Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2.5 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. Overhead Width (cm) 7.6 7.4 9.8 8.8 9.3 7.3 Weight (kg) 142 163 256 188 231 156 The regression equation is y= + x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 2.5 cm is kg? (Round to one decimal place as needed.) Can the prediction be correct? What is wrong with predicting the weight in this case? A. The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data. B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. C. The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data. D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is

2.5

cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of

0.05.

Overhead Width (cm)	7.6	7.4	9.8	8.8	9.3	7.3
Weight (kg)	142	163	256	188	231	156

The regression equation is

y= + x.

(Round to one decimal place as needed.)

The best predicted weight for an overhead width of

2.5

cm is kg?

(Round to one decimal place as needed.)

Can the prediction be correct? What is wrong with predicting the weight in this case?

The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data.

The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation.

The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data.

The prediction can be correct. There is nothing wrong with predicting the weight in this case.

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