In a survey of 3470 adults age
In a survey of 3470 adults aged 57 through 85 years, it wasfound that 82.3% of them used at least one prescriptionmedication. Complete parts (a) through (c) below. a. How many ofthe 3470 subjects used at least one prescription medication?nothing (Round to the nearest integer as needed.) b. Construct a90% confidence interval estimate of the percentage of adults aged57 through 85 years who use at least one prescription medication.nothing%less thanpless than nothing% (Round to one decimal placeas needed.) c. What do the results tell us about the proportion ofcollege students who use at least one prescription medication? A.The results tell us that there is a 90% probability that the trueproportion of college students who use at least one prescriptionmedication is in the interval found in part (b). B. The resultstell us nothing about the proportion of college students who use atleast one prescription medication. C. The results tell us that,with 90% confidence, the probability that a college student usesat least one prescription medication is in the interval found inpart (b). D. The results tell us that, with 90% confidence, thetrue proportion of college students who use at least oneprescription medication is in the interval found in part (b).
Answer:
Solution:
Sample size = n = 3470
Sample proportion = 
= probability of adults aged 57 through 85 years used at leastone prescription medication = 0.823
Part a. How many of the 3470 subjects used atleast one prescription medication?
Number of subjects used at least one prescription medication =3470 x 0.823
Number of subjects used at least one prescription medication =2855.81
Number of subjects used at least one prescription medication =2856
Part b. Construct a 90% confidence intervalestimate of the percentage of adults aged 57 through 85 years whouse at least one prescription medication.
Formula:
where
Zc is z critical value for c = 90% confidencelevel.
Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 =0.9500
Look in z table for Area = 0.9500 or its closest area and findcorresponding z value.
Area 0.9500 is in between 0.9495 and 0.9505 and both the areaare at same distance from 0.9500
Thus we look for both area and find both z values
Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to1.65
Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645
Thus Zc = 1.645
Thus
Thus
Part c . What do the results tell us about theproportion of college students who use at least one prescriptionmedication?
Since the study is about adults aged 57 through 85 years whoused at least one prescription medication , confidence intervalshould say about adults aged 57 through 85 years , not about thecollege students.
Thus correct option is:
B. The results tell us nothing about the proportion ofcollege students who use at least one prescriptionmedication.
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
