1. A new drug for migraine headaches has been introduced to the market. You would
like to know if migraine patients prefer the old drug or the new drug. So, you sample
190 patients that have used both drugs and you find that 52% of the sampled patients
prefer the new drug, 48% of the patients prefer the old drug. Find the 90% confidence
limit for the proportion of all patients that prefer the new drug. Can you be 90%
confident that a majority of all the patients prefer the new drug?
p=.52 n = 190 Based on a confidence limit of 90 %, we find in table 6.1 that z=1.645
So, the 90% confidence limit is:
Notice that the proportion that like the new drug may be as small as .46 which is less
than the .5. So, we cannot be 90% confident that a majority of all the patients will prefer
the new drug.
, 2. Suppose that you are a nurse and you are assigned to do checkups of people one day per week
in a certain village. You have a total of 300 patients in the village. You have the option of doing the
checkups in the mornings or in the afternoons. Therefore, you ask 35 patients and find that 62%
prefer afternoon appointments while 38% prefer morning appointments. Find the 95% confidence
limit for the proportion of all patients that prefer afternoon appointments.
Since 62% prefer afternoon, we set P = .62. As we mentioned previously, we estimate p by P.
So, p=.62. The total population is 300, so set N=300. A total of 35 patients were surveyed, so
Based on a confidence limit of 95
%, we find in table 6.1 that z=1.96. Now, we can substitute all of these values into our
equation:
So the proportion of the total who prefer afternoon appointments is between .469 and .771.