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 1012 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 = 1012

x = 260

Point estimate = sample proportion = = x / n = 260/1012=0.2569

1 - = 1-0.2569=0.7431

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

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

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

= 1.96 (((0.2569*0.7431) / 1012)

E= 0.027

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.2569 -0.027 < p <0.2569+ 0.027

0.2299< p < 0.2839

The 95% confidence interval for the population proportion p is;

lower limit 0.230 , upper limit 0.284