Question

It is known that ibuprofen reduces pain for 85% of all sufferers. Researchers for a particular brand plan to survey 200 users to see how their formulation works; they will sample the n=200 subjects and calculate the sample proportion who have less pain. Over time, we would expect the sample proportions to fall around 0.85, since that is a known parameter value. Of course, they will be different each time, since different subjects varying ailments are selected each time.

1.) Draw a density curve of all possible sample proportions of size 200 that could occur. What is the mean and the standard deviation of your graph?

2.) The scientist survey 200 random subjects and get a P̂ of 0.87 - 87% of these particular subjects experience pain relief. This P̂ of 0.87 makes the researchers excited, it is higher than the national average of 85%, so maybe this means that their ibuprofen brand works better than the other ones! Using your graph and your parameter values from the previous problem, what is the probability of getting a P̂ of 0.87 or higher? (hint: shade in the graph, calculate a z-score, etc) Should the researchers be concerned or is this sample proportion a common variation?

Answer 1

Page No.: * Solution! - Let Po= Percentage of sufferers for whome pain is reduced. Po=0.85 So=1-Po= 0.15 , n= 200 ay We know that for large n sample proportions. p has Normal disth with parametes Pot Polo - Po=0.85 Pogo - (0.85 (0.15) - 0.000 6375 200 pa N(M=0.85 02 = 0.0006375) :.0=0.0252 Probability density curve for p is 0.8258 0.85 0.8752 (uto). M = 0.85 & S.D.= 0 = 0.0252 CM-8) Here mean = b) Here P = p = 0.87, we can shade this region in graph as 0.85 0-87 & Zscore = þ- Po V popoln - 0. 87-0.85 0.0252 Z sore. = 0.7936 The problem is to test Ho: P = Pa=0.85 Ils Hi P >0.85 Test statistic. Z = 0.7936 Table value - At st. d.o.s. Za = 1.64 Decision - As Zcal < Zx , we accept Ho at 5%lo.s. Conclusion - There is no sufficient avidence to say that there is increase in percestage of pain relief.