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ROZKŁAD POISSONA:
$$P\left( X = k \right) = \frac{\lambda^{k}}{k!}e^{- \lambda};\ \ \ \ \lambda > 0$$
EX = λ; λ = EX = D2X = np
D2X = λ e = 2, 7182818 |
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FUNKCJA GĘSTOŚCI:
f(x) ≥ 0,
∫−∞+∞f(x)dx = 1;
∫abf(x)dx = P(a<X≤b)
F(x) = ∫−∞xf(t)dt = P(X<x)
P(X≤a) = F(a)
EX = ∫−∞+∞xf(x)dx
D2X = ∫−∞+∞[x−EX]2f(x)dx
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ROZKŁAD χ2:
z k stopniami swobody:
χ12 + χ22+…+χk2
EX = k
D2X = 2k
1 − α |
$$EX = \sum_{}^{}{x_{i}p_{i}}$$
$$D^{2}X = \sum_{}^{}{\left( x_{i} - EX \right)^{2}p_{i}} = EX^{2} - \left( \text{EX} \right)^{2}$$
$$DX = \sqrt{D^{2}X}$$
$$\sum_{}^{}{p_{i} = 1}$$
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ROZKŁAD JEDNOSTAJNY:
$$f\left( x \right) = \left\{ \begin{matrix}
0\ , \\
\frac{1}{\begin{matrix}
b - a \\
0, \\
\end{matrix}} \\
\end{matrix}\ , \right.\ \text{\ \ \ \ \ }\begin{matrix}
x < a \\
a \leq x \leq b \\
x > b \\
\end{matrix}$$
$$F\left( x \right) = \left\{ \begin{matrix}
0\ , \\
\frac{x - a}{\begin{matrix}
b - a \\
1, \\
\end{matrix}} \\
\end{matrix}\ , \right.\ \text{\ \ \ \ \ }\begin{matrix}
x < a \\
a \leq x \leq b \\
x > b \\
\end{matrix}$$
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ROZKŁAD T-STUDENTA:
z k stopniami swobody:
$$T_{k} = \frac{T}{\sqrt{\chi_{\alpha}^{2}}}\sqrt{k}$$
EX = 0
$$D^{2}X = \frac{k}{k - 2}$$
$$\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{\left( 1 - \alpha \right)}{2}$$
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DYSTRYBUANTA:
F(X) = P(X≤x)
P(X≤a) = F(a)
P(X>a) = 1 − F(a)
P(a<X<b) = F(b) − F(a)
P(X<a) = F(a) − P(X=a)
P(a≤X<b) = F(b) − F(a) − P(X=b) + P(X=a)
F(−1) = P(X≤−1) = 1 − F(1)
F(1) = P(X≤1)
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$$EX = \frac{a + b}{2}$$
$$D^{2}X = \frac{\left( b - a \right)^{2}}{12}$$
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TESTY NIEZALEŻNOŚCI:
H0 : P(X=xi, Y=yi) = P(X=xi)P(Y=yi)
H1 : nie
$\chi^{2} = \frac{\sum_{}^{}\left( n_{\text{ij}} - np_{\text{ij}} \right)^{2}}{np_{\text{ij}}}$ H0 − odrzucic : χ2 ≥ χα2
α = (s−1)(k−1)
$$T = \sqrt{\frac{\chi^{2}}{n\left( s - 1 \right)\left( k - 1 \right)}}$$
T = 0; T = 1; T ∈ ⟨0;1⟩
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ROZKŁAD WYKŁADNICZY:
$$f\left( x \right) = \left\{ \begin{matrix}
0, \\
\lambda e^{- \lambda x}, \\
\end{matrix} \right.\ \text{\ \ \ \ }\begin{matrix}
x < 0 \\
x \geq 0 \\
\end{matrix}$$
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ROZKŁAD BERNOULLIEGO:
$P\left( X = k \right) = \begin{pmatrix}
n \\
k \\
\end{pmatrix}p^{k}\left( 1 - p \right)^{n - k}$ $\begin{pmatrix}
n \\
k \\
\end{pmatrix} = \frac{n!}{k!\left( n - k \right)!}$
$\begin{pmatrix}
n \\
0 \\
\end{pmatrix} = 1$ 0!=1
$\begin{pmatrix}
n \\
n \\
\end{pmatrix} = 1$ 1!=1
$$\begin{pmatrix}
