A recent survey on drug use and health measured the prevalen

A recent survey on drug use and health measured the prevalence of drug use in the United States. This survey is designed to provide estimates on the use of illicit drugs, alcohol, and tobacco among members of United States households aged 12 and older. In addition, numerous demographic variables like gender, age, race, etc. are recorded. Assume that this is a legitimate statistical survey and that valid inferences for the US population may be drawn from its results. In the 2007 survey, 54 out of 880 respondents aged 12 to 15 years old indicated that they had tried marijuana. Investigators would like to estimate the true proportion for the population of 12 to 15 year olds in the US that have tried marijuana. To do this, they wish to construct a 95% confidence interval. The point estimate for the true proportion of 12 to 15 year olds that have tried marijuana is: (Round the answer to three decimal places.) Calculate the 95% margin of error: (Round the answer to three decimal places.) Find the 95% confidence interval for the proportion of 12 to 15 year olds that have tried marijuana. (Round the answer to three decimal places.)

Solution

A)

Note that              
              
p^ = point estimate of the population proportion = x / n = 54/880 =    0.061363636   [ANSWER]

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B)      
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.008090268          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.015856633 [ANSWER]

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C)
          
lower bound = p^ - z(alpha/2) * sp =   0.045507003          
upper bound = p^ + z(alpha/2) * sp =    0.07722027          
              
Thus, the confidence interval is              
              
(   0.045507003   ,   0.07722027   ) [ANSWER]