Diszkrét illeszkedésvizsgálat
Adott egy n n n r r r x 1 , x 2 , … x r x_1, x_2, \dots x_r x 1 , x 2 , … x r
Az x i x_i x i v i ( i = 1 , 2 , … , r ) v_i \; (i=1,2,\dots,r) v i ( i = 1 , 2 , … , r )
Pr ( X j = x j ) = p j ( j = 1 , … , r ) \Pr(X_j = x_j) = p_j \quad (j=1,\dots,r) Pr ( X j = x j ) = p j ( j = 1 , … , r ) Ellenhipotézisünk, hogy a minta nem a nullhipotézisben feltett eloszlást követi, azaz van olyan j j j
A próbastasztikát a következő módon definiáljuk:
T n = ∑ j = 1 r ( v j − n ⋅ p j ) 2 n ⋅ p j T_n = \sum_{j=1}^{r}{\frac{\left( v_j - n \cdot p_j \right)^2}{n \cdot p_j}} T n = j = 1 ∑ r n ⋅ p j ( v j − n ⋅ p j ) 2 ahol:
v j v_j v j p j p_j p j n ⋅ p j n \cdot p_j n ⋅ p j
Ha a nullhipotézis teljesül a próbastasztika χ 2 \chi^2 χ 2 ( r − 1 ) (r-1) ( r − 1 )
Egy gyárban egy termék minőségét 4 elemű mintákat véve ellenőrzik, havonta 300 mintavétellel. Megszámolták, hogy a
legutóbbi hónapban hány alkalommal volt a mintában j j j
( 80 , 113 , 77 , 27 , 3 ) \left( 80, \; 113, \; 77, \; 27, \; 3 \right) ( 80 , 113 , 77 , 27 , 3 ) Modellezhető-e a minta ( 4 ; 0 , 25 ) (4; 0,25) ( 4 ; 0 , 25 )
H 0 : H_0: H 0 : B i n ( 4 ; 0 , 25 ) \mathrm{Bin}{(4; 0,25)} Bin ( 4 ; 0 , 25 )
A nullhipotézisről α = 0 , 05 \alpha = 0,05 α = 0 , 05
Ahhoz hogy a hipotézist vizsgálni tudjuk meg kell határoznunk a T n T_n T n
A valószínűségek:
p 0 = Pr ( X = 0 ) = ( 4 0 ) ⋅ ( 1 4 ) 0 ⋅ ( 3 4 ) 4 = ( 3 4 ) 4 = 81 256 ≈ 0 , 3164 p 1 = Pr ( X = 1 ) = ( 4 1 ) ⋅ ( 1 4 ) 1 ⋅ ( 3 4 ) 3 = 4 ⋅ 1 4 ⋅ ( 3 4 ) 3 = 108 256 ≈ 0 , 4219 p 2 = Pr ( X = 2 ) = ( 4 2 ) ⋅ ( 1 4 ) 2 ⋅ ( 3 4 ) 2 = 54 256 ≈ 0 , 2109 p 3 = Pr ( X = 3 ) = ( 4 3 ) ⋅ ( 1 4 ) 3 ⋅ ( 3 4 ) 1 = 12 256 ≈ 0 , 04688 p 4 = Pr ( X = 4 ) = ( 4 4 ) ⋅ ( 1 4 ) 4 ⋅ ( 3 4 ) 0 = ( 1 4 ) 4 = 1 256 ≈ 0 , 0039 \begin{align*}
p_0 = \Pr(X = 0) &= {4 \choose 0} \cdot \left(\frac{1}{4}\right)^0 \cdot \left(\frac{3}{4}\right)^4
= \left(\frac{3}{4}\right)^4
= \frac{81}{256}
\approx 0,3164 \\
p_1 = \Pr(X = 1) &= {4 \choose 1} \cdot \left(\frac{1}{4}\right)^1 \cdot \left(\frac{3}{4}\right)^3
= 4 \cdot \frac{1}{4} \cdot \left(\frac{3}{4}\right)^3
