Chapter 12: Q. 12. 7 (page 694)

For a given line of best fit, you computed that $r = 0.6501$ using $n = 12$ data points and the critical value is $0.576$ . Can the line be used for prediction? Why or why not?

Short Answer

Expert verified

Yes. the line can be used for the prediction because $r > c r i t i c a l v a l u e$

Step by step solution

Given information

For a given line of best fit, we computed that $r = 0.6501$ using $n = 12$ data points and the critical value is $0.576$ .

Explanation

Suppose you computed $r = 0.6501$ using $n = 12$ data points.

$d f = n - 2 = 12 - 2 = 10 .$

The critical values associated with $d f = 12$ are $- 0.576$ and $+ 0.576$ . If $r <$ negative critical value or $r >$ positive critical value, then $r$ is significant.

Since $r = 0.6501$ and $0.6501 > 0.576$ , $r$ is significant and the line may be used for prediction.

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For a given line of best fit, you computed that $r = 0.6501$ using $n = 12$ data points and the critical value is $0.576$ . Can the line be used for prediction? Why or why not?

Short Answer

Step by step solution

Given information

Explanation

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For a given line of best fit, you computed that r=0.6501using n=12data points and the critical value is 0.576. Can the line be used for prediction? Why or why not?

Short Answer

Step by step solution

Given information

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Applied Mathematics

Logic and Functions

Probability and Statistics

Calculus

Decision Maths

Study anywhere. Anytime. Across all devices.

For a given line of best fit, you computed that $r = 0.6501$ using $n = 12$ data points and the critical value is $0.576$ . Can the line be used for prediction? Why or why not?