Question

1. A CBS News/N.Y. Times poll found that 329 out of 763 randomly selected adults said they would travel to outer space in their lifetime, given the chance. Construct and interpret a 96% confidence interval for the population proportion of adults who would said they would travel to outer space in their lifetime, given the chance.

1. Identify the proper Test or Confidence interval:

2. Complete the Test or Confidence Interval.

Answer 1

Solution :

n = 763

x = 329

= x / n = 329 / 763 = 0.431

1 - = 1 - 0.431 = 0.569

At 96% confidence level the z is ,

= 1 - 96% = 1 - 0.96 = 0.04

/ 2 = 0.04 / 2 = 0.02

Z/2 = Z0.02 = 2.050

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

= 2.050 * (((0.431 * 0.569) / 763)

= 0.037

A 96 % confidence interval for population proportion p is ,

- E < P < + E

0.431 - 0.037 < p < 0.431 + 0.037

0.394 < p < 0.468