Question

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 favour the proposal. Of these, 89 say “Yes.”

(a) Construct and interpret a 99% confidence interval for the proportion of all visitors to Yellowstone who favours the restrictions.

(b) Based on your work in part (a), can the U.S. Forest Service conclude that more than half of visitors to Yellow-stone National Park favour the proposal? Justify your answer.

Short answer

a. the confidence interval is$(0.49,0.70)$

b. No

Step-by-step solution

introduction

Given, the service asks a random sample n =$150$

people who say yes x = $89$

Using,

$\widehat{p}-z_{\frac{a}{2}}\times \sqrt{\frac{\widehat{p}\times (1-\widehat{p})}{n}}<p<\widehat{p}-z_{\frac{a}{2}}\times \sqrt{\frac{\widehat{p}\times (1-\widehat{p})}{n}}$

$99\%$confidence interval for the proportion of all visitors to Yellowstone who favours the restrictions is$(0.49,0.70)$

explanation

From the above part, it is realized that the population extent is somewhere in the range of$0.49$and$0.70$. No, the U.S. Forest Service cant conclude that more than half of visitors to Yellow-stone National Park favour the proposal since the interval has the range$0.49$. Accordingly, it can't be inferred that the greater part of the guests are supportive of the proposition.