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.

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

, where z* = Critical value.

= 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

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.:

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

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 .

idk

step-by-step explanation:

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!