Respuesta :
Answer:
The 90% confidence interval estimate for the proportion of all the voters who are in favor of stricter gun control measures is (0.318, 0.382).
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 interval [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:
A poll of 600 voters showed 210 that were in favor of stricter gun control measures. This means that [tex]n = 600, p = \frac{210}{600} = 0.35[/tex]
90% confidence interval
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{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}} = 0.35 - 1.645\sqrt{\frac{0.35*0.65}{600}} = 0.3180[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.35 + 1.645\sqrt{\frac{0.35*0.65}{600}} = = 0.382[/tex]
The 90% confidence interval estimate for the proportion of all the voters who are in favor of stricter gun control measures is (0.318, 0.382).
