Respuesta :
Answer:
(72460, 97540)
Step-by-step explanation:
Given:-
- The sample size, n = 8
- The sample parameters of normal distributed data:
mean u = $85000
standard deviation s.d = $15000
Find:-
Construct a 95% confidence interval for the average starting salary of all computer scientists in gainesville.
Solution:-
- A random variable (X) denotes average starting salary of all computer scientists in gainesville has a sample normally distribution as follows:
X ~ N ( $85000 , $15000^2 )
- The general formulation for CI for known standard deviation with significance α = 1 - 0.95 = 0.05:
( u - Z_a/2*(s.d/√n) < X < u + Z_a/2*(s.d/√n) )
- The Z_a/2 = Z_0.025 = 1.96:
( 85,000 - 1.96*(15,000/√8) < X < 85,000 + 1.96*(15,000/√8) )
(72460, 97540)
The confidence interval for the average starting salary is (72460, 97540)
Calculation of the confidence interval:
Since The sample size, n = 8, the mean is $85,000, and the standard deviation is $15,000
Now the random variable should be in
X ~ N ( $85000 , $15000^2 )
Now standard deviation is
= 1 - 0.95
= 0.05
So, the confidence interval is
[tex]= ( 85,000 - 1.96\times (15,000\div \sqrt8) < X < 85,000 + 1.96\times (15,000\div \sqrt8) )[/tex]
= (72460, 97540)
hence, The confidence interval for the average starting salary is (72460, 97540)
Learn more about mean here: https://brainly.com/question/1863752
