Question

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. Create a 95% confidence interval for the population proportion of adults who ate fast food at least once in the past month. Use a TI-83, TI-83 plus, or TI-84 calculator, rounding your answers to three decimal places.

Answer 1

95% confidence interval for the population proportion of adults who ate fast food at least once in the past month

( 0.6226 , 0.6814)

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

`p^(-) = (x)/(n) = (652)/(1000) = 0.652`

Step(ii):-

95% confidence interval for the population proportion of adults who ate fast food at least once in the past month

`( p^(-) - Z_(0.05) \sqrt{(p^(-) (1-p^(-) ))/(n) } , p^(-) + Z_(0.05) \sqrt{(p^(-) (1-p^(-) ))/(n) })`

`( 0.652 - 1.96 \sqrt{(0.652 (1-0.652 ))/(1000) } , 0.652 + 1.96 \sqrt{(0.652 (1-0.652))/(1000) })`

( 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)