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) Weight (kg) 7.4 144 7.8 8.7 9.6 7.9 8.8 D 190 224 237 184 229 9 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =D+x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 2 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? O 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. O B. 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. O 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. O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.

MATLAB: An Introduction with Applications
6th Edition
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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

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

0.05.
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
7.8
8.7
9.6
7.9
8.8
Weight (kg)
144
190
224
237
184
229
E Click the icon to view the critical values of the Pearson correlation coefficient r.
The regression equation is y =D+x.
(Round to one decimal place as needed.)
The best predicted weight for an overhead width of 2 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?
O 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.
O B. 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.
O 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.
O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
Transcribed Image Text: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 7.8 8.7 9.6 7.9 8.8 Weight (kg) 144 190 224 237 184 229 E Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =D+x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 2 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? O 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. O B. 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. O 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. O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
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