Question

A poll is taken in which 350 out of 525 randomly selected voters indicated their preference for a certain candidate.

(a) Find a 99% confidence interval for p.:_______ ≤p≤________

(b) Find the margin of error for this 99% confidence interval for p.:___________

Answer 1

Solution :

Given that,

n = 525

x = 350

= x / n = 350 /525 = 0.667

1 - = 1 - 0.667 = 0.333

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 * (((0.667 * 0.333) / 525) = 0.0530

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.667 - 0.0530 < p < 0.667 + 0.0530

( A )0.6140 < p < 0.7200

(b) E = 0.0530