= \frac{108}{256}
\approx 0,4219 \\
p_2 = \Pr(X = 2) &= {4 \choose 2} \cdot \left(\frac{1}{4}\right)^2 \cdot \left(\frac{3}{4}\right)^2
= \frac{54}{256}
\approx 0,2109 \\
p_3 = \Pr(X = 3) &= {4 \choose 3} \cdot \left(\frac{1}{4}\right)^3 \cdot \left(\frac{3}{4}\right)^1
= \frac{12}{256}
\approx 0,04688 \\
p_4 = \Pr(X = 4) &= {4 \choose 4} \cdot \left(\frac{1}{4}\right)^4 \cdot \left(\frac{3}{4}\right)^0
= \left(\frac{1}{4}\right)^4
= \frac{1}{256}
\approx 0,0039
\end{align*} p 0 = Pr ( X = 0 ) p 1 = Pr ( X = 1 ) p 2 = Pr ( X = 2 ) p 3 = Pr ( X = 3 ) p 4 = Pr ( X = 4 ) = ( 0 4 ) ⋅ ( 4 1 ) 0 ⋅ ( 4 3 ) 4 = ( 4 3 ) 4 = 256 81 ≈ 0 , 3164 = ( 1 4 ) ⋅ ( 4 1 ) 1 ⋅ ( 4 3 ) 3 = 4 ⋅ 4 1 ⋅ ( 4 3 ) 3 = 256 108 ≈ 0 , 4219 = ( 2 4 ) ⋅ ( 4 1 ) 2 ⋅ ( 4 3 ) 2 = 256 54 ≈ 0 , 2109 = ( 3 4 ) ⋅ ( 4 1 ) 3 ⋅ ( 4 3 ) 1 = 256 12 ≈ 0 , 04688 = ( 4 4 ) ⋅ ( 4 1 ) 4 ⋅ ( 4 3 ) 0 = ( 4 1 ) 4 = 256 1 ≈ 0 , 0039 Ebből az elméleti gyakoriságok
( 0 , 3164 0 , 4219 0 , 2109 0 , 0469 0 , 0039 ) ⋅ 300 = ( 94 , 9 126 , 6 63 , 3 14 , 1 1 , 2 ) \begin{pmatrix}
0,3164 \\
0,4219 \\
0,2109 \\
0,0469 \\
0,0039
\end{pmatrix}
\cdot
300
=
\begin{pmatrix}
94,9 \\
126,6 \\
63,3 \\
14,1 \\
1,2
\end{pmatrix} 0 , 3164 0 , 4219 0 , 2109 0 , 0469 0 , 0039 ⋅ 300 = 94 , 9 126 , 6 63 , 3 14 , 1 1 , 2 Ekkor a próbastatisztika:
T 300 = ∑ j = 1 r ( v j − n p j ) 2 n p j = ( 80 − 94 , 9 ) 2 94 , 9 + ( 113 − 126 , 6 ) 2 126 , 6 + ( 77 − 63 , 3 ) 2 63 , 3 + ( 27 − 14 , 1 ) 2 14 , 1 + ( 3 − 1 , 2 ) 2 1 , 2 = 21 , 5 \begin{align*}
T_{300} &= \sum_{j=1}^{r}{\frac{\left( v_j - n p_j \right)^2}{n p_j}} \\
&= \frac{ \left( 80 - 94,9 \right)^2 }{94,9} +
\frac{ \left( 113 - 126,6 \right)^2 }{126,6} +
\frac{ \left( 77 - 63,3 \right)^2 }{63,3} +
\frac{ \left( 27 - 14,1 \right)^2 }{14,1} +
\frac{ \left( 3 - 1,2 \right)^2 }{1,2} \\
&= 21,5
\end{align*} T 300 = j = 1 ∑ r n p j ( v j − n p j ) 2 = 94 , 9 ( 80 − 94 , 9 ) 2 + 126 , 6 ( 113 − 126 , 6 ) 2 + 63 , 3 ( 77 − 63 , 3 ) 2 + 14 , 1 ( 27 − 14 , 1 ) 2 + 1 , 2 ( 3 − 1 , 2 ) 2 = 21 , 5 χ 2 \chi^2 χ 2 ( r − 1 ) = ( 5 − 1 ) = 4 (r-1)=(5-1)=4 ( r − 1 ) = ( 5 − 1 ) = 4 χ 4 ; 0 , 05 2 \chi^2_{4; \; 0,05} χ 4 ; 0 , 05 2 táblázatból :
