Question

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.

Short answer

a) Based on limited information, there are no immediate concerns about fitting a linear model. b) The least squares line equation is MatingActivity \(= 0.480 - 0.323\) * FemalesHiding and the predicted mating activity when females spend 50% of the time in hiding is 0.3185. c) Hypotheses; H0: Beta = 0, H1: Beta ≠ 0. T-statistic is -2.56 and p-value is 0.033. Based on p-value < 0.05, the time in hiding is an effective predictor of mating activity. d) Hypotheses; H0: All Betas = 0, H1: At least one Beta ≠ 0. F-statistic is 6.58 and p-value is 0.033. Based on p-value < 0.05, the regression model is effective at predicting mating activity. e) The p-values from parts c and d are both 0.033 — identical because it's a simple linear regression with one predictor variable. f) \(R^{2}\) is 45.1% or in other words, 45.1% of the variation in mating activity can be explained by the linear regression model.

Step-by-step solution

The Condition for Fitting a Linear Model

Since no information is provided in the exercise regarding non-linearity, outliers, non-constant variation, or high influence observations, the conditions for fitting the linear model seem satisfactory. This question requires more information to make a definitive conclusion.

Equation of the Least Squares Line and Predicting Mating Activity

The regression equation given is MatingActivity \(= 0.480 - 0.323\) * FemalesHiding. Using this equation and substituting FemalesHiding with 0.5 gives an estimate for the MatingActivity = \(0.480 - 0.323 * 0.5 = 0.3185\). Hence, the predicted mating activity is 0.3185 when females spend 50% of the time in hiding.

Hypotheses, t-statistic, p-value, Conclusion of the t-test of the Slope

Null Hypothesis (H0): The slope parameter (Beta) is zero, which would mean that time in hiding is not an effective predictor of mating activity (Beta = 0). Alternative Hypothesis (H1): The slope parameter (Beta) is not zero (Beta ≠ 0). The provided output gives t-statistic of -2.56 and a p-value of 0.033. Given that the p-value is less than 0.05, the null hypothesis can be rejected with a conclusion that the time in hiding is an effective predictor of mating activity.

Hypotheses, F-statistic, p-value, Conclusion of the ANOVA test

Null Hypothesis (H0): All regression coefficients (Beta) are zero, which would mean that the regression model is not effective at predicting mating activity. The alternative hypothesis (H1): At least one Beta ≠ 0. The output provides: F-statistic of 6.58 and a p-value of 0.033, given the p-value < 0.05, the null hypothesis can be rejected. Thus the regression model is effective at predicting mating activity.

Comparing the Two P-values

The p-values from the t-test for the slope (part c) and the F-test from the ANOVA (part d) are both 0.033 — they are identical. This is expected in a simple linear regression model where there's only one predictor variable.

Interpretation of \(R^{2}\)

\(R^{2}\) (coefficient of determination) for this model is given as 45.1%. This implies that approximately 45.1% of the variation in mating activity is explained by the linear model considering female water striders' time in hiding.

Key concepts

T-test for Slope

In linear regression analysis, the t-test for the slope is essential for understanding the relationship between the independent (predictor) and dependent (response) variables. By testing the slope, we aim to ascertain whether changes in the predictor variable have a statistically significant effect on the response variable. The null hypothesis (\(H_0\)) for this test usually states that the slope coefficient (\beta) is zero, implying no effect. If the p-value obtained from the test is less than the significance level (often 0.05), we reject the null hypothesis and infer that the predictor variable does have a significant effect on the response.

For instance, with the water striders' data, the slope of the regression line representing the relationship between 'FemalesHiding' and 'MatingActivity' is tested. A t-statistic of -2.56 and a p-value of 0.033 suggest that 'FemalesHiding' is significantly related to 'MatingActivity', and this variable is an effective predictor within the context of the study.

ANOVA Test

The Analysis of Variance, or ANOVA test, in regression, is used to evaluate the overall significance of the model. It compares the model with one that has no predictors and essentially checks if your regression model is better than just using the mean as the prediction. This involves comparing the variance explained by the model against the variance within the residuals.

With only one predictor, as in our water striders example, the ANOVA test provides an F-statistic which indicates the ratio of model variance to residual variance. A low p-value associated with this F-statistic, as in our exercise (0.033), leads to rejecting the null hypothesis that the model with the predictors is no better than a model without them. Consequently, one could conclude that the regression model does provide a valuable prediction of 'MatingActivity'.

Coefficient of Determination

The coefficient of determination, denoted as \(R^2\), is the proportion of the variance in the dependent variable that is predictable from the independent variable(s). An \(R^2\) of 0% indicates that the model explains none of the variability of the response data around its mean, while 100% indicates that the model explains all the variability.

In our case study, with an \(R^2 = 45.1%\) it means that approximately 45.1% of the variation in 'MatingActivity' can be explained by 'FemalesHiding'. In educational content, we might analogize \(R^2\) to the percentage of questions on a test that were answered correctly because of studying a specific book. If the book was about the test topic, you'd expect a high percentage—just as you would expect a high \(R^2\) if your model variables are closely related.

Predictive Modeling

In statistics, predictive modeling uses mathematical techniques to predict future outcomes. In the context of linear regression, the model, derived from historical data, forecasts values by applying the regression equation to new data.

The water striders study is an example of a predictive model where the mating activity of the insects is predicted based on the observed data about females' hiding times. By carefully choosing predictors and assessing model fit through measures like \(R^2\) and p-values from hypothesis testing, researchers use predictive modeling to draw inferences and make decisions based on the derived statistical relationships.

Least Squares Line

The least squares line is the heart of a linear regression model, providing the best fit line through a set of data points. This line minimizes the sum of the squares of the residuals—the vertical distances between the actual data points and the line. Mathematically, if we're given the regression equation as MatingActivity = 0.480 - 0.323 FemalesHiding, we can use this line to make predictions.

For example, if females are observed hiding 50% of the time, we insert 0.50 in place of 'FemalesHiding' in our equation to predict 'MatingActivity'. The resulting calculation, MatingActivity = 0.480 - 0.323 * 0.50, enables researchers to predict mating activity based on this linear relationship. By introducing students to this concept, we equip them with a tool to make informed predictions in a variety of disciplines such as biology, economics, and engineering.