SOLUTION: In a random sample of 500 people of a ctty, it was found that 160 preferred seafood. Find a 95% confidence interval for the actual proportion of people who preferred seafood.

Question 371232: In a random sample of 500 people of a ctty, it was
found that 160 preferred seafood. Find a 95%
confidence interval for the actual proportion of
people who preferred seafood.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In a random sample of 500 people of a city, it was
found that 160 preferred seafood.
Find a 95% confidence interval for the actual proportion of
people who preferred seafood.
-----
p-hat = 160/500 = 0.32
---
ME = 1.96*sqrt(0.32*0.68/500) = 0.0410
----
95% CI: 0.32-0.0410 < p < 0.32+0.0410
-----
Cheers,
Stan H.

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