We are given the following information.
Proportion = p = 83.7%
Sample size = n = 3467
Confidence level = 90%
b. Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication
The confidence interval is given by

Where MoE is the margin of error and is given by
![MoE=z_{(\alpha)/(2)}\cdot\sqrt[]{(p(1-p))/(n)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/owhvnlaugmrniy9vw0k24yitbgwonbm6on.png)
Where

From the normal distribution table, the value of z_0.05 corresponding to a 90% confidence interval is found to be 1.65
So, the margin of error is
![\begin{gathered} MoE=z_{(\alpha)/(2)}\cdot\sqrt[]{(p(1-p))/(n)} \\ MoE=1.65_{}\cdot\sqrt[]{(0.837(1-0.837))/(3467)} \\ MoE=1.65_{}\cdot\sqrt[]{(0.1364)/(3467)} \\ MoE=1.65_{}\cdot0.00627 \\ MoE=0.0103 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zulgd6ofebw5f5s759t3qaikqtkgnq62mq.png)
So, the confidence interval is
[tex]\begin{gathered} CI=(p-MoE,\: p+MoE) \\ CI=(0.837-0.0103,\: 0.837+0.0103) \\ CI=(0.8267,\: 0.8473) \\ CI=(82.7\%,\: 84.7\%) \\ or \\ 82.7\%
What does it mean?
It means that we are 90% confident that the percentage of adults aged 57 through 85 years who use at least one prescription medication lies within the interval of (82.7%, 84.7%)