9 , 488 9,488 9 , 488
Függetlenségvizsgálat
Adott egy n n n
az 1. szempont szerint legyen r r r
az 2. szempont szerint legyen s s s
Ezekből r × s r \times s r × s v i j v_{ij} v ij i i i j j j
Az adott sorok (1. kategória elemei) és oszlopok( 2. kategória elemei) összegeire speciális jelöléseket vezetünk be:
v i ∙ = ∑ j = 1 s v i j e ˊ s v ∙ j = ∑ i = 1 r v i j v_{i\bullet} = \sum_{j=1}^{s}{v_{ij}} \qquad \text{és} \qquad v_{\bullet j} = \sum_{i=1}^{r}{v_{ij}} v i ∙ = j = 1 ∑ s v ij e ˊ s v ∙ j = i = 1 ∑ r v ij Hasonlóan jelöljük a kategóriák valószínűségét:
p i ∙ = ∑ j = 1 s p i j e ˊ s p ∙ j = ∑ i = 1 r p i j p_{i\bullet} = \sum_{j=1}^{s}{p_{ij}} \qquad \text{és} \qquad p_{\bullet j} = \sum_{i=1}^{r}{p_{ij}} p i ∙ = j = 1 ∑ s p ij e ˊ s p ∙ j = i = 1 ∑ r p ij Annak (elméleti) valószínűsége, hogy egy elem az i i i j j j p i j p_{ij} p ij
A teljes kontingenciatáblázat a következő módon néz ki:
2. szempont 1 ⋯ j ⋯ s Sor o ¨ sszegek 1 v 11 ⋯ v 1 j ⋯ v 1 s v 1 ∙ ⋮ ⋮ ⋮ ⋮ ⋮ 1. szempont i v i 1 ⋯ v i j ⋯ v i s v i ∙ ⋮ ⋮ ⋮ ⋮ ⋮ r v r 1 ⋯ v r j ⋯ v r s v r ∙ Oszlop o ¨ sszegek v ∙ 1 ⋯ v ∙ j ⋯ v ∙ s n \begin{array}{cc|ccccc|c}
& & & & \text{2. szempont} & & \\
& & 1 & \cdots & j & \cdots & s & \text{Sorösszegek} \\
\hline
& 1 & v_{11} & \cdots & v_{1j} & \cdots & v_{1s} & v_{1\bullet} \\
& \vdots & \vdots & & \vdots & & \vdots & \vdots \\
\text{1. szempont} & i & v_{i1} & \cdots & v_{ij} & \cdots & v_{is} & v_{i\bullet} \\
& \vdots & \vdots & & \vdots & & \vdots & \vdots \\
& r & v_{r1} & \cdots & v_{rj} & \cdots & v_{rs} & v_{r\bullet} \\
\hline
\text{Oszlopösszegek} & & v_{\bullet 1} & \cdots & v_{\bullet j} & \cdots & v_{\bullet s} & n
\end{array} 1. szempont Oszlop o ¨ sszegek 1 ⋮ i ⋮ r 1 v 11 ⋮ v i 1 ⋮ v r 1 v ∙ 1 ⋯ ⋯ ⋯ ⋯ ⋯ 2. szempont j v 1 j ⋮ v ij ⋮ v r j v ∙ j ⋯ ⋯ ⋯ ⋯ ⋯ s v 1 s ⋮ v i s ⋮ v rs v ∙ s Sor o ¨ sszegek v 1 ∙ ⋮ v i ∙ ⋮ v r ∙ n Nullhipotézisünk (H 0 H_0 H 0 p i j = p i ∙ ⋅ p ∙ j ( i ∈ [ 1 , r ] , j ∈ [ 1 , s ] ) p_{ij} = p_{i\bullet} \cdot p_{\bullet j} \; \left( i \in [1,r], \; j \in [1,s] \right) p ij = p i ∙ ⋅ p ∙ j ( i ∈ [ 1 , r ] , j ∈ [ 1 , s ] )
