Question

4. Comparing the fit of the regression lines for two sets of data Aa Aa E Examine each of the following scatter diagrams and
Assignment 14 5.6 9.6 6.6 9 = -0.25 + 1.44x Calculate the missing predicted values of y, residuals, and squared residuals to
The following are the six pairs of data values for Graph II, along with the regression equation: 3.4 5.6 8.8 9.4 y = 1.44x +
0 0
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Answer #1

The 2nd image is a better fit, as the observed points are very close to regression line as opposed to the 1st image.

For Graph I,
The regression line is given as:
\widehat{y}=-0.25+1.44x

We predict on the given data, using the given regression estimate.
Using the above equation, we get the following table:

xi yi \widehat{y_i} y_i-\widehat{y_i} (y_i-\widehat{y_i})^2
1 2.6 1.19 1.41 1.9881
2.2 0.2 2.92 -2.72 7.3984
3.4 7 4.65 2.35 5.5225
4.4 4.4 6.09 -1.69 2.8561
5.6 8.2 7.81 0.39 0.1521
6.6 9.6 9.25 0.35 0.1225
Total 18.0397

Thus, the SSE is 18.0397

The corrosponding degrees of freedom is 6-2 = 4.

The Standard Error of the Estimate =\sqrt{\frac{SSE}{N}}=\sqrt{\frac{18.0397}{6}}=1.73396

For Graph II,
The regression line is given as:
\widehat{y}=1.44x+0.23

We predict on the given data, using the given regression estimate.
Using the above equation, we get the following table:

xi yi \widehat{y_i} y_i-\widehat{y_i} (y_i-\widehat{y_i})^2
1 1.6 1.67 -0.07 0.0049
2.2 3.4 3.4 0 0
3.4 5.2 5.13 0.07 0.0049
4.4 6.4 6.57 -0.17 0.0289
5.6 8.8 8.29 0.51 0.2601
6.6 9.4 9.73 -0.33 0.1089
Total 0.4077



Thus, the SSE is 0.4077

The corrosponding degrees of freedom is 6-2 = 4.


The Standard Error of the Estimate =\sqrt{\frac{SSE}{N}}=\sqrt{\frac{0.4077}{6}}=0.06795



Thus, the Standard Error of estimate for Graph I is more than that of Graph II.
Thus, the least square lie for Graph I fits the data less efficiently than the least square line for Graph II fits its' data.
The magnitude of the correlation cofficient for Graph i to be less than the correlation coefficient for Graph II.



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