Answer:
95% confidence interval for the population proportion of adults who ate fast food at least once in the past month
( 0.6226 , 0.6814)
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 1000
Given data in a survey of 1,000 adults in a country, 652 said that they had eaten fast food at least once in the past month
Sample proportion
[tex]p^{-} = \frac{x}{n} = \frac{652}{1000} = 0.652[/tex]
Step(ii):-
95% confidence interval for the population proportion of adults who ate fast food at least once in the past month
[tex]( p^{-} - Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-} )}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-} )}{n} })[/tex]
[tex]( 0.652 - 1.96 \sqrt{\frac{0.652 (1-0.652 )}{1000} } , 0.652 + 1.96 \sqrt{\frac{0.652 (1-0.652)}{1000} })[/tex]
( 0.652 - 0.0294 ,0.652 + 0.0294 )
( 0.6226 , 0.6814)
Final answer:-
95% confidence interval for the population proportion of adults who ate fast food at least once in the past month
( 0.6226 , 0.6814)
