Mathematical Methods for Physics and Engineering - Matematica.NET

STATISTICS31.3.5 Confidence limits for a Gaussian sampling distributionAn important special case occurs when the sampling distribution is Gaussian; ifthe mean is a and the standard deviation is σâ then]1 (â − a)2P (â|a, σâ) = √ exp[−2πσâ2 2σâ2 . (31.30)For almost any (consistent) estimator â, the sampling distribution will tend tothis form in the large-sample limit N →∞, as a consequence of the central limittheorem. For a sampling distribution of the form (31.30), the above procedurefor determining confidence intervals becomes straightforward. Suppose, from oursample, we obtain the value â obs for our estimator. In this case, equations (31.28)and (31.29) become)(âobs − a +Φ= α,σâ)(âobs − a −1 − Φ= β,σâwhere Φ(z) is the cumulative probability function for the standard Gaussian distribution,discussed in subsection 30.9.1. Solving these equations for a − and a + givesa − = â obs − σâΦ −1 (1 − β), (31.31)a + = â obs + σâΦ −1 (1 − α); (31.32)we have used the fact that Φ −1 (α) =−Φ −1 (1−α) to make the equations symmetric.The value of the inverse function Φ −1 (z) can be read off directly from table 30.3,given in subsection 30.9.1. For the normally used central confidence interval onehas α = β. In this case, we see that quoting a result using the standard error, asa = â ± σâ, (31.33)is equivalent to taking Φ −1 (1 − α) = 1. From table 30.3, we find α =1− 0.8413 =0.1587, and so this corresponds to a confidence level of 1 − 2(0.1587) ≈ 0.683.Thus, the standard error limits give the 68.3% central confidence interval.◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with standard deviation σ =1. The sample values are as follows (to two decimalplaces):2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Find the 90% central confidence interval on the population mean µ.Our estimator ˆµ is the sample mean ¯x. As shown towards the end of section 31.3, thesampling distribution of ¯x is Gaussian with mean E[¯x] and variance V [¯x] =σ 2 /N. Sinceσ = 1 in this case, the standard error is given by σˆx = σ/ √ N =0.32. Moreover, insubsection 31.3.3, we found the mean of the above sample to be ¯x =1.11.1238