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 127 athletes at Wahtonka and 20 of 119 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?

(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 90% confidence interval for the difference between the proportion of athletes using drugs at schools with and without testing.

Interval: ____ to ____

Answer 1

Answer :

a) There were not enough successes at Wahtonka High School. (8<10), so the large sample confidence interval should not be use

(b)

Wahtonka sample size: 127+2 =129

Wahtonka drug users: 8+1 =9

Warrenton sample size: 119+2=121

Warrenton drug users: 20+1=21

c) the z-critical value for 95% confidence interval is 1.96.

P1= sample 1 proportion X1/n1= 9/129=0.06976

P2= sample 2 proportion X2/n2= 21/121=0.17355

P1-P2=0.06976-0.17355=-0.10379

Standard error of difference in proportion = √[((P1*(1-P1))/n1)+((P2*(1-P2)/n2)]

=√[((0.06976*0.93024)/129)+((0.17355*0.82645)/121)]

=0.04109

Limit=(P1-P2)+-z*error

Lower limit=(P1-P2)-z*error

= -0.10379-(1.96)*(0.04109)

=-0.18432

Upper limit=(P1-P2)+z*error

= -0.10379+(1.96)*(0.04109)

=-0.02325

A 95% confidence interval for the difference is given by (0.02325,0.18432).