Consider the data. 3 12 6 20 14 Y, 65 40 60 10 20 The estimated regression equation for these data is ý = 77.5 - 3.5x. (a) Compute SE, SST, and SSR using equations SSE = E(y, - ý)?, SST = E(y, - y)?, and SSR = E(ŷ, - y)?. SSE = SST = SSR = (b) Compute the coefficient of determination 2. (Round your answer to three decimal places.) 12 = Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line. O The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squares line. O The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares line. O The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squares line. (c) Compute the sample correlation coefficient. (Round your answer to three decimal places.)

Consider the data. 3 12 6 20 14 Y, 65 40 60 10 20 The estimated regression equation for these data is ý = 77.5 - 3.5x. (a) Compute SE, SST, and SSR using equations SSE = E(y, - ý)?, SST = E(y, - y)?, and SSR = E(ŷ, - y)?. SSE = SST = SSR = (b) Compute the coefficient of determination 2. (Round your answer to three decimal places.) 12 = Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line. O The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squares line. O The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares line. O The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squares line. (c) Compute the sample correlation coefficient. (Round your answer to three decimal places.)

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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Concept explainers

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

Consider the data.
12
20
14
65
40
60
10
20
The estimated regression equation for these data is ŷ = 77.5 - 3.5x.
(a) Compute SSE, SST, and SSR using equations SSE = E(y, - ý,)2, sST = E(y, - y)?, and SSR = E(9, - v)?.
SSE =
SST =
SSR =
(b) Compute the coefficient of determination 2. (Round your answer to three decimal places.)
Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.)
O The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line.
O The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squares line.
O The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares line.
O The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squares line.
(c) Compute the sample correlation coefficient. (Round your answer to three decimal places.)

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