Option B. ($99,242.00, $100,758.01)
a) For n - 1 = 49 degrees of freedom, we have form t distribution tables:
P( t < 2.010) = 0.975
Therefore, P(-2.010 < t < 2.010) = 0.95
Therefore the confidence interval here is obtained as:
X +₋ t 8/√ⁿ
100 \pm 2.010*\frac{2}{\sqrt{50}}
100 \pm 0.5685
(99431.61, 100568.39)
Therefore A is the correct interval here.
Q2) From t distribution tables, we have here:
P(-2.680 < t < 2.680) = 0.99
Therefore the confidence interval here is obtained as:
100 \pm 2.010*\frac{2}{\sqrt{50}}
100 \pm 0.7580
(99242, 100785)
1) A university wants to determine the average salary of its graduating computer science majors. The university asks the graduating classes of 50 students to share their starting salaries when hired. The 50 students had an average salary of $100,000 with a standard deviation of $2,000. Find a 95% confidence interval for the average starting salary of computer science majors from the university.
a - ($99,431.61, $100,568.39)
b - ($99,445.64, $100,554.36)
c- ($99,982.26, $100,017.74)
d - The salaries are not normally distributed so we cannot compute a confidence interval
2) Find a 99% confidence interval for the previous problem.
a - ($99,271.44, $100,728.56)
b - ($99,242.00, $100,758.01)
c - ($99,996.46, $100,003.55)
d - The salaries are not normally distributed so we cannot compute a confidence interval
Learn more about confidence intervals at
https://brainly.com/question/15712887
#SPJ4
