Respuesta :
Answer:
The 95% confidence interval for the population proportion of Americans over 42 who smoke is (0.265, 0.325).
Step-by-step explanation:
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]1-\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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
897 people, of which 897 - 632 = 265 smoke.
Then [tex]n = 897, \pi = \frac{265}{897} = 0.295[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{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}} = 0.295 - 1.96\sqrt{\frac{0.295*0.705}{897}} = 0.265[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.295 - 1.96\sqrt{\frac{0.295*0.705}{897}} = 0.325[/tex]
The 95% confidence interval for the population proportion of Americans over 42 who smoke is (0.265, 0.325).
