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 [tex]\r p = 43\% = 0.43[/tex]
The standard error is [tex]SE = 0.017[/tex]
Given that the confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha = (100-95)\%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * SE[/tex]
=> [tex]E = 1.96 * 0.017[/tex]
=> [tex]E = 0.03332 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\r p -E < p < \r p +E[/tex]
=> [tex]0.43 -0.03332 < p < 0.43 + 0.03332[/tex]
=> [tex] 0.39668 < p < 0.46332[/tex]
Converting to percentage
[tex] (0.39668 * 100)< p < (0.46332*100)[/tex]
[tex] 39.7\% < p < 46.33 \%[/tex]
