Question

ch 5 predicting the response (y) using the regression equation to a value which lies outside of the range of the values of the predictor variable is called extrapolation. prediction. interpolation. estimation. question 20 1 pts ch 5 for a set of data, r is -0.8. one of the following statements is incorrect. which is it? sixty - four percent of the variability in y is explained by the regression of y on x. eighty percent of the variability in y is explained by the regression of y on x. a change of one standard deviation in x corresponds to a decrease of 0.8 standard deviations in y.

Explanation:

• For the first question:
• Extrapolation is defined as predicting the response \(y\) using the regression equation for a value outside the range of the predictor variable \(x\).
• Prediction is a general term for estimating \(y\) but not specifically for values outside the range.
• Interpolation is for predicting \(y\) within the range of \(x\) values.
• Estimation is also a general term and not specific to the context of being outside the range.
• For the second question:
• The coefficient of determination \(R^{2}=r^{2}\). Given \(r = - 0.8\), then \(R^{2}=(-0.8)^{2}=0.64\) or \(64\%\). So, \(64\%\) of the variability in \(y\) is explained by the regression of \(y\) on \(x\), not \(80\%\).
• The interpretation of the correlation coefficient \(r\): A change of one standard deviation in \(x\) corresponds to a change of \(r\) standard deviations in \(y\). Here \(r=-0.8\), so a change of one standard deviation in \(x\) corresponds to a decrease of \(0.8\) standard deviations in \(y\).

Answer:

• First question: extrapolation.
• Second question: Eighty percent of the variability in \(y\) is explained by the regression of \(y\) on \(x\).