Statistic Exercise (2)

[brockjames4]

I am in a statistics class at the University of Phoenix. I have really been struggling with it and it is the final week of the course and am lost on 2 statistics exercises need to be completed by Monday Sept 27th. It is Thursday Sept 23rd today and I would greatly appreciate any help to solve the 2 exercise statistic problems below. If someone does have time to to solve these 2 exercise problems by Monday Sept 27th I would be very grateful for you help. If you could show each step used to solve them also I would appreciate that as well. Thank you very much in advance for all the help and time anyone spent to help me with this!


8.62 In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive.

(a) Construct a 95 percent confidence interval for the population proportion of positive drug tests.
(b) Why is the normality assumption not a problem, despite the very small value of p?
(Data are from Flying 120, no.11 [November 1993], p.31.)

 

[galactus]

I am in a statistics class at the University of Phoenix. I have really been struggling with it and it is the final week of the course and am lost on 2 statistics exercises need to be completed by Monday Sept 27th. It is Thursday Sept 23rd today and I would greatly appreciate any help to solve the 2 exercise statistic problems below. If someone does have time to to solve these 2 exercise problems by Monday Sept 27th I would be very grateful for you help. If you could show each step used to solve them also I would appreciate that as well. Thank you very much in advance for all the help and time anyone spent to help me with this!

In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive.

(a) Construct a 95 percent confidence interval for the population proportion of positive drug tests.

Use the formula \(\displaystyle E=z\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

Remember, for a 95% CI, z=1.96

The proportion who tested positive is obviously \(\displaystyle \hat{p}=\frac{1143}{86991}=.0131\)

Add and Subtract the value of E from p to set up the confidence interval.

\(\displaystyle \hat{p}-E<p<\hat{p}+E\)

This tells us, with 95% confidence, the proportion of adults who will test positive will fall in that interval.

 

[brockjames4]

Thanks for the help galactus I appreciate it! Is there still some calculating I need to do to finish it up or is that the answer?

 

[Deleted member 4993]

I need to do to finish it up or is that the answer?

Come to think about it - you will pass this course and claim you know statistics - without knowing the answer to the question above.

And now we encourage more on line courses...............

And we cry why the better paid jobs are flying overseas......

 

[brockjames4]

I never claimed I know statistics first of all. How could I become an expert on the subject after 5 weeks of being introduced to it? How much of statistics did you understand in your 5th week of being introduced to statistics? I will be getting a Bachelors Degree in IT and don't even know why statistics is a required course. In your opinion you would much rather have someone who never went to class on campus somewhere and cheated his entire way through 4 years of college and spent the entire time getting wasted doing keg stands than someone like me fixing your networking or security issues? I don't think so. If you think jobs are going over seas because a lack of quality Americans to do the job effectively then you don't know much about outsourcing and why it is done. It's done because business don't have to worry about paying minimum wages or providing any benefits or pay overtime over seas because countries like India do have human right laws. Maybe you should take an online course in business.