Respuesta :
Answer:
The margin of error for a 96% confidence interval is of 0.0114 = 1.14%.
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].
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
For this problem, we have that:
[tex]n = 400, p = \frac{395}{400} = 0.9875[/tex]
96% confidence level
So [tex]\alpha = 0.04[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.056[/tex].
Calculate the margin of error for a 96% confidence interval.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 2.056\sqrt{\frac{0.9875(0.0125)}{400}}[/tex]
[tex]M = 0.0114[/tex]
The margin of error for a 96% confidence interval is of 0.0114 = 1.14%.