n \\
1 \\
\end{pmatrix} = n$$
$$\begin{pmatrix}
n \\
n - k \\
\end{pmatrix} = \begin{pmatrix}
n \\
k \\
\end{pmatrix} = n$$
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$$F\left( x \right) = \left\{ \begin{matrix}
0, \\
1 - e^{- \lambda x},\ \ \ \ \ \\
\end{matrix}\begin{matrix}
x < 0 \\
x \geq 0 \\
\end{matrix} \right.\ $$
$$EX = \frac{1}{\lambda};\ \ \ \lambda = \frac{1}{\text{EX}}$$
$$D^{2}X = \frac{1}{\lambda^{2}}$$
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MODEL LINIOWY:
y = ax + b y = axb lna = x
$a = r\frac{\text{Sy}}{\text{Sx}}$ y = aebx a = ex
$b = \overset{\overline{}}{y} - a\overset{\overline{}}{x}$ y = ax2 + b
$$r = \frac{\sum_{}^{}{\left( x_{i} - \overset{\overline{}}{x} \right)\left( y_{i} - \overset{\overline{}}{y} \right)}}{\text{nSxSy}};\ \ \ \ \ \ \ \ r \in \left\langle - 1;1 \right\rangle$$
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LOSOWANIE NIEZALEŻNE:
PRZEDZIAŁ UFNOŚCI DLA ŚREDNIEJ
I model
$$P\left\{ \overset{\overline{}}{x} - u_{\alpha}\frac{\sigma}{\sqrt{n}} < m < \overset{\overline{}}{x} + u_{\alpha}\frac{\sigma}{\sqrt{n}} \right\} = 1 - \alpha$$
II model
$$P\left\{ \overset{\overline{}}{x} - t_{\alpha}\frac{s}{\sqrt{n - 1}} < m < \overset{\overline{}}{x} + t_{\alpha}\frac{s}{\sqrt{n - 1}} \right\} = 1 - \alpha$$
III model
$$P\left\{ \overset{\overline{}}{x} - u_{\alpha}\frac{s}{\sqrt{n}} < m < \overset{\overline{}}{x} + u_{\alpha}\frac{s}{\sqrt{n}} \right\} = 1 - \alpha$$
$\mathbf{t}_{\mathbf{\alpha,n - 1}}\mathbf{\text{\ \ \ \ \ \ \ \ \ \ }}\mathbf{u}_{\mathbf{\alpha}}\mathbf{= 1 -}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\alpha}$ |
LOSOWANIE ZALEŻNE:
ESTYMACJA ŚREDNIEJ:
I model
$$P\left\{ \overset{\overline{}}{x} - u_{\alpha}\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}} < m < \overset{\overline{}}{x} + u_{\alpha}\frac{s}{\sqrt{n}}\sqrt{1 - \frac{n}{N}} \right\} = 1 - \alpha$$
II model
$$P\left\{ \overset{\overline{}}{x} - t_{\alpha}\frac{\hat{s}}{\sqrt{n}}\sqrt{1 - \frac{n}{N}} < m < \overset{\overline{}}{x} + t_{\alpha}\frac{\hat{s}}{\sqrt{n}}\sqrt{1 - \frac{n}{N}} \right\} = 1 - \alpha$$
$$n = \frac{N}{1 + \frac{Nd^{2}}{{u_{\alpha}^{2}s}^{2}}}$$
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PRZEDZIAŁ UFNOŚCI DLA PROCENTÓW/WSKAŹNIKA STRUKTURY:
$$P\left\{ \frac{m}{n} - u_{\alpha}\sqrt{\frac{\frac{m}{n}\left( 1 - \frac{m}{n} \right)}{n}} < p < \frac{m}{n} + u_{\alpha}\sqrt{\frac{\frac{m}{n}\left( 1 - \frac{m}{n} \right)}{n}} \right\} = 1 - \alpha$$
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LOSOWANIE WARSTWOWE:
ESTYMACJA ŚREDNIEJ POPULACJI:
$$P\left\{ \overset{\overline{}}{x_{w}} - u_{\alpha}D\left( \overset{\overline{}}{x_{w}} \right) < m < \overset{\overline{}}{x_{w}} + u_{\alpha}D\left( \overset{\overline{}}{x_{w}} \right) \right\} = 1 - \alpha$$
