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 133 athletes at Wahtonka and 27 of 115 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 99.5% confidence interval for the difference between the proportion of athletes using drugs at schools with and without testing. Interval: to

please show your work and what function to use on the calculator . thank you !

Answer 1

Solution :

a)at least one sample has too few success

b) Wahtonka sample size =113 , drug use =8

Warrenton sample size =115 , drug use =27

c)

	Wahtonka:		Warranton:
x1                =	8	x2                =	27
p̂1=x1/n1 =	0.0708	p̂2=x2/n2 =	0.2348
n1                       =	113	n2                       =	115
estimated difference in proportion   =p̂1-p̂2   =	-0.164
std error Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) =	0.0463
for 99. 5 % CI value of z=	2.807
margin of error E=z*std error =	0.1299
lower bound=(p̂1-p̂2)-E=	-0.2939
Upper bound=(p̂1-p̂2)+E=	-0.0341

from above 99. 5% confidence interval for difference in population proportion =(-0.2939, -0.0341)

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