SOLUTION: 1. A regeneration survey has indicated that 75 of the 90 plots examined were stocked. a) Find the 99% confidence limits for the fraction of the total area being stocked. b) Ho

Algebra ->  Probability-and-statistics -> SOLUTION: 1. A regeneration survey has indicated that 75 of the 90 plots examined were stocked. a) Find the 99% confidence limits for the fraction of the total area being stocked. b) Ho      Log On


   

Question 1054795: 1. A regeneration survey has indicated that 75 of the 90 plots examined were stocked.
a) Find the 99% confidence limits for the fraction of the total area being stocked.
b) How large a sample is needed if we wish to be 95% confident that our sample proportion will be within 0.05 of the true fraction?
2. Another survey from another area indicated 75% stocking from 100 plots. Is this different from the area described in Question 1? Use 90% and 95% confidence limits.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
p = 75/90 = 5/6 and q = 1/6
99%
a. ME = z%2Asqrt%28%28p%281-p%29%29%2Fn%29 = ± 2.576%2Asqrt%28%28%285%2F6%29%281%2F6%29%29%2F90%29
5/6 - ME < mu < 5/6 + ME
b. n = %28z%2FME%29%5E2+%28p%281-p%29%29%29 = %281.645%2F.05%29%5E2+%28%285%2F6%29%281%2F6%29%29%29
2. Yes: p = 75/100 = 3/4 and q = 1/4
follow steps in a above for z = 1.645 and z = 1.96
|
= CI z = value
90% z =1.645
92% z = 1.751
95% z = 1.96
98% z = 2.326
99% z = 2.576