Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal ift overhead width measured from a photograph is 1.6 cm. Can the prediction be correct? What is wrong with predicting the we in this case? Use a significance level of 0.05. Overhead Width (cm) 7.8 7.4 9.7 7.3 8.2 9.4 Weight (kg) 153 275 177 190 286 216 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =+ x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 1.6 cm is O 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 there is not sufficient evidence of a linear correlation. The width in this cas is beyond the scope of the available sample data. O B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficien evidence of a linear correlation. O 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. O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal ift overhead width measured from a photograph is 1.6 cm. Can the prediction be correct? What is wrong with predicting the we in this case? Use a significance level of 0.05. Overhead Width (cm) 7.8 7.4 9.7 7.3 8.2 9.4 Weight (kg) 153 275 177 190 286 216 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =+ x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 1.6 cm is O 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 there is not sufficient evidence of a linear correlation. The width in this cas is beyond the scope of the available sample data. O B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficien evidence of a linear correlation. O 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. 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
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
Question
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 1.6 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)
7.8
7.4
9.7
7.3
8.2
9.4
Weight (kg)
177
190
286
153
216
275
Click the icon to view the critical values of the Pearson correlation coefficient r.
The regression equation is y =+x.
(Round to one decimal place as needed.)
The best predicted weight for an overhead width of 1.6 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 there is not sufficient evidence of a linear correlation. 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.
OC. 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 D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
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