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 984 supermarket shoppers revealed that 262 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.)

(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

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

Answer 1

n = 984

x = 262

= x / n = 262 / 984=0.2663

1 - = 1 - 0.2663 = 0.7337

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.05 = 1.960

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n) = 1.960 * (((0.2663 * 0.7337) / 984) = 0.028

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.266 - 0.028 < p < 0.266 + 0.028

0.238 < p < 0.294

Lower limit = 0.238

Upper limit = 0.294