1 pt) In 2002 the Supreme Court ruled that schools could require random drug tes

Question

1 pt) In 2002 the Supreme Court ruled that schools could require random drug tests of students participating in competitive after-school activities such as athletics. Does drug testing reduce use of illegal drugs? A study compared two similar high schools in Oregon. Wahtonka High School tested athletes at random and Warrenton High School did not. In a confidential survey, 5 of 102 athletes at Wahtonka and 28 of 146 athletes at Warrenton said they were using drugs. Regard these athletes as SRSs from the populations of athletes at similar schools with and without drug testing.

(b) The plus four method adds two observations, a success and a failure, to each sample. What are the sample sizes and the numbers of drug users after you do this?

Wahtonka sample size: ________ Wahtonka drug users: ______________
Warrenton sample size: ________ Warrenton drug users: ______________

(c) Give the plus four 99.5% confidence interval for the difference between the proportion of athletes using drugs at schools with and without testing.
Interval: _______ to_____________

Explanation / Answer

(b) The plus four method adds two observations, a success and a failure, to each sample. What are the sample sizes and the numbers of drug users after you do this?

We add 2 to the drug users, and add 4 to the sample sizes.

Wahtonka sample size: 106
Wahtonka drug users: 7
Warrenton sample size: 150   
Warrenton drug users: 30 [answers]

(c) Give the plus four 99.5% confidence interval for the difference between the proportion of athletes using drugs at schools with and without testing.
Interval: _______ to_____________

Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.066037736      
p2 = x2/n2 =    0.2      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.040602005      
  
          
For the   99.5%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.0025      
z(alpha/2) =    2.807033768      
          
lower bound = p1^ - p2^ - z(alpha/2) * sd =    -0.247933464      
upper bound = p1^ - p2^ + z(alpha/2) * sd =    -0.019991064      
          
Thus, the confidence interval is          
          
(   -0.247933464   ,   -0.019991064 ) [ANSWER]

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