Simple random sampling examples
Example 1:
In this section, SRS estimators are illustrated with an example. This example will be pursued through the entire Lecture notes and illustrates that different perform differently for the same population and with the same sample size. The example population has \(N = 30\) individual elements; we may imagine 30strip plots that cover a forest area. This dataset will also be used in the further chapters for comparison among the performance of different sampling techniques. Table 11 lists the values of the 30 units. Here, for SRS, we are only interested in the y values. The \(x\) values are a measure for the size (area) of the strips; this will later be used in the context of other estimators. From this population we get the following parametric values:
\[\mu= \frac{\sum_{i=1}^N y_i}{N} = 7.0667\] and\(\sigma^2 = \frac{\sum_{i=1}^N (y_i - \mu)^2}{N} =7.1289\)
If we take samples of size \(n=10\), then the parametric error variance of the estimated mean is:
\[var (\bar {y}) = \frac {N-n}{N-1} * \frac {\sigma^2}{n} =0.491645\]
Table 1:Example Population of N = 30 individual elements
| Number | y | x |
|---|---|---|
| 1 | 2 | 50 |
| 2 | 3 | 50 |
| 3 | 6 | 100 |
| 4 | 5 | 100 |
| 5 | 6 | 125 |
| 6 | 8 | 130 |
| 7 | 6 | 130 |
| 8 | 7 | 140 |
| 9 | 8 | 140 |
| 10 | 6 | 130 |
| 11 | 7 | 140 |
| 12 | 7 | 150 |
| 13 | 9 | 160 |
| 14 | 8 | 170 |
| 15 | 10 | 180 |
| 16 | 9 | 200 |
| 17 | 12 | 210 |
| 18 | 8 | 210 |
| 19 | 14 | 210 |
| 20 | 7 | 200 |
| 21 | 12 | 200 |
| 22 | 9 | 180 |
| 23 | 8 | 160 |
| 24 | 6 | 140 |
| 25 | 7 | 120 |
| 26 | 4 | 90 |
| 27 | 5 | 90 |
| 28 | 6 | 100 |
| 29 | 4 | 100 |
| 30 | 3 | 80 |
| Mean | 7.0667 | 13950 |
| Pop. variance | 7.1289 | 2087.25 |
This value will be compared in the subsequent chapters with the error variances produced by other with the same sample size \(n=10\). The square root of the error variance is the standard error; and this is the true parametric standard error which we strive to estimate from a single sample then.To recap: the parametric standard error is the standard deviation ofall possible samples of size \(n=10\). In a concrete sampling study, we have only one single sample of size\(n=10\) and from this sample the standard error can only be estimated.
Example 2:
Let´stake one single sample of \(n=10\) from thepopulation of \(N=30\) given in Figure 1 and Table1. Assume that the following elements were randomly selected:
| Number | \(y_i\) |
|---|---|
| 3 | 6 |
| 5 | 6 |
| 9 | 8 |
| 11 | 7 |
| 15 | 10 |
| 16 | 9 |
| 21 | 12 |
| 26 | 4 |
| 27 | 5 |
| 29 | 4 |
We now take these ten selected sampling elements to produce estimates of the population parameters of interest. The estimated mean, variance in the population and error variance are, respectively:
\[\bar y = \hat {\mu} = 7.1 m^3\]
\[s^2 = \hat \sigma^2 = 6.9889\]
\[v\hat {a}r (\bar y) = 0.4659 = \frac {N-n}{N} * \frac {s^2}{n} = fpc \frac {s^2}{n}\]
Note that all estimated values differ from the true parametric values. In practice, however, we will never come to know how much this deviation actually is because the parametric values remain unknown.
However,we can make a probabilistic statement about the range in which we expect the true value to be; which is the confidence interval. For the estimated mean, the confidence interval is calculated from the estimated standard error and an assumption about the distribution of the sample means which is reflected in the value of thet-distribution. Accepting an error probability of\(\alpha = 5%\) that our statement is wrong, the width of one side of the confidence interval is:
\[t_{\alpha,v}S_{\bar y} = 2.262*0.6826 = 1.5440\] and then \(P(5.5560< \mu < 8.6440)\,\).
This reads: the probability that the true parametric mean is in the interval between 5.556 and 8.644 is 0.95; it may however be that the true parametric mean is smaller or larger; this is the error probability of\(\alpha = 5%\). The given t-value can be readfrom tables or calculated from functions that usually every statistical software has built in. The actual values are depending on the degrees of freedom and the chosen error probability.
One may also calculate the confidence interval for the estimated variance in the population; which we do not here.
The above sample has a size of n=10 and a sampling intensity of f= 10/30*100 = 33%. Here, sampling intensity can be calculated in terms of number of sampling units. Usually, however, the sampling intensity in forest inventories is calculated in terms of
\[f= \frac {total\, area\, of\, all\, sample\, plots}{total\, area\, of\,inventory\, region}\]
Estimation of the total: The estimator of the total is
\[\hat\tau = N * \bar y\], here\[\hat \tau = N * \bar y = 30 * 7.1 m^3 = 213\],
and the estimated variance of the estimated total is
\[\hat{var}(\hat \tau) = N^2\,v{\hat a}r(\bar y) = 30^2 * 0.4659 =419.31\].
