A school administrator wants to know what proportion of the approximately 80,000 teachers in their state
have a Master's degree. The administrator takes an SRS of 100 teachers from a statewide database
containing every teacher, and they find 55 teachers in the sample have a Master's degree.
Based on this sample, which of the following is a 99% confidence interval for the proportion of teachers
in the state with a Master's degree?
Choose 1
~
0.55 +0.128
Ⓡ
0.55 = 0.098
©
55 +0.128
©
55 = 0.098

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