The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal, Of these, 89 say "Yes."

a. Construct and interpret a $99\%$confidence interval for the proportion of all visitors to Yellowstone who favor the restrictions.

b. Based on your work in part (a), is there convincing evidence that more than half of all visitors to Yellowstone National Park favor the proposal? Justify your answer.

The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal, Of these, 89 say "Yes."

The formula to compute the confidence interval for population proportion is:

$CI=\hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Here,

$\text{Number of trials}\left(n\right)=150$

$\text{Number of events}\left(x\right)=89$

The obtained Minitab output is:

Therefore,

There is $99\%$probability that the proportion of visitors who are favor of restrictions lies between $0.485$and $0.696$

The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal, Of these, 89 say "Yes."

The confidence interval computed in above part lies between $0.486$and $0.696$. Since, it contains the value (0.50) and upper limit of the confidence interval is higher than $0.50$. Thus, it could be concluded that more than half of visitors are favor of restriction.