Question 460767: A doctor says that less than 25% of US adults chew tobacco. In a random sample of 170 US adults, 18.5% say they chew tobacco. At alpha = .05, is there enough evidence to reject the doctor's claim?
Since I don't have a standard deviation, do I use t-distribution? Or do I use binomial distribution and find it by square root of npq?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A doctor says that less than 25% of US adults chew tobacco.
In a random sample of 170 US adults, 18.5% say they chew tobacco.
At alpha = .05, is there enough evidence to reject the doctor's claim?
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Since I don't have a standard deviation, do I use t-distribution? Or do I use --binomial distribution and find it by square root of npq?
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It is a proportion test so you use the z-distribution.
The standard deviation is sqrt[pq/n]. You have all
the information you need.
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Cheers,
Stan H.
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