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For this problem, camy 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 994 supermarket shoppers revealed that 268 always stock up on an item when they find that item at a real bargain price. (b) Find a $95%$ confidence interval for p. (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. 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^ $. Report $p^ $ along with the margin of error. Report the confidence interval. Report the margin of error. What is the margin of error based on a $95%$ confidence interval? (Round your answer to three decimal places.)

To calculate the proportion (p) of all supermarket shoppers who always stock up on an item when they find a real bargain:

1) Identify the given values:

> The number of shoppers who always stock up on an item when they find a real bargain (x) is 268.

> The total number of supermarket shoppers in the sample (n) is 994.

2) Use the formula: