Listed below are the overhead widths (cm) of seals measured from photographs and weights (kg) of the seals. Find the regression equation, letting the overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is1.8cm, using the regression equation. Can the prediction be correct? If not, what is wrong? Use a significance level of 0.05. Overhead_Width_(cm) Weight_(kg) 7.2 128 7.4 167 9.8 261 9.5 221 8.7 211 8.3 201 The regression equation is y=enter your response here+enter your response herex. (Round the y-intercept to the nearest integer as needed. Round the slope to one decimal place as needed.) Part 2 The best predicted weight for an overhead width of 1.8 cm, based on the regression equation, is enter your response here kg. (Round to one decimal place as needed.) Part 3 Can the prediction be correct? If not, what is wrong? A. The prediction cannot be correct because a weight of zero does not make sense and because there is not sufficient evidence of a linear correlation. B. 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. C. 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. D. The prediction can be correct.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section4.5: Correlation And Causation

Problem 2AGP

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Part 2

The best predicted weight for an overhead width of

1.8

cm, based on the regression equation, is

enter your response here

kg.

(Round to one decimal place as needed.)

Part 3

Can the prediction be correct? If not, what is wrong?

The prediction cannot be correct because a weight of zero 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 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 can be correct.

Expert Solution