Question

In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”.39 However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the confidence interval in the context of the study.

Answer 1

Answer: (0.42648,0.47352)

It means the interval (0.42648,0.47352) that we can be 95% sure contains the population proportion.

Step-by-step explanation:

Given : The proportion of U.S. adult report that they live with one or more chronic conditions : $p=0.45$

Standard error : $s.=0.012$

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=z_{0.025}=1.96$

The confidence interval for population proportion is given by :-

$p\pm\ z_{\alpha/2}\ s\\\\=0.45\pm(1.96)(0.012)\\\\\approx0.45\pm0.02352=(0.45-0.02352,0.45+0.02352)=(0.42648,0.47352)$

Hence,  a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions =(0.42648,0.47352)

It means the interval (0.42648,0.47352) that we can be 95% sure contains the population proportion.