Respuesta :
Answer:0.55 ± 0.128
Step-by-step explanation:
Confidence interval is written as
Sample proportion ± margin of error
Margin of error = z × √pq/n
Where
z represents the z score corresponding to the confidence level
p = sample proportion. It also means probability of success
q = probability of failure
q = 1 - p
p = x/n
Where
n represents the number of samples
x represents the number of success
From the information given,
n = 100
x = 55
p = 55/1)00 = 0.55
q = 1 - 0.55 = 0.45
To determine the z score, we subtract the confidence level from 100% to get α
α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005
This is the area in each tail. Since we want the area in the middle, it becomes
1 - 0.005 = 0.995
The z score corresponding to the area on the z table is 2.58. Thus, confidence level of 99% is 2.58
Therefore, the 99% confidence interval is
0.55 ± 2.58√(0.55)(0.45)/100
= 0.55 ± 0.128
Using the z-distribution, it is found that the 99% confidence interval for the proportion of teachers in the state with a Master's degree is:
[tex]0.55 \pm 0.128[/tex]
Confidence interval of proportions
- 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:
- 55 out of 100 teachers in the sample have a Master's degree, hence [tex]n = 100, \pi = \frac{55}{100} = 0.55[/tex].
- 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].
Then, the margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}} = 2.575\sqrt{\frac{0.55(0.45)}{100}} = 0.128[/tex]
Hence, the interval is:
[tex]0.55 \pm 0.128[/tex]
To learn more about the z-distribution, you can take a look at https://brainly.com/question/25730047
