Question

The Monterey Bay Aquarium, founded in 1984, is situated on the beautiful coast of Monterey Bay in historic Cannery Row district. From 2009 to 2013, the aquarium conducted a survey of a random sample of visitors as they exited the museum. The survey includes visitor demographic information, use of social media, and opinions on their aquarium visit. In 2013, the survey included 192 visitors over 65, of which 98 used a mobile device during their visit. Find the margin of error of the large-sample 90% confidence interval for the proportion of visitors over 65 who used a mobile device during their visit?

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 98 / 192 = 0.510

1 - = 1 - 0.510 = 0.490

Z/2 = Z0.05 = 1.645

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

E = 1.645 (((0.510 * 0.490) / 192)

E = 0.059

answered by: anonymous