1. The Standard Deviation of a population of numbers can be estimated from the s

Question

1. The Standard Deviation of a population of numbers can be estimated from the standard deviation of a random sample drawn from the population. This is a biased estimator of the population standard deviation. We know that s2 is an unbiased estimator for the variance 2 of the underlying population provided the population variance exists and the sample values are drawn independently with replacement. Suppose the population has a normal distribution. Then (n – 1)s2 / 2 has a chi-square distribution with n 1 degrees of freedom and thus its square root has a chi distribution with n 1 degrees of freedom. Thereby the mean of s, the sample standard deviation, is E[s] = bn , where bn is a function of n, the sample size:

            bn = sqrt(2/(n-1))*((n/2)/((n-1)/2)= 1 - (1/4n) - (7/32n2) - (19/128n3) + O(n-4) . Note: (x) is the gamma function. (n+1) = n!.

Run your R code and find bn for n = 2, n = 3, n = 4, n = 5 and n = 10.

2. Other approximations to bnare cn = 1 - (0.75/(n-1)) and dn = (3.5n - 3.62)/(3.5n-1).       

Run your R code and find cn and dn for n = 2, n = 3, n = 4, n = 5 and n = 10.

Which adjustment method seems to be better? bn          cn         dn ?

Explanation / Answer

R-code

1.

bn=function(n){sqrt(2/(n-1))*(gamma(n/2)/gamma((n-1)/2))}
n=c(2,3,4,5,10)
b=bn(n)
b

[1] 0.7978846 0.8862269 0.9213177 0.9399856 0.9726593

2.

cn = function(n){1 - (0.75/(n-1))}
c=cn(n)
c

[1] 0.2500000 0.6250000 0.7500000 0.8125000 0.9166667
dn =function(n){(3.5*n - 3.62)/(3.5*n-1)}
d=dn(n)
d

[1] 0.5633333 0.7242105 0.7984615 0.8412121 0.9229412

dn adjustment method seems to be better

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