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.

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

Step(ii):-

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

( 0.652 -   0.0294 ,0.652 + 0.0294 )

( 0.6226  , 0.6814)

Final -

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

( 0.6226  , 0.6814)

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;