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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.

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JeanaShupp

Answer with explanation:

Formula to find the confidence interval for population proportion (p) is given by :-

[tex]\hat{p}\pm z^* SE[/tex]

, where z* = Critical value.

[tex]\hat{p}[/tex] = Sample proportion.

SE= Standard error.

Let p be the true population proportionof U.S. adults who live with one or more chronic conditions.

As per given , we have

[tex]\hat{p}=0.45[/tex]

SE=0.012

By z-table , the critical value for 95% confidence interval : z* = 1.96

Now , a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions.:

[tex]0.45\pm (1.96) (0.012)[/tex]

[tex]0.45\pm (0.02352)[/tex]

[tex]=(0.45-0.02352,\ 0.45+0.02352)=(0.42648,\ 0.47352)[/tex]

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

Interpretation : Pew Research Foundation can be 95% confident that the true population proportion (p) of U.S. adults who live with one or more chronic conditions lies between 0.42648 and 0.47352 .

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