Respuesta :
Answer:
The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)
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:
632 American, 462 don't smoke, 632 - 462 = 170 smoke.
Proportion who smoke, so
[tex]n = 632, \pi = \frac{170}{632} = 0.269[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.269 - 1.645\sqrt{\frac{0.269*0.731}{632}} = 0.24[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.269 + 1.645\sqrt{\frac{0.269*0.731}{632}} = 0.298[/tex]
The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)
