2 Finally, in Figure 103, we superimpose the 4 distribution onour histogram of sample variances

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Note that exponent 2 carries no meaning whatsoever; it is simply a symbol and alerts us to the fact that the random variable is nonnegative It is possible, but probably useless, to nd the probability distribution for , as we could nd the probability distribution for the square root of a normal random variable We close this section with an admonition: while the central limit theorem affords us the luxury of sampling from any probability distribution, the probability distribution of 2 highly depends on the fact that the samples arise from a normal distribution

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Probability

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Statistical Inference II: Continuous Probability Distributions II

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STATISTICAL INFERENCE ON THE VARIANCE

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While the mean value of the diameter of a manufactured part is very important for the part to t into a mechanism, the variance of the diameter is also crucial so that parts do not vary widely from their target value A sample of 12 parts showed a sample variance s2 = 00025 Is this the evidence that the true variance 2 exceeds 00010 To solve the problem, we must calculate some probabilities One dif culty with the chi-squared distribution, and indeed with almost all practical continuous probability distributions, is the fact that areas, or probabilities, are very dif cult to compute and so we rely on computers to do that work for us The computer system Mathematica and the statistical program Minitab r have both been used in this book for these calculations and the production of graphs 2 Here are some examples where we need some points on 11 :

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2 We nd that P( 11 < 457) = 005 and so P((n 1)s2 / 2 < 457) = 005, which means that P( 2 > (n 1)s2 /457) = 005 and in this case we nd P( 2 > 11 00025/457) = P( 2 > 00060175) = 005 and so we have a con dence interval for 2 Hypothesis tests are carried out in a similar manner In this case, as in many other industrial examples, we are concerned that the variance may be too large; small variances are of course desirable For example, consider the hypotheses from the previous example, H0 : 2 = 00010 and HA : 2 > 00010 From our data, where s2 = 00025, and from the example above, we see that this value for s2 is in the rejection region We also nd that 2 11 = (n 1)s2 / 2 = 11 00025/00010 = 275 and we can calculate that 2 P( 11 > 275) = 000385934, and so we have a p value for the test Without a computer this could only be done with absolutely accurate tables The situation is shown in Figure 104

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Chi square, df = 11 009 008 007 006 Density 005 004 003 002 001 000 0 X 275 000386

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Statistical Inference on the Variance

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Distribution plot

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Now we look at some graphs with other degrees of freedom in Figures 105 and 106 The chi-squared distribution becomes more and more normal-like as the degrees of freedom increase Figure 107 shows a graph of the distribution with 30 degrees of freedom

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Statistical Inference II: Continuous Probability Distributions II

Distribution plot

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2 2 We see that P( 30 > 438) = 005 It can be shown that E( n ) = n and that 2 ) = 2n If we approximate 2 by a normal distribution with mean 30 and Var( n 30 standard deviation 60 = 7746,we nd that the point with 5% of the curve in the right-hand tail is 30 + 1645 60 = 427, so the approximation is not too bad The approximation is not very good, however, for small degrees of freedom Now we are able to consider inferences about the sample mean when is unknown