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 is 1.8cm, using the regression equation. Can the prediction be correct? If not, what is wrong? Use a significance level of 0.05. Overhead Width (cm) 7.1 7.3 9.9 9.3 8.8 8.3 Weight (kg) 137 176 282 230 230 214 The regression equation is y=+x. (Round the constant to the nearest integers needed. Round the coefficient to one decimal place as needed.) The best-predicted weight for an overhead width of 1.8 cm, based on the regression equation, is: ____ kg. (Round to one decimal place as needed.) 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 a negative weight does not make sense. The width, in this case, is beyond the scope of the available sample data. 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.
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 is 1.8cm, using the regression equation. Can the prediction be correct? If not, what is wrong? Use a significance level of 0.05. Overhead Width (cm) 7.1 7.3 9.9 9.3 8.8 8.3 Weight (kg) 137 176 282 230 230 214 The regression equation is y=+x. (Round the constant to the nearest integers needed. Round the coefficient to one decimal place as needed.) The best-predicted weight for an overhead width of 1.8 cm, based on the regression equation, is: ____ kg. (Round to one decimal place as needed.) 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 a negative weight does not make sense. The width, in this case, is beyond the scope of the available sample data. 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.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
Related questions
Question
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 is 1.8cm, using the regression equation. Can the prediction be correct? If not, what is wrong? Use a significance level of
0.05.
Overhead Width (cm)
|
7.1
|
7.3
|
9.9
|
9.3
|
8.8
|
8.3
|
|
---|---|---|---|---|---|---|---|
Weight (kg)
|
137
|
176
|
282
|
230
|
230
|
214
|
|
The regression equation is
y=+x.
(Round the constant to the nearest integers needed. Round the coefficient to one decimal place as needed.)
The best-predicted weight for an overhead width of 1.8 cm, based on the regression equation, is: ____ kg.
(Round to one decimal place as needed.)
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 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 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.
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