Answer:
a) CI = ( 5,1 ; 5,7 )
b) SE = 0,1
Step-by-step explanation:
a) Sample random n = 100
Mean = μ = 5,4
Standard deviation s = 1,3
CI = 99 % α = 1 % α = 0,01 α/2 = 0,005
z(c) for 0,005 is from z-table z(c) = 2,575
z(c) = ( X - μ ) /s/√n CI = μ ± z(c) * s/√n
CI = 5,4 ± 2,575* 1,3/10
CI = 5,4 ± 0,334
CI = ( 5,1 ; 5,7 )
b) SE = Standard deviation / √n
SE = 1,3 /10 SE = 0,1
We can support that with 99 % of probability our random variable will be in the CI.
