A professor obtains SAT scores and freshman grade point averages (GPAs) for a group of n=15 college students. The SAT scores have a mean of M=580 with SS = 22,400, and the GPAs have a mean of 3.10 with SS = 1.26, and SP = 84. Find the regression equation for predicting GPA from SAT scores. What percentage of the variance in GPAs is accounted for by the regression equation? (Compute the correlation, r, then find r2.) C) Does the regression equation account for a significant portion of the variance in GPA? Use alpha = 0.05 to evaluate the F-ratio. A) b=0.00375; a=0.925; Y (hat) = 0.00375X+0.925 B) r= 0.5 r^2= 0.25
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.
A professor obtains SAT scores and freshman grade point averages (GPAs) for a group of n=15 college students. The SAT scores have a mean of M=580 with SS = 22,400, and the GPAs have a mean of 3.10 with SS = 1.26, and SP = 84. Find the regression equation for predicting GPA from SAT scores. What percentage of the variance in GPAs is accounted for by the regression equation? (Compute the
A) b=0.00375; a=0.925; Y (hat) = 0.00375X+0.925
B) r= 0.5 r^2= 0.25
Given:
n = 15
Mx = 580
My = 3.10
SSx = 22400
SSy = 1.26
SSxy = 84
∝ = 0.05
Part a:
We compute the slope and intercept as,
Slope = b = SSxy/SSx= 84/22400 = 0.00375
Intercept = a = My - Slope×Mx
= 3.10 – 0.00375-580
= 0.925.
Thus, the regression equation is,
= a + bx
= 0.925 + 0.00375x
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