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An Example (Simplified with a Few Numbers)

Suppose there is a deposit $V$, which is divided by $n=5$ blocks $v_{1}$ to $v_{5}$. The true (averaged) values $z_{i}$ of $Z$ in $v_{i}$ and the estimated ones $/hat{Z}$ are given in Table 1.1.


Table 1.1: Values of a Fictive Deposit.
$i$ $z_{i}$ $/hat{z}_{i}$ $(z_{i}-/bar{z})^{2}$ $(/hat{z}_{i}-/bar{/hat{z}})^{2}$ $z_{i}-/hat{z}_{i}$ $(z_{i}-/hat{z}_{i})^2$
1 5 6 0 1 -1 1
2 7 6 4 1 1 1
3 6 4 1 1 2 4
4 2 4 9 1 -2 4
5 5 5 0 0 0 0
$/sum$ 25 25 14 4 0 10

The mean value of $z$ obviously is $/bar{/hat{z}}=25/5=5.$ The variance of the spatial distribution is

/begin{displaymath}/sigma^2_D (v/V)=/frac{1}{n-1} /sum(z_{i}-/bar{z})^{2}=14/4=3.5/end{displaymath}

and the variance of the estimated values

/begin{displaymath}/sigma^2_D(/hat{v}/V)=/frac{1}{n-1}/sum(/hat{z}_{i}-/bar{/hat{z}})^{2}=4/4=1.0/ / / ./end{displaymath}

The estimation variance therefore is

/begin{displaymath}/sigma^2_E=/frac{1}{n-1}/sum(/hat{z_{i}}-z_{i})^{2}=10/4=2.5/ / / ./end{displaymath}

Empirically we can see that the approximate relation between the three variances is

/begin{displaymath}/sigma^2_D(v/V)/cong /sigma_{D}^2 (/hat{v}/V) + /sigma^2_E/ / / ./end{displaymath}

In order to construct a tolerance interval for the true values of $z_{i}$ we assume an approximate normal distribution of the associated random variable. The approximate (95 %) tolerance interval for $z_{i}$ then is $/hat{z}_{i} /pm 2 /times /sigma_{E}$ or, in detail,

/begin{displaymath}/hat{z}_{i}-2 /times /sigma_{E} /leq z_{i} /leq /hat{z}_{i}+2 /times /sigma_{E}/ / / ,/end{displaymath}


/begin{displaymath}/hat{z}_{i}-2 /times /sqrt{2.5} /leq z_{i} /leq /hat{z}_{i}+2 /times /sqrt{2.5}/ / / ,/end{displaymath}


/begin{displaymath}/hat{z}_{i}-3.16 /leq z_{i} /leq /hat{z}_{i}+3.16/ / / ./end{displaymath}


next up previous contents
Next: Some Typical Problems and Up: The Geostatistical Language Previous: Case Study, After J.P.   Contents
Rudolf Dutter 2003-03-13