I am not able to figure out all the answers to this problem. Please help with the formula and steps to get the answer. Thanks.

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 992 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.) .2661 (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. We are 95% confident that the true proportion of shoppers who stock up on bargains falls within this interval. We are 5% confident that the true proportion of shoppers who stock up on bargains falls above this interval. O We are 5% confident that the true proportion of shoppers who stock up on bargains falls within this interval. We are 95% confident that the true proportion of shoppers who stock up on bargains falls outside this interval. 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 along with the margin of error. O Report the margin of error. Report p. O 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.) 0.026 X