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 740 women shoppers revealed that 626 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores).

(a) Let p represent the proportion of all women shoppers who remain loyal to their favorite supermarket. 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 = 740

x = 626

Point estimate = sample proportion = = x / n = 626/740=0.8459

1 -   = 1-0.8459 =0.1541

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 (((0.8459*0.1541) / 740)

E = 0.033

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.8459- 0.033< p <0.8459+ 0.033

0.813< p < 0.879

The 95% confidence interval for the population proportion p is : lower limit=0.813,upper limit= 0.879