Mathematical Methods for Physics and Engineering - Matematica.NET

STATISTICS31.4.7 A worked exampleTo conclude our discussion of basic estimators, we reconsider the set of experimentaldata given in subsection 31.2.4. We carry the analysis as far as calculatingthe standard errors in the estimated population parameters, including the populationcorrelation.◮Ten UK citizens are selected at random and their heights and weights are found to be asfollows (to the nearest cm or kg respectively):Person A B C D E F G H I JHeight (cm) 194 168 177 180 171 190 151 169 175 182Weight (kg) 75 53 72 80 75 75 57 67 46 68Estimate the means, µ x and µ y , and standard deviations, σ x and σ y , of the two-dimensionaljoint population from which the sample was drawn, quoting the standard error on the estimatein each case. Estimate also the correlation Corr[x, y] of the population, and quote thestandard error on the estimate under the assumption that the population is a multivariateGaussian.In subsection 31.2.4, we calculated various sample statistics for these data. In particular,we found that for our sample of size N = 10,¯x = 175.7, ȳ =66.8,s x =11.6, s y =10.6, r xy =0.54.Let us begin by estimating the means µ x and µ y . As discussed in subsection 31.4.1, thesample mean is an unbiased, consistent estimator of the population mean. Moreover, thestandard error on ¯x (say) is σ x / √ N. In this case, however, we do not know the true valueof σ x and we must estimate it using ̂σ x = √ N/(N − 1)s x . Thus, our estimates of µ x andµ y , with associated standard errors, areˆµ x = ¯x ±ˆµ y = ȳ ±s x√N − 1= 175.7 ± 3.9,s y√N − 1=66.8 ± 3.5.We now turn to estimating σ x and σ y . As just mentioned, our estimate of σ x (say)is ̂σ x = √ N/(N − 1)s x . Its variance (see the final line of subsection 31.4.3) is givenapproximately byV [ ˆσ] ≈ 1 (ν 4 − N − 3 )4Nν 2 N − 1 ν2 2 .Since we do not know the true values of the population central moments ν 2 and ν 4 ,wemust use their estimated values in this expression. We may take ˆν 2 = ̂σ x 2 =(ˆσ) 2 ,whichwehave already calculated. It still remains, however, to estimate ν 4 .Asimpliedneartheendof subsection 31.4.5, it is acceptable to take ˆν 4 = n 4 . Thus for the x i and y i values, we have(ˆν 4 ) x = 1 N(ˆν 4 ) y = 1 NN∑(x i − ¯x) 4 = 53 411.6i=1N∑(y i − ȳ) 4 = 27 732.5i=11254