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 93% accuracy round your answer to three decimal places.

Answer 1

Let be the population proportion.

The population proportion is = 329/763 = 0.431

Confidence interval is given by

                                        

Two sided critical z value at 93% alpha level = Z_α/2 = Z_0.035 = 1.812

(From the standard normal statistical table)

93% Confidence Interval = 0.431 ± 1.812 * sqrt( 0.431*(1 – 0.431)/763) = (0.399 , 0.463)

The true proportion of adults who would like to travel to outer space with 93% accuracy is (0.399, 0.463).

0.399 < p < 0.463.