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(75887) (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
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cheers,
Stan H.
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