120k views
20 votes
From a random sample of workers at a large corporation you find that 58% of 200 went on a vacation last year away from home for at least a week. What is the margin of error for the 95% confidence interval of (0.50, 0.66) of coworkers who went on a vacation last year away from home for at least a week

User Max Pronko
by
3.0k points

1 Answer

2 votes

Answer:

The margin of error for the confidence interval is of 0.08 = 8%.

Explanation:

A confidence interval has two bounds, an upper bound and a lower bound. The margin of error is half the subtraction between these two bounds.

In this question:

Lower bound: 0.5

Upper bound: 0.66

Margin of error: (0.66 - 0.5)/2 = 0.16/2 = 0.08

The margin of error for the confidence interval is of 0.08 = 8%.

User Japanjot Singh
by
3.2k points