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Solutions for odd number questions in 12.1-12.2 12.7 A deterministic model does not allow for random error or variation, whereas a probabilistic model does. An example where a deterministic model would be appropriate is: Let y =cost of a 2 × 4 piece of lumber and x =length(in feet) The model would be y = β1 x. There should be no variation in price for the same length of wood. An example where a probabilistic model would be appropirate is: Let y = sales per month of a commodity and x =amount of money spent advertising The model would be y = β0 + β1 x + ². The sales per month will probably vary even if the amount of money spent on advertising remains the same. 12.11 ˆ 1 = −.78. From Exercise 12.10, β̂0 = 7.10 and beta The fitted line is ŷ = 7.10 − .78x. To obtain predicted values of y, we substitute values of x into the prediction equation(fitted line). a. x y ŷ = 7.10 − .78x (y − ŷ) (y − ŷ)2 7 2 1.64 .36 .1296 4 4 3.98 .02 .0004 6 2 2.42 -.42 .1764 2 5 5.54 -.54 .2916 1 7 6.32 .68 .4624 1 6 6.32 -.32 .1024 3 5 4.76 .24 SE = P (y − ŷ) = .02 c. 1 .0576 SSE = P (y − ŷ)2 = 1.2204 x y ŷ = 14 − 2.5x (y − ŷ) (y − ŷ)2 7 2 -3.5 5.5 30.25 4 4 4 0 0 6 2 -1 3 9 2 5 9 -4 16 1 7 11.5 -4.5 20.25 1 6 11.5 -5.5 30.25 3 5 6.5 -1.5 SE = P (y − ŷ) = −7 2.25 SSE = P (y − ŷ)2 = 108.00 The line in part(b) has SSE = 108.00, which is obviously greater than the SSE(= 1.2204) for the least squares line. 12.13 b. Looking at the scattergram, x and y appear to have a negative linear relationship. c. The least squares estimates of β1 , β̂1 = −.9939; the least squares estimates of β0 , β̂0 = 8.543. d. The least squares line is ŷ = 8.543 − .9939x. After plotting the line on the scattergram, we can see it fits the data well. ˆ 0 has no meaning other than the e. β̂0 = 8.543, since x = 0 is not in the observed range, beta y-intercept. β̂1 = −.9939. The estimated change in the mean value of y for each unit change in x is -.0039. These interpretations are valid only for values of x in the range from 2 to 8. 2