式-003
\( S(xy) = \displaystyle \sum_{ i=1 }^n ( x_{ i } – \bar{ x })( y_{ i } – \bar{ y }) \)
\( S(xy) = \displaystyle \sum_{ i=1 }^n x_{ i }y_{ i } – n \sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i } \)
\( S(xy) = \displaystyle \sum_{ i=1 }^n x_{ i }y_{ i } – \displaystyle\frac{\sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i } }{ n } \)
\( S(xy) = – \displaystyle\frac{\sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i } }{ n } \)
\( S(xy) = \displaystyle \sum_{ i=1 }^n x_{ i }y_{ i } – \displaystyle\frac{ 1 }{ n } \sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i } \)
\( \bar{ x } \) \( V_{ x } \) \( s_{ x } \) \( z = ax + b \) \( \bar{ z } = a \bar{ x } + b \)
\( V_{ z } = a^2 \cdot V_{ x } \) \( s_{ z } = | a | s_{ x } \)
\( \tilde{ x } , \tilde{ y } \)
\( t = \displaystyle\frac{ r \sqrt{ n – 2 } }{ \sqrt{ 1 – r^2 }} \)
\( t = 0.631 \times \sqrt{ (20 – 2 ) }/ \sqrt{ 1 – 0.631^2 } = 3.451 \)
\( t = 0.234 \times \sqrt{ (30 – 2 ) }/ \sqrt{ 1 – 0.234^2 } = 1.274 \)
抜取検査
\( P(x, n | p, N ) = \displaystyle\frac{ \displaystyle{{N-p・N} \choose {n-x}} \cdot \displaystyle{{p・N} \choose x }}{ \displaystyle{ N \choose n}} \)
\( P(x, n | p, N ) \) \( P(x, n \vert p, N ) \)
\( n/N \leq 1/10 \)
\( P( x, n | p ) = \displaystyle{ n \choose x } \cdot p^x \cdot (1 – p ) ^ {n – x} \)
\( n/N \leq 1/10 \) \( p \leq 0.1 \)
\( P( x, np ) = \displaystyle\frac{ e^{ -np } \cdot (np)^x }{ x! } \)
\( L ( p ) = \displaystyle \sum_{ x=0 }^c \displaystyle{ n \choose x } \cdot p^x \cdot (1 – p ) ^ { n – x } \)
\( L ( p ) = \displaystyle \sum_{ x=0 }^c \displaystyle{ n \choose x } \displaystyle\frac{ e^{ -np } \cdot (np)^x }{ x! } \)
\( \displaystyle{ {N-x} \choose {Y-n}} \)
\( P(x, n | p, N ) = \displaystyle\frac{ \displaystyle{{N-p・N} \choose {n-x}} \cdot \displaystyle{{p・N} \choose x }}{ \displaystyle{ N \choose n}} \)
統計的方法の基礎
\( \bar{ x } \)
\( E( \bar{ x }) = \mu \) \( D( \bar{ x } = \sigma / \sqrt{ n }) \)
\( N( \mu, \sigma^2 / n) \)
\( E( x_{ i } ) = \mu \) \( V( x_{ i } ) = \sigma^2 \)
\( n \rightarrow \infty \)
\( \bar{ X_{ n }} = \displaystyle\frac{ X_{ 1 } + X_{ 2 } + \cdots + X_{ n }}{ n } \rightarrow \mu \)
\( n \rightarrow \infty \) \( P( | \bar{ X_{ n}} – \mu | \gt \epsilon ) \rightarrow 0 \)
\( u = \displaystyle\frac{ \bar{ x } – \mu }{ \displaystyle\frac{ \sigma }{ \sqrt{ n }}} \)