Respuesta :
Answer:
Step-by-step explanation:
From the information given,
Number of sample, n = 200
Probability of success, p = 43/100 = 0.43
q = 1 - p = 1 - 0.43
q = 0.57
For a confidence level of 95%, the corresponding z value is 1.96.
The formula for determining the error bound for the proportion is
z × √pq/n
= 1.96 ×√(0.43 × 0.57)/200
= 1.96 × 0.035 = 0.0686
The upper boundary of the population proportion is
0.43 + 0.0686 = 0.5
The lower boundary of the population proportion is
0.43 - 0.0686 = 0.4
The error in the solution is
phat = 0.43/200 = 0.00215
Also,
[-0.0043, 0.0086] is wrong
From the information given, the 95% confidence interval for the population proportion is: (0.3614, 0.4986).
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]\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]\frac{1+\alpha}{2}[/tex].
In this problem:
- Sample of 200 people, thus [tex]n = 200[/tex].
- Proportion of 43%, thus [tex]p = 0.43[/tex].
95% confidence level
So [tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{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}} = 0.43 - 1.96\sqrt{\frac{0.43*0.57}{200}} = 0.3614[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.43 + 1.96\sqrt{\frac{0.43*0.57}{200}} = 0.4986[/tex]
The 95% confidence interval for the population proportion is: (0.3614, 0.4986).
A similar problem is given at https://brainly.com/question/16807970
