Answer:
a. 2,727
b. (85.6%, 87.6%)
Step-by-step explanation:
The percentage of the adults aged 57 through 85 that used at least one prescription medication = 86.6%
a. The expected number of the 3,149 subjects aged 57 through 85 that used at least one prescription medication = 3,149 × 86.6/100 = 2,727.034 ≈ 2,727 (subjects)
b. The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is given as follows;
![CI=\hat{p}\pm z\times \sqrt{\dfrac{\hat{p} \cdot (1-\hat{p})}{n}}](https://tex.z-dn.net/?f=CI%3D%5Chat%7Bp%7D%5Cpm%20z%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B%5Chat%7Bp%7D%20%5Ccdot%20%281-%5Chat%7Bp%7D%29%7D%7Bn%7D%7D)
Where;
= 86.6/100= 0.866
n = 3,149
z = The z-value at 90% confidence level = 1.645
Therefore, we get the following confidence interval of the percentage of adults (rounded to one decimal place as required);
![\left (0.866 - 1.645\times \sqrt{\dfrac{0.866 \times (1-0.866)}{3,149}}\right) \times 100 \% \approx 85.6 \%](https://tex.z-dn.net/?f=%5Cleft%20%280.866%20-%201.645%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B0.866%20%5Ctimes%20%281-0.866%29%7D%7B3%2C149%7D%7D%5Cright%29%20%5Ctimes%20%20100%20%5C%25%20%20%5Capprox%2085.6%20%5C%25)
![\left( 0.866 + 1.645\times \sqrt{\dfrac{0.866 \times (1-0.866)}{3,149}} \right) \times 100 \% \approx 87.6 \%](https://tex.z-dn.net/?f=%5Cleft%28%200.866%20%2B%201.645%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B0.866%20%5Ctimes%20%281-0.866%29%7D%7B3%2C149%7D%7D%20%5Cright%29%20%5Ctimes%20100%20%5C%25%20%20%5Capprox%2087.6%20%5C%25)
The 90% confidence interval, of the percentage C.I. ≈ (85.6%, 87.6%).