Please tell if I am doing things correctly. Question 1: What is the Pearson correlation r? (report to 2 decimal places) The Pearson correlation coefficient rr is given by the square root of R2R2. Since the provided R2R2 (coefficient of determination) is 89.89%, r=0.8989=0.9481r=0.8989 r =0.9481 The slope sign in the regression equation will determine the r sign. Since the slope is negative, rr will also be negative. r=−0.9481r=−0.9481 Answer: r=−0.95r=−0.95 (rounded to two decimal places) What is the strength of this relationship? Given that the absolute value of rr is 0.95, which is very close to 1, this indicates a very strong linear relationship. Answer: Strong If you use the regression equation to make a prediction, what error or residual would you expect? (report to 3 decimal places) The residual standard error SS represents the typical amount by which the data points deviate from the regression line. It gives a measure of the spread of the residuals about the fitted regression line. The given value of SS is 0.345210. Answer: S=0.345S=0.345 (rounded to three decimal places) Find the residual or error for the specific point with Natural growth rate = -1.39 and Median age = 44.5. (report to 2 decimal places) Using the regression equation: Natural growth rate=4.2464−0.10808×Median ageNatural growth rate=4.2464−0.10808×Median age Plug in the value for Median age: Predicted growth rate=4.2464−0.10808×44.5=4.2464−4.80956=−0.56316Predicted growth rate=4.2464−0.10808×44.5=4.2464−4.80956=−0.56316 Now, find the residual: Residual=Observed growth rate−Predicted growth rateResidual=Observed growth rate−Predicted growth rate Residual=−1.39+0.56316=−0.82684Residual=−1.39+0.56316=−0.82684 Answer: Residual = -0.83 (rounded to two decimal places) Now, compiling all answers: r=−0.95r=−0.95 Strong S=0.345S=0.345 Residual = -0.83

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.6: Applications And The Perron-frobenius Theorem

Problem 22EQ

See similar textbooks

Related questions

Question

Please tell if I am doing things correctly.

Question 1: What is the Pearson correlation r? (report to 2 decimal places) The Pearson correlation coefficient rr is given by the square root of R2R2. Since the provided R2R2 (coefficient of determination) is 89.89%, r=0.8989=0.9481r=0.8989

r =0.9481 The slope sign in the regression equation will determine the r sign. Since the slope is negative, rr will also be negative. r=−0.9481r=−0.9481

Answer: r=−0.95r=−0.95 (rounded to two decimal places)

What is the strength of this relationship? Given that the absolute value of rr is 0.95, which is very close to 1, this indicates a very strong linear relationship.

Answer: Strong

If you use the regression equation to make a prediction, what error or residual would you expect? (report to 3 decimal places) The residual standard error SS represents the typical amount by which the data points deviate from the regression line. It gives a measure of the spread of the residuals about the fitted regression line. The given value of SS is 0.345210.

Answer: S=0.345S=0.345 (rounded to three decimal places)

Find the residual or error for the specific point with Natural growth rate = -1.39 and Median age = 44.5. (report to 2 decimal places) Using the regression equation: Natural growth rate=4.2464−0.10808×Median ageNatural growth rate=4.2464−0.10808×Median age Plug in the value for Median age: Predicted growth rate=4.2464−0.10808×44.5=4.2464−4.80956=−0.56316Predicted growth rate=4.2464−0.10808×44.5=4.2464−4.80956=−0.56316 Now, find the residual: Residual=Observed growth rate−Predicted growth rateResidual=Observed growth rate−Predicted growth rate Residual=−1.39+0.56316=−0.82684Residual=−1.39+0.56316=−0.82684

Answer: Residual = -0.83 (rounded to two decimal places)

Now, compiling all answers: