Question

A poll found that out of 723 randomly selected adults 315 said that, given the chance, they would travel to Mars. Estimate the true proportion of adults who would like to travel to Mars with 99% confidence.

Answer 1

Solution :

Given that,

n = 723

x = 315

Point estimate = sample proportion = = x / n = 315/723=0.436

1 -   = 1- 0.436 =0.564

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.436*0.564) / 723)

= 0.0475

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.436 -0.0475 < p < 0.436+0.0475

0.3885< p < 0.4835

The 99% confidence interval for the population proportion p is 0.3885,0.4835