Question

In: Statistics and Probability

I am working problem 11.32 on page 610 of Ott and Longnecker's 'Statistical Methods and Data...

I am working problem 11.32 on page 610 of Ott and Longnecker's 'Statistical Methods and Data Analysis. I cannot figure out how the answer for the regression equation is:

y = -1.73 + 1.32x and not y = 3.21 + 0.46x if you would include all points. If you exclude some outliers you can approach the former solution, but I can never attain the solution. Also, the narrative provided by Chegg to accompany the solution is too small to be read.

"Can you please give me some guidance on the technique and/or points you eliminated from the regression. The problem even indicates you are using least-squares regression and I cannot seem to get your numbers, -1.73 and 1.31 for intercept and slope respectively

Solutions

Expert Solution

The calculations are:

y x (y - ybar) (x - xbar) (x - xbar)*(y - ybar) (y - ybar)² (x - xbar)²
4.3 4 -1.21 -1.5 1.81 1.46 2.25
5.5 5 -0.01 -0.5 0.00 0.00 0.25
6.8 6 1.29 0.5 0.65 1.67 0.25
8 7 2.49 1.5 3.74 6.21 2.25
4 4 -1.51 -1.5 2.26 2.28 2.25
5.2 5 -0.31 -0.5 0.15 0.10 0.25
6.6 6 1.09 0.5 0.55 1.19 0.25
7.5 7 1.99 1.5 2.99 3.97 2.25
2 4 -3.51 -1.5 5.26 12.31 2.25
4 5 -1.51 -0.5 0.75 2.28 0.25
5.7 6 0.19 0.5 0.10 0.04 0.25
6.5 7 0.99 1.5 1.49 0.98 2.25
ybar xbar Σ(y - ybar) Σ(x - xbar) Σ(x - xbar)*(y - ybar) Σ(y - ybar)² Σ(x - xbar)²
5.51 5.5 0 0 19.75 32.47 15
b1 = Σ(x - xbar)*(y - ybar)/Σ(x - xbar)² 1.32
bo = ybar - b1*xbar -1.73

where ybar = average of the y terms

xbar = average of the x terms

The regression equation is:

y = -1.73 + 1.32*x


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