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
![\begin{gathered} CI=(p\pm MoE) \\ CI=(p-MoE,\: p+MoE) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/tghv8rel70q089w097ufi16ffv52uzoaxm.png)
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
![z_{(\alpha)/(2)}=1-0.90=(0.10)/(2)=0.05](https://img.qammunity.org/2023/formulas/mathematics/high-school/k71ev0y4ytj2eh6fngqv3cphgo6ks0og8c.png)
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%)