Answer:
Following are the solution to this question:
Explanation:
Please find the complete question in the attached file.
Let p become a screen positive for substance use throughout California
To test

In California,
is a good test for medication and n seems to be the random sample.
The stats of the examination are given

They reject
if it's relevant:


We, therefore, refuse
level and we can infer that such a test supports an argument which Upwards of 10% of California jobseekers screen positive for use of medicines.