Explore a world of knowledge and get your questions answered on IDNLearn.com. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
Sagot :
Using the z-distribution, it is found that the 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have that:
- 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].
- 49% out of 2001 U.S. adults play video games, hence [tex]\pi = 0.49, n = 2001[/tex].
The lower bound of the interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 - 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.4681[/tex]
The upper bound of the interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 + 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.5119[/tex]
The 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.
To learn more about the z-distribution, you can take a look at https://brainly.com/question/25730047
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
