Mathematical Methods for Physics and Engineering : A Comprehensive Guide

STATISTICS

31.4.7 A worked example

To conclude our discussion of basic estimators, we reconsider the set of experi-

mental data given in subsection 31.2.4. We carry the analysis as far as calculating

the standard errors in the estimated population parameters, including the popu-

lation correlation.

Ten UK citizens are selected at random and their heights and weights are found to be as

follows (to the nearestcmorkgrespectively):

Person ABCDEFGH I J

Height (cm) 194 168 177 180 171 190 151 169 175 182

Weight (kg) 75 53 72 80 75 75 57 67 46 68

Estimate the means,μxandμy, and standard deviations,σxandσy, of the two-dimensional

joint population from which the sample was drawn, quoting the standard error on the esti-

mate in each case. Estimate also the correlationCorr[x, y]of the population, and quote the

standard error on the estimate under the assumption that the population is a multivariate

Gaussian.

In subsection 31.2.4, we calculated various samplestatistics for these data. In particular,
we found that for our sample of sizeN= 10,

̄x= 175. 7 , ̄y=66. 8 ,

sx=11. 6 ,sy=10. 6 ,rxy=0. 54.

Let us begin by estimating the meansμxandμy. As discussed in subsection 31.4.1, the
sample mean is an unbiased, consistent estimator of the population mean. Moreover, the
standard error onx ̄(say) isσx/

√

N. In this case, however, we do not know the true value
ofσxand we must estimate it usinĝσx=

√

N/(N−1)sx. Thus, our estimates ofμxand
μy, with associated standard errors, are

μˆx= ̄x±

sx

√

N− 1

= 175. 7 ± 3. 9 ,

μˆy= ̄y±

sy

√

N− 1

=66. 8 ± 3. 5.

We now turn to estimatingσxandσy. As just mentioned, our estimate ofσx(say)

iŝσx=

√

N/(N−1)sx. Its variance (see the final line of subsection 31.4.3) is given
approximately by

V[σˆ]≈

1

4 Nν 2

(

ν 4 −

N− 3

N− 1

ν 22

)

.

Since we do not know the true values of the population central momentsν 2 andν 4 ,we
must use their estimated values in this expression. We may takeνˆ 2 =σ̂^2 x=(σˆ)^2 ,whichwe
have already calculated. It still remains, however, to estimateν 4 .Asimpliedneartheend
of subsection 31.4.5, it is acceptable to takeνˆ 4 =n 4. Thus for thexiandyivalues, we have

(ˆν 4 )x=

1

N

∑N

i=1

(xi−x ̄)^4 = 53 411. 6

(νˆ 4 )y=

1

N

∑N

i=1

(yi− ̄y)^4 = 27 732. 5