It could be any value
It is a linear model.
There is not enough information to say much. To start with, the correlation may not be significant. Furthermore, a linear relationship may not be an appropriate model. If you assume that a linear model is appropriate and if you assume that there is evidence to indicate that the correlation is significant (by this time you might as well assume anything you want!) then you could say that the dependent variable increases by 1.67 for every unit change in the independent variable - within the range of the independent variable.
Linear functions are used to model situations that show a constant rate of change between 2 variables. For example, the relation between feet and inches is always 12 inches/foot. so a linear function would be y = 12 x where y is the number of inches and x is the number of feet. y = 24 x models the number of hours in any given number of days {x}. Business applications abound. If a cell phone company charges a start-up fee of $50 and then $.05 for every minute used, the function is y = .05 x + 50.
The LPP is a class of mathematical programming where the functions representing the objectives and the constraints are linear. Optimisation refers to the maximisation or minimisation of the objective functions. The following are the characteristics of this form. • All decision variables are non-negative. • All constraints are of = type. • The objective function is of the maximisation type.
When the correlation coefficient isn't equal to 1 you have any number of choices. Contrary to what a maths syllabus might tell you, there is no right or wrong answer here. Do whatever you think best! Maths does have a creative element to it (this isn't it though...) If its close to zero though, a regression line is probably a poor choice. There aren't many ways to draw a nice fitting curve in this case, but you might be able to model it with a random (e.g. bivariate normal) distribution.
Linear regression can be used in statistics in order to create a model out a dependable scalar value and an explanatory variable. Linear regression has applications in finance, economics and environmental science.
I want to develop a regression model for predicting YardsAllowed as a function of Takeaways, and I need to explain the statistical signifance of the model.
Ridge regression is used in linear regression to deal with multicollinearity. It reduces the MSE of the model in exchange for introducing some bias.
The value depends on the slope of the line.
in general regression model the dependent variable is continuous and independent variable is discrete type. in genral regression model the variables are linearly related. in logistic regression model the response varaible must be categorical type. the relation ship between the response and explonatory variables is non-linear.
There are many possible reasons. Here are some of the more common ones: The underlying relationship is not be linear. The regression has very poor predictive power (coefficient of regression close to zero). The errors are not independent, identical, normally distributed. Outliers distorting regression. Calculation error.
One of the main reasons for doing so is to check that the assumptions of the errors being independent and identically distributed is true. If that is not the case then the simple linear regression is not an appropriate model.
A correlation coefficient close to 0 makes a linear regression model unreasonable. Because If the correlation between the two variable is close to zero, we can not expect one variable explaining the variation in other variable.
George Portides has written: 'Robust regression with application to generalized linear model'
O. A. Sankoh has written: 'Influential observations in the linear regression model and Trenkler's iteration estimator' -- subject(s): Regression analysis, Estimation theory
+ Linear regression is a simple statistical process and so is easy to carry out. + Some non-linear relationships can be converted to linear relationships using simple transformations. - The error structure may not be suitable for regression (independent, identically distributed). - The regression model used may not be appropriate or an important variable may have been omitted. - The residual error may be too large.
You can conclude that there is not enough evidence to reject the null hypothesis. Or that your model was incorrectly specified. Consider the exact equation y = x2. A regression of y against x (for -a < x < a) will give a regression coefficient of 0. Not because there is no relationship between y and x but because the relationship is not linear: the model is wrong! Do a regression of y against x2 and you will get a perfect regression!