Respuesta :
Answer:
a) The 99% confidence interval for the proportion of all students who had use of a computer at home and give it in interval notation is (0.709, 0.911).
b) 0.81
c) 0.039.
d) 0.101
Step-by-step explanation:
Question a:
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home.
This means that [tex]n = 100, \pi = \frac{81}{100} = 0.81[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.81 - 2.575\sqrt{\frac{0.81*0.19}{100}} = 0.709[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.81 + 2.575\sqrt{\frac{0.81*0.19}{100}} = 0.911[/tex]
The 99% confidence interval for the proportion of all students who had use of a computer at home and give it in interval notation is (0.709, 0.911).
b) Give the value of the point estimate described in this scenario.
Sample proportion of [tex]\pi = 0.81[/tex]
c) Give the value of the standard error for the point estimate.
This is:
[tex]s = \sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.81*0.19}{100}} = 0.039[/tex]
The standard error is of 0.039.
d) Give the value of the margin of error if you were to calculate a 99% confidence interval.
This is:
[tex]M = zs = 2.575*0.039 = 0.101[/tex]
