Prob 13.5

# Where y is the mean weekly gross revenue expected for

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where Y is the mean weekly gross revenue expected for a weekly TV advertising budget of x1 and a weekly newspaper advertising budget of x2 (all in \$K). The R^2 is 0.9190 (in cell B5). This means that 91.9% of the variation in the actual values of y is due to varying x1 and x2 according to the regression equation. The adjusted R^2 is 0.8866 (in cell B6). This means that after discounting for using two independent variables in the multiple regression, 88.66% of the variation in the actual values of y is due to varying x1 and x2 according to the regression equation. This is a truer representation of the variation in y accounted for by the model. (However, 12.34% [= 100% - 88.66%] is not accounted for by the multiple regression. Some of this unaccounted-for variation is due to random error. Most of this unaccounted-for variation is probably due to a still missing or “omitted” variable.) The adjusted R^2 of the multiple regression is significantly higher than the R^2 of either simple linear regression. Therefore the multiple regression is the best of all the regressions performed. Moreover, the multiple regression is a good predictive tool, because the adjusted R^2 is high. Suppose in some week, \$3500 is spent on TV advertising and \$1800 is spent on newspaper advertising. How much is expected to gross revenue is expected to be earned for that week? (This is asked for part d of the problem.) Let x1 = 3.5 and x2 = 1.8. Then Y = 83.2301 + .(2.2902*3.5) + (1.3010*1.8) = 83.2301 + 8.0157 + 2.3418 = 93.5876 So the gross revenue is expected to be \$ 93,587.60. ASW 5e Prob 13.5 (p. 541) 3 © 2008 Harvey Singer

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(New question) Suppose in some week, \$5000 is spent on TV advertising and \$1500 is spent on newspaper advertising. How much is expected to gross revenue is expected to be earned for that week?
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