Question

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, 8 of 132 athletes at Wahtonka and 29 of 111 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.

(a) You should not use the large-sample confidence interval. Why not?
Choose a reason. The sample sizes are too small. The sample sizes are not identical. The sample proportions are too small. At least one sample has too few failures. At least one sample has too few successes.

(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 99.9% confidence interval for the difference between the proportion of athletes using drugs at schools with and without testing.
Interval:  to

Answer 1

a)

At least one sample has too few successes.

b)

Wahtonka sample size:136 Wahtonka drug users: 10
Warrenton sample size:115 Warrenton drug users: 31

c)

sample #1   ----->  
first sample size,     n1=   136
number of successes, sample 1 =     x1=   10
proportion success of sample 1 , p̂1=   x1/n1=   0.0735294
      
sample #2   ----->  
second sample size,     n2 =    115
number of successes, sample 2 =     x2 =    31
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.269565

level of significance, α =   0.001   
Z critical value =   Z α/2 =    3.291   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.04704          
margin of error , E = Z*SE =    3.291   *   0.0470   =   0.15480
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.196   -   0.1548   =   -0.3508332
upper limit = (p̂1 - p̂2) + E =    -0.196   +   0.1548   =   -0.0412385
                  
so, confidence interval is (   -0.3508   < p1 - p2 <   -0.0412   )