Теория вероятностей и математическая статистика. Билялов Р.Ф. - 91 стр.

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α = P
H
0
(S
C
) =
1
√
2π
∞
Z
C+1.377
1.469
e
−
t
2
2
dt.
H
1
: m
2
∼ N(15p
(1)
2
, 15p
(1)
2
q
(1)
2
) = N(6.3, 1.912
2
),
m
3
∼ N(15p
(1)
3
, 15p
(1)
3
q
(1)
3
) = N(2.7, 1.488
2
). η
2
=
m
2
−6.3
1.912
,
η
2
=
m
3
−2.7
1.488
. P (m
2
+ m
3
> 15) = P (0.789η
2
+ 0.614η
3
> 2.248) =
0.007.
β = P
H
1
(
¯
S
C
) = P (0.049(1.912η
2
+ 6.3) − 0.798(1.488η
3
+ 2.7) ≤ C) =
= (0.079η
2
− 0.997η
3
≤
C − 1.846
1.191
) =
1
√
2π
∞
Z
−
C−1.846
1.191
e
−
t
2
2
dt.
α = β
C+1.377
1.469
= −
C−1.846
1.191
, C = −1.63.
0.049m
2
− 0.798m
3
H
0
−2.2 −3.6 −3.70 −6.3 −6.30 −6.3 −1.30 −2.10 −4.50 −5.6
H
1
−0.5 −1.2 −2.15 −1.5 −0.31 −2.1 −0.65 −2.25 −1.15 −1.20
n = 16
¯
X = 20
σ = 4
σ =
4
(
¯
X−m)
√
n
σ
,
α =
0.05 α = 0.005.
1
√
2π
∞
R
u
α
e
−
u
2
2
du = α
u
0.05
= 1.65 u
0.005
= 2.58
P (20 − 1.65 < m < 20 + 1.65) = 0.9 P (20 − 2.26 < m < 20 + 2.26) =
0.99.
Îøèáêà ïåðâîãî ðîäà
                                              Z∞
                                   1                  t2
                  α = PH0 (SC ) = √                e− 2 dt.
                                   2π
                                        C+1.377
                                         1.469

                                        (1)        (1) (1)
    Äëÿ ãèïîòåçû H1 : m2 ∼ N (15p2 , 15p2 q2 ) = N (6.3, 1.9122 ),
             (1)   (1) (1)
m3 ∼ N (15p3 , 15p3 q3 ) = N (2.7, 1.4882 ). Ïîëîæèì η2 = m1.9122 −6.3
                                                                       ,
     m3 −2.7
η2 = 1.488 . Òîãäà P (m2 + m3 > 15) = P (0.789η2 + 0.614η3 > 2.248) =
0.007. Ýòîé âåðîÿòíîñòüþ òîæå ïðåíåáðåãàåì.

β = PH1 (S̄C ) = P (0.049(1.912η2 + 6.3) − 0.798(1.488η3 + 2.7) ≤ C) =
                                                             Z∞
                              C − 1.846     1                          t2
       = (0.079η2 − 0.997η3 ≤           )= √                       e− 2 dt.
                                1.191       2π
                                                       − C−1.846
                                                          1.191

Èç α = β ñëåäóåò C+1.377     C−1.846
                    1.469 = − 1.191 , îòêóäà C = −1.63. Çíà÷åíèÿ
0.049m2 − 0.798m3 :
  H0     −2.2|−3.6|−3.70|−6.3|−6.30|−6.3|−1.30|−2.10|−4.50|−5.6
                                                                .
  H1    −0.5|−1.2|−2.15|−1.5|−0.31|−2.1|−0.65|−2.25|−1.15|−1.20
×àñòîòà îøèáêè ïåðâîãî ðîäà: 0.1, âòîðîãî ðîäà  0.3.
    Çàäà÷à 15.170.  ðåçóëüòàòå n = 16 íàáëþäåíèé äëÿ åìêî-
ñòè êîíäåíñàòîðà íàéäåíî X̄ = 20 ìêô. Íàéòè 90%-íûå è 99%-íûå
äîâåðèòåëüíûå èíòåðâàëû äëÿ ìàòåìàòè÷åñêîãî îæèäàíèÿ åìêîñòè.
Ñðåäíåêâàäðàòè÷íîå îòêëîíåíèå σ = 4 ìêô.
    Ðåøåíèå. Òàê êàê ñðåäíåêâàäðàòè÷íîå îòêëîíåíèå
                                                √      èçâåñòíî, σ =
                                          (X̄−m) n
4 ìêô, òî íóæíî èñïîëüçîâàòü ñòàòèñòèêó       σ    , èìåþùóþ íîð-
ìàëüíîå ñòàíäàðòíîå ðàñïðåäåëåíèå. Ðàññìàòðèâàåì äâà ñëó÷àÿ: α =
                                                      R∞ u2
0.05 è α = 0.005. Â òàáëèöå ðåøåíèé óðàâíåíèÿ √12π e− 2 du = α
                                                                  uα
íàõîäèì ñîîòâåòñòâåííî çíà÷åíèÿ u0.05 = 1.65 è u0.005 = 2.58. Òîãäà
P (20 − 1.65 < m < 20 + 1.65) = 0.9 è P (20 − 2.26 < m < 20 + 2.26) =
0.99.
    Çàäà÷à 15.196. Èç áîëüøîé ïàðòèè òðàíçèñòîðîâ îäíîãî òè-
ïà áûëè ñëó÷àéíûì îáðàçîì îòîáðàíû è ïðîâåðåíû 100 øòóê. Ó

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