You intend to estimate a population mean μ from the following sample. 26.2 27.7 8.6 3.8 11.6 You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places.
Answer:
The Confidence interval = (8.98 , 22.18)
Step-by-step explanation:
From the given information:
mean = 
mean = 15.58
the standard deviation
= 
the standard deviation = 
standard deviation = 9.62297
Degrees of freedom df = n-1
Degrees of freedom df = 5 - 1
Degrees of freedom df = 4
For df at 4 and 80% confidence level, the critical value t from t table = 1.533
The Margin of Error M.O.E = 
The Margin of Error M.O.E = 
The Margin of Error M.O.E = 
The Margin of Error M.O.E = 6.60
The Confidence interval = (
)
The Confidence interval = (
,
)
The Confidence interval = ( 15.58 - 6.60 , 15.58 + 6.60)
The Confidence interval = (8.98 , 22.18)