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H0H1
H1
H0
H0H1
H0H0H1
H0
α
H0H1
5%; 1%; 0,1%
H0
H0
H0
H0
H1α H0
H0
H0
H0
H1
H1

α
α= 5% 5%
H0
αobs
H1
H0
H0
H1
H0
α
αobs ≤α H0H1α
αobs > α H0H1β
α H0H1H1
β H0H1H1
= 1 −β H1H1

T1
T2X=T1−T2
X=T1−T2
X
µ
n= 24 <30
X
X N(µ;σ)
µ σ
H0µ= 0
H1µ > 0
α= 1 %
x=t1−t2>0
H1
n= 24 xi
¯x= 4,257 s∗= 12,08.
Xn
¯x= 4,257
H0
X H0
X
H0XnN(0 ; σ
√n)

σ
Xn
Tn=Xn−0
S∗
n
√n
H0Tnn−1 = 23
Tn, α = 1 %
Tn
H0Xn
H1Xn
XnH1
¯
Xn.
Tn
Tn
IA = ]−∞;t1−α] = ]−∞; 2,5] t1−α= 2,5 1 −α= 0,99
Tn
t=4,257 −0
12,08
√24
= 1,726
t∈IA, H0H1.
β
α= 1 %
α=PH0¯ Xn≥4,257=PH0(Tn≥1,726) .
α
α= 0,0489 = 4,89 %.
α= 4,89% > α = 1% H0
T1−T2T2−T1
X
X
X