Question

A cbs news/new york times poll found that 329 out of 763 randomly selected adults said they would travel to outer space in their lifetime, given the chance. estimate the true proportion of adults who would like to travel to outer space with 92% confidence

Answer 1

Answer:

0.339 < p < 0.461

Explanation:

Given data:

confidence interval is 92%

Randomly selected adults = 329

Total number of adults is 763

$\alpha = 1 - 0.92 = 0.08$

$\frac{\alpha}{2} = \frac{0.08}{2} = 0.04$

for alpha = 0.04

z value is = 1.75

$p = \frac{329}{763} = 0.43$

$= p \pm z \times \sqrt{\frac{p \times (1-p)}{N}$

$= 0.43 \pm 1.75 \tiimes \sqrt{\frac{0.43 \times0.57)}{763}$

$=0.43 \pm 0.031$

0.339 < p < 0.461