Question

problem set 2 questions:

1)Calculate the mean of the estimates at each 'effort' using data from the entire class.

2)Calculate the standard error of the mean and 95% CI at each ‘effort” using data from the entire class.

3)What happens to each of the parameters when sampling effort is first doubled (two scoops) and then increased again by 50% (three scoops)?

4)Assuming cost is an issue, which of the three sample sizes would be the best? (In the sampling of actual wildlife, cost usually is an issue).

Group	Effort	m	r	t	N
1	1	54	2	49	1323
1	2	113	12	110	1035.83
1	3	154	41	179	672.34
2	1	64	4	69	1104
2	2	77	11	100	700
2	3	131	20	141	923.55
3	1	60	4	65	975
3	2	118	15	115	904.67
3	3	198	80	211	522.23
4	1	56	2	41	1148
4	2	97	19	110	561.58
4	3	136	20	131	890.80
5	1	39	6	49	318.5
5	2	101	19	128	680.42
5	3	137	33	188	780.48
6	1	52	2	58	1508
6	2	125	19	128	842.11
6	3	180	40	186	837.00
7	1	31	2	43	666.5
7	2	69	4	90	1552.5
7	3	138	29	149	709.034483

Answer 1

Ans-1)- To calculate mean: 54+113+154+64+77+131+60+118+198+56+97+136+39+101+137+52+125+180+31+69+138=2130/21=101.43.

Ans-2)- Standard deviation: 2249.60+133.86+2763.6+1401+596.82+874.38+1716.45+274.56+9325.76+2063.88+19.62+1195.08+3897.50+0.18+1265.22+2443.325+555.54+6173.24+4960.38+1051.70+1337.36=37154.23/21=1769.24

Standard error: Standard deviation/sq. root. of N= 1769.24/ Sq. root of 18665.54 =12.95

If 95% confidence level then, N=Sq. (z*s/m)= 34.85

Estimate margin of error: 34.85/136.62=0.26

Ans-3)- If the sampling effort is first doubled, then standard deviation decrease by half.

If the sampling effort is 50% increased then standard deviation decreases by 1/50%.

Ans-4)- The three sample sizes would be 318.5, 522.23 and 561.58 as smaller sample size gives less cost and more accurate results.