Prove that the average ssd is minimized

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To calculate variance and standard deviation, we take the deviations from the mean. At times, we need to consider the deviations from a target value rather than the mean. Consider the case of a machine that bottles cola into 2-liter (2,000- cm3) bottles. The target is thus 2,000 cm3. The machine, however, may be bottling 2,004 cm3 on average into every bottle. Call this 2,004 cm3 the process mean. The damage from process errors is determined by the deviations from the target rather than from the process mean. The variance, though, is calculated with deviations from the process mean, and therefore is not a measure of the damage. Suppose we want to calculate a new variance using deviations from the target value. Let "SSD(Target)" denote the sum of the squared deviations from the target.

[For example, SSD(2,000) denotes the sum of squared deviations when the deviations are taken from 2,000.] Dividing the SSD by the number of data points gives the Average SSD(Target). The following spreadsheet is set up to calculate the deviations from the target, SSD(Target), and the Average SSD(Target). Column B contains the data, showing a process mean of 2,004. (Strictly speaking, this would be sample data. But to simplify matters, let us assume that this is population data.) Note that the population variance (VARP) is 3.5 and the Average SSD(2,000) is 19.5. In the range G5:H13, a table has been created to see the effect of changing the target on Average SSD(Target). The offset refers to the difference between the target and the process mean

1. Study the table and find an equation that relates the Average SSD to VARP and the Offset. [Hint: Note that while calculating SSD, the deviations are squared, so think in squares.]

2. Using the equation you found in part 1, prove that the Average SSD(Target) is minimized when the target equals the process mean.

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Reference no: EM131369462

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