Question 923201
18 venture-capital investments in the fiber optics business sector
Mean = 6.33 ,  standard deviation = 2.04  (also Pop sd)
 90% confidence interval 
ME = 1.74 (2.04/sqrt(18))
CI: 6.33-ME < u < 6.33 + ME
.....
50% confidence interval ...
ME = .689 (2.04/sqrt(18))  
CI: 6.33-ME < u < 6.33 + ME
.......
 minimum sample size required to have a margin of error of $0.05 million with a 90% confidence interval.
.05 = 1.645((2.04/sqrt(n))  
n = {{{ (1.645*2.04/.05)^2}}}
.......
 <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
DF	Probability, p 2-tailed			
	0.1	0.05	0.01	0.001
1	6.31	12.71	63.66	636.62
2	2.92	4.3	9.93	31.6
3	2.35	3.18	5.84	12.92
4	2.13	2.78	4.6	8.61
5	2.02	2.57	4.03	6.87
6	1.94	2.45	3.71	5.96
7	1.89	2.37	3.5	5.41
8	1.86	2.31	3.36	5.04
9	1.83	2.26	3.25	4.78
10	1.81	2.23	3.17	4.59
11	1.8	2.2	3.11	4.44
12	1.78	2.18	3.06	4.32
13	1.77	2.16	3.01	4.22
14	1.76	2.14	2.98	4.14
15	1.75	2.13	2.95	4.07
16	1.75	2.12	2.92	4.02
17	{{{ highlight_green(1.74)}}} 	2.11	2.9	3.97
18	1.73	2.1	2.88	3.92