Get comprehensive answers to your questions with the help of IDNLearn.com's community. Discover reliable and timely information on any topic from our network of knowledgeable professionals.
Sagot :
Using the z-distribution, it is found that the 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
195 out of 500 people said they would take more vacations this year than last year, hence:
[tex]n = 500, \pi = \frac{195}{500} = 0.39[/tex]
95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 - 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.347[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 + 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.433[/tex]
The 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).
A similar problem is given at https://brainly.com/question/15850972
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
