Respuesta :
Answer:
The 95% confidence interval for the true population proportion of adults with at least a high school education is (0.6819, 0.7381). This means that we are 95% sure that the true proportion of adults in the entire population surveyed with at least a high school education is (0.6819, 0.7381).
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 = 1000, \pi = \frac{710}{1000} = 0.71[/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}} = 0.71 - 1.96\sqrt{\frac{0.71*0.29}{1000}} = 0.6819[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 + 1.96\sqrt{\frac{0.71*0.29}{1000}} = 0.7381[/tex]
The 95% confidence interval for the true population proportion of adults with at least a high school education is (0.6819, 0.7381). This means that we are 95% sure that the true proportion of adults in the entire population surveyed with at least a high school education is (0.6819, 0.7381).
