SOLUTION: In 1990, 5.8% of job applicants who were tested for drugs failed the test. A random sample of 1640 current job applicants results in 64 failures (based loosely on data from the A

Question 1150143: In 1990, 5.8% of job applicants who were tested for drugs failed the test.
A random sample of 1640 current job applicants results in 64 failures (based loosely on
data from the American Management Association).
At the α = 0.01 significance level, test the below claim (i.e., that the failure rate is now
lower)
HO: p = 0.058
HA: p < 0.058
a) Is this an upper tail, lower tail, or two-tail test? (5 points)
b) Are we testing means or proportions? (5 points)
c) State the rule of rejection (in terms of p-value and level of significance) (5 points)
d) Find the p-value (5 points)
e) Should you reject or not reject HO? (5 points)
f) Does the result suggest that fewer job applicants now use drugs? (5 points)

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Lower tail test, because looking at size of the failures being below a number.
Testing a proportion.
reject if z<-2.33
p-value <0.01
test statistic is z=(phat-p)/sqrt (.058*.942/1640), where p hat is 64/1640 or 0.0390
this is z= (0.0390-0.058)/0.00577
z=-3.29
Reject Ho
p-value is 0.0005
The results do suggest that fewer applicants are using drugs
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