Question 33051: I've tried working this but i'm pretty sure I am really wrong!
A drug company had previously reported that 1% of patients have a negative reaction to drug X. After the drug hits the market concerns are raised that the actual probability of a negative reaction is higher than the drug company reported. In an investegatory trial 500 patients are given the drug and 9 have a negative reaction.
a) what would be the expected number of negative reactions in the study of 500 patients given the drug company's rate is correct? .01* 500=5
b) What would be the variance of negative reactions in this study of 500 patients given that the drug company's rate is correct? I have no idea!
c) How likely would observing 9 negative reactions in 500 patients have been given the drug company's intial claim was correct. I'm not sure how to type in what I did but i'll try (500!/9!)*.01^9*.99^491. my answer is .035 which means very unlikely.
d) How likely would observing 1 negative reaction in 500 patients have been given the drug company's initial claim was correct? I did the same sort of thing (500!/1!)*.01^1*.99^499=.033.
e) How likely would it be to observe 9 or more negative reactions. The answer is the same as part c.
Any help is greatly appreciated!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a. np=500(0.01)=5
b) It's binomial. The standard deviation is sqrt(npq)
= sqrt(500.01(0.9)=0.671
c) I get 0.036. , but it's 500!/[491!*9!](0.01)^9(0.9)^491 or binompdf(500,0.01,9)
d) I get 0.33 also; it's 500!/[499!*1!](0.01)(0.99)^499
e) I get 0.067 or [1 - binomcdf(500,0.01,8)]
Cheers,
Stan H.
|
|
|
