I just need help with explaining how I find the margin of error.

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 1.010 supermarket shoppers revealed that 252 always stock up on an item when they find that item at a real bargain price. (a) Letp 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 ,0. (Enter a number. Round your answer to four decimal places.) .2594 g (b) Find a 95% condence interval for p. {For each answer, enter a number. Round your answers to three decimal places.) lower limit 2324 q] upper limit .2664 ./ Give a brief explanation of the meaning of the interval. (9 We are 95% condent that the true proportion of shoppers who stock up on bargains falls within this interval. 0 We are 5% confident that the true proportion of shoppers who stock up on bargains falls above this interval. 0 We are 5% confident that the true proportion of shoppers who stock up on bargains falls within this interval. 0 We are 95% condent that the true proportion of shoppers who stock up on bargains falls outside this interval. 1' (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 nd the item is a real bargain? 0 Report 33. 0 Report the condence interval. 0 Report the margin of error. @ Report g along with the margin of error. I What is the margin of error based on a 95% condence interval? (Enter a number. Round your answer to three decimal places.)