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DochEvi [55]
3 years ago
5

In a survey of adults aged 57 through 85​ years, it was found that 86.6​% of them used at least one prescription medication. Com

plete parts​ (a) through​ (c) below.
a. How many of the 3149 subjects used at least one prescription​ medication?

​(Round to the nearest integer as​ needed.)

b. Construct a​ 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication.


​(Round to one decimal place as​ needed.)
Mathematics
1 answer:
OleMash [197]3 years ago
7 0

Answer:

a) 272 used at least one prescription​ medication.

b) The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is (85.6%, 87.6%).

Step-by-step explanation:

Question a:

86.6% out of 3149, so:

0.866*3149 = 2727.

272 used at least one prescription​ medication.

Question b:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 3149, \pi = 0.866

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.866 - 1.645\sqrt{\frac{0.866*0.134}{3149}} = 0.856

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.866 + 1.645\sqrt{\frac{0.866*0.134}{3149}} = 0.876

For the percentage:

0.856*100% = 85.6%

0.876*100% = 87.6%.

The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is (85.6%, 87.6%).

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