Ha teljesül a nullhipotézis a
T n = ∑ i = 1 r ∑ j = 1 s ( ( v i j − v i ∙ ⋅ v ∙ j n ) 2 ⋅ n v i ∙ ⋅ v ∙ j ) T_n = \sum_{i=1}^{r}{\sum_{j=1}^{s}{\left( \left( v_{ij} - \frac{v_{i\bullet} \cdot v_{\bullet j}}{n} \right)^2 \cdot \frac{n}{v_{i\bullet} \cdot v_{\bullet j}} \right)}} T n = i = 1 ∑ r j = 1 ∑ s ( ( v ij − n v i ∙ ⋅ v ∙ j ) 2 ⋅ v i ∙ ⋅ v ∙ j n ) próbastasztika χ 2 \chi^2 χ 2 ( r − 1 ) ⋅ ( s − 1 ) (r-1) \cdot (s-1) ( r − 1 ) ⋅ ( s − 1 )
Adott a következő kontingenciatáblázat:
Hőmérséklet / Csapadék kevés átlagos sok hűvös 15 10 5 átlagos 10 10 20 magas 5 20 5
α = 0 , 05 \alpha = 0,05 α = 0 , 05
H 0 H_0 H 0
A kiegészített kontingenciatáblázat:
Hőmérséklet / Csapadék kevés átlagos sok Összeg hűvös 15 10 5 v 1 ∙ = 30 v_{1 \bullet} = 30 v 1 ∙ = 30 átlagos 10 10 20 v 2 ∙ = 40 v_{2 \bullet} = 40 v 2 ∙ = 40 magas 5 20 5 v 3 ∙ = 30 v_{3 \bullet} = 30 v 3 ∙ = 30 Összeg v ∙ 1 = 30 v_{\bullet 1} = 30 v ∙ 1 = 30 v ∙ 2 = 40 v_{\bullet 2} = 40 v ∙ 2 = 40 v ∙ 3 = 30 v_{\bullet 3} = 30 v ∙ 3 = 30 n = 100 n = 100 n = 100
T 100 = ∑ i = 1 3 ∑ j = 1 3 ( ( v i j − v i ∙ ⋅ v ∙ j n ) 2 ⋅ n v i ∙ ⋅ v ∙ j ) = ( v 11 − v 1 ∙ ⋅ v ∙ 1 100 ) 2 ⋅ 100 v 1 ∙ ⋅ v ∙ 1 + ( v 12 − v 1 ∙ ⋅ v ∙ 2 100 ) 2 ⋅ 100 v 1 ∙ ⋅ v ∙ 2 + ( v 13 − v 1 ∙ ⋅ v ∙ 3 100 ) 2 ⋅ 100 v 1 ∙ ⋅ v ∙ 3 + ( v 21 − v 2 ∙ ⋅ v ∙ 1 100 ) 2 ⋅ 100 v 2 ∙ ⋅ v ∙ 1 + ( v 22 − v 2 ∙ ⋅ v ∙ 2 100 ) 2 ⋅ 100 v 2 ∙ ⋅ v ∙ 2 + ( v 23 − v 2 ∙ ⋅ v ∙ 3 100 ) 2 ⋅ 100 v 2 ∙ ⋅ v ∙ 3 + ( v 31 − v 3 ∙ ⋅ v ∙ 1 100 ) 2 ⋅ 100 v 3 ∙ ⋅ v ∙ 1 + ( v 32 − v 3 ∙ ⋅ v ∙ 2 100 ) 2 ⋅ 100 v 3 ∙ ⋅ v ∙ 2 + ( v 33 − v 3 ∙ ⋅ v ∙ 3 100 ) 2 ⋅ 100 v 3 ∙ ⋅ v ∙ 3 = ( 15 − 30 ⋅ 30 100 ) 2 ⋅ 100 30 ⋅ 30 + ( 10 − 30 ⋅ 40 100 ) 2 ⋅ 100 30 ⋅ 40 + ( 5 − 30 ⋅ 30 100 ) 2 ⋅ 100 30 ⋅ 30 + ( 10 − 40 ⋅ 30 100 ) 2 ⋅ 100 40 ⋅ 30 + ( 10 − 40 ⋅ 40 100 ) 2 ⋅ 100 40 ⋅ 40 + ( 20 − 40 ⋅ 30 100 ) 2 ⋅ 100 40 ⋅ 30 + ( 5 − 30 ⋅ 30 100 ) 2 ⋅ 100 30 ⋅ 30 + ( 20 − 30 ⋅ 40 100 ) 2 ⋅ 100 30 ⋅ 40 + ( 5 − 30 ⋅ 30 100 ) 2 ⋅ 100 30 ⋅ 30 = ( 15 − 9 ) 2 ⋅ 1 9 + ( 10 − 12 ) 2 ⋅ 1 12 + ( 5 − 9 ) 2 ⋅ 1 9 + ( 10 − 12 ) 2 ⋅ 1 12 + ( 10 − 16 ) 2 ⋅ 1 16 + ( 20 − 12 ) 2 ⋅ 1 12 + ( 5 − 9 ) 2 ⋅ 1 9 + ( 20 − 12 ) 2 ⋅ 1 12 + ( 5 − 9 ) 2 ⋅ 1 9 = 4 + 1 3 + 16 9 + 1 3 + 36 16 + 64 12 + 16 9 + 64 12 + 16 9 = 22 , 916 \begin{align*}
