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For this problem, carry at least four digits after the decimal in your calculati

ID: 3053829 • Letter: F

Question

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 990 supermarket shoppers revealed that 270 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. (Round your answer to four decimal places.)

(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.
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.   
95% 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 p?.
Report the margin of error.   
Report the confidence interval.
Report p? along with the margin of error.

What is the margin of error based on a 95% confidence interval? (Round your answer to three decimal places.)

Explanation / Answer

a)
Point estimate for p = 270/990 = 0.2727

b)
z value at 95% = 1.96
CI = p +/- z *sqrt(p * (1-p)/n)
= 0.2727 +/- 1.96 * sqrt(0.2727 * (1-0.2727)/990)
= (0.2450 , 0.3005)

lower limit = 0.2450
upper limit = 0.3005
95% of all confidence intervals would include the true proportion of shoppers who stock up on bargains.

c)
p =0.2727

ME = z * sqrt(p * (1-p)/n)
= 1.96 * sqrt(0.2727 * (1-0.2727)/990)
= 0.028

Report p along with ME

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