Question

A certain transportation system of buses and commuter trains is heavily utilired so that it is not practical to check every traveler's ticket. Rather, only a small, randomly selected group of travelers on any given trip will be asked to show their tickets. Suppose that in a random sample of 928 train travelers is selected and 91 of them admitted they did not buy a ticket. Find the upper bound of a 95% confidence interval for the true proportion of all train travelers who do not buy tickets. Take all calculations to three decimal places and report your answer to three decimal places

Answer 1

solution:-

given that n = 928 ,

x = 91

proportion p = x/n = 91/928 = 0.098

the value of 95% confidence from z table is 1.96

confidence interval

formula for upper bound

=> p + z * sqrt(p*(1-p)/n)

=> 0.098 + 1.96 * sqrt(0.098*(1-0.098)/928)

=> 0.117