SOLUTION: A random sample of 1100 car owners in a particular city found 220 car owners who recieved a speeding ticket this year. Find a 95% confidence interval for the true percent of car ow

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Question 1039096: A random sample of 1100 car owners in a particular city found 220 car owners who recieved a speeding ticket this year. Find a 95% confidence interval for the true percent of car owners in this city who recieved a speeding ticket this year. Express your results to the nearest hundredth of a percent.
Answer:

______ to ____%
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
1 sample proportion
ci is 1.96 sqrt{p(1-p)/n}
1.96*sqrt (1/5)(4/5)/1100}; the sqrt term is 0.0121
the interval is 0.0236
The CI is 0.20+/-0.0236
(0.1764,0.2236)
(17.64%,22.36%)

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A random sample of 1100 car owners in a particular city found 220 car owners who recieved a speeding ticket this year. Find a 95% confidence interval for the true percent of car owners in this city who recieved a speeding ticket this year. Express your results to the nearest hundredth of a percent.
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p-hat = 220/1100 = 0.2
ME = 1.96*sqrt[0.2*0.8/1100] = 0.024
Answer:
0.2 - 0.024% < p < 0.2 + 0.024%
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0.2024% < p < 0.2024%
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Cheers,
Stan H.
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