In 2002 the Supreme Court ruled that schools could require r

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.

(a) 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: 104 Wahtonka drug users: 6
Warrenton sample size: 148 Warrenton drug users: 29

(b) 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

Solution

Given that,

Wahtonka sample size: 104 Wahtonka drug users: 6
Warrenton sample size: 148 Warrenton drug users: 29

First we are finding two proportions.

p1^ = Wahtonka drug users / Wahtonka drug users = 6 / 104 = 0.0577

p2^ = Warrenton drug users / Warrenton sample size = 29 / 148 = 0.1959

q1^ = 1 - p1^ = 1 - 0.0577 = 0.9423

q2^ = 1 - p2^ = 1 - 0.1959 = 0.8041

Now we have to check the condition that,

n1*p1^ = 104*0.0577 = 6

n1*q1^ = 104 * 0.9423 = 98

n2*p2^ = 148*0.1959 = 29

n2*q2^ = 148 * 0.8041 = 119

All are greator than 5.

sample statistic = p1^ - p2^ = 0.0577 - 0.1959 = -0.1383

99.5% confidence interval for difference of proportion is,

(p1^ - p2^) - E < p1 - p2 < (p1^ - p2^) + E

E is margin of error.

E = Zc * sqrt [ p1^q1^/n1 + p2^q2^/n2 ]

Zc is the critical value for normal distribution.

c = confidence level = 99.5% = 99.5/100 = 0.995

a = 1 - c = 1 - 0.995 = 0.005

We can find this critical value by using EXCEL.

syntax :

=normsinv(probability)

where probability = a/2 = 0.005 / 2 =0.0025

Zc = 2.8070 (take positive sign to the value of Zc)

E = 2.8070 * sqrt [0.0577*0.9423/104 + 0.0577*0.9423/148 ]

E = 2.8070 * sqrt (0.0016) = 0.1118

lower limit = (p1^-p2^) - E = -0.1383 - 0.1118 = -0.2501

upper limit = (p1^ + p2^) + E = -0.1383 + 0.1118 = -0.0264

We see that CI contains only negative values so we conclude that p1 - p2 < 0.

We are 99.5% confident that p1<p2.