Reply
Wed 11 Mar, 2015 11:53 am
I'm not sure how to do this... help?
The probability of flu symptoms for a person not receiving any treatment is 0.019. In a clinical trial of Lipitor, a common drug used to lower cholesterol, 863 patients were given a treatment of 10-mg Lipitor tablets, and 19 of those patients experienced flu symptoms.
a) Assuming that lipitor has no effect on flu symptoms, find the mean and standard deviation for the numbers of people in groups of 863 that can be expected to have flu symptoms.
b) Based on the result from part (a), is it unusual to find that among 863 people, there are 19 who experience flu symptoms? why or why not?
@aegarnham,
aegarnham wrote:
I'm not sure how to do this... help?
The probability of flu symptoms for a person not receiving any treatment is 0.019. In a clinical trial of Lipitor, a common drug used to lower cholesterol, 863 patients were given a treatment of 10-mg Lipitor tablets, and 19 of those patients experienced flu symptoms.
a) Assuming that lipitor has no effect on flu symptoms, find the mean and standard deviation for the numbers of people in groups of 863 that can be expected to have flu symptoms.
Possible values of the number of people with flu symptoms are binomially distributed, so μ = pn, and σ^2 = np(1-p).
Quote:
b) Based on the result from part (a), is it unusual to find that among 863 people, there are 19 who experience flu symptoms? why or why not?
Using the normal approximation to the binomial distribution, we pretend the possible value of the number of people with flu symptoms are normally distributed. You now know μ and σ, so use them to calculate the z-score of 19. Then comment on whether the z-score is reasonably close to 0.
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace