Question

A poll is taken in which 324 out of 550 randomly selected voters indicated their preference for a certain candidate.

(a) Find a 99% confidence interval for ?.

≤?≤

(b) Find the margin of error for this 99% confidence interval for ?.



(c) Without doing any calculations, indicate whether the margin of error is larger or smaller or the same for an 80% confidence interval.
A. smaller
B. larger
C. same

Answer 1

Solution :

Given that,

n = 550

x = 324

Point estimate = sample proportion = = x / n = 324/550=0.589

1 -   = 1- 0.589 =0.411

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 ( Using z table )

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

= 2.576* (((0.589*0.411) / 550)

= 0.0540

A 99% confidence interval for population proportion p is ,

- E ≤?≤ + E

0.589-0.0540 ≤?≤ 0.589+0.0540

0.5350≤?≤ 0.6430

The 99% confidence interval for the population proportion p is : 0.5350≤?≤< 0.6430