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? 
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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.