Chpt09

# Chpt09 - Chapter 9 The Analysis of Variance Section 9.1 1(a...

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Chapter 9 The Analysis of Variance Section 9.1 1. (a) H 0 : μ A = μ B = μ B C ; where μ i = average strength of wood of Type i. (b) When a null hypothesis is not rejected in an ANOVA test it can often be good news. When there is no significant difference between two populations (types of beams in this case), you are then free to use other factors when making decisions about the populations. For instance, in this exercise, the choice is narrowed to Type A or B (both of which were shown by the ANOVA test to be superior in strength to Type C). To make the final decision (between A and B), we might use another factor, such as the cost of each type of beam, as a factor in deciding between them (since the ANOVA test shows no significant difference in their strengths). (c) This is similar to pat (b), except now there is no significant difference between the strengths of any of the three beams, so another factor (e.g., cost) can be used to make a decision about which one to use. 2. In general the decision about how to allocate samples depends upon what results you expect. Suppose that you expect the variation "within samples" to be large as compared to the variation "between sample means." That is, when you subject several plastic parts to the same chemical treatment level, the results you obtain for each part are quite different from one another, as compared to the differences you expect between the different chemical treatment levels you are experimenting with. In such a case, you would want to use fewer experimental levels of chemical concentration and, correspondingly, more plastic parts at each concentration level. The reason many replicates would be desirable is so that the mean effect of the treatment level can be better estimated. If you recall from earlier chapters, as the sample size increases, the sampling variability of the sample mean decreases. Conversely, if each time you subject a plastic part to the same chemical treatment level, the results you obtain are very similar, there is no need to "waste" plastic parts in conducting replicates. Instead, you can use a larger number of concentration levels with only a few replicates at each level. 3. The two ANOVA tests will give identical conclusions. The reason for this is that an ANOVA test is based on comparing variances , which will not be affected by a calibration error. The calibration will certainly cause the mean of the measurements to shift, but the variation of the measurements around the mean will be the same as the variation of accurate measurements around the true mean. For example, if x i denotes the true measured strength of an alloy sample, then x i + 2.5 would be the reading given by the instrument. Letting x denote the mean of the true measurements, then x +2.5 would be the mean of the instrument's measurements. Because (x i - x ) = (x i +2.5) - ( x +2.5), the sample variances of the true measurements and the instrument's measurements will be identical.

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