Respuesta :
Answer:
The margin of error is of 3.97 percentage points.
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].
The confidence interval has the following margin of error
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence interval
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].
We also have that
[tex]n = 600, \pi = 0.56[/tex]
So, the margin of error is:
[tex]M = 1.96*\sqrt{\frac{0.56*0.44}{600}} = 0.0397[/tex]
So the margin of error is of 3.97 percentage points.
The margin of error for the 56% point estimate using a 95% confidence level is; 3.97%
Formula for margin of error here is;
M = z√(p(1 - p)/n)
Where;
z is critical value at given confidence level
p is sample proportion
n is sample size
We are given;
p = 56% = 0.56
n = 600
Confidence level = 95%
Now,from tables, the critical value at a confidence level of 95% is; z = 1.96
Thus;
M = 1.96√(0.56(1 - 0.56)/600)
M = 0.0397 or 3.97%
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