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 1020 supermarket shoppers revealed that 276 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.

5% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.

95% 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 all confidence intervals 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 p.

Report the confidence interval.

Report p along with the margin of error.

Report the margin of error.

What is the margin of error based on a 95% confidence interval? (Enter a number. Round your answer to three decimal places.)