# SOLUTION: (a) In a random sample of 150 persons having their lunch at the University cafeteria on meatless day it was observed that 20 percent preferred vegetable dishes. • Find 95% confide

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 Question 466310: (a) In a random sample of 150 persons having their lunch at the University cafeteria on meatless day it was observed that 20 percent preferred vegetable dishes. • Find 95% confidence interval for P (proportion of those who preferred vegetables) • How large a sample is needed, if we want to be 98% confident that our estimate of P is within 0.01 Answer by stanbon(60771)   (Show Source): You can put this solution on YOUR website!(a) In a random sample of 150 persons having their lunch at the University cafeteria on meatless day it was observed that 20 percent preferred vegetable dishes. • Find 95% confidence interval for P (proportion of those who preferred vegetables) --- p-hat = 0.2 ME = 1.96*sqrt[0.2*0.8/sqrt(150)] = 0.0640 --- 95% CI: 0.2-0.0640 < p < 0.2640 95% CI: 0.1360 < p < 0.2640 ============================================ • How large a sample is needed, if we want to be 98% confident that our estimate of P is within 0.01 --- n = [z/E]^2*pq --- n = [2.32633/0.01]^2*0.2*0.8 ----- n = 8660 when rounded up ============================ cheers, Stan H. ============