The Manufacturer of a Light Fixture Believes That the Dollars

Essay

The manufacturer of a light fixture believes that the dollars spent on advertising,the price of the fixture,and the number of retail stores selling the fixture in a particular month influence the light fixture sales.The manufacturer randomly selects 10 months and collects the following data: $$\begin{array} { l l l l } \text { Sales } & \text { Advertising } & \text { Price } & \text { \# of stores } \\41 & 20 & 40 & 1 \\42 & 40 & 60 & 3 \\59 & 40 & 20 & 4 \\60 & 50 & 80 & 5 \\81 & 50 & 10 & 6 \\80 & 60 & 40 & 6 \\100 & 70 & 20 & 7 \\82 & 70 & 60 & 8 \\101 & 80 & 30 & 9 \\110 & 90 & 40 & 10\end{array}$$ The sales are in thousands of units per month,the advertising is given in hundreds of dollars per month,the price is the unit retail price for the particular month.Using this data,the following computer output is obtained.
The regression equation is
Sales = 31.0 + 0.820 Advertising - 0.325 Price + 1.84 Stores $$\begin{array} { l l l l l } \text { Predictor } & \text { Coef } & \text { StDev } & \text { T } & \text { P } \\\text { Constant } & 30.992 & 7.728 & 4.01 & 0.007 \\\text { Advertising } & 0.8202 & 0.5023 & 1.63 & 0.154 \\\text { Price } & - 0.32502 & 0.08935 & - 3.64 & 0.011 \\\text { Stores } & 1.841 & 3.855 & 0.48 & 0.650\end{array}$$ S = 5.465R-Sq = 96.7%R-Sq(adj)= 95.0%
Analysis of Variance $$\begin{array} { l l l l l l } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Regression } & 3 & 5179.2 & 1726.4 & 57.81 & 0.000 \\\text { Residual Error } & 6 & 179.2 & 29.9 & & \\\text { Total } & 9 & 5358.4 & & &\end{array}$$
-Calculate the 95% prediction interval for this point estimate.