Respuesta :
Answer:
A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).
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:
[tex]n = 150, \pi = 0.3[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 - 2.327\sqrt{\frac{0.3*0.7}{150}} = 0.213[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 + 2.327\sqrt{\frac{0.3*0.7}{150}} = 0.387[/tex]
A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).
