Question

The world's smallest mammal is the bumblebee bat. The mean weight of 50 randomly selected bumblebee bats is 1.659 grams, with a standard deviation of 0.264 grams.
a) a 99.9% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal):
( , )
b) Find a 99% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal):
( ,   )
c) Find a 95% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal):
( ,  )
d) Find an 80% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal):
( ,   )

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.001 /2, 50- 1 ) = 3.5
1.659 ± t(0.001/2, 50 -1) * 0.264/√(50)
Lower Limit = 1.659 - t(0.001/2, 50 -1) 0.264/√(50)
Lower Limit = 1.53
Upper Limit = 1.659 + t(0.001/2, 50 -1) 0.264/√(50)
Upper Limit = 1.79
99.9% Confidence interval is ( 1.53 , 1.79 )

Part  b)
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 50- 1 ) = 2.68
1.659 ± t(0.01/2, 50 -1) * 0.264/√(50)
Lower Limit = 1.659 - t(0.01/2, 50 -1) 0.264/√(50)
Lower Limit = 1.56
Upper Limit = 1.659 + t(0.01/2, 50 -1) 0.264/√(50)
Upper Limit = 1.76
99% Confidence interval is ( 1.56 , 1.76 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 50- 1 ) = 2.01
1.659 ± t(0.05/2, 50 -1) * 0.264/√(50)
Lower Limit = 1.659 - t(0.05/2, 50 -1) 0.264/√(50)
Lower Limit = 1.58
Upper Limit = 1.659 + t(0.05/2, 50 -1) 0.264/√(50)
Upper Limit = 1.73
95% Confidence interval is ( 1.58 , 1.73 )

Part d)

  Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.2 /2, 50- 1 ) = 1.299
1.659 ± t(0.2/2, 50 -1) * 0.264/√(50)
Lower Limit = 1.659 - t(0.2/2, 50 -1) 0.264/√(50)
Lower Limit = 1.61
Upper Limit = 1.659 + t(0.2/2, 50 -1) 0.264/√(50)
Upper Limit = 1.71
80% Confidence interval is ( 1.61 , 1.71 )