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Example 15.1b Prediction
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15.115.1 | 15.2 | 15.1a | 15.2a | 15.2b | 15.3 | 15.4 | 15.5 | 15.615.215.1a15.2a15.2b15.315.415.515.6 The Problem n Besides the 50 regions in the data set, Pharmex, does business in five other regions, which have promotional expenses indexes of 114, 98, 84, 122, and 101. n Find the predicted Sales and a 95% prediction interval for each of these regions.
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15.115.1 | 15.2 | 15.1a | 15.2a | 15.2b | 15.3 | 15.4 | 15.5 | 15.615.215.1a15.2a15.2b15.315.415.515.6 PHARMEX.XLS n This example cannot be solved with StatPro but it is relatively easy with Excels built-in functions. n We illustrate the procedure in this file shown here on the next slide. n The original data appear in Column B and C. We use the range names SalesOld and PromoteOld for the data in these columns. n The new regions appear in rows 8-12. Their given values of Promote are in the range G8:G12, which we name PromoteNew.
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15.115.1 | 15.2 | 15.1a | 15.2a | 15.2b | 15.3 | 15.4 | 15.5 | 15.615.215.1a15.2a15.2b15.315.415.515.6
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15.115.1 | 15.2 | 15.1a | 15.2a | 15.2b | 15.3 | 15.4 | 15.5 | 15.615.215.1a15.2a15.2b15.315.415.515.6 Solution n To obtain the predicted sales for these regions, we use Excels TREND function by highlighting the range H8:H12, typing the formula =TREND(SalesOld,PromoteOld,PromoteNew) and pressing Ctrl-Shift-Enter. n This substitutes the new values of the explanatory variable (in the third argument) into the regression equation based on the data from the first two arguments.
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15.115.1 | 15.2 | 15.1a | 15.2a | 15.2b | 15.3 | 15.4 | 15.5 | 15.615.215.1a15.2a15.2b15.315.415.515.6 Solution -- continued n We then calculate the limits of a 95% prediction interval for each new region in columns I and J by going out two standard errors on each side of the predicted values in column H. n We see from the wide predication intervals how much uncertainty remains. n This implies that the point predication for this region, 112.03, contains a considerable amount of uncertainty. The reason is the relatively large standard error of estimate, s e.
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15.115.1 | 15.2 | 15.1a | 15.2a | 15.2b | 15.3 | 15.4 | 15.5 | 15.615.215.1a15.2a15.2b15.315.415.515.6 Solution -- continued n If we reduce the s e, the length of the prediction interval would be reduced accordingly. n Contrary to what you might expect, this is not a sample size problem. A larger sample size would almost surely not produce a smaller value of s e. n The problem is that Promote is not highly correlated with Sales. n The only way to decrease se and obtain more accurate predictions is to find other explanatory variables that are more closely related to Sales.
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