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Sagot :
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
[tex]\begin{gathered} CI=(p\pm MoE) \\ CI=(p-MoE,\: p+MoE) \end{gathered}[/tex]Where MoE is the margin of error and is given by
[tex]MoE=z_{\frac{\alpha}{2}}\cdot\sqrt[]{\frac{p(1-p)}{n}}[/tex]Where
[tex]z_{\frac{\alpha}{2}}=1-0.90=\frac{0.10}{2}=0.05[/tex]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
[tex]\begin{gathered} MoE=z_{\frac{\alpha}{2}}\cdot\sqrt[]{\frac{p(1-p)}{n}} \\ MoE=1.65_{}\cdot\sqrt[]{\frac{0.837(1-0.837)}{3467}} \\ MoE=1.65_{}\cdot\sqrt[]{\frac{0.1364}{3467}} \\ MoE=1.65_{}\cdot0.00627 \\ MoE=0.0103 \end{gathered}[/tex]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%)
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