Using the information given, it is found that the 95% confidence interval for the proportion of all U.S. adults who experienced depression in that year is (0.075, 0.081).
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].
3156 out of 40467 adults experienced depression, hence:
[tex]n = 40467, \pi = \frac{3156}{40467} = 0.078[/tex]
95% confidence level
So [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}} = 0.078 - 1.96\sqrt{\frac{0.078(0.922)}{40467}} = 0.075[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.078 + 1.96\sqrt{\frac{0.078(0.922)}{40467}} = 0.081[/tex]
The 95% confidence interval for the proportion of all U.S. adults who experienced depression in that year is (0.075, 0.081).
A similar problem is given at https://brainly.com/question/16807970
