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Answer:
Following are the solution to this question:
Step-by-step explanation:
Please find the complete question in the attached file.
Let p become a screen positive for substance use throughout California
To test [tex]H_0 : p = 0.1_{against} \ H_0:p>0.1[/tex]
In California, [tex]\hat{P}[/tex] is a good test for medication and n seems to be the random sample.
The stats of the examination are given
[tex]\to z=\frac{\hat{p} -0.1}{\sqrt{\frac{0.1 \times 0.9}{n} }}_{\ that \ is \ under H_o \ follows \ a \ normal \ distribution}.[/tex]
They reject [tex]H_o \ at \ 5\%[/tex] if it's relevant:
[tex]\to |Z_{obs} |> Z_{0.05}\\\\Here\\\\ \to \hat{p} = \frac{145}{1200} = 0.121 \\\\ \to n = 1200\\\\ \to z_{obs} = 2.42487\\\\ \to Z_{0.05} = 1.64485[/tex]
[tex]\to z_{obs} = 2.42487 > z_{0.05}= 1.64485[/tex]
We, therefore, refuse [tex]H_o \ at \ 5\%[/tex] 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.
