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 1020 supermarket shoppers revealed that 260 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

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

Answer 1

Solution :

Given that,

n = 1020

x = 260

Point estimate = sample proportion = = x / n = 260/1020=0.2550

1 -   = 1- 0.2550 =0.7450

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.2550*0.7450) / 1020)

= 0.027

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.2550- 0.027< p <0.2550+ 0.027

0.228< p < 0.282

The 95% confidence interval for the population proportion p is : lower limit =0.228 , upper limit= 0.282