Respuesta :
Answer:
b
Step-by-step explanation:
just took the test and got it right
Given that:
Number of samples from each high school = 30
Proportion of students from high school A = 20%
Proportion of students from high school B = 18%
Number of sample A ; n(A) = 30
Proportion of sample A ; p(A) = 0.20
Number of sample B ; n(B) = 30
Proportion of sample B ; p(B) = 0.18
To Prove:
10% condition met
The standard deviation of ˆ p A − ˆ p B
The difference (high school A – high school B) in the ________ that participate on a school athletic team _______ varies by about ____ from the true difference in proportions of _____.
Solution:
School A
Multiply Number of sample A to proportion of students at school A= nA×pA = 30 × 0.20 = 6.
So out of 30 students sample in School A, only 6 were the students who participated in Athletic team, and remained 24 students who were not the part of athletic.
School B
Multiply Number of sample B to proportion of students at school B= nB×pB = 30×0.18 = 5.4
So out of 30 students sample in School B, only 5.4 were the students who participated in Athletic team, and remained 24.6 students who were not the part of athletic.
The standard deviation of pA - pB is not approximately normal.
The value of nA.pA and nB.pB are less than 10.
Learn More: https://brainly.com/question/475676