$D^{2}\left( \overset{\overline{}}{x_{w}} \right)\mathbf{=}\left( \frac{\mathbf{1}}{\mathbf{n}}{\mathbf{-}\frac{\mathbf{1}}{\mathbf{N}}} \right)\sum_{h = 1}^{L}{W_{h}s_{h}^{2}}$ $\overset{\overline{}}{x_{w}}\mathbf{=}\sum_{h = 1}^{L}{W_{h}\overset{\overline{}}{x_{h}}}$
$n_{h} = \frac{N_{h}}{N}n = W_{h}n$ $W_{h} = \frac{N_{h}}{N}$ $n = \frac{\sum_{}^{}{W_{h}s_{h}^{2}}}{\frac{d^{2}}{u_{\alpha}^{2}} + \frac{\mathbf{1}}{\mathbf{N}}\sum_{}^{}{W_{h}s_{h}^{2}}}$ |
PRZEDZIAŁ UFNOŚCI DLA WARIANCJI:
I model
$$P\left\{ \frac{ns^{2}}{c_{2}} < \sigma^{2} < \frac{ns^{2}}{c_{1}} \right\} = 1 - \alpha;\ \ \ \ \left\{ \begin{matrix}
c_{1}\ \ \ \ dla\ \ \ \ n - 1\ \ \ i\ \ \ 1 - \frac{1}{2}\alpha \\
c_{2}\ \ dla\ \ \ \ \ n - 1\ \ \ i\ \ \ \frac{1}{2}\alpha \\
\end{matrix} \right.\ $$
$$s^{2} = \frac{\frac{\sum_{}^{}\left( x_{i} - \overset{\overline{}}{x} \right)^{2}}{n}}{\bullet n};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ns^{2} = \sum_{}^{}\left( x_{i} - \overset{\overline{}}{x} \right)^{2}$$
II model
$$P\left\{ \frac{s}{1 + \frac{u_{\alpha}}{\sqrt{2n}}} < \sigma < \frac{s}{1 - \frac{u_{\alpha}}{\sqrt{2n}}} \right\} = 1 - \alpha$$
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TESTY PARAMETRYCZNE:
TEST DLA WARTOŚCI ŚREDNIEJ POPULACJI:
H0 : m = m0 H0 : m = m0 H0 : m = m0 ∖ nH1 : m ≠ m0 H1 : m > m0 H1 : m < m0
I, III model (H0 − odrzucic)
|u| ≥ uα u ≥ uα u ≤ uα
$1 - \frac{1}{2}\alpha$ 1 − α α
II model (H0 − odrzucic)
|t| ≥ tα t ≥ tα t ≤ −tα
α 2α 2α |
WYZNACZENIE NIEZBĘDNEJ LICZBY POMIARU DO PRÓBY:
I model
$$n = \frac{u_{\alpha}^{2}\sigma^{2}}{d^{2}}$$
II model
$$n = \frac{t_{\alpha}^{2}{\hat{s}}^{2}}{d^{2}};\ \ \ \ \ \ \ \ \ {\hat{s}}^{2} = \frac{1}{n - 1}\sum_{}^{}{\left( x_{i} - \overset{\overline{}}{x} \right)^{2}n_{i}}$$
III model
$$n = \frac{u_{\alpha}^{2}p\left( 1 - p \right)}{d^{2}};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{u_{\alpha}^{2}}{{4d}^{2}}$$
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I model
$$u = \frac{\overset{\overline{}}{x} - m_{0}}{\sigma}\sqrt{n}$$
II model
$$t = \frac{\overset{\overline{}}{x} - m_{0}}{\sigma}\sqrt{n - 1} = \frac{\overset{\overline{}}{x} - m_{0}}{\hat{s}}\sqrt{n}\text{\ \ \ \ \ \ \ \ \ \ \ \ }\hat{s} = \sqrt{\frac{\sum_{}^{}\left( x_{i} - \overset{\overline{}}{x} \right)^{2}}{n - 1}}$$
III model
$$u = \frac{\overset{\overline{}}{x} - m_{0}}{s}\sqrt{n}$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = \frac{\frac{m}{n} - p}{\sqrt{\frac{p\left( 1 - p \right)}{n}}}$ $\overline{p} = \frac{m_{1} + m_{2}}{n_{1} + n_{2}}$ $n = \frac{n_{1}*\ n_{2}}{n_{1} + n_{2}}$ $u = \ \frac{\frac{m_{1}}{n_{1}} - \frac{m_{2}}{n_{2}}}{\sqrt{\frac{\overline{p}(1 - \overline{p})}{n}}}$ |
ANALIZA WARIANCJI – TEST:
H0 : m1 = m2 = … = mk $\left. \ \begin{matrix}
\alpha \\
k - 1\ oraz\ n - k \\
\end{matrix} \right\} F_{\alpha}$
H1 : nie F ≥ Fα; H0 − odrzucic,
ma wplyw
(czynnik) między grupami |
