over other refractory brick of the same classlower production and application costs, versatility, and performance at high temperatures. The accompanying data on \(x=\) viscosity \((\mathrm{MPa} \cdot \mathrm{s})\) and \(y=\) free-flow \((\%)\) was read from a graph in the article "Processing of Zero-Cement Self-Flow Alumina Castables" (The Amer. Ceramic Soc. Bull., 1998: 60-66): \begin{tabular}{c|ccccccc} \(x\) & 351 & 367 & 373 & 400 & 402 & 456 & 484 \\ \hline\(y\) & 81 & 83 & 79 & 75 & 70 & 43 & 22 \end{tabular} The authors of the cited paper related these two variables using a quadratic regression model. The estimated regression function is \(y=-295.96+2.1885 x-.0031662 x^{2}\). a. Compute the predicted values and residuals, and then SSE and \(s^{2}\). b. Compute and interpret the coefficient of multiple determination. c. The estimated SD of \(\hat{\beta}_{2}\) is \(s_{\hat{\beta}_{3}}=.0004835\). Does the quadratic predictor belong in the regression model? d. The estimated SD of \(\hat{\beta}_{1}\) is \(.4050\). Use this and the information in (c) to obtain joint CIs for the linear and quadratic regression coefficients with a joint confidence level of (at least) \(95 \%\). e. The estimated SD of \(\hat{\mu}_{y, 200}\) is 1.198. Calculate a 95\% CI for true average free-flow when viscosity \(=400\) and also a \(95 \%\) PI for free- flow resulting from a single observation made when viscosity \(=400\), and compare the intervals.

An investigation of a die casting process resulted in the accompanying data on \(x_{1}=\) furnace temperature, \(x_{2}=\) die close time, and \(y=\) temperature difference on the die surface ("A Multiple-Objective Decision-Making Approach for Assessing Simultaneous Improvement in Die Life and Casting Quality in a Die Casting Process," Quality Engineering, 1994: 371-383). a. Carry out the model utility test. b. Calculate and interpret a \(95 \%\) confidence interval for \(\beta_{2}\), the population regression coefficient of \(x_{2}\). c. When \(x_{1}=1300\) and \(x_{2}=7\), the estimated standard deviation of \(\hat{y}\) is \(s_{\bar{Y}}=.353\). Calculate a \(95 \%\) confidence interval for true average temperature difference when furnace temperature is 1300 and die close time is \(7 .\) d. Calculate a \(95 \%\) prediction interval for the temperature difference resulting from a single experimental run with a furnace temperature of 1300 and a die close time of 7 .

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The article -Creep and Fatigue Characteristics of Ferrocement Slabs" (U. Ferrvcement, 1984: 399-322) reported data on \(y=\) tensile strength (MPa), \(x_{1}=\) slab thickness \((c m), x_{2}=\) load (kg), \(x_{3}=\) age at loading (days), and \(x_{4}=\) time under test (days) resulting from stress tests of \(n=9\) reinforced concrete slabs. The results of applying the BE elimination method of variable selection are summarized in the accompanying tabular format. Explain what occurred at each step of the procedure. \begin{tabular}{lccc} Step & 1 & 2 & 3 \\ Constant & \(8.496\) & \(12.670\) & \(12.989\) \\ \hline\(x_{1}\) & \(-.29\) & \(-.42\) & \(-.49\) \\ T-RATIO & \(-1.33\) & \(-2.89\) & \(-3.14\) \\ \(x_{2}\) & \(.0104\) & \(.0110\) & \(.0116\) \\ T-RATIO & \(6.30\) & \(7.40\) & \(7.33\) \\ \(x_{3}\) & \(.00159\) & & \\ T-RATIO & \(.83\) & & \\ \(x_{4}\) & \(-.023\) & \(-.023\) & \(.570\) \\ T-RATIO & \(-1.48\) & \(-1.53\) & \(92.82\) \\ 5 & 533 & \(.516\) & \\ R-5Q & \(95.81\) & \(95.10\) & \\ \hline \end{tabular}

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