Prob 13.5

# Let x1 50 and x2 15 then y 832301 2290250 1301015

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Let x1 = 5.0 and x2 = 1.5. Then Y = 83.2301 + .(2.2902*5.0) + (1.3010*1.5) = 83.2301 + 11.4510+ 1.9515 = 96.6326 But according to the first line of the table, the gross revenue for this week was 96, so the equation very slightly over-predicted. The error is 96 – 96.6326 = –0.6326 (all in \$K). Why the difference (small as it is)? Because of the calculated result is the expected value, and there will be variation around this expected value with a normal distribution. (New question) For a special promotion running through an upcoming week, the gross revenue is targeted to be \$100,000. It has already been decided to allocate \$2000 for newspaper advertising. How much money should be budgeted for TV advertising? Here, Y = 100 and x2 = 2. The problem is calculate the value of x1. First, solve the multiple regression equation for x1 in terms of Y and x2: 2.2902x1 = Y – 83.2301– 1.3010x2 so x1 = (Y – 83.2301– 1.3010x2) / 2.2902 Substituting Y = 100 and x2 = 2, x1 = (100 – 83.2301 – [1.3010*2]) / 2.2902 so x1 = 6.1863 As a result, \$6.1863K = \$6,186.30 should be used as a planning value for the TV advertising budget. ASW 5e Prob 13.5 (p. 541) 4 © 2008 Harvey Singer
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