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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
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