T_{100} &= \sum_{i=1}^{3}{\sum_{j=1}^{3}{\left( \left( v_{ij} - \frac{v_{i\bullet} \cdot v_{\bullet j}}{n} \right)^2 \cdot \frac{n}{v_{i\bullet} \cdot v_{\bullet j}} \right)}} \\
&=
\left( v_{11} - \frac{v_{1\bullet} \cdot v_{\bullet 1}}{100} \right)^2 \cdot \frac{100}{v_{1\bullet} \cdot v_{\bullet 1}} +
\left( v_{12} - \frac{v_{1\bullet} \cdot v_{\bullet 2}}{100} \right)^2 \cdot \frac{100}{v_{1\bullet} \cdot v_{\bullet 2}} +
\left( v_{13} - \frac{v_{1\bullet} \cdot v_{\bullet 3}}{100} \right)^2 \cdot \frac{100}{v_{1\bullet} \cdot v_{\bullet 3}} \\
&+ \left( v_{21} - \frac{v_{2\bullet} \cdot v_{\bullet 1}}{100} \right)^2 \cdot \frac{100}{v_{2\bullet} \cdot v_{\bullet 1}} +
\left( v_{22} - \frac{v_{2\bullet} \cdot v_{\bullet 2}}{100} \right)^2 \cdot \frac{100}{v_{2\bullet} \cdot v_{\bullet 2}} +
\left( v_{23} - \frac{v_{2\bullet} \cdot v_{\bullet 3}}{100} \right)^2 \cdot \frac{100}{v_{2\bullet} \cdot v_{\bullet 3}} \\
&+ \left( v_{31} - \frac{v_{3\bullet} \cdot v_{\bullet 1}}{100} \right)^2 \cdot \frac{100}{v_{3\bullet} \cdot v_{\bullet 1}} +
\left( v_{32} - \frac{v_{3\bullet} \cdot v_{\bullet 2}}{100} \right)^2 \cdot \frac{100}{v_{3\bullet} \cdot v_{\bullet 2}} +
\left( v_{33} - \frac{v_{3\bullet} \cdot v_{\bullet 3}}{100} \right)^2 \cdot \frac{100}{v_{3\bullet} \cdot v_{\bullet 3}} \\
&=
\left( 15 - \frac{30 \cdot 30}{100} \right)^2 \cdot \frac{100}{30 \cdot 30} +
\left( 10 - \frac{30 \cdot 40}{100} \right)^2 \cdot \frac{100}{30 \cdot 40} +
\left( 5 - \frac{30 \cdot 30}{100} \right)^2 \cdot \frac{100}{30 \cdot 30} \\
&+ \left( 10 - \frac{40 \cdot 30}{100} \right)^2 \cdot \frac{100}{40 \cdot 30} +
\left( 10 - \frac{40 \cdot 40}{100} \right)^2 \cdot \frac{100}{40 \cdot 40} +
\left( 20 - \frac{40 \cdot 30}{100} \right)^2 \cdot \frac{100}{40 \cdot 30} \\
&+ \left( 5 - \frac{30 \cdot 30}{100} \right)^2 \cdot \frac{100}{30 \cdot 30} +
\left( 20 - \frac{30 \cdot 40}{100} \right)^2 \cdot \frac{100}{30 \cdot 40} +
