Question

g In 2013, the Pew Research Foundation reported that 43% of U.S. adults report that they live with one or more chronic conditions. 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 1.7%, 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. Report your answer as a percentage rounded to 3 decimal places. We are 95% confident that the true proportion of U.S. adults who live with one or more chronic conditions is between % and % Submit All Parts Question 6

Answer 1

Answer:

We are 95% confident that the true proportion of U.S. adults who live with one or more chronic conditions is between 39.7% and 46.33%

Step-by-step explanation:

From the question we are told that

  The  sample  proportion is $\r p = 43\% = 0.43$

   The  standard error is $SE = 0.017$

Given that the confidence level is 95%  then the level of significance is mathematically represented as  

      $\alpha = (100-95)\%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

     $E = Z_{\frac{\alpha }{2} } * SE$

=>    $E = 1.96 * 0.017$

=>     $E = 0.03332$

Generally 95% confidence interval is mathematically represented as

      $\r p -E < p < \r p +E$

=>   $0.43 -0.03332 < p < 0.43 + 0.03332$

=>    $0.39668 < p < 0.46332$

Converting to percentage

$(0.39668 * 100)< p < (0.46332*100)$        

$39.7\% < p < 46.33 \%$