20 CRR-NY 8186-26.7NY-CRR
20 CRR-NY 8186-26.7
20 CRR-NY 8186-26.7
8186-26.7 Determination of intervals and final sample size.
(a) Determine the number of “S” intervals in a sample class.
(1) For survey units exclusive of those in the special assessing unit of Nassau County determine the initial number of “S” intervals in a sample class from Table 1 below.
(i) If the number of parcels in the sample class is less than 16 parcels, the initial number of intervals equals the number of intervals indicated in the table.
(ii) If the number of parcels in the sample class is greater than 15 parcels, the initial number of intervals is equal to the number of intervals indicated in the table minus one.
TABLE 1
| Number of parcels in sample class | Number of intervals | Minimum interval sample size |
|---|---|---|
| less than 7 | 1 | Number of parcels |
| 7 to 15 | 2 | 3 |
| 16 to 79 | 3 | 3 |
| 80 to 499 | 4 | 3 |
| 500 to 999 | 4 | 4 |
| 1,000 to 3,999 | 5 | 4 |
| 4,000 to 9,999 | 6 | 4 |
| 10,000 to 19,999 | 7 | 4 |
| 20,000 to 29,999 | 8 | 4 |
| 30,000 to 39,999 | 9 | 4 |
| equal to or greater than 40,000 | 10 | 4 |
(2) For the special assessing unit of Nassau County, determine the initial number of “S” intervals in a sample class from Table 2 below.
(i) If the number of parcels in the sample class is less than 20, the initial number of intervals equals the number of intervals in the table.
(ii) If the number of parcels in the sample class is greater than 19, the initial number of intervals equals the number of intervals indicated in the table minus one.
TABLE 2
| Number of parcels in sample class | Number of intervals | Minimum interval sample size |
|---|---|---|
| less than 13 | 1 | Number of parcels |
| 13 to 19 | 2 | 6 |
| 20 to 79 | 3 | 4 |
| 80 to 499 | 4 | 4 |
| 500 to 999 | 4 | 5 |
| 1,000 to 3,999 | 5 | 4 |
| 4,000 to 9,999 | 6 | 4 |
| 10,000 to 19,999 | 7 | 4 |
| 20,000 to 29,999 | 8 | 4 |
| 30,000 to 39,999 | 9 | 4 |
| equal to or greater than 40,000 | 10 | 4 |
(b) Determine the boundaries and contents of “S” intervals.
(1) Sum the parcel assessed values in the sample class from high to low, and test the sum after each parcel is added.
(2) At each step in the process, test the cumulative assessed value of the interval being constructed against the result of the sample class total assessed value, minus the cumulative assessed value of previously completed intervals, divided by the number of intervals left to be completed.
(3) When the test indicates that, including the latest parcel added, the cumulative assessed value for the interval exceeds the test amount, the latest parcel is determined to be included in the interval, and the interval being processed is complete.
(4) When an interval is complete, the interval total assessed value is subtracted from the sample class test amount remaining to be assigned, and the count of intervals remaining to be constructed is reduced by one.
(5) This procedure continues until all parcels in the sample class have been assigned to a value interval.
(6) For identification purposes, the value interval which contains the parcel with the lowest assessed value is numbered “001.” Subsequent intervals are identified from high to low by adding one to the previous interval identifying number.
(c) Determine boundaries and contents of subintervals.
(1) Examine the first value interval (S001) and determine the number of parcels that it contains.
(i) If the interval contains less than 15 parcels, or all the parcels in the original interval have the same assessed value, the boundaries and contents of the interval remain unchanged.
(ii) If the interval contains 15 or more parcels with different assessed values, create two subintervals by comparing the assessed values of the parcels in the original interval to the mean assessed value of the original interval and assigning parcels, highest assessed value to lowest, to the higher subinterval until an assessed value not greater than the mean assessed value is found.
(iii) The first parcel found with an assessed value not greater than the mean assessed value of the original interval, and all remaining parcels from the original interval, are assigned to the lower subinterval.
(2) For identification purposes, the subinterval which contains the parcel with the lowest assessed value is numbered “001.” The second subinterval is numbered “002.” Subsequent intervals are identified from low to high by adding one to the previous interval identifying number.
(d) Allocate class tentative sample size to “S” intervals.
(1) The class tentative sample size, rounded to the nearest integer, is divided by the number of intervals, resulting in an integer and a remainder. Interval sample sizes are then set, from the lowest value to the highest, equal to the integer plus one for the number of intervals equal to the remainder, and equal to the integer for any additional intervals.
(2) Compare the interval sample sizes from paragraph (1) of this subdivision to the minimum interval sample sizes shown in subdivision (a) of this section; the larger sample size is the interval sample size.
(3) A sample class in the special assessing unit of Nassau County which has greater than 19 parcels and two intervals has a minimum interval sample size of six. A sample class in the special assessing unit of Nassau County which has greater than 19 parcels and three intervals has a minimum interval sample size of four.
(4) Where the interval sample size would be greater than the number of parcels in the interval, the interval sample size is equal to the number of parcels.
(5) The sum of the interval sample sizes is the sample class sample size.
(6) Sample class sample size cannot exceed the number of parcels in the class.
(e) Randomly select the sample parcels and alternates for each “S” interval.
(f) Assign to each large unit exactly one “T” interval and select the large unit as the sample parcel for that interval.
20 CRR-NY 8186-26.7
Current through February 28, 2023
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