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For the following dataset, enter the data into a calculator and find the least squares regression line, correlation coefficient, and coefficient of determination. Remember to show your thinking through your work. . . X1: 10, 8, 13, 9, 11, 14, 6, 4, 12, 7, 5. y1: 8.04, 6.95, 7.58, 8.81, 8.33, 9.96, 7.24, 4.26, 10.84, 4.82, 5.68

Respuesta :

meerkat18
We input simply the data to the calculator using the stat mode on the right x and y columns. The coefficient of lnear regression, r2 is equal to 0.67. The equation in the calculator resulted to y = 0.5 x + 3. The low r^2 means the relationship is not linear and the equation is not accurate to use.

The correct answers are:

y = 0.5x+3; r = 0.816; and r² = 0.667.

Explanation:

Plotting the x-values into List 1 and the y-values into List 2, then running the regression, we get an equation of the form y=ax+b, where a = 0.500 and b = 3.000.  This means the rate of change, or slope, of the regression line is 0.5 (up one and over 2 between points) and the y-intercept is 3 (the data crosses the y-axis at (0, 3)).

The r-value, or correlation coefficient, tells us how closely the regression fits the line.  Since it is 0.816, which is fairly close to 1, this is a good fit.

The r²-value, or coefficient of determination, shows percentage variation in y which is explained by all the x variables together.  The value of r² is always between 0 and 100%;  0% indicates that the model explains none of the variability of the response data around its mean, while 100% indicates that the model explains all the variability of the response data around its mean.  Since the value of r² is 0.667, or 66.7%, this means that it is a fairly good fit.