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 = [tex]\dfrac{ 26.2+ 27.7+ 8.6+ 3.8 +11.6 }{5}[/tex]
mean = 15.58
the standard deviation [tex]\sigma[/tex] = [tex]\sqrt{\dfrac{\sum(x_i - \mu)^2 }{n}}[/tex]
the standard deviation = [tex]\sqrt{\dfrac{(26.2 - 15.58)^2 +(27.7 - 15.58)^2 +(8.6 - 15.58)^2 + (3.8 - 15.58)^2 + (11.6 - 15.58)^2 }{5 } }[/tex]
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 = [tex]t \times \dfrac{\sigma}{\sqrt{n}}[/tex]
The Margin of Error M.O.E = [tex]1.533 \times \dfrac{9.62297}{\sqrt{5}}[/tex]
The Margin of Error M.O.E = [tex]1.533 \times 4.3035[/tex]
The Margin of Error M.O.E = 6.60
The Confidence interval = ( [tex]\mu \pm M.O.E[/tex] )
The Confidence interval = ( [tex]\mu + M.O.E[/tex] , [tex]\mu - M.O.E[/tex] )
The Confidence interval = ( 15.58 - 6.60 , 15.58 + 6.60)
The Confidence interval = (8.98 , 22.18)
