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 988 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. (Enter a number. Round your answer to four decimal places.) Correct: Your answer is correct. (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 within this interval. We are 5% confident that the true proportion of shoppers who stock up on bargains falls above 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 p̂ along with the margin of error. Report the margin of error. Report the confidence interval. Report p̂. What is the margin of error based on a 95% confidence interval? (Enter a number. Round your answer to three decimal places.)

Answer 1

Solution :

Given that,

n = 988

x = 270

Point estimate = sample proportion = = x / n = 270 / 988 = 0.2733

1 - = 1 - 0.2733 = 0.7267

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.2733 * 0.7267) / 988)

= 0.0277

A 95% confidence interval for population proportion p is ,

± E

= 0.2733 ± 0.0277

= ( 0.246, 0.301 )

lower limit = 0.246

upper limit = 0.301

We are 95% confident that the true proportion of shoppers who stock up on bargains falls within this interval.

c) Report p̂ along with the margin of error

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

E = 1.96 (((0.2733 * 0.7267) / 988)

E = 0.028