Question

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 $$ \begin{aligned} \hat{y}=& 3.89+.033 x_{1}+.024 x_{2}+.023 x_{3} \\ &+.008 x_{4}-.13 x_{5}-.72 x_{6} \end{aligned} $$ where \(y=\) species richness, \(x_{1}=\) watershed area, \(x_{2}=\) shore width, \(x_{3}=\) drainage \((\%), x_{4}=\) water color (total color units), \(x_{5}=\) sand \((\%)\), and \(x_{6}=\) alkalinity. The coefficient of multiple determination was reported as \(R^{2}=.83\). Use a test with significance level \(.01\) to decide whether the chosen model is useful.

Short answer

Given that the \( R^2 = 0.83 \) is high and exceeds the 0.7 - 0.75 benchmark, the model can be considered as significantly useful for predicting species richness based on the given factors. However, a proper F-test would confirm this, which cannot be done here due to unavailability of needed data.

Step-by-step solution

Understand the Model

First, familiarize with the provided multiple linear regression model. Here, the species richness (\(y\)) is predicted based on the variables: watershed area (\(x_1\)), shore width (\(x_2\)), drainage (\(x_3\)), water color (\(x_4\)), sand content (\(x_5\)), and alkalinity (\(x_6\)). The model is considered significantly useful if it explains a significant amount of the variance in \(y\). This is measured by \(R^2\).

Null and Alternative Hypothesis

The null hypothesis (\(H_0\)) is that the regression model is not useful, meaning all regression coefficients except the constant are zero. The alternative hypothesis (\(H_a\)) is that at least one regression coefficient is not zero, making the model useful. Formally, these can be written as: \(H_0\): All \(\beta_i = 0\), except perhaps the constant (\(\beta_0\)). \(H_a\): At least one \(\beta_i ≠ 0\) where \(i = 1, 2, ..., 6\).

Test Statistic Calculation

In multiple linear regression, the usefulness of the model is often tested using the F-statistic. However, as the F-statistic value is not provided, it cannot be calculated here. Instead, the given coefficient of multiple determination (\(R^2\)) value can be used to gauge the fit of the model

Decision Rule

A common practice in stats is to consider a model significantly useful if more than 70-75% of the variance is explained, which is indicated by an \(R^2\) value larger than 0.7-0.75. Hence, with an \(R^2\)value of 0.83, it can be anticipated that model is significantly useful.

Conclusion

As the coefficient of multiple determination (\( R^2 = 0.83\)) is high and exceeds the 0.7 - 0.75 benchmark, the model can be considered as significantly useful for predicting species richness based on the given variables. However, a proper F-test should be done to formally test these hypotheses, which unfortunately cannot be done here due to unavailability of needed data.