$$\sum_{}^{}\left( {\overset{\overline{}}{x}}_{i} - \overset{}{x} \right)^{2}n_{i}$$
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k − 1
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$${\hat{s}}_{1}^{2}$$
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$$F = \frac{{\hat{s}}_{1}^{2}}{{\hat{s}}_{2}^{2}}$$
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wewnątrz grup (składnik losowy) |
$$\sum_{}^{}{\sum_{}^{}\left( x_{\text{ij}} - \overset{\overline{}}{x_{i}} \right)^{2}}$$
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n − k
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$${\hat{s}}_{2}^{2}$$
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H0 : m1 = m2 = … = mk FA ≥ FαA; H0 − odrzucic, ma wplyw
H1 : nie FB ≥ FαB; H0 − odrzucic, ma wplyw
między grupami
(czynnik A) |
SKA
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r − 1
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$${\hat{s}}_{1}^{2}$$
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$$F_{A} = \frac{{\hat{s}}_{1}^{2}}{{\hat{s}}_{3}^{2}}$$
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między grupami
(czynnik B) |
SKB
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k − 1
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$${\hat{s}}_{2}^{2}$$
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$$F_{B} = \frac{{\hat{s}}_{2}^{2}}{{\hat{s}}_{3}^{2}}$$
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wewnątrz grup |
SKR
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(r−1)(k−1)
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$${\hat{s}}_{3}^{2}$$
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TEST LUKI:
$\overset{\overline{}}{x_{1}} < \overset{\overline{}}{x_{2}} < \ldots < \overset{\overline{}}{x_{n}}$ $\overset{\overline{}}{x_{2}} - \overset{\overline{}}{x_{1}} > L - istotne$
$$L = t_{\alpha}\sqrt{\frac{2{\hat{s}}_{2}^{2}}{d}};\ \ \ \ \ \ d = \frac{n}{k}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }t_{\alpha}\ \ \ \ \ dla\ \ \ \ n - k$$
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$$SKC = \sum_{}^{}{\sum_{}^{}\left( x_{\text{ij}} - \overset{}{x} \right)^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ SKR = SKC - SKA - SKB$$
$$SKA = k\sum_{}^{}\left( \overset{\overline{}}{x_{i}} - \overset{}{x} \right)^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\left. \ \begin{matrix}
\alpha \\
r - 1\ oraz\ \left( r - 1 \right)\left( k - 1 \right) \\
\end{matrix} \right\} F_{\text{αA}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$$
$$SKB = r\sum_{}^{}\left( \overset{\overline{}}{x_{j}} - \overset{}{x} \right)^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ }\left. \ \begin{matrix}
\alpha \\
\ k - 1\ oraz\ \left( r - 1 \right)\left( k - 1 \right) \\
\end{matrix} \right\} F_{\text{αB}}\ $$
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