The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park.

Question

To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal. Of these, 89 say “Yes.” 

1.Construct and interpret a 99% confidence interval for the proportion of all visitors to Yellowstone who favor the restrictions. (Use PANIC or State-Plan-Do-Conclude)

2.Based on your work in part (1), can the U.S. Forest Service conclude that more than half of visitors to Yellowstone National Park favor the proposal? Justify your answer.

Answer 1

1) Using PANIC

Parameter: p = the true proportion of all visitors to Yellowstone who favor the restrictions.

Assumptions: Random - The problem states that the visitors in the sample were selected at random. 10% condition - n=150 is less than 10% of all visitors to Yellowstone. Success/Failure - n(p-hat) = 89 successes; n(1 - p-hat) = 150-89 = 61; Both are greater than 10.

Name the procedure: 1-sample z-interval for p.

Interval: p-hat +/- z* {sqrt[p-hat*(1 - p-hat)/n]}

.593 +/- 2.576 * sqrt(.593 * .407 / 150)

(0.490, 0.696)

Conclude: We are 99% confident that the true proportion of all visitors to Yellowstone who favor the restrictions is between .490 and .696.

2) Any value in the interval is a plausible value for the true proportion. The Forest Service cannot conclude that more than 50% of the visitors favor the restrictions because there are values in the interval that are less than 50%.