Question

A tourism agency surveyed a random sample of 400 European tourists to see if they have plan to visit Toronto or Vancouver. Of the 400 surveyed, 280 said they would visit Toronto. Using the 99% confidence level, what are the confidence interval limits for the proportion that plan to visit Toronto?

A) 0.647 and 0.753

B) 0.660 and 0.740

C) 0.666 and 0.734

D) 0.671 and 0.719

Answer 1

Solution :

n = 400

x = 280

= x / n = 280 / 400 = 0.700

1 - = 1 - 0.700 = 0.300

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.700 * 0.300) / 400)

= 0.059

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.700 - 0.059 < p < 0.700 + 0.059

0.641 < p < 0.759