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 88% accuracy. Round your answers to at least three decimal places.

Answer 1

The true proportion of adults is in the interval 0.4026< P< 0.4594.

Here we have to estimate the true population of adults who would like to travel to outer space with 88% accuracy.

Data provided:

selected (x) = 329

Total(n) = 763

Sample proportion (p') = x/n

                                       = 329/ 763

                                       = 0.431

Critical z-value = 1.59

margin of error(E) = z$\sqrt{p'( 1-p')/n}$

                             = 1.59 $\sqrt{0.431(1-0.431) / 763}$

                             = 1.59 × 0.0179

                             = 0.0284

88% confidence interval is:

(p' ± E) = ( 0.431 ± 0.0284)

          = 0.4594 and 0.4026

Therefore the estimate is 0.4026< P< 0.4594.

To know more about critical z-value refer to the link given below:

https://brainly.com/question/16029738

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