I need assistance with the group section that is highlighted below - must be 1-2 slides
peta.c
unit_4_gp_2.docxPage | 3
A.
The Reliability Factor is being determined by using the reliability factor table, with the risk of incorrect acceptance of 5% and 0 overstatements so that the sample size and sampling interval are being determined easily and effectively. Therefore,
|
Number of Overstatement Misstatements |
Risk of Incorrect Acceptance |
||||||||
|
|
1% |
5% |
10% |
15% |
20% |
25% |
30% |
37% |
50% |
|
0* |
4.61 |
3.00 |
2.31 |
1.90 |
1.61 |
1.39 |
1.21 |
1.00 |
0.70 |
|
1 |
6.64 |
4.75 |
3.89 |
3.38 |
3.00 |
2.70 |
2.44 |
2.14 |
1.68 |
|
2 |
8.41 |
6.30 |
5.33 |
4.72 |
4.28 |
3.93 |
3.62 |
3.25 |
2.68 |
|
3 |
10.05 |
7.76 |
6.69 |
6.02 |
5.52 |
5.11 |
4.77 |
4.34 |
3.68 |
|
4 |
11.61 |
9.16 |
8.00 |
7.27 |
6.73 |
6.28 |
5.90 |
5.43 |
4.68 |
|
5 |
13.11 |
10.52 |
9.28 |
8.50 |
7.91 |
7.43 |
7.01 |
6.49 |
5.68 |
|
6 |
14.57 |
11.85 |
10.54 |
9.71 |
9.08 |
8.56 |
8.12 |
7.56 |
6.67 |
|
7 |
16.00 |
13.15 |
11.78 |
10.90 |
10.24 |
9.69 |
9.21 |
8.63 |
7.67 |
|
8 |
17.41 |
14.44 |
13.00 |
12.08 |
11.38 |
10.81 |
10.31 |
9.68 |
8.67 |
|
9 |
18.79 |
15.71 |
14.21 |
13.25 |
12.52 |
11.92 |
11.39 |
10.74 |
9.67 |
|
10 |
20.15 |
16.97 |
15.41 |
14.42 |
13.66 |
13.02 |
12.47 |
11.79 |
10.67 |
|
*The 0 row is always used for the reliability factor in the sample size formula and for basic precision. |
Considering the table given above the Reliability factor = 3.0
B. The Expansion Factor is being determined using an expansion factor table, given risk of incorrect Given acceptance of 5% Expansion factor is = 1.6 C. Sample size = (Book value of population * Reliability Factor)/(Total misstatement-(Expected misstatement*Expansion factor)) where:
|
Book value of the population = $3,460,000 Reliability Factor = 3.0 (calculated and showed in part A) Total misstatement (or tolerable error) = $63,460 Expected misstatement (or anticipated error) = $13,000 Expansion factor = 1.6 (calculated and shown in part B) |
|
So, sample size = (3,460,000*3.0)/(63460-(13000*1.6) |
|
Sample size =243.319 |
D. Sampling interval = Book value of population/Sample size where:
|
Book value of population = $3,460,000 Sample size = 243.319 |
|
So, Sampling interval = 3460000/243.319 |
|
= 14220.01 |
