In a marketing survey, a random sample of1012supermarket shoppers revealed that270always 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 your answer to four decimal places.)

0.2668 ok

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

lower limit0.207 wrong

upper limit0.326 wrong

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. xx

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 the confidence interval.

Reportp.

Reportpalong with the margin of error. x

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.)

1.96 I got wrong