Question 450368: In 1992, the FAA conducted 30,000 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1100 were positive. (a) Construct a 95 percent confidence interval for the population proportion of positive drug tests. ..
I have this as the answer but not sure if I did it correctly..
N= positive tests / O=number of applicants tested / P= proportioned sample
P= n/o
P=1100/30,000=11/300=0.036
A standard error of p = SE(p) =(0.036*(1-0.036)/30000) = 0.11566
and I have to also answer...(b) May you assume that this is a normal distribution? (Check the normality of your proportion!)
Can you advise what formula to use if I have this wrong, please. I have only one more day to figure this///and thank you!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In 1992, the FAA conducted 30,000 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1100 were positive. (a) Construct a 95 percent confidence interval for the population proportion of positive drug tests. ..
I have this as the answer but not sure if I did it correctly..
N= positive tests / O=number of applicants tested / P= proportioned sample
P= n/o
P=1100/30,000=11/300=0.036
A standard error of p = SE(p) =(0.036*(1-0.036)/30000) = 0.11566
and I have to also answer...(b) May you assume that this is a normal distribution? (Check the normality of your proportion!)
Can you advise what formula to use if I have this wrong, please. I have only one more day to figure this///and thank you!
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If you are rounding to 2 decimal places you probably should
use p-hat = 0.37 for your sample proportion.
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SE = z*sqrt[pq/n]
SE = 1.96*sqrt[0.37*0.63/30000] = 0.00546..
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95% CI: 0.37-0.00546 < p < 0.37+0.00546
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Also: I ran a 1-PropZInt on my TI-84 and got this same answer.
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Cheers,
Stan H.
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