# 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

Z_{c} 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 Z_{c} = 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.**