\left( 5 - \frac{30 \cdot 30}{100} \right)^2 \cdot \frac{100}{30 \cdot 30} \\
&=
\left( 15 - 9 \right)^2 \cdot \frac{1}{9} +
\left( 10 - 12 \right)^2 \cdot \frac{1}{12} +
\left( 5 - 9 \right)^2 \cdot \frac{1}{9} \\
&+ \left( 10 - 12 \right)^2 \cdot \frac{1}{12} +
\left( 10 - 16 \right)^2 \cdot \frac{1}{16} +
\left( 20 - 12 \right)^2 \cdot \frac{1}{12} \\
&+ \left( 5 - 9 \right)^2 \cdot \frac{1}{9} +
\left( 20 - 12 \right)^2 \cdot \frac{1}{12} +
\left( 5 - 9 \right)^2 \cdot \frac{1}{9} \\
&= 4 + \frac{1}{3} + \frac{16}{9} + \frac{1}{3} + \frac{36}{16} + \frac{64}{12} + \frac{16}{9} + \frac{64}{12} + \frac{16}{9} \\
&= 22,916
\end{align*} T 100 = i = 1 ∑ 3 j = 1 ∑ 3 ( ( v ij − n v i ∙ ⋅ v ∙ j ) 2 ⋅ v i ∙ ⋅ v ∙ j n ) = ( v 11 − 100 v 1 ∙ ⋅ v ∙ 1 ) 2 ⋅ v 1 ∙ ⋅ v ∙ 1 100 + ( v 12 − 100 v 1 ∙ ⋅ v ∙ 2 ) 2 ⋅ v 1 ∙ ⋅ v ∙ 2 100 + ( v 13 − 100 v 1 ∙ ⋅ v ∙ 3 ) 2 ⋅ v 1 ∙ ⋅ v ∙ 3 100 + ( v 21 − 100 v 2 ∙ ⋅ v ∙ 1 ) 2 ⋅ v 2 ∙ ⋅ v ∙ 1 100 + ( v 22 − 100 v 2 ∙ ⋅ v ∙ 2 ) 2 ⋅ v 2 ∙ ⋅ v ∙ 2 100 + ( v 23 − 100 v 2 ∙ ⋅ v ∙ 3 ) 2 ⋅ v 2 ∙ ⋅ v ∙ 3 100 + ( v 31 − 100 v 3 ∙ ⋅ v ∙ 1 ) 2 ⋅ v 3 ∙ ⋅ v ∙ 1 100 + ( v 32 − 100 v 3 ∙ ⋅ v ∙ 2 ) 2 ⋅ v 3 ∙ ⋅ v ∙ 2 100 + ( v 33 − 100 v 3 ∙ ⋅ v ∙ 3 ) 2 ⋅ v 3 ∙ ⋅ v ∙ 3 100 = ( 15 − 100 30 ⋅ 30 ) 2 ⋅ 30 ⋅ 30 100 + ( 10 − 100 30 ⋅ 40 ) 2 ⋅ 30 ⋅ 40 100 + ( 5 − 100 30 ⋅ 30 ) 2 ⋅ 30 ⋅ 30 100 + ( 10 − 100 40 ⋅ 30 ) 2 ⋅ 40 ⋅ 30 100 + ( 10 − 100 40 ⋅ 40 ) 2 ⋅ 40 ⋅ 40 100 + ( 20 − 100 40 ⋅ 30 ) 2 ⋅ 40 ⋅ 30 100 + ( 5 − 100 30 ⋅ 30 ) 2 ⋅ 30 ⋅ 30 100 + ( 20 − 100 30 ⋅ 40 ) 2 ⋅ 30 ⋅ 40 100 + ( 5 − 100 30 ⋅ 30 ) 2 ⋅ 30 ⋅ 30 100 = ( 15 − 9 ) 2 ⋅ 9 1 + ( 10 − 12 ) 2 ⋅ 12 1 + ( 5 − 9 ) 2 ⋅ 9 1 + ( 10 − 12 ) 2 ⋅ 12 1 + ( 10 − 16 ) 2 ⋅ 16 1 + ( 20 − 12 ) 2 ⋅ 12 1 + ( 5 − 9 ) 2 ⋅ 9 1 + ( 20 − 12 ) 2 ⋅ 12 1 + ( 5 − 9 ) 2 ⋅ 9 1 = 4 + 3 1 + 9 16 + 3 1 + 16 36 + 12 64 + 9 16 + 12 64 + 9 16 = 22 , 916 Ha a nullhipotézis teljesül a próbastasztika χ 2 \chi^2 χ 2 ( r − 1 ) ⋅ ( s − 1 ) = 2 ⋅ 2 = 4 (r-1) \cdot (s-1) = 2 \cdot 2 = 4 ( r − 1 ) ⋅ ( s − 1 ) = 2 ⋅ 2 = 4 χ 4 ; 0 , 05 2 \chi^2_{4; 0,05} χ 4 ; 0 , 05 2 táblázatból : 9 , 49 9,49 9 , 49