Question

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 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) 7.4 7.3 9.8 9.6 8.7 7.5 Weight​ (kg) 143 168 270 235 218 173 LOADING... Click the icon to view the critical values of the Pearson correlation coefficient r.

The regression equation is ModifyingAbove y with caretequals nothingplus nothingx. ​(Round to one decimal place as​ needed.)

The best predicted weight for an overhead width of 2 cm is nothing 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 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. There is nothing wrong with predicting the weight in this case.

Answer 1

The regression equation is : y^ =-136.3+40.3x

The best predicted weight for an overhead width of 2 cm =-136.3+40.3*2= -55.7

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.