In a marketing survey, a random sample of994supermarket shoppers revealed that272always stock up on an item when they find that item at a real bargain price.

(a)Letprepresent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain.

Find a point estimate forp. (Enter a number. Round up to four decimal places.)

(b)Find a 95% confidence interval forp. (For each answer, enter a number. Round up to three decimal places.)

lower limit

upper limit

Give a brief explanation of the meaning of the interval.

95% 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.

5% 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 the confidence interval.

Reportp.

Reportpalong 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 up to three decimal places.)