For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.
In a marketing survey, a random sample of 980 supermarket shoppers revealed that 264 always stock up on an item when they find that item at a real bargain price.
(a)
Let
p represent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain. Find a point estimate for
p. (Enter a number. Round your answer to four decimal places.)
(b)
Find a 95% confidence interval for
p. (For each answer, enter a number. Round your answers to three decimal places.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
95% of all confidence intervals would include the true proportion of shoppers who stock up on bargains.5% of all confidence intervals would include the true proportion of shoppers who stock up on bargains. 95% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.5% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.
(c)
As a news writer, how would you report the survey results on the percentage of supermarket shoppers who stock up on items when they find the item is a real bargain?
Report p̂ along with the margin of error.Report p̂. Report the margin of error.Report the confidence interval.
What is the margin of error based on a 95% confidence interval? (Enter a number. Round your answer to three decimal places.)