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Does this sample support a claim that more than 10% of job applicants in California test positive for drug use

User Cyprian
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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
H_0 : p = 0.1_(against) \ H_0:p>0.1

In California,
\hat{P} is a good test for medication and n seems to be the random sample.

The stats of the examination are given


\to z=\frac{\hat{p} -0.1}{\sqrt{(0.1 * 0.9)/(n) }}_(\ that \ is \ under H_o \ follows \ a \ normal \ distribution).

They reject
H_o \ at \ 5\% if it's relevant:


\to |Z_(obs) |> Z_(0.05)\\\\Here\\\\ \to \hat{p} = (145)/(1200) = 0.121 \\\\ \to n = 1200\\\\ \to z_(obs) = 2.42487\\\\ \to Z_(0.05) = 1.64485


\to z_(obs) = 2.42487 > z_(0.05)= 1.64485

We, therefore, refuse
H_o \ at \ 5\% 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.

User Naruil
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