Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 1 - Section 1.4 - Linear Regression - Exercises - Page 103: 22d

Answer

The linear model is reasonable, as the graph of the regression line is a good fit for the data points. .

Work Step by Step

Plotting the data points and graphing the regression line, we would say that the regression line seems to be a good fit. Calculating the correlation coefficient, we add another column to the table, $\left[\begin{array}{llllll} & S & E & SE & S^{2} & E^{2}\\ & & & & & \\ \hline & 10 & 2 & 20 & 100 & 4\\ & 15 & 2.5 & 37.5 & 225 & 6.25\\ & 20 & 3 & 60 & 400 & 9\\ & 25 & 4.5 & 112.5 & 625 & 20.25\\ \hline & & & & & \\ \sum & 70 & 12 & 230 & 1350 & 39.5\\ & & & & & \end{array}\right]$ and calculate $r=\displaystyle \frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{n(\sum x^{2})-(\sum x)^{2}}\cdot\sqrt{n(\sum y^{2})-(\sum y)^{2}}}\approx 0.956182887468$, which is very close to $1$, indicating